Chemistry for the IB Diploma (second edition)

for the IB Diploma Coursebook with free additional online material Second edition The second edition of this well-received coursebook is fully updated for the 2014 IB Chemistry syllabus, comprehensively covering all requirements. Get the best coverage of the syllabus with clear assessment statements, and links to Theory of Knowledge, International-mindedness and Nature of Science themes. Exam preparation is supported with plenty of sample exam questions, online test questions and exam tips. Clear, simple language makes the text accessible to students of all abilities. Easy navigation is ensured with Standard and Higher Level material clearly marked in all chapters.

The coursebook contains: • test yourself questions throughout the chapters to test knowledge, incorporating command terms as used in IB examinations to cultivate familiarity with the terms and develop skill in answering questions appropriately • worked examples to illustrate how to tackle complex problems • exam-style questions at the end of each chapter, offering thorough practice for the examination • definitions of key terms displayed alongside the text for easy reference • links to Theory of Knowledge concepts alongside appropriate topics, to stimulate thought and discussion • a chapter on the Nature of Science • clear, well-labelled illustrations and photos to help make concepts easy to understand.

Steve Owen teaches at Sevenoaks School, Kent, one of the leading IB schools in the UK, and has over 14 years’ experience in teaching IB Chemistry.

Chemistry for the IB Diploma SECOND EDITION

Steve Owen

9781107622708 Owen Chemistry for the IB Diploma Cover C M Y K

Steve Owen

Chemistry for the IB Diploma

Chemistry

The free additional online student material contains: • all four Option chapters • guidance for the Internal and External Assessment and Nature of Science • additional practice questions.

See inside for details

Steve Owen Other titles available: Biology for the IB Diploma

ISBN 978-1-107-65460-0 ISBN 978-1-107-62819-9

with additional online material

Chemistry for the IB Diploma Second edition Steve Owen with Caroline Ahmed Chris Martin Roger Woodward

Cambridge University Press’s mission is to advance learning, knowledge and research worldwide.

Our IB Diploma resources aim to: t FODPVSBHF MFBSOFST UP FYQMPSF DPODFQUT JEFBT BOE topics that have local and global significance t IFMQ TUVEFOUT EFWFMPQ B QPTJUJWF BUUJUVEF UP MFBSOJOH JO QSFQBSBUJPO for higher education t BTTJTU TUVEFOUT JO BQQSPBDIJOH DPNQMFY RVFTUJPOT BQQMZJOH critical-thinking skills and forming reasoned answers.

University Printing House, Cambridge cb2 8bs, United Kingdom Cambridge University Press is part of the University of Cambridge.

It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781107622708 Š Cambridge University Press 2011, 2014 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

First published 2011 Second edition 2014 Printed in the United Kingdom by Latimer Trend A catalogue record for this publication is available from the British Library isbn 978-1-107-62270-8 Paperback Additional resources for this publication at education.cambridge.org/ibsciences Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables, and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter.

The material has been developed independently by the publisher and the content is in no way connected with nor endorsed by the International Baccalaureate Organization.

notice to teachers in the uk It is illegal to reproduce any part of this book in material form (including photocopying and electronic storage) except under the following circumstances: (i) where you are abiding by a licence granted to your school or institution by the Copyright Licensing Agency; (ii) where no such licence exists, or where you wish to exceed the terms of a licence, and you have gained the written permission of Cambridge University Press; (iii) where you are allowed to reproduce without permission under the provisions of Chapter 3 of the Copyright, Designs and Patents Act 1988, which covers, for example, the reproduction of short passages within certain types of educational anthology and reproduction for the purposes of setting examination questions.

The website accompanying this book contains further resources to support your IB Chemistry studies.Visit education.cambridge.org/ibsciences and register for access.

Separate website terms and conditions apply.

Contents Introduction

v

1 Stoichiometric relationships

1

1.1 Introduction to the particulate nature of matter and chemical change 1.2 The mole concept 1.3 Reacting masses and volumes Exam-style questions

2 Atomic structure 2.1 The nuclear atom 2.2 Electron configuration 2.3 Electrons in atoms (HL) Exam-style questions

3 The periodic table 3.1 3.2 3.3 3.4

The periodic table Physical properties First-row d-block elements (HL) Coloured complexes (HL) Exam-style questions

1 8 18 51

56 56 62 72 80

85 85 88 104 110 115

4 Chemical bonding and structure 119 4.1 4.2 4.3 4.4 4.5 4.6

Ionic bonding and structure Covalent bonding Covalent structures Intermolecular forces Metallic bonding Covalent bonding and electron domains and molecular geometries (HL) 4.7 Hybridisation (HL) Exam-style questions

5 Energetics/thermochemistry 5.1 5.2 5.3 5.4 5.5

Measuring energy changes Hess’s law Bond enthalpies Energy cycles (HL) Entropy and spontaneity (HL) Exam-style questions

119 125 130 148 160 162 176 180

185 185 196 207 215 225 235

6 Chemical kinetics 6.1 Collision theory and rate of reaction 6.2 Rate expression and reaction mechanism (HL) 6.3 Activation energy (HL) Exam-style questions

7 Equilibrium 7.1 Equilibrium 7.2 The equilibrium law (HL) Exam-style questions

8 Acids and bases 8.1 8.2 8.3 8.4 8.5 8.6 8.7

Theories of acids and bases Lewis acids and bases (HL) Properties of acids and bases The pH scale Strong and weak acids and bases Acid deposition Calculations involving acids and bases (HL) 8.8 pH curves (HL) Exam-style questions

9 Redox processes 9.1 Oxidation and reduction 9.2 Electrochemical cells 9.3 Electrochemical cells (HL) Exam-style questions

10 Organic chemistry 10.1 10.2 10.3 10.4 10.5

Fundamentals of organic chemistry Functional group chemistry Types of organic reactions (HL) Synthetic routes (HL) Stereoisomerism (HL) Exam-style questions

241 241 252 268 272

278 278 293 303

308 308 310 312 314 319 325 328 341 362

368 368 386 393 416

422 422 447 466 484 489 501

iii

11 Measurement and data processing

507

11.1 Uncertainties and errors in measurements and results 507 11.2 Graphical techniques 519 11.3 Spectroscopic identification of organic compounds 524 11.4 Spectroscopic identification of organic compounds (HL) 536 Exam-style questions 548

Appendix: the periodic table

Answers to test yourself questions 558 Glossary

576

Index

585

Acknowledgements

593

Free online material The website accompanying this book contains further resources to support your IB Chemistry studies. Visit education.cambridge.org/ibsciences and register to access these resources:

Options

Self-test questions

Option A Materials

Assessment guidance

Option B Biochemistry

Model exam papers

Option C Energy

Nature of Science

Option D Medicinal chemistry

Answers to exam-style questions Answers to Options questions

iv

557

Introduction This second edition of Chemistry for the IB Diploma is fully updated to cover the content of the IB Chemistry Diploma syllabus that will be examined in the years 2016–2022.

Chemistry may be studied at Standard Level (SL) or Higher Level (HL). Both share a common core, and at HL the core is extended with additional HL material. In addition, at both levels, students then choose one Option to complete their studies. Each Option consists of common core and additional HL material. All common core and additional HL material is covered in this print book. The Options are included in the free online material that is accessible with the code available in this book.

The content is arranged in topics that match the syllabus topics, with core and additional HL material on each topic combined in the book topics. The HL content is identified by ‘HL’ included in relevant section titles, and by a yellow page border.

Each section in the book begins with learning objectives as starting and reference points. Test yourself questions appear throughout the text so students can check their progress and become familiar with the style and command terms used, and exam-style questions appear at the end of each topic. Many worked examples appear throughout the text to help students understand how to tackle different types of questions.

Theory of Knowledge (TOK) provides a cross-curricular link between different subjects. It stimulates thought about critical thinking and how we can say we know what we claim to know. Throughout this book, TOK features highlight concepts in Chemistry that can be considered from a TOK perspective. These are indicated by the ‘TOK’ logo, shown here.

Science is a truly international endeavour, being practised across all continents, frequently in international or even global partnerships. Many problems that science aims to solve are international, and will require globally implemented solutions. Throughout this book, InternationalMindedness features highlight international concerns in Chemistry. These are indicated by the ‘International-Mindedness’ logo, shown here.

Nature of Science is an overarching theme of the Chemistry course.

The theme examines the processes and concepts that are central to scientific endeavour, and how science serves and connects with the wider community. Throughout the book, there are ‘Nature of Science’ paragraphs that discuss particular concepts or discoveries from the point of view of one or more aspects of Nature of Science. A chapter giving a general introduction to the Nature of Science theme is available in the free online material.

INTRODUCTION

v

Free online material Additional material to support the IB Chemistry Diploma course is available online.Visit education.cambridge.org/ibsciences and register to access these resources.

Besides the Options and Nature of Science chapter, you will find a collection of resources to help with revision and exam preparation. This includes guidance on the assessments, interactive self-test questions and model exam papers. Additionally, answers to the exam-style questions in this book and to all the questions in the Options are available.

vi

Stoichiometric relationships 1 1.1 Introduction to the particulate nature of matter and chemical change

Learning objectives

1.1.1 The particulate nature of matter

NBUUFS r 6OEFSTUBOE UIF DIBOHFT JOWPMWFE XIFO UIFSF JT B DIBOHF JO TUBUF

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5IF QSPQFSUJFT PG UIF UISFF TUBUFT PG NBUUFS BSF TVNNBSJTFE JO 5BCMF 1.1

r %FTDSJCF UIF UISFF TUBUFT PG

heating ďšť energy is supplied particles gain energy sublimation

deposition

melting

solid

liquid

freezing particles vibrating about mean positions

particles moving around each other

boiling evaporating

gas

condensing particles moving at high speeds in all directions

cooling ďšť energy taken out particles lose energy

Figure 1.1 The arrangement of particles in solids, liquids and gases and the names of the changes of state. Note that evaporation can occur at any temperature – boiling occurs at a fixed temperature.

Solids

Liquids

Gases

Distance between particles

close together

close but further apart than in solids

particles far apart

Arrangement

regular

random

random

Shape

fixed shape

no fixed shape – take up the shape of the container

no fixed shape – fill the container

Volume

fixed

fixed

not fixed

Movement

vibrate

move around each other

move around in all directions

Speed of movement

slowest

faster

fastest

Energy

lowest

higher

highest

Forces of attraction

strongest

weaker

weakest

Table 1.1 The properties of the three states of matter.

1 STOICHIOMETRIC RELATIONSHIPS

1

*G B QVSF TVCTUBODF JT IFBUFE TMPXMZ GSPN CFMPX JUT NFMUJOH QPJOU UP BCPWF JUT boiling point B HSBQI PG UFNQFSBUVSF BHBJOTU UJNF DBO CF PCUBJOFE 'JHVSF 1.2

100

Temperature / °C

90 boiling point = 80 °C 80

boiling

70 60 melting point = 50 °C 50 melting 40 solid 30

gas

liquid

20 10 0

0

5

10

15 20 25 Time / minutes

30

35

Figure 1.2 A heating curve showing changes of state.

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1.1.2 Chemical change

Learning objectives

Elements and compounds

r 6OEFSTUBOE UIBU DPNQPVOET

$IFNJTUSZ JT QBSUMZ B TUVEZ PG IPX DIFNJDBM FMFNFOUT DPNCJOF UP NBLF UIF XPSME BOE UIF 6OJWFSTF BSPVOE VT (PME JT BO element BOE BMM TBNQMFT PG QVSF HPME DPOUBJO POMZ HPME BUPNT An element is a pure substance that contains only one type of atom (but see isotopes in Topic 2).

An atom is the smallest part of an element that can still be recognised as that element.

The physical and chemical properties of a compound are very different to those of the elements from which it is formed.

4PEJVN BOE DIMPSJOF BSF FMFNFOUT m XIFO UIFZ BSF NJYFE BOE IFBUFE UIFZ DPNCJOF DIFNJDBMMZ UP GPSN B DPNQPVOE DBMMFE TPEJVN DIMPSJEF 4PEJVN JT B HSFZ SFBDUJWF NFUBM XJUI B MPX NFMUJOH QPJOU BOE DIMPSJOF JT B ZFMMPX HSFFO QPJTPOPVT HBT m CVU TPEJVN DIMPSJEF DPNNPO TBMU JT B OPO UPYJD DPMPVSMFTT DPNQPVOE XJUI B IJHI NFMUJOH QPJOU 4JNJMBSMZ XIFO JSPO B NBHOFUJD NFUBM JT IFBUFE XJUI TVMGVS B OPO NBHOFUJD ZFMMPX TPMJE B HSFZ OPO NFUBMMJD TPMJE DBMMFE JSPO TVMê EF JT GPSNFE 'JHVSF 1.3

Chemical properties EJDUBUF IPX TPNFUIJOH SFBDUT JO B DIFNJDBM SFBDUJPO Physical properties BSF CBTJDBMMZ BMM UIF PUIFS QSPQFSUJFT PG B TVCTUBODF m TVDI BT NFMUJOH QPJOU EFOTJUZ IBSEOFTT FMFDUSJDBM DPOEVDUJWJUZ FUD

IBWF EJÄ„ FSFOU QSPQFSUJFT UP UIF FMFNFOUT UIFZ BSF NBEF GSPN r 6OEFSTUBOE IPX UP CBMBODF DIFNJDBM FRVBUJPOT r 6OEFSTUBOE IPX UP VTF TUBUF TZNCPMT JO DIFNJDBM FRVBUJPOT r %FTDSJCF UIF EJÄ„ FSFODFT CFUXFFO FMFNFOUT DPNQPVOET BOE NJYUVSFT r 6OEFSTUBOE UIF EJÄ„ FSFODFT CFUXFFO IPNPHFOFPVT BOE IFUFSPHFOFPVT NJYUVSFT A compound is a pure substance formed when two or more elements combine chemically.

Figure 1.3 Iron (left) combines with sulfur (centre) to form iron sulfide (right).

The meaning of chemical equations 8IFO FMFNFOUT DPNCJOF UP GPSN DPNQPVOET UIFZ BMXBZT DPNCJOF JO fixed ratios EFQFOEJOH PO UIF OVNCFST PG BUPNT SFRVJSFE 8IFO TPEJVN BOE DIMPSJOF DPNCJOF UIFZ EP TP JO UIF NBTT SBUJP TP UIBU H PG TPEJVN SFBDUT FYBDUMZ XJUI H PG DIMPSJOF 4JNJMBSMZ XIFO IZESPHFO BO FYQMPTJWF HBT DPNCJOFT XJUI PYZHFO B IJHIMZ SFBDUJWF HBT

UP GPSN XBUFS MJRVJE BU SPPN UFNQFSBUVSF H PG IZESPHFO DPNCJOFT XJUI H PG PYZHFO PS H PG IZESPHFO SFBDUT XJUI H PG PYZHFO VTJOH SPVOEFE SFMBUJWF BUPNJD NBTTFT m UIBU JT UIFZ BMXBZT DPNCJOF JO B NBTT SBUJP PG

1 STOICHIOMETRIC RELATIONSHIPS

3

C C

+

C O

O O

C O

Figure 1.4 Two carbon atoms react with one oxygen molecule to form two molecules of carbon monoxide.

C

C

C

C

+

O O

C O

C O

O O

C O

C O

Figure 1.5 Four carbon atoms react with two oxygen molecules to form four molecules of carbon monoxide.

C

C

O O

C O

C O

C

C

O O

C O

C O

C

C

O O

C O

C O

C

C

O O

C O

C O

+

Figure 1.6 Eight carbon atoms react with four oxygen molecules to form eight molecules of carbon monoxide.

Mass is conserved in all chemical reactions.

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Balancing equations *G B SFBDUJPO JOWPMWFT H PG POF TVCTUBODF SFBDUJOH XJUI H PG BOPUIFS TVCTUBODF JO B DMPTFE DPOUBJOFS OPUIJOH DBO CF BEEFE PS DBO FTDBQF UIFO BU UIF FOE PG UIF SFBDUJPO UIFSF XJMM TUJMM CF FYBDUMZ H PG TVCTUBODF QSFTFOU 5IJT H NBZ CF NBEF VQ PG POF PS NPSF QSPEVDUT BOE TPNF SFBDUBOUT UIBU IBWF OPU GVMMZ SFBDUFE CVU UIF LFZ QPJOU JT UIBU UIFSF XJMM OP NPSF BOE OP MFTT UIBO H QSFTFOU " DIFNJDBM SFBDUJPO JOWPMWFT BUPNT KPJOJOH UPHFUIFS JO EJÄ„ FSFOU XBZT BOE FMFDUSPOT SFEJTUSJCVUJOH UIFNTFMWFT CFUXFFO UIF BUPNT CVU JU JT OPU QPTTJCMF GPS UIF SFBDUJPO UP JOWPMWF BUPNT PS FMFDUSPOT CFJOH DSFBUFE PS EFTUSPZFE 8IFO B DIFNJDBM SFBDUJPO JT SFQSFTFOUFE CZ B DIFNJDBM FRVBUJPO UIFSF NVTU CF FYBDUMZ UIF TBNF OVNCFS BOE UZQF PG BUPNT PO FJUIFS TJEF PG UIF FRVBUJPO SFQSFTFOUJOH UIF TBNF OVNCFS PG BUPNT CFGPSF BOE BGUFS UIJT SFBDUJPO

BUPNT

$ H 0 reactants $

)

0

→

$0 ) 0 products $

)

0

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4

State symbols are often used to indicate the physical state of substances involved in a reaction: (s) = solid (l) = liquid (g) = gas (aq) = aqueous (dissolved in water)

Worked examples 1.1 #BMBODF UIF GPMMPXJOH FRVBUJPO w / H

w ) H → w /) H

BOE XPSL PVU UIF TVN PG UIF DPFą DJFOUT JO UIF FRVBUJPO *O UIF VOCBMBODFE FRVBUJPO UIFSF BSF UXP / BUPNT BOE UXP ) BUPNT PO UIF MFGU IBOE TJEF PG UIF FRVBUJPO CVU POF / BUPN BOE UISFF ) BUPNT PO UIF SJHIU IBOE TJEF *U JT OPU QPTTJCMF GPS UXP / BUPNT UP SFBDU XJUI UXP ) BUPNT UP QSPEVDF POF / BUPN BOE UISFF ) BUPNT UIFSFGPSF UIJT FRVBUJPO JT OPU CBMBODFE *U DBO CF CBMBODFE JO UXP TUBHFT BT GPMMPXT w / w ) → /) BUPNT / /

) ) w / ) → /) BUPNT / /

) ) 5IJT FRVBUJPO JT OPX CBMBODFE CFDBVTF UIFSF JT UIF TBNF OVNCFS PG FBDI UZQF PG BUPN PO CPUI TJEFT PG UIF FRVBUJPO 5IF TVN PG UIF DPFÄ… DJFOUT JO UIJT FRVBUJPO JT

5IF DPFÄ… DJFOU PG / JT BMUIPVHI XF EP OPU VTVBMMZ XSJUF UIJT JO BO FRVBUJPO 1.2 #BMBODF UIF GPMMPXJOH FRVBUJPO w $ H H

w 0 H → w $0 H

w ) 0 M

$PNQPVOET BSF CBMBODFE ê STU UIFO FMFNFOUT w $ H H

w 0 H → $0 H

) 0 M

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¤ > w $ H H

0 H â&#x2020;&#x2019; $0 H

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0 H â&#x2020;&#x2019; $0 H

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1 STOICHIOMETRIC RELATIONSHIPS

5

?

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Mixtures

A mixture contains two or more substances mixed together.

A homogeneous mixture has the same (uniform) composition throughout the mixture and consists of only one phase.

A heterogeneous mixture does not have uniform composition and consists of separate phases.

Heterogeneous mixtures can be separated by mechanical means.

6

&MFNFOUT BOE DPNQPVOET BSF QVSF TVCTUBODFT CVU NPTU UIJOHT BSPVOE VT BSF OPU QVSF TVCTUBODFT CVU NJYUVSFT 8F CSFBUIF JO BJS XIJDI JT B NJYUVSF BMM UIF GPPET XF FBU BSF NJYUVSFT PYZHFO JT DBSSJFE BSPVOE PVS CPEZ CZ CMPPE BOPUIFS NJYUVSF 5IF DPNQPOFOUT PG B NJYUVSF DBO CF FMFNFOUT PS DPNQPVOET m PS NJYUVSFT "JS JT B NJYUVSF PG NPTUMZ FMFNFOUT OJUSPHFO PYZHFO BSHPO

XJUI TNBMMFS BNPVOUT PG DPNQPVOET DBSCPO EJPYJEF XBUFS WBQPVS FUD

5IF DPNQPOFOUT PG B NJYUVSF BSF OPU DIFNJDBMMZ CPOEFE UPHFUIFS BOE TP SFUBJO UIFJS JOEJWJEVBM QSPQFSUJFT *O B NJYUVSF PG JSPO BOE TVMGVS UIF JSPO JT TIJOZ BOE NBHOFUJD UIF TVMGVS JT ZFMMPX BOE CVSOT JO BJS UP GPSN TVMGVS EJPYJEF 8IFO UIF NJYUVSF JT IFBUFE BOE GPSNT UIF DPNQPVOE JSPO TVMê EF UIJT JT OPU TIJOZ PS NBHOFUJD PS ZFMMPX m JU JT EVMM BOE HSFZ BOE IBT DPNQMFUFMZ EJÄ&#x201E; FSFOU QSPQFSUJFT UP JUT FMFNFOUT "T ZPV TBX FBSMJFS XIFO BUPNT DPNCJOF UP GPSN DPNQPVOET UIFZ EP TP JO ê YFE SBUJPT CVU UIF DPNQPOFOUT PG B NJYUVSF DBO CF NJYFE UPHFUIFS JO BOZ QSPQPSUJPO 'PS FYBNQMF FUIBOPM BOE XBUFS DBO CF NJYFE UPHFUIFS JO BOZ SBUJP 4PMVUJPOT BSF NJYUVSFT BOE B TPMVUJPO PG TPEJVN DIMPSJEF DPVME CF NBEF CZ EJTTPMWJOH H PG TPEJVN DIMPSJEF JO cm PG XBUFS PS H PG TPEJVN DIMPSJEF JO DN XBUFS PS H PG TPEJVN DIMPSJEF JO cm PG XBUFS PS NBOZ PUIFS BNPVOUT 5IF DPNQPOFOUT PG B NJYUVSF DBO CF TFQBSBUFE GSPN FBDI PUIFS CZ QIZTJDBM NFBOT m GPS FYBNQMF B NJYUVSF PG TBOE BOE TBMU DPVME CF TFQBSBUFE CZ EJTTPMWJOH UIF TBMU JO XBUFS ê MUFSJOH PÄ&#x201E; UIF TBOE BOE UIFO IFBUJOH UIF TBMU TPMVUJPO UP ESJWF PÄ&#x201E; UIF XBUFS

Homogeneous and heterogeneous mixtures 0OF FYBNQMF PG B homogeneous mixture JT B TPMVUJPO /P JOEJWJEVBM QBSUJDMFT DBO CF TFFO JO UIF TPMVUJPO BOE JUT DPODFOUSBUJPO JT UIF TBNF UISPVHIPVU *G TFWFSBM cm TBNQMFT PG B TPMVUJPO PG TPEJVN DIMPSJEF BSF UBLFO GSPN B CFBLFS BOE FWBQPSBUFE TFQBSBUFMZ UP ESZOFTT UIF TBNF NBTT PG TPEJVN DIMPSJEF XJMM CF GPSNFE CZ FBDI TBNQMF $MFBO BJS XJUI OP QBSUJDVMBUFT JT BMTP B IPNPHFOFPVT NJYUVSF 0OF FYBNQMF PG B heterogeneous mixture JT TBOE JO B CFBLFS PG XBUFS 5IF TBOE BOE XBUFS DBO CF EJTUJOHVJTIFE GSPN FBDI PUIFS BOE DBO BMTP CF TFQBSBUFE CZ ê MUFSJOH

.JYUVSFT PG EJÄ&#x201E; FSFOU TPMJET BSF BMTP IFUFSPHFOFPVT 'PS FYBNQMF FWFO UIPVHI B NJYUVSF PG JSPO BOE TVMGVS NBZ IBWF CFFO NBEF WFSZ DBSFGVMMZ TP UIBU UIFSF BSF UIF TBNF NBTTFT PG JSPO BOE TVMGVS JO FBDI DVCJD DFOUJNFUSF UIF DPNQPTJUJPO JT OPU VOJGPSN CFDBVTF UIFSF BSF EJTUJODU QBSUJDMFT PG JSPO BOE TVMGVS BOE FBDI QBSUJDMF PG JSPO BOE TVMGVS SFQSFTFOUT B EJÄ&#x201E; FSFOU QIBTF 5IF DPNQPOFOUT PG UIF NJYUVSF DPVME CF TFQBSBUFE GSPN FBDI PUIFS VTJOH B NBHOFU m PS FWFO B QBJS PG UXFF[FST UP QJDL PVU FBDI JOEJWJEVBM QJFDF PG JSPO BOE TVMGVS 4FB XBUFS JT B NJYUVSF BOE UIF QSPDFTT PG PCUBJOJOH GSFTI XBUFS GSPN TFB XBUFS JT DBMMFE EFTBMJOBUJPO %FTBMJOBUJPO JT WFSZ JNQPSUBOU JO TPNF QBSUT PG UIF XPSME XIFSF TVÄ&#x2026; DJFOU GSFTI XBUFS JT OPU BWBJMBCMF GSPN PUIFS TPVSDFT GPS FYBNQMF JO UIF .JEEMF &BTU 'SFTI XBUFS PCUBJOFE CZ EFTBMJOBUJPO DBO CF VTFE GPS IVNBO DPOTVNQUJPO BHSJDVMUVSF PS JO JOEVTUSZ

Nature of science %BUB DPMMFDUJPO JT FTTFOUJBM JO TDJFODF 5IF EJTDVTTJPO BCPWF IBT VTFE CPUI RVBOUJUBUJWF SFHBSEJOH SFBDUJOH NBTTFT BOE RVBMJUBUJWF EBUB BCPVU UIF QSPQFSUJFT PG TVCTUBODFT "DDVSBUF RVBOUJUBUJWF EBUB BSF FTTFOUJBM GPS UIF BEWBODFNFOU PG TDJFODF BOE TDJFOUJTUT BOBMZTF TVDI EBUB UP NBLF IZQPUIFTFT BOE UP EFWFMPQ UIFPSJFT 5IF MBX PG EFê OJUF QSPQPSUJPOT HPWFSOJOH IPX FMFNFOUT DPNCJOF NBZ TFFN PCWJPVT OPXBEBZT JO UIF MJHIU PG UIF BUPNJD UIFPSZ CVU JO UIF TFWFOUFFOUI BOE FJHIUFFOUI DFOUVSJFT JU XBT UIF TVCKFDU PG NVDI EFCBUF

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Test yourself 2 $MBTTJGZ FBDI PG UIF GPMMPXJOH BT BO FMFNFOU B DPNQPVOE PS B NJYUVSF a XBUFS b PYZHFO c QPUBTTJVN JPEJEF d PSBOHF KVJDF e DSVEF PJM f WBOBEJVN g BNNPOJB h BJS i IZESPHFO DIMPSJEF j NBHOFTJVN PYJEF

3 $MBTTJGZ FBDI PG UIF EJBHSBNT CFMPX VTJOH BT NBOZ XPSET BT BQQSPQSJBUF GSPN UIF MJTU element compound mixture solid liquid gas

a

b

c

d

1 STOICHIOMETRIC RELATIONSHIPS

7

Learning objectives

1.2 The mole concept

r %Fê OF SFMBUJWF BUPNJD NBTT BOE

1.2.1 Relative masses

SFMBUJWF NPMFDVMBS NBTT r 6OEFSTUBOE XIBU JT NFBOU CZ POF NPMF PG B TVCTUBODF r $BMDVMBUF UIF NBTT PG POF NPMF PG B TVCTUBODF r $BMDVMBUF UIF OVNCFS PG NPMFT QSFTFOU JO B TQFDJê FE NBTT PG B TVCTUBODF r 8PSL PVU UIF OVNCFS PG QBSUJDMFT JO B TQFDJê FE NBTT PG B TVCTUBODF BOE BMTP UIF NBTT PG POF NPMFDVMF

The relative atomic mass (Ar) of an element is the average of the masses of the isotopes in a naturally occurring sample of the 1 element relative to the mass of 12 of an atom of carbon-12.

5IF Ar PG DBSCPO JT OPU CFDBVTF DBSCPO DPOUBJOT JTPUPQFT PUIFS UIBO DBSCPO TFF QBHF 58

.PTU DIFNJDBM SFBDUJPOT JOWPMWF UXP PS NPSF TVCTUBODFT SFBDUJOH XJUI FBDI PUIFS 4VCTUBODFT SFBDU XJUI FBDI PUIFS JO DFSUBJO SBUJPT BOE TUPJDIJPNFUSZ JT UIF TUVEZ PG UIF SBUJPT JO XIJDI DIFNJDBM TVCTUBODFT DPNCJOF *O PSEFS UP LOPX UIF FYBDU RVBOUJUZ PG FBDI TVCTUBODF UIBU JT SFRVJSFE UP SFBDU XF OFFE UP LOPX UIF OVNCFS PG BUPNT NPMFDVMFT PS JPOT JO B TQFDJê D BNPVOU PG UIBU TVCTUBODF )PXFWFS UIF NBTT PG BO JOEJWJEVBM JPO BUPN PS NPMFDVMF JT TP TNBMM BOE UIF OVNCFS PG QBSUJDMFT UIBU NBLF VQ FWFO B WFSZ TNBMM NBTT TP MBSHF UIBU B NPSF DPOWFOJFOU NFUIPE PG XPSLJOH PVU SFBDUJOH RVBOUJUJFT IBE UP CF EFWFMPQFE

Relative atomic mass (Ar)

5IF NBTT PG B IZESPHFO BUPN JT BQQSPYJNBUFMZ ¤ â&#x2C6;&#x2019; H 4VDI TNBMM OVNCFST BSF OPU DPOWFOJFOU UP VTF JO FWFSZEBZ MJGF TP XF VTF TDBMFT PG SFMBUJWF NBTT 5IFTF DPNQBSF UIF NBTTFT PG BUPNT BOE NPMFDVMFT FUD UP UIF NBTT PG POF BUPN PG DBSCPO XIJDI JT BTTJHOFE B NBTT PG FYBDUMZ "T UIFTF RVBOUJUJFT BSF SFMBUJWF UIFZ IBWF OP VOJUT 5IF relative atomic mass (Ar) PG TJMWFS JT " OBUVSBMMZ PDDVSSJOH TBNQMF PG TJMWFS DPOUBJOT UIF JTPUPQFT "H BOE "H 5IF JTPUPQF JT TMJHIUMZ NPSF BCVOEBOU UIBO UIF JTPUPQF 5BLJOH JOUP BDDPVOU UIF BNPVOU PG FBDI JTPUPQF QSFTFOU JO B TBNQMF UIF XFJHIUFE NFBO JU JT GPVOE UIBU PO BWFSBHF UIF NBTT PG B TJMWFS BUPN JT UJNFT

PG B DBSCPO BUPN /P TJMWFS BUPNT BDUVBMMZ FYJTU XJUI UIF NBTT PG UIF NBTT PG UIJT JT KVTU UIF BWFSBHF SFMBUJWF BUPNJD NBTT PG TJMWFS

Relative molecular mass (Mr) The relative molecular mass (Mr) of a compound is the mass of a molecule of that compound 1 of an relative to the mass of 12 atom of carbon-12.

"O relative molecular mass (Mr) JT UIF TVN PG UIF SFMBUJWF BUPNJD NBTTFT PG UIF JOEJWJEVBM BUPNT NBLJOH VQ B NPMFDVMF 5IF SFMBUJWF NPMFDVMBS NBTT PG NFUIBOF $) JT Ar PG $

¤ Ar PG )

5IF SFMBUJWF NPMFDVMBS NBTT PG FUIBOPJD BDJE $) $00) JT

¤

¤

The relative formula mass is the mass of one formula unit relative 1 of an atom of to the mass of 12 carbon-12.

8

*G B DPNQPVOE JT NBEF VQ PG JPOT BOE UIFSFGPSF EPFT OPU DPOUBJO EJTDSFUF NPMFDVMFT XF TIPVME SFBMMZ UBML BCPVU relative formula mass )PXFWFS SFMBUJWF NPMFDVMBS NBTT JT VTVBMMZ VTFE UP SFGFS UP UIF NBTT PG UIF GPSNVMB VOJU PG BO JPOJD DPNQPVOE BT XFMM

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Test yourself 4 8PSL PVU UIF SFMBUJWF NPMFDVMBS NBTTFT PG UIF GPMMPXJOH DPNQPVOET NH3 C2H5OH MgCl2 Ca(NO3)2 CH3(CH2)5CH3 SO2 PCl5 Mg3(PO4)2 Na2S2O3 CH3CH2CH2COOCH2CH3

Moles 0OF NPMF JT UIF BNPVOU PG TVCTUBODF UIBU DPOUBJOT UIF TBNF OVNCFS PG QBSUJDMFT BUPNT JPOT NPMFDVMFT FUD BT UIFSF BSF DBSCPO BUPNT JO H PG DBSCPO 5IJT OVNCFS JT DBMMFE Avogadroâ&#x20AC;&#x2122;s constant IBT TZNCPM L PS NA BOE IBT UIF WBMVF ¤ NPM¢ 4P H PG DBSCPO DPOUBJOT ¤ DBSCPO BUPNT :PV DBO IBWF B NPMF PG BCTPMVUFMZ BOZUIJOH 8F VTVBMMZ DPOTJEFS B NPMF PG BUPNT ¤ BUPNT PS B NPMF PG NPMFDVMFT ¤ NPMFDVMFT CVU XF DPVME BMTP IBWF GPS JOTUBODF B NPMF PG QJOH QPOH CBMMT ¤ QJOH QPOH CBMMT

5IF Ar PG PYZHFO JT XIJDI NFBOT UIBU PO BWFSBHF FBDI PYZHFO BUPN JT UJNFT BT IFBWZ BT B DBSCPO BUPN 5IFSFGPSF H PG PYZHFO BUPNT NVTU DPOUBJO UIF TBNF OVNCFS PG BUPNT BT H PG DBSCPO J F POF NPMF PS ¤ BUPNT 4JNJMBSMZ POF NBHOFTJVN BUPN JT PO BWFSBHF UJNFT BT IFBWZ BT B DBSCPO BUPN BOE UIFSFGPSF H PG NBHOFTJVN BUPNT DPOUBJOT ¤ NBHOFTJVN BUPNT 5IF OVNCFS PG NPMFT QSFTFOU JO B DFSUBJO NBTT PG TVCTUBODF DBO CF XPSLFE PVU VTJOH UIF FRVBUJPO OVNCFS PG NPMFT n

The molar mass (M) of a substance is its Ar or Mr in grams.

The units of molar mass are g molâ&#x2C6;&#x2019;1. For example, the Ar of silicon is 28.09, and the molar mass of silicon is 28.09 g molâ&#x2C6;&#x2019;1.

This means that 28.09 g of silicon contains 6.02 Ã&#x2014; 1023 silicon atoms.

8IFO DBMDVMBUJOH UIF OVNCFS PG NPMFT QSFTFOU JO B DFSUBJO NBTT PG B TVCTUBODF UIF NBTT NVTU CF JO HSBNT

NBTT PG TVCTUBODF NPMBS NBTT

5IF USJBOHMF JO 'JHVSF 1.7 JT B VTFGVM TIPSUDVU GPS XPSLJOH PVU BMM UIF RVBOUJUJFT JOWPMWFE JO UIF FRVBUJPO *G BOZ POF PG UIF TFDUJPOT PG UIF USJBOHMF JT DPWFSFE VQ UIF SFMBUJPOTIJQ CFUXFFO UIF PUIFS UXP RVBOUJUJFT UP HJWF UIF DPWFSFE RVBOUJUZ JT SFWFBMFE 'PS FYBNQMF JG ANBTT PG TVCTUBODF JT DPWFSFE XF BSF MFGU XJUI OVNCFS PG NPMFT NVMUJQMJFE CZ NPMBS NBTT NBTT PG TVCTUBODF OVNCFS PG NPMFT ¤ NPMBS NBTT *G ANPMBS NBTT JT DPWFSFE XF BSF MFGU XJUI NBTT PG TVCTUBODF EJWJEFE CZ OVNCFS PG NPMFT NBTT PG TVCTUBODF NPMBS NBTT

OVNCFS PG NPMFT

mass of substance

number of moles

molar mass

Figure 1.7 The relationship between the mass of a substance, the number of moles and molar mass.

1 STOICHIOMETRIC RELATIONSHIPS

9

0OF NPMF JT BO FOPSNPVT OVNCFS BOE CFZPOE UIF TDPQF PG PVS OPSNBM FYQFSJFODF )PX EP XF VOEFSTUBOE B OVNCFS UIJT MBSHF 0OF XBZ JT UP EFTDSJCF UIF OVNCFS JO UFSNT PG UIJOHT XF BSF GBNJMJBS XJUI GSPN FWFSZEBZ MJGF 'PS JOTUBODF POF NPMF PG QJOH QPOH CBMMT XPVME DPWFS UIF TVSGBDF PG UIF &BSUI UP BCPVU UJNFT UIF IFJHIU PG .PVOU &WFSFTU 8F LOPX XIBU B QJOH QPOH CBMM MPPLT MJLF BOE XF NBZ IBWF B SPVHI JEFB PG UIF IFJHIU PG .PVOU &WFSFTU TP QFSIBQT UIJT EFTDSJQUJPO HJWFT VT B DPOUFYU JO XIJDI XF DBO VOEFSTUBOE ¤ "OPUIFS EFTDSJQUJPO TPNFUJNFT VTFE JT JO UFSNT PG B NPMF PG DPNQVUFS QBQFS POF NPMF PG DPNQVUFS QSJOUFS QBQFS TIFFUT JG TUBDLFE POF PO UPQ PG FBDI PUIFS XPVME

TUSFUDI PWFS MJHIU ZFBST POF MJHIU ZFBS JT UIF EJTUBODF UIBU MJHIU USBWFMT JO POF ZFBS m UIJT JT PWFS UXJDF UIF UIJDLOFTT PG PVS HBMBYZ *T UIJT EFTDSJQUJPO CFUUFS PS XPSTF UIBO UIF QSFWJPVT POF *U DFSUBJOMZ TPVOET NPSF JNQSFTTJWF CVU EPFT JU TVÄ&#x201E; FS GSPN UIF GBDU UIBU XF IBWF OP SFBM DPODFQU PG UIF TJ[F PG PVS HBMBYZ $BO ZPV UIJOL PG BOZ PUIFS XBZT PG EFTDSJCJOH UIJT OVNCFS JO UFSNT PG UIJOHT ZPV BSF GBNJMJBS XJUI GSPN FWFSZEBZ MJGF 5IJT JT BO FYBNQMF PG B XJEFS JEFB UIBU XF UFOE UP VOEFSTUBOE UIJOHT UIBU BSF CFZPOE PVS OPSNBM FYQFSJFODF CZ SFGFSFODF UP UIJOHT XJUI XIJDI XF BSF NPSF GBNJMJBS

Worked examples 1.3 $BMDVMBUF UIF OVNCFS PG NPMFT PG NBHOFTJVN BUPNT JO H PG NBHOFTJVN NBTT PG TVCTUBODF OVNCFS PG NPMFT n

NPMBS NBTT

n NPM

H PG NBHOFTJVN JT NPM

/PUF UIF VOJU GPS NPMFT JT NPM

5IF BOTXFS JT HJWFO UP UISFF TJHOJê DBOU ê HVSFT CFDBVTF UIF NBTT PG TVCTUBODF JT HJWFO UP UISFF TJHOJê DBOU ê HVSFT

1.4 $BMDVMBUF UIF NBTT PG NPM $) $00) NBTT PG TVCTUBODF OVNCFS PG NPMFT ¤ NPMBS NBTT NBTT PG TVCTUBODF ¤ H 5IF NBTT PG NPM $) $00) JT H

10

5IF BOTXFS JT HJWFO UP GPVS TJHOJê DBOU ê HVSFT CFDBVTF UIF OVNCFS PG NPMFT BOE UIF NPMBS NBTT BSF HJWFO UP GPVS TJHOJê DBOU ê HVSFT

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Test yourself 5 $PQZ BOE DPNQMFUF UIF UBCMF 5IF ê STU SPX IBT CFFO EPOF GPS ZPV

Compound H2O

Molar mass / g molâ&#x2C6;&#x2019;1 18.02

CO2

Mass / g

Number of moles / mol

9.01

0.500

5.00

H2S

0.100

NH3

3.50

Q

1.00

0.0350

Z

0.0578

1.12 Ã&#x2014; 10â&#x2C6;&#x2019;3

Mg(NO3)2

1.75

C3H7OH

2500 5.68 Ã&#x2014; 10â&#x2C6;&#x2019;5

Fe2O3

The mass of a molecule 5IF NBTT PG POF NPMF PG XBUFS JT H 5IJT DPOUBJOT ¤ NPMFDVMFT PG XBUFS 5IF NBTT PG POF NPMFDVMF PG XBUFS DBO CF XPSLFE PVU CZ EJWJEJOH UIF NBTT PG POF NPMF H CZ UIF OVNCFS PG NPMFDVMFT JU DPOUBJOT ¤

NBTT PG POF NPMFDVMF

Exam tip 3FNFNCFS m UIF NBTT PG B NPMFDVMF JT B WFSZ TNBMM OVNCFS %P OPU DPOGVTF UIF NBTT PG B TJOHMF NPMFDVMF XJUI UIF NBTT PG POF NPMF PG B TVCTUBODF XIJDI JT B OVNCFS HSFBUFS UIBO

¤ ¢ H ¤

mass of one molecule =

molar mass Avogadroâ&#x20AC;&#x2122;s constant

The number of particles 8IFO XF XSJUF A NPM 0 JU NFBOT POF NPMF PG 0 NPMFDVMFT UIBU JT ¤ 0 NPMFDVMFT &BDI 0 NPMFDVMF DPOUBJOT UXP PYZHFO BUPNT UIFSFGPSF POF NPMF PG 0 NPMFDVMFT DPOUBJOT ¤ ¤ ¤ BUPNT 5IBU JT POF NPMF PG 0 molecules JT NBEF VQ PG UXP NPMFT PG PYZHFO atoms 8IFO XF UBML BCPVU A NPM ) 0 XF NFBO NPM ) 0 NPMFDVMFT J F ¤ ¤ ) 0 NPMFDVMFT J F ¤ ) 0 NPMFDVMFT &BDI H 0 NPMFDVMF DPOUBJOT UXP IZESPHFO BUPNT BOE POF PYZHFO BUPN 5IF UPUBM OVNCFS PG IZESPHFO BUPNT JO NPM ) 0 JT ¤ ¤ J F ¤ IZESPHFO BUPNT J F NPM IZESPHFO BUPNT &BDI ) 0 NPMFDVMF DPOUBJOT UISFF BUPNT 5IFSFGPSF UIF UPUBM OVNCFS PG BUPNT JO NPM ) 0 JT ¤ ¤ J F ¤ BUPNT PS NPM BUPNT *G ZPV MPPL BU 5BCMF 1.2 ZPV DBO TFF UIF DPOOFDUJPO CFUXFFO UIF OVNCFS PG NPMFT PG NPMFDVMFT BOE UIF OVNCFS PG NPMFT PG B QBSUJDVMBS BUPN JO UIBU NPMFDVMF 'JHVSF 1.8 JMMVTUSBUFT UIF SFMBUJPOTIJQ CFUXFFO OVNCFS PG QBSUJDMFT OVNCFS PG NPMFT BOE "WPHBESP T DPOTUBOU

O O oxygen

Exam tip :PV NVTU CF DMFBS XIJDI UZQF PG QBSUJDMF ZPV BSF DPOTJEFSJOH %P ZPV IBWF POF NPMF PG BUPNT NPMFDVMFT PS JPOT

H

O

H

water

1 STOICHIOMETRIC RELATIONSHIPS

11

Compound number of particles

number of moles

Avogadroâ&#x20AC;&#x2122;s constant

Figure 1.8 The relationship between the number of moles and the number of particles.

Moles of molecules

Moles of O atoms

H2O

0.1

0.1

SO2

0.1

0.2

SO3

0.1

0.3

H3PO4

0.1

0.4

O3

0.5

1.5

CH3COOH

0.2

0.4

Table 1.2 The relationship between the number of moles of molecules and the number of moles of particular atoms.

*G XF NVMUJQMZ UIF OVNCFS PG NPMFT PG NPMFDVMFT CZ UIF OVNCFS PG B QBSUJDVMBS UZQF PG BUPN JO B NPMFDVMF J F CZ UIF TVCTDSJQU PG UIF BUPN

XF HFU UIF OVNCFS PG NPMFT PG UIBU UZQF PG BUPN 5IVT JO NPM H 40 UIFSF BSF ¤ J F NPM PYZHFO BUPNT

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Test yourself 6 8PSL PVU UIF NBTT PG POF NPMFDVMF PG FBDI PG UIF GPMMPXJOH a ) 0 b /) c $0

8 $BMDVMBUF UIF UPUBM OVNCFS PG BUPNT JO FBDI PG UIF GPMMPXJOH a NPM /) b NPM $ H c NPM $ H 0)

7 8PSL PVU UIF UPUBM OVNCFS PG IZESPHFO BUPNT JO FBDI PG UIF GPMMPXJOH a NPM ) b NPM $) c NPM /)

9 $BMDVMBUF UIF OVNCFS PG NPMFT PG PYZHFO BUPNT JO FBDI PG UIF GPMMPXJOH a NPM ) 40 b NPM $M 0 c NPM 9F0

Learning objectives

1.2.2 Empirical and molecular formulas

r %FUFSNJOF UIF QFSDFOUBHF

Percentage composition of a compound

DPNQPTJUJPO CZ NBTT PG B TVCTUBODF r 6OEFSTUBOE XIBU JT NFBOU CZ FNQJSJDBM BOE NPMFDVMBS GPSNVMBT r $BMDVMBUF FNQJSJDBM BOE NPMFDVMBS GPSNVMBT

12

5IF QFSDFOUBHF CZ NBTT PG FBDI FMFNFOU QSFTFOU JO B DPNQPVOE DBO CF XPSLFE PVU VTJOH UIF GPSNVMB % by mass of number of atoms of the element Ã&#x2014; relative atomic mass = an element relative molecular mass

Worked examples 1.5 'JOE UIF QFSDFOUBHF CZ NBTT PG FBDI FMFNFOU QSFTFOU JO $ H /0 5IF SFMBUJWF NPMFDVMBS NBTT PG $ H /0 JT 1FSDFOUBHF PG DBSCPO UIF SFMBUJWF BUPNJD NBTT PG DBSCPO JT BOE UIFSF BSF TJY DBSCPO BUPNT QSFTFOU TP UIF UPUBM NBTT PG UIF DBSCPO BUPNT JT ¤ J F

DBSCPO ¤

1FSDFOUBHF PG UIF PUIFS FMFNFOUT QSFTFOU

IZESPHFO ¤ ¤

OJUSPHFO ¤

PYZHFO ¤ ¤

1.6 $BMDVMBUF UIF NBTT PG PYZHFO QSFTFOU JO H PG $0 5IF SFMBUJWF NPMFDVMBS NBTT PG $0 JT 0G UIJT UIF BNPVOU DPOUSJCVUFE CZ UIF UXP PYZHFO BUPNT JT ¤

4P UIF GSBDUJPO PG UIF NBTT PG UIJT DPNQPVOE UIBU JT DPOUSJCVUFE CZ PYZHFO JT

5IFSFGPSF JO H PG $0 UIF BNPVOU PG PYZHFO JT

¤ H

1.7 8IBU NBTT PG )/0 DPOUBJOT H PG PYZHFO 5IF SFMBUJWF NPMFDVMBS NBTT PG )/0 JT &BDI NPMFDVMF DPOUBJOT UISFF PYZHFO BUPNT XJUI B UPUBM NBTT PG ¤ J F 5IF PYZHFO BOE UIF )/0 BSF JO UIF SBUJP 5IFSFGPSF UIF NBTT PG )/0 DPOUBJOJOH H PG PYZHFO JT

¤ H

Alternative method

¤ 5IF QFSDFOUBHF PG PYZHFO JO )/0 JT ¤

4P PG UIJT TBNQMF JT PYZHFO BOE IBT B NBTT PG H 8F OFFE UIFSFGPSF UP ê OE UIF NBTT PG XIJDI JT HJWFO CZ

¤ H

/PUF JO PSEFS UP PCUBJO UIJT BOTXFS NPSF ê HVSFT XFSF DBSSJFE UISPVHI PO UIF DBMDVMBUPS 1 STOICHIOMETRIC RELATIONSHIPS

13

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Test yourself 10 $BMDVMBUF UIF QFSDFOUBHF CZ NBTT PG PYZHFO JO FBDI PG UIF GPMMPXJOH DPNQPVOET a $ H 0) b $) $) $00) c $M 0 11 $BMDVMBUF UIF NBTT PG PYZHFO JO FBDI PG UIF GPMMPXJOH TBNQMFT a H PG $ H 0) b H PG 40 c H PG 1 0

12 'PS FBDI PG UIF GPMMPXJOH DPNQPVOET XPSL PVU UIF NBTT PG TVCTUBODF UIBU XJMM DPOUBJO H PG PYZHFO a $) 0) b 40 c 1 0

Empirical and molecular formulas " NPMFDVMBS GPSNVMB JT B XIPMF OVNCFS NVMUJQMF PG UIF FNQJSJDBM GPSNVMB 5IFSFGPSF JG UIF FNQJSJDBM GPSNVMB PG B DPNQPVOE JT $) UIF NPMFDVMBS GPSNVMB JT $) n J F $ H PS $ H PS $ H FUD

Empirical formula: the simplest whole number ratio of the elements present in a compound.

Molecular formula: the total number of atoms of each element present in a molecule of the compound. (The molecular formula is a multiple of the empirical formula.)

Worked examples 1.8 *G UIF NPMFDVMBS GPSNVMBT PG UXP DPNQPVOET BSF b 3F $M a $ H 0 XIBU BSF UIF FNQJSJDBM GPSNVMBT a 8F OFFE UP ê OE UIF TJNQMFTU SBUJP PG UIF FMFNFOUT QSFTFOU BOE UIFSFGPSF OFFE UP ê OE UIF IJHIFTU OVNCFS UIBU EJWJEFT FYBDUMZ JOUP UIF TVCTDSJQU PG FBDI FMFNFOU *O UIJT DBTF FBDI TVCTDSJQU DBO CF EJWJEFE CZ UXP BOE TP UIF FNQJSJDBM GPSNVMB JT $ H 0 b *O UIJT DBTF FBDI TVCTDSJQU JT EJWJTJCMF CZ UISFF BOE TP UIF FNQJSJDBM GPSNVMB JT 3F$M 1.9 5IF FNQJSJDBM GPSNVMB PG CFO[FOF JT $) (JWFO UIBU UIF NPMBS NBTT JT H NPM¢ XPSL PVU JUT NPMFDVMBS GPSNVMB 5IF NBTT PG UIF FNQJSJDBM GPSNVMB VOJU $) JT 5IF OVNCFS PG UJNFT UIBU UIF FNQJSJDBM GPSNVMB VOJU PDDVST JO UIF BDUVBM NPMFDVMF n JT HJWFO CZ SFMBUJWF NPMFDVMBS NBTT n

FNQJSJDBM GPSNVMB NBTT 5IFSFGPSF UIF NPMFDVMBS GPSNVMB JT $) XIJDI JT NPSF DPNNPOMZ XSJUUFO BT $ H

14

$IFNJDBM BOBMZTJT PG B TVCTUBODF DBO QSPWJEF UIF DPNQPTJUJPO CZ NBTT PG UIF DPNQPVOE 5IF FNQJSJDBM GPSNVMB DBO UIFO CF DBMDVMBUFE GSPN UIFTF EBUB *O PSEFS UP XPSL PVU UIF NPMFDVMBS GPSNVMB UIF SFMBUJWF NPMFDVMBS NBTT PG UIF DPNQPVOE JT BMTP SFRVJSFE

Worked examples 1.10 " DPNQPVOE IBT UIF GPMMPXJOH DPNQPTJUJPO CZ NBTT $ H ) H 0 H a $BMDVMBUF UIF FNQJSJDBM GPSNVMB PG UIF DPNQPVOE b *G UIF SFMBUJWF NPMFDVMBS NBTT PG UIF DPNQPVOE JT DBMDVMBUF UIF NPMFDVMBS GPSNVMB a 5IJT JT NPTU FBTJMZ EPOF CZ MBZJOH FWFSZUIJOH PVU JO B UBCMF C mass / g

H 0.681

O 0.137

0.181

divide by relative atomic mass to give number of moles

0.681 / 12.01

0.137 / 1.01

0.181 / 16.00

number of moles / mol

0.0567

0.136

0.0113

divide by smallest to get ratio

0.0567 / 0.0113

0.136 / 0.0113

0.0113 / 0.0113

ratio

5

12

1

5IFSFGPSF UIF FNQJSJDBM GPSNVMB JT $ H 0 b 5IF FNQJSJDBM GPSNVMB NBTT PG UIF DPNQPVOE JT 5IJT JT UIF TBNF BT UIF SFMBUJWF NPMFDVMBS NBTT BOE TP UIF NPMFDVMBS GPSNVMB JT UIF TBNF BT UIF FNQJSJDBM GPSNVMB $ H 0

1.11 *G B ë VPSJEF PG VSBOJVN DPOUBJOT VSBOJVN CZ NBTT XIBU JT JUT FNQJSJDBM GPSNVMB " VSBOJVN ë VPSJEF DPOUBJOT POMZ VSBOJVN BOE ë VPSJOF ë VPSJOF ¢ *U NBLFT OP EJÄ&#x201E; FSFODF IFSF UIBU UIF QFSDFOUBHF DPNQPTJUJPO JT HJWFO JOTUFBE PG UIF NBTT PG FBDI FMFNFOU QSFTFOU BT UIF QFSDFOUBHF JT UIF TBNF BT UIF NBTT QSFTFOU JO H U

F

percentage

67.62

32.38

mass in 100 g / g

67.62

32.38

divide by relative atomic mass to give number of moles

67.62 / 238.03

32.38 / 19.00

number of moles

0.2841

1.704

divide by smallest to get ratio

0.2841 / 0.2841

1.704 / 0.2841

ratio

1

6

5IFSF BSF UIFSFGPSF TJY ë VPSJOF BUPNT GPS FWFSZ VSBOJVN BUPN BOE UIF FNQJSJDBM GPSNVMB JT 6'

1 STOICHIOMETRIC RELATIONSHIPS

15

1.12 5IF FYQFSJNFOUBM TFU VQ TIPXO JO UIF ê HVSF DBO CF VTFE UP EFUFSNJOF UIF FNQJSJDBM GPSNVMB PG DPQQFS PYJEF 5IF GPMMPXJOH FYQFSJNFOUBM SFTVMUT XFSF PCUBJOFE Mass of empty dish / g

24.58

Mass of dish + copper oxide / g

30.12

Mass of dish + copper at end of experiment / g

29.00

$BMDVMBUF UIF FNQJSJDBM GPSNVMB PG UIF DPQQFS PYJEF BOE XSJUF BO FRVBUJPO GPS UIF SFBDUJPO

copper oxide power

excess hydrogen gas burning

stream of hydrogen gas

Hydrogen gas is passed over the heated copper oxide until all the copper oxide is reduced to copper.

NBTT PG DPQQFS PYJEF BU TUBSU ¢ H NBTT PG DPQQFS BU FOE ¢ H 5IF EJÄ&#x201E; FSFODF JO NBTT JT EVF UP UIF PYZHFO GSPN UIF DPQQFS PYJEF DPNCJOJOH XJUI UIF IZESPHFO NBTT PG PYZHFO JO DPQQFS PYJEF ¢ H 'SPN OPX PO UIF RVFTUJPO JT B TUSBJHIUGPSXBSE FNQJSJDBM GPSNVMB RVFTUJPO

NPM OVNCFS PG NPMFT PG DPQQFS

OVNCFS PG NPMFT PG PYZHFO NPM

*G FBDI OVNCFS PG NPMFT JT EJWJEFE CZ UIF TNBMMFS OVNCFS

Cu

O

UIF SBUJP PG DPQQFS UP PYZHFO JT BOE UIF FNQJSJDBM GPSNVMB JT $V0 5IF FRVBUJPO GPS UIF SFBDUJPO JT $V0 ) â&#x2020;&#x2019; $V ) 0

Composition by mass from combustion data Worked examples 1.13 "O PSHBOJD DPNQPVOE A DPOUBJOT POMZ DBSCPO BOE IZESPHFO 8IFO H PG A CVSOT JO FYDFTT PYZHFO H PG DBSCPO EJPYJEF BOE H PG XBUFS BSF GPSNFE $BMDVMBUF UIF FNQJSJDBM GPSNVMB y y 5IF FRVBUJPO GPS UIF SFBDUJPO JT PG UIF GPSN $xHy x

0 â&#x2020;&#x2019; x$0 H 0

"MM UIF $ JO UIF $0 DPNFT GSPN UIF IZESPDBSCPO A

OVNCFS PG NPMFT PG $0 NPM

&BDI $0 NPMFDVMF DPOUBJOT POF DBSCPO BUPN 5IFSFGPSF UIF OVNCFS PG NPMFT PG DBSCPO JO H PG UIF IZESPDBSCPO JT NPM 16

"MM UIF IZESPHFO JO UIF XBUFS DPNFT GSPN UIF IZESPDBSCPO A

NPM OVNCFS PG NPMFT PG ) 0

.PSF TJHOJê DBOU ê HVSFT BSF DBSSJFE UISPVHI JO TVCTFRVFOU DBMDVMBUJPOT

&BDI ) 0 NPMFDVMF DPOUBJOT UXP IZESPHFO BUPNT TP UIF OVNCFS PG NPMFT PG IZESPHFO JO H PG ) 0 JT ¤ NPM 5IFSFGPSF UIF OVNCFS PG NPMFT PG IZESPHFO JO H PG UIF IZESPDBSCPO JT NPM 5IF FNQJSJDBM GPSNVMB BOE NPMFDVMBS GPSNVMB DBO OPX CF DBMDVMBUFE C

H

number of moles

0.184

0.293

divide by smaller

0.184 / 0.184

0.293 / 0.184

ratio

1.00

1.60

5IF FNQJSJDBM GPSNVMB NVTU CF B SBUJP PG XIPMF OVNCFST BOE UIJT DBO CF PCUBJOFE CZ NVMUJQMZJOH FBDI OVNCFS CZ ê WF 5IFSFGPSF UIF FNQJSJDBM GPSNVMB JT $ H 1.14 "O PSHBOJD DPNQPVOE B DPOUBJOT POMZ DBSCPO IZESPHFO BOE PYZHFO 8IFO H PG B CVSOT JO FYDFTT PYZHFO H PG DBSCPO EJPYJEF BOE H PG XBUFS BSF GPSNFE a 8IBU JT UIF FNQJSJDBM GPSNVMB PG B b *G UIF SFMBUJWF NPMFDVMBS NBTT JT XIBU JT UIF NPMFDVMBS GPSNVMB PG B a 5IF EJÄ&#x2026; DVMUZ IFSF JT UIBU UIF NBTT PG PYZHFO JO B DBOOPU CF XPSLFE PVU JO UIF TBNF XBZ BT UIF QSFWJPVT FYBNQMF BT POMZ TPNF PG UIF PYZHFO JO UIF $0 BOE ) 0 DPNFT GSPN UIF PYZHFO JO B UIF SFTU DPNFT GSPN UIF PYZHFO JO XIJDI JU JT CVSOU

¤ H NBTT PG DBSCPO JO H PG $0

¤ H NBTT PG IZESPHFO JO H PG ) 0

/PUF BT UIFSF BSF ) BUPNT JO B XBUFS NPMFDVMF

NBTT PG PYZHFO JO H PG B JT ¢ ¢ H 5IF FNQJSJDBM GPSNVMB DBO OPX CF DBMDVMBUFE C

H

O

0.76

0.19

0.51

moles / mol 0.063

0.19

0.032

ratio

6

1

mass / g

2

5IFSFGPSF UIF FNQJSJDBM GPSNVMB JT $ H 0 b 5IF FNQJSJDBM GPSNVMB NBTT JT

5IFSFGPSF UIF NPMFDVMBS GPSNVMB JT $ H 0 J F $ H 0

1 STOICHIOMETRIC RELATIONSHIPS

17

?

Test yourself 13 8IJDI PG UIF GPMMPXJOH SFQSFTFOU FNQJSJDBM GPSNVMBT CO2 CH C2H4 C4H10 HO C3H8 H2O2 N2H4 H 2O CH3COOH C6H5CH3 PCl5 14 $PQZ UIF UBCMF CFMPX BOE DPNQMFUF JU XJUI UIF NPMFDVMBS GPSNVMBT PG UIF DPNQPVOET HJWFO UIF FNQJSJDBM GPSNVMBT BOE SFMBUJWF NPMFDVMBS NBTTFT Empirical formula

Relative molecular mass

HO

34.02

ClO3

166.90

CH2

84.18

BNH2

80.52

Molecular formula

16 *G BO PYJEF PG DIMPSJOF DPOUBJOT DIMPSJOF DBMDVMBUF JUT FNQJSJDBM GPSNVMB 17 " DPNQPVOE DPOUBJOT JPEJOF BOE PYZHFO $BMDVMBUF UIF FNQJSJDBM GPSNVMB PG UIF DPNQPVOE 18 8IFO H PG BO PSHBOJD DPNQPVOE D XIJDI DPOUBJOT POMZ DBSCPO IZESPHFO BOE PYZHFO JT CVSOU JO FYDFTT PYZHFO H PG DBSCPO EJPYJEF BOE H PG XBUFS BSF QSPEVDFE 8IBU JT UIF FNQJSJDBM GPSNVMB PG D 19 8IFO H PG BO JSPO PYJEF JT IFBUFE XJUI DBSCPO H PG JSPO JT QSPEVDFE $BMDVMBUF UIF FNQJSJDBM GPSNVMB PG UIF JSPO PYJEF

15 "OBMZTJT PG B TBNQMF PG BO PSHBOJD DPNQPVOE QSPEVDFE UIF GPMMPXJOH DPNQPTJUJPO C: 0.399 g H: 0.101 g a $BMDVMBUF UIF FNQJSJDBM GPSNVMB b (JWFO UIBU UIF SFMBUJWF NPMFDVMBS NBTT JT EFUFSNJOF UIF NPMFDVMBS GPSNVMB

Learning objectives

r 4PMWF QSPCMFNT JOWPMWJOH NBTTFT PG TVCTUBODFT r $BMDVMBUF UIF UIFPSFUJDBM BOE QFSDFOUBHF ZJFME JO B SFBDUJPO r 6OEFSTUBOE UIF UFSNT limiting reactant BOE reactant in excess BOE TPMWF QSPCMFNT JOWPMWJOH UIFTF

1.3 Reacting masses and volumes 1.3.1 Calculations involving moles and masses Conservation of mass 5IF GBDU UIBU NBTT JT DPOTFSWFE JO B DIFNJDBM SFBDUJPO DBO TPNFUJNFT CF VTFE UP XPSL PVU UIF NBTT PG QSPEVDU GPSNFE 'PS FYBNQMF JG H PG JSPO SFBDUT exactly BOE completely XJUI H PG TVMGVS H PG JSPO TVMê EF JT GPSNFE 'F T

4 T â&#x2020;&#x2019; 'F4 T

18

Worked example 1.15 $POTJEFS UIF DPNCVTUJPO PG CVUBOF $ H H

0 H â&#x2020;&#x2019; $0 H

) 0 M

H PG CVUBOF SFBDUT FYBDUMZ XJUI H PG PYZHFO UP QSPEVDF H PG DBSCPO EJPYJEF 8IBU NBTT PG XBUFS XBT QSPEVDFE 5IF NBTTFT HJWFO SFQSFTFOU BO FYBDU DIFNJDBM SFBDUJPO TP XF BTTVNF UIBU BMM UIF SFBDUBOUT BSF DPOWFSUFE UP QSPEVDUT 5IF UPUBM NBTT PG UIF SFBDUBOUT H 5IF UPUBM NBTT PG UIF QSPEVDUT NVTU BMTP CF H 5IFSFGPSF UIF NBTT PG XBUFS ¢ H

Using moles 8F PGUFO XBOU UP XPSL PVU UIF NBTT PG POF SFBDUBOU UIBU SFBDUT FYBDUMZ XJUI B DFSUBJO NBTT PG BOPUIFS SFBDUBOU m PS IPX NVDI QSPEVDU JT GPSNFE XIFO DFSUBJO NBTTFT PG SFBDUBOUT SFBDU 5IJT DBO CF EPOF CZ DBMDVMBUJOH UIF OVNCFST PG FBDI NPMFDVMF PS BUPN QSFTFOU JO B QBSUJDVMBS NBTT PS NVDI NPSF TJNQMZ CZ VTJOH UIF NPMF DPODFQU "T XF IBWF TFFO POF NPMF PG BOZ TVCTUBODF BMXBZT DPOUBJOT UIF TBNF OVNCFS PG QBSUJDMFT TP JG XF LOPX UIF OVNCFS PG NPMFT QSFTFOU JO B DFSUBJO NBTT PG SFBDUBOU XF BMTP LOPX UIF OVNCFS PG QBSUJDMFT BOE DBO UIFSFGPSF XPSL PVU XIBU NBTT PG BOPUIFS SFBDUBOU JU SFBDUT XJUI BOE IPX NVDI QSPEVDU JT GPSNFE

There are three main steps in a moles calculation.

1 Work out the number of moles of anything you can.

2 Use the chemical (stoichiometric) equation to work out the number of moles of the quantity you require.

3 Convert moles to the required quantity â&#x20AC;&#x201C; volume, mass etc.

Questions involving masses of substances

Worked examples 1.16 $POTJEFS UIF SFBDUJPO PG TPEJVN XJUI PYZHFO /B T

0 H â&#x2020;&#x2019; /B 0 T

a )PX NVDI TPEJVN SFBDUT FYBDUMZ XJUI H PG PYZHFO b 8IBU NBTT PG /B 0 JT QSPEVDFE a 4UFQ m UIF NBTT PG PYZHFO JT HJWFO TP UIF OVNCFS PG NPMFT PG PYZHFO DBO CF XPSLFE PVU ZPV DPVME VTF UIF USJBOHMF TIPXO IFSF

NPM OVNCFS PG NPMFT PG PYZHFO

/PUF UIF NBTT PG PYZHFO XBT HJWFO UP UISFF TJHOJê DBOU ê HVSFT TP BMM TVCTFRVFOU BOTXFST BSF BMTP HJWFO UP UISFF TJHOJê DBOU ê HVSFT

mass of substance

number of moles

molar mass

1 STOICHIOMETRIC RELATIONSHIPS

19

4UFQ m UIF DPFÄ&#x2026; DJFOUT JO UIF DIFNJDBM TUPJDIJPNFUSJD FRVBUJPO UFMM VT UIBU NPM 0 SFBDUT XJUI NPM TPEJVN 5IFSFGPSF NPM 0 SFBDUT XJUI ¤ NPM TPEJVN J F NPM TPEJVN 4UFQ m DPOWFSU UIF OVNCFS PG NPMFT UP UIF SFRVJSFE RVBOUJUZ NBTT JO UIJT DBTF NBTT PG TPEJVN ¤ H /PUF UIF NBTT PG TPEJVN JT XPSLFE PVU CZ NVMUJQMZJOH UIF NBTT PG POF NPMF CZ UIF OVNCFS PG NPMFT m UIF OVNCFS PG NPMFT JT not NVMUJQMJFE CZ UIF NBTT PG GPVS TPEJVN BUPNT m UIF GPVS XBT BMSFBEZ UBLFO JOUP BDDPVOU XIFO NPM XBT NVMUJQMJFE CZ UP HJWF UIF OVNCFS PG NPMFT PG TPEJVN b 'SPN UIF DPFÄ&#x2026; DJFOUT JO UIF FRVBUJPO XF LOPX UIBU NPM 0 SFBDUT XJUI NPM TPEJVN UP QSPEVDF NPM /B 0 5IFSFGPSF NPM 0 SFBDUT XJUI NPM TPEJVN UP HJWF ¤ NPM /B 0 J F NPM /B 0 5IF NPMBS NBTT PG /B 0 H NPM¢ 4P NBTT PG /B 0 ¤ H "MUFSOBUJWFMZ UIF NBTT PG /B 0 DBO CF XPSLFE PVU VTJOH UIF JEFB PG DPOTFSWBUJPO PG NBTT J F NBTT PG /B 0 NBTT PG 0 NBTT PG /B

Exam tip .BTTFT NBZ BMTP CF HJWFO JO LJMPHSBNT PS UPOOFT LH H UPOOF ¤ H #FGPSF XPSLJOH PVU UIF OVNCFS PG NPMFT ZPV NVTU DPOWFSU UIF NBTT UP HSBNT 5P DPOWFSU LJMPHSBNT UP HSBNT NVMUJQMZ CZ UP DPOWFSU UPOOFT UP HSBNT NVMUJQMZ UIF NBTT CZ ¤

1.17 $POTJEFS UIF GPMMPXJOH FRVBUJPO /) $V0 â&#x2020;&#x2019; / ) 0 $V *G H PG BNNPOJB /) JT SFBDUFE XJUI FYDFTT $V0 DBMDVMBUF UIF NBTT PG DPQQFS QSPEVDFE 5IF $V0 JT JO FYDFTT TP UIFSF JT NPSF UIBO FOPVHI UP SFBDU XJUI BMM UIF /) 5IJT NFBOT UIBU XF EP OPU OFFE UP XPSSZ BCPVU UIF OVNCFS PG NPMFT PG $V0 4UFQ m UIF OVNCFS PG NPMFT PG /) DBO CF DBMDVMBUFE

NPM /)

4UFQ m UXP NPMFT PG /) QSPEVDF UISFF NPMFT PG DPQQFS TP NPM /) QSPEVDFT ¤ NPM DPQQFS J F NPM DPQQFS 5IF OVNCFS PG NPMFT PG DPQQFS JT UIFSFGPSF UJNFT UIF OVNCFS PG NPMFT PG /) 4UFQ m UIF NBTT PG NPM DPQQFS H TP UIF NBTT PG DPQQFS QSPEVDFE ¤ H

20

Formula for solving moles questions involving masses "O BMUFSOBUJWF XBZ PG EPJOH UIFTF RVFTUJPOT JT UP VTF B GPSNVMB m1 m2 = n1M1 n2M2 where m1 = mass of first substance n1 = coefficient of first substance (number in front in the chemical equation) M1 = molar mass of first substance

Worked example 1.18 5IF GPMMPXJOH FRVBUJPO SFQSFTFOUT UIF DPNCVTUJPO PG CVUBOF $ H H

0 H â&#x2020;&#x2019; $0 H

) 0 M

*G H PG CVUBOF JT VTFE DBMDVMBUF UIF NBTT PG PYZHFO SFRVJSFE GPS BO FYBDU SFBDUJPO 8F XJMM DBMM CVUBOF TVCTUBODF BOE PYZHFO TVCTUBODF JU EPFTO U NBUUFS XIJDI ZPV DBMM XIBU CVU ZPV IBWF UP CF DPOTJTUFOU

m H n M H NPM¢

m n M H NPM¢

m1 m2

n1M1 n2M2

m

¤ ¤ 5IF FRVBUJPO DBO CF SFBSSBOHFE ¤ ¤

m ¤ 5IFSFGPSF UIF NBTT PG PYZHFO SFRVJSFE GPS BO FYBDU SFBDUJPO JT H

1 STOICHIOMETRIC RELATIONSHIPS

21

?

Test yourself 20 a )PX NBOZ NPMFT PG IZESPHFO HBT BSF QSPEVDFE XIFO NPM TPEJVN SFBDU XJUI FYDFTT XBUFS /B ) 0 â&#x2020;&#x2019; /B0) ) b )PX NBOZ NPMFT PG 0 SFBDU XJUI NPM $ H $ H 0 â&#x2020;&#x2019; $0 ) 0 c )PX NBOZ NPMFT PG ) 4 BSF GPSNFE XIFO NPM )$M SFBDU XJUI FYDFTT 4C S Sb S )$M â&#x2020;&#x2019; 4C$M ) S d )PX NBOZ NPMFT PG PYZHFO BSF GPSNFE XIFO NPM ,$M0 SFBDU ,$M0 T â&#x2020;&#x2019; ,$M T

0 H

e )PX NBOZ NPMFT PG JSPO BSF GPSNFE XIFO NPM $0 SFBDU XJUI FYDFTT JSPO PYJEF 'F 0 $0 â&#x2020;&#x2019; 'F $0 f )PX NBOZ NPMFT PG IZESPHFO XPVME CF SFRVJSFE UP NBLF ¤ ¢ NPM /) / ) â&#x2020;&#x2019; /)

5IF GBDU UIBU B UIFPSZ DBO FYQMBJO FYQFSJNFOUBM PCTFSWBUJPOT EPFT OPU OFDFTTBSJMZ NBLF JU DPSSFDU 5IF FYQMBOBUJPOT QSFTFOUFE JO UIJT CPPL ê U JO XJUI FYQFSJNFOUBM PCTFSWBUJPOT CVU UIJT EPFT OPU NFBO UIBU UIFZ BSF AUSVF m UIFZ KVTU SFQSFTFOU PVS JOUFSQSFUBUJPO PG UIF EBUB BU UIJT TUBHF JO UJNF &BDI HFOFSBUJPO PG TDJFOUJTUT CFMJFWFT UIBU UIFZ BSF QSFTFOUJOH B USVF EFTDSJQUJPO PG SFBMJUZ CVU JT JU QPTTJCMF GPS NPSF UIBO POF FYQMBOBUJPO UP ê U UIF GBDUT :PV PS JOEFFE * NBZ OPU CF BCMF UP UIJOL PG B CFUUFS FYQMBOBUJPO UP ê U NBOZ PG UIF FYQFSJNFOUBM PCTFSWBUJPOT JO NPEFSO TDJFODF CVU UIBU EPFT OPU NFBO UIBU UIFSF JTO U POF $POTJEFS UIF GPMMPXJOH USJWJBM FYBNQMF &YQFSJNFOUBMMZ XIFO LH PG DBMDJVN DBSCPOBUF $B$0 JT IFBUFE LH PG DBSCPO EJPYJEF $0 JT PCUBJOFE 5IF GPMMPXJOH DBMDVMBUJPO DBO CF DBSSJFE PVU UP FYQMBJO UIJT

22

21 a $BMDVMBUF UIF NBTT PG BSTFOJD *** DIMPSJEF QSPEVDFE XIFO H PG BSTFOJD SFBDUT XJUI FYDFTT DIMPSJOF BDDPSEJOH UP UIF FRVBUJPO "T $M â&#x2020;&#x2019; "T$M b 8IBU NBTT PG TVMGVS JT QSPEVDFE XIFO H PG JSPO *** TVMê EF JT SFBDUFE XJUI FYDFTT PYZHFO 'F S 0 â&#x2020;&#x2019; 'F 0 4 c $BMDVMBUF UIF NBTT PG JPEJOF UIBU NVTU CF SFBDUFE XJUI FYDFTT QIPTQIPSVT UP QSPEVDF H PG QIPTQIPSVT *** JPEJEF BDDPSEJOH UP UIF FRVBUJPO CFMPX 1 * â&#x2020;&#x2019; 1* d $POTJEFS UIF SFBDUJPO TIPXO CFMPX 8IBU NBTT PG 4$M NVTU CF SFBDUFE XJUI FYDFTT /B' UP QSPEVDF H PG /B$M 4$M /B' â&#x2020;&#x2019; 4 $M 4' /B$M

5IF FRVBUJPO GPS UIF SFBDUJPO JT $B$0 â&#x2020;&#x2019; $B0 $0

OVNCFS PG NPMFT PG $B$0

¤

NPM

5XP NPMFT PG $B$0 QSPEVDFT UXP NPMFT PG $0 5IF NBTT PG UXP NPMFT PG $0 JT ¤ ¤

LH )PQFGVMMZ ZPV DBO TFF TPNF NJTUBLFT JO UIJT DBMDVMBUJPO CVU UIF SFTVMU JT XIBU XF HPU FYQFSJNFOUBMMZ *U JT BMTP JOUFSFTUJOH UP OPUF UIBU JG JO ZPVS *# FYBNJOBUJPO ZPV IBE KVTU XSJUUFO EPXO UIF ê OBM BOTXFS ZPV XPVME QSPCBCMZ IBWF HPU GVMM NBSLT

Calculating the yield of a chemical reaction *O BOZ DPNNFSDJBM QSPDFTT JU JT WFSZ JNQPSUBOU UP LOPX UIF yield UIF BNPVOU PG EFTJSFE QSPEVDU PG B DIFNJDBM SFBDUJPO 'PS JOTUBODF JG B QBSUJDVMBS QSPDFTT GPS UIF QSFQBSBUJPO PG B ESVH JOWPMWFT GPVS TFQBSBUF TUFQT BOE UIF ZJFME PG FBDI TUFQ JT JU JT QSPCBCMZ RVJUF B QSPNJTJOH TZOUIFUJD SPVUF UP UIF ESVH *G IPXFWFS UIF ZJFME PG FBDI TUFQ JT POMZ UIFO JU JT MJLFMZ UIBU UIF DPNQBOZ XPVME MPPL GPS B NPSF FÄ&#x2026; DJFOU TZOUIFUJD QSPDFTT 5IF ZJFME PG B DIFNJDBM SFBDUJPO JT VTVBMMZ RVPUFE BT B QFSDFOUBHF m UIJT HJWFT NPSF JOGPSNBUJPO UIBO KVTU RVPUJOH UIF ZJFME PG UIF QSPEVDU BT B NBTT $POTJEFS UIF QSFQBSBUJPO PG EJCSPNPFUIBOF $ H #S

$ H H

#S M â&#x2020;&#x2019; $ H #S M

H PG FUIFOF $ H XJMM SFBDU FYBDUMZ XJUI H PG CSPNJOF 5IF theoretical yield GPS UIJT SFBDUJPO JT H m UIJT JT UIF NBYJNVN QPTTJCMF ZJFME UIBU DBO CF PCUBJOFE 5IF actual yield PG $ H #S NBZ CF H % yield =

ZJFME ¤

actual yield Ã&#x2014; 100 theoretical yield

Worked example 1.19 $ H 0) M

$) $00) M â&#x2020;&#x2019; $) $00$ H M

) 0 M

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OVNCFS PG NPMFT PG $ H 0) NPM

5IF $) $00) JT JO FYDFTT J F NPSF UIBO FOPVHI JT QSFTFOU UP SFBDU XJUI BMM UIF $ H 0) 5IJT NFBOT UIBU XF EP OPU OFFE UP XPSSZ BCPVU UIF OVNCFS PG NPMFT PG $) $00)

5IF DIFNJDBM FRVBUJPO UFMMT VT UIBU NPM $ H 0) QSPEVDFT NPM $) $00$ H 5IFSFGPSF NPM $ H 0) QSPEVDFT NPM $) $00$ H 5IF NPMBS NBTT PG $) $00$ H H NPM¢ 5IF NBTT PG FUIZM FUIBOPBUF $) $00$ H ¤ H 4P UIF UIFPSFUJDBM ZJFME JT H 5IF BDUVBM ZJFME JT H HJWFO JO UIF RVFTUJPO

ZJFME ¤

5IF QFSDFOUBHF ZJFME PG $) $00$ H JT

1 STOICHIOMETRIC RELATIONSHIPS

23

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Test yourself 22 $BMDVMBUF UIF QFSDFOUBHF ZJFME JO FBDI PG UIF GPMMPXJOH SFBDUJPOT a 8IFO H PG 40 JT IFBUFE XJUI FYDFTT PYZHFO H PG 40 JT PCUBJOFE 40 0 â&#x2020;&#x2019; 40 b 8IFO H PG BSTFOJD JT IFBUFE JO FYDFTT PYZHFO H PG "T 0 JT QSPEVDFE "T 0 â&#x2020;&#x2019; "T 0

c 8IFO H PG $ H SFBDUT XJUI FYDFTT CSPNJOF H PG $) #S$) #S JT QSPEVDFE $ H #S â&#x2020;&#x2019; $) #S$) #S

Limiting reactant 7FSZ PGUFO XF EP OPU VTF FYBDU RVBOUJUJFT JO B DIFNJDBM SFBDUJPO CVU SBUIFS XF VTF BO FYDFTT PG POF PS NPSF SFBDUBOUT 0OF SFBDUBOU JT UIFSFGPSF VTFE VQ CFGPSF UIF PUIFST BOE JT DBMMFE UIF limiting reactant 8IFO UIF MJNJUJOH SFBDUBOU JT DPNQMFUFMZ VTFE VQ UIF SFBDUJPO TUPQT 'JHVSF 1.9 JMMVTUSBUFT UIF JEFB PG B MJNJUJOH SFBDUBOU BOE TIPXT IPX UIF QSPEVDUT PG UIF SFBDUJPO EFQFOE PO XIJDI SFBDUBOU JT MJNJUJOH

Figure 1.9 The reaction between magnesium and hydrochloric acid. In each test tube a small amount of universal indicator has been added. a In this test tube, the magnesium is in excess and the reaction finishes when the hydrochloric acid runs out.

There is still magnesium left over at the end, and the solution is no longer acidic. b In this test tube, the hydrochloric acid is in excess.

The magnesium is the limiting reactant, and the reaction stops when the magnesium has been used up. The solution is still acidic at the end.

HCI(aq)

HCI(aq)

Mg a

b

Worked examples 1.20 $POTJEFS UIF SFBDUJPO CFUXFFO NBHOFTJVN BOE OJUSPHFO .H T

/ H â&#x2020;&#x2019; .H / T

H PG NBHOFTJVN JT SFBDUFE XJUI H PG OJUSPHFO 8IJDI JT UIF MJNJUJOH SFBDUBOU OVNCFS PG NPMFT PG NBHOFTJVN

NPM

OVNCFS PG NPMFT PG / NPM

5IF FRVBUJPO UFMMT VT UIBU NPM NBHOFTJVN SFBDUT XJUI NPM / 4P NPM NBHOFTJVN SFBDUT XJUI

NPM / J F NPM /

24

5IFSFGPSF GPS BO FYBDU SFBDUJPO NPM / BSF SFRVJSFE UP SFBDU XJUI NPM NBHOFTJVN )PXFWFS NPM / BSF VTFE XIJDI JT NPSF UIBO FOPVHI UP SFBDU 5IJT NFBOT UIBU / JT JO FYDFTT CFDBVTF UIFSF JT NPSF UIBO FOPVHI UP SFBDU XJUI BMM UIF NBHOFTJVN QSFTFOU .BHOFTJVN JT UIFSFGPSF UIF MJNJUJOH SFBDUBOU 5IJT DBO BMTP CF TFFO GSPN XPSLJOH XJUI UIF OVNCFS PG NPMFT PG / m NPM / XBT VTFE JO UIJT SFBDUJPO 5IJT OVNCFS PG NPMFT PG / XPVME SFRVJSF ¤ NPM NBHOFTJVN GPS BO FYBDU SFBDUJPO J F NPM NBHOFTJVN )PXFWFS POMZ NPM NBHOFTJVN BSF QSFTFOU UIFSFGPSF UIF NBHOFTJVN XJMM SVO PVU CFGPSF BMM UIF / IBT SFBDUFE Exam tip *G UIF OVNCFS PG NPMFT PG FBDI SFBDUBOU JT EJWJEFE CZ JUT DPFÄ&#x2026; DJFOU JO UIF TUPJDIJPNFUSJD FRVBUJPO UIF TNBMMFTU OVNCFS JOEJDBUFT UIF MJNJUJOH SFBDUBOU 1.21 $POTJEFS UIF SFBDUJPO CFUXFFO TVMGVS BOE ë VPSJOF 4 T

' H â&#x2020;&#x2019; 4' H

H PG TVMGVS SFBDUT XJUI H PG ë VPSJOF a 8IJDI JT UIF MJNJUJOH SFBDUBOU b 8IBU NBTT PG TVMGVS 7* ë VPSJEF JT GPSNFE c 8IBU NBTT PG UIF SFBDUBOU JO FYDFTT JT MFGU BU UIF FOE a OVNCFS PG NPMFT PG TVMGVS

NPM

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"MUFSOBUJWFMZ XF DBO SFBTPO GSPN UIF DIFNJDBM FRVBUJPO UIBU NPM ' TIPVME SFBDU XJUI NPM TVMGVS J F NPM EJWJEFE CZ UISFF 5IFSF JT NPSF UIBO NPM TVMGVS QSFTFOU TP TVMGVS JT QSFTFOU JO FYDFTT BOE ' JT UIF MJNJUJOH SFBDUBOU 'PS UIF SFTU PG UIF RVFTUJPO XF NVTU XPSL XJUI UIF MJNJUJOH SFBDUBOU b 8IFO UIF MJNJUJOH SFBDUBOU JT VTFE VQ DPNQMFUFMZ UIF SFBDUJPO TUPQT 5IJT NFBOT UIBU UIF BNPVOU PG QSPEVDU GPSNFE JT EFUFSNJOFE CZ UIF BNPVOU PG UIF MJNJUJOH SFBDUBOU XF TUBSUFE XJUI 'SPN UIF DIFNJDBM FRVBUJPO NPM ' QSPEVDFT NPM 4' J F NPM EJWJEFE CZ UISFF

NPMBS NBTT PG 4' H NPM¢ NBTT PG 4' GPSNFE ¤ H c 'SPN UIF DIFNJDBM FRVBUJPO NPM ' SFBDUT XJUI NPM TVMGVS J F NPM TVMGVS EJWJEFE CZ UISFF 0SJHJOBMMZ UIFSF XFSF NPM TVMGVS QSFTFOU UIFSFGPSF UIF OVNCFS PG NPMFT PG TVMGVS MFGU BU UIF FOE PG UIF SFBDUJPO JT â&#x2C6;&#x2019; 5IF NBTT PG TVMGVS MFGU BU UIF FOE PG UIF SFBDUJPO JT ¤ H Exam tip 5P EP B NPMFT RVFTUJPO ZPV OFFE UP LOPX UIF NBTT PG KVTU POF PG UIF SFBDUBOUT *G ZPV BSF HJWFO UIF NBTTFT PG NPSF UIBO POF SFBDUBOU ZPV NVTU DPOTJEFS UIBU POF PG UIFTF SFBDUBOUT XJMM CF UIF MJNJUJOH SFBDUBOU BOE ZPV NVTU VTF UIJT POF GPS BMM DBMDVMBUJPOT

1 STOICHIOMETRIC RELATIONSHIPS

25

1.22 'PS UIF SFBDUJPO 'F$S 0 /B $0 0 â&#x2020;&#x2019; /B $S0 'F 0 $0 UIFSF BSF H PG FBDI SFBDUBOU BWBJMBCMF 8IJDI JT UIF MJNJUJOH SFBDUBOU 5IJT RVFTUJPO DPVME CF EPOF CZ XPSLJOH PVU UIF OVNCFS PG NPMFT PG FBDI SFBDUBOU BOE UIFO DPNQBSJOH UIFN CVU UIFSF JT B TIPSUDVU m UP XPSL PVU UIF NBTTFT PG FBDI TVCTUBODF JG NPMBS RVBOUJUJFT SFBDUFE

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¤

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Nature of science 4DJFODF JT B DPOTUBOUMZ DIBOHJOH CPEZ PG LOPXMFEHF 4DJFOUJTUT UBLF FYJTUJOH LOPXMFEHF BOE USZ UP CVJME PO JU UP JNQSPWF UIFPSJFT TP UIBU UIFZ BSF NPSF XJEFMZ BQQMJDBCMF BOE IBWF CFUUFS FYQMBOBUPSZ QPXFS 5IF DPODFQU PG UIF NPMF EFWFMPQFE GSPN UIF DPODFQU PG FRVJWBMFOU XFJHIU

?

26

Test yourself 23 8IBU JT UIF MJNJUJOH SFBDUBOU JO FBDI PG UIF GPMMPXJOH SFBDUJPOT a NPM 4C 0 SFBDU XJUI NPM ) 40 Sb 0 ) 40 â&#x2020;&#x2019; 4C 40

) 0 b NPM "T$M SFBDU XJUI NPM ) 0 "T$M ) 0 â&#x2020;&#x2019; "T 0 )$M c NPM DPQQFS SFBDU XJUI NPM EJMVUF )/0 BDDPSEJOH UP UIF FRVBUJPO $V )/0 â&#x2020;&#x2019; $V /0

) 0 /0 d NPM /B$M SFBDU XJUI NPM .O0 BOE NPM ) 40 /B$M .O0 ) 40 â&#x2020;&#x2019; /B 40 .O40 ) 0 $M

24 #PSPO DBO CF QSFQBSFE CZ SFBDUJOH # 0 XJUI NBHOFTJVN BU IJHI UFNQFSBUVSFT # 0 .H â&#x2020;&#x2019; # .H0 8IBU NBTT PG CPSPO JT PCUBJOFE JG H # 0 BSF SFBDUFE XJUI H NBHOFTJVN 25 *SPO *** PYJEF SFBDUT XJUI DBSCPO UP QSPEVDF JSPO 'F 0 $ â&#x2020;&#x2019; 'F $0 8IBU NBTT PG JSPO JT PCUBJOFE JG UPOOFT PG 'F 0 BSF SFBDUFE XJUI UPOOF PG DBSCPO

1.3.2 Calculations involving volumes of gases

Learning objectives

Real gases and ideal gases

r 6OEFSTUBOE "WPHBESP T MBX

"O AJEFBM HBT JT B DPODFQU JOWFOUFE CZ TDJFOUJTUT UP BQQSPYJNBUF NPEFM

UIF CFIBWJPVS PG SFBM HBTFT 6OEFS OPSNBM DPOEJUJPOT BSPVOE L1B <BQQSPYJNBUFMZ BUNPTQIFSF> QSFTTVSF BOE $ SFBM HBTFT TVDI BT IZESPHFO CFIBWF QSFUUZ NVDI MJLF JEFBM HBTFT BOE UIF BQQSPYJNBUJPOT XPSL WFSZ XFMM 5XP BTTVNQUJPOT XF NBLF XIFO EFê OJOH UIF ideal gas BSF UIBU UIF NPMFDVMFT UIFNTFMWFT IBWF OP WPMVNF UIFZ BSF QPJOU NBTTFT BOE UIBU OP GPSDFT FYJTU CFUXFFO UIFN FYDFQU XIFO UIFZ DPMMJEF *G XF JNBHJOF DPNQSFTTJOH B SFBM HBT UP B WFSZ IJHI QSFTTVSF UIFO UIF QBSUJDMFT XJMM CF NVDI DMPTFS UPHFUIFS BOE VOEFS UIFTF DPOEJUJPOT UIF GPSDFT CFUXFFO NPMFDVMFT BOE UIF WPMVNFT PDDVQJFE CZ UIF NPMFDVMFT XJMM CF TJHOJê DBOU 5IJT NFBOT UIBU XF DBO OP MPOHFS JHOPSF UIFTF GBDUPST BOE UIF CFIBWJPVS PG UIF HBT XJMM EFWJBUF TJHOJê DBOUMZ GSPN PVS JEFBM HBT NPEFM 5IJT XJMM BMTP CF UIF DBTF BU WFSZ MPX UFNQFSBUVSFT XIFO UIF NPMFDVMFT BSF NPWJOH NPSF TMPXMZ 6OEFS DPOEJUJPOT PG WFSZ MPX UFNQFSBUVSF BOE WFSZ IJHI QSFTTVSF B HBT JT BQQSPBDIJOH UIF MJRVJE TUBUF BOE XJMM CF MFBTU MJLF PVS QSFEJDUJPOT GPS BO JEFBM HBT 5IF JEFB UIBU UIF WPMVNFT PG UIF JOEJWJEVBM HBT NPMFDVMFT BSF [FSP TP JU NBLFT OP EJĄ FSFODF JG UIF HBT JT ) PS /) BOE UIBU UIFSF BSF OP GPSDFT CFUXFFO UIF NPMFDVMFT BHBJO OP EJĄ FSFODF CFUXFFO /) BOE H NFBOT UIBU BMM JEFBM HBTFT NVTU CFIBWF JO UIF TBNF XBZ 5IJT NFBOT UIBU UIF WPMVNF PDDVQJFE CZ B HBT BU B DFSUBJO UFNQFSBUVSF BOE QSFTTVSF EFQFOET POMZ PO UIF OVNCFS PG NPMFDVMFT QSFTFOU BOE OPU PO UIF OBUVSF PG UIF HBT *O PUIFS XPSET BU B DFSUBJO UFNQFSBUVSF BOE QSFTTVSF UIF WPMVNF PG B HBT JT QSPQPSUJPOBM UP UIF OVNCFS PG NPMFT PG UIF HBT

BOE VTF JU UP DBMDVMBUF SFBDUJOH WPMVNFT PG HBTFT r 6TF UIF NPMBS WPMVNF PG B HBT JO DBMDVMBUJPOT BU TUBOEBSE UFNQFSBUVSF BOE QSFTTVSF r 6OEFSTUBOE UIF SFMBUJPOTIJQT CFUXFFO QSFTTVSF WPMVNF BOE UFNQFSBUVSF GPS BO JEFBM HBT r 4PMWF QSPCMFNT VTJOH UIF FRVBUJPO P V P V T T r 4PMWF QSPCMFNT VTJOH UIF JEFBM HBT FRVBUJPO Gases deviate most from ideal behaviour at high pressure and low temperature.

Volume of gas ∝ number of moles of the gas

Using volumes of gases Avogadro’s law: equal volumes of ideal gases measured at the same temperature and pressure contain the same number of molecules. *O PUIFS XPSET DN PG ) DPOUBJOT UIF TBNF OVNCFS PG NPMFDVMFT BU $ BOE L1B BT DN PG /) JG XF BTTVNF UIBU UIFZ CPUI CFIBWF BT JEFBM HBTFT 6OEFS UIF TBNF DPOEJUJPOT DN PG $0 XPVME DPOUBJO IBMG BT NBOZ NPMFDVMFT 5IJT NFBOT UIBU WPMVNFT DBO CF VTFE EJSFDUMZ JOTUFBE PG NPMFT JO FRVBUJPOT JOWPMWJOH HBTFT H H

$M H → )$M H

5IF BCPWF FRVBUJPO UFMMT VT UIBU POF NPMF PG ) SFBDUT XJUI POF NPMF PG $M UP HJWF UXP NPMFT PG )$M 0S POF WPMVNF PG ) SFBDUT XJUI POF WPMVNF PG $M UP HJWF UXP WPMVNFT PG )$M J F DN PG ) SFBDUT XJUI DN PG $M UP HJWF DN PG )$M

DN JT UIF TBNF BT NM

5IF JEFBM HBT DPODFQU JT BO BQQSPYJNBUJPO XIJDI JT VTFE UP NPEFM UIF CFIBWJPVS PG SFBM HBTFT 8IZ EP XF MFBSO BCPVU JEFBM HBTFT XIFO UIFZ EP OPU FYJTU 8IBU JNQMJDBUJPOT EPFT UIF JEFBM HBT DPODFQU IBWF PO UIF MJNJUT PG LOPXMFEHF HBJOFE GSPN UIJT DPVSTF

1 STOICHIOMETRIC RELATIONSHIPS

27

Worked examples *O CPUI PG UIFTF XPSLFE FYBNQMFT BTTVNF UIBU BMM HBTFT CFIBWF BT JEFBM HBTFT BOE UIBU BMM NFBTVSFNFOUT BSF NBEF VOEFS UIF TBNF DPOEJUJPOT PG UFNQFSBUVSF BOE QSFTTVSF 1.23 $POTJEFS UIF GPMMPXJOH SFBDUJPO GPS UIF TZOUIFTJT PG NFUIBOPM $0 H

) H â&#x2020;&#x2019; $) 0) H

a 8IBU WPMVNF PG ) SFBDUT FYBDUMZ XJUI EN PG $0 b 8IBU WPMVNF PG $) 0) JT QSPEVDFE a 'SPN UIF FRVBUJPO XF LOPX UIBU NPM $0 SFBDUT XJUI NPM ) 5IFSFGPSF POF WPMVNF PG $0 SFBDUT XJUI UXP WPMVNFT PG ) m EN PG $0 SFBDUT XJUI ¤ J F EN PG ) b 0OF WPMVNF PG $0 QSPEVDFT POF WPMVNF PG $) 0) 5IFSFGPSF UIF WPMVNF PG $) 0) QSPEVDFE JT EN 1.24 *G DN PG PYZHFO SFBDUT XJUI DN PG NFUIBOF JO UIF GPMMPXJOH SFBDUJPO IPX NVDI PYZHFO XJMM CF MFGU BU UIF FOE PG UIF SFBDUJPO $) H

0 H â&#x2020;&#x2019; $0 H

) 0 M

'SPN UIF FRVBUJPO XF LOPX UIBU NPM $) SFBDUT XJUI NPM 0 5IFSFGPSF POF WPMVNF PG $) SFBDUT XJUI UXP WPMVNFT PG 0 m TP DN PG $) SFBDUT XJUI ¤ J F DN PG 0 5IF PSJHJOBM WPMVNF PG 0 XBT DN UIFSFGPSF JG DN SFBDUFE UIF WPMVNF PG PYZHFO HBT MFGU PWFS BU UIF FOE PG UIF SFBDUJPO XPVME CF â&#x2C6;&#x2019; DN

STP = standard temperature and pressure = 273 K, 100 kPa (1 bar) L1B Ã&#x2014; 1B

volume in dm3

number of moles

molar volume (22.7 dm3 molâ&#x20AC;&#x201C;1)

Figure 1.10 The relationship between the number of moles of a gas and its volume.

28

Converting volumes of gases to number of moles #FDBVTF UIF WPMVNF PDDVQJFE CZ BO JEFBM HBT EFQFOET POMZ PO UIF OVNCFS PG QBSUJDMFT QSFTFOU BTTVNJOH UIBU QSFTTVSF BOE UFNQFSBUVSF BSF DPOTUBOU BOE OPU PO UIF OBUVSF PG UIF QBSUJDMFT UIF WPMVNF PDDVQJFE CZ POF NPMF PG BOZ JEFBM HBT VOEFS B DFSUBJO TFU PG DPOEJUJPOT XJMM BMXBZT CF UIF TBNF 5IF WPMVNF PDDVQJFE CZ POF NPMF PG B HBT VOEFS DFSUBJO DPOEJUJPOT JT DBMMFE UIF molar volume molar volume of an ideal gas at STP = 22.7 dm3 molâ&#x2C6;&#x2019;1 or 2.27 Ã&#x2014; 10â&#x2C6;&#x2019;2 m3 molâ&#x2C6;&#x2019;1 5IJT NFBOT UIBU VOEFS UIF TBNF TFU PG DPOEJUJPOT UIF WPMVNF PDDVQJFE CZ POF NPMF PG /) JT UIF TBNF BT UIF WPMVNF PDDVQJFE CZ POF NPMF PG $0 BOE POF NPMF PG ) BOE UIJT WPMVNF JT EN BU 451 5IF SFMBUJPOTIJQ CFUXFFO UIF OVNCFS PG NPMFT PG B HBT BOE JUT WPMVNF JT

OVNCFS PG NPMFT

WPMVNF NPMBS WPMVNF

5IJT JT TVNNBSJTFE JO 'JHVSF 1.10

5IF BCTPMVUF PS ,FMWJO TDBMF PG UFNQFSBUVSF TUBSUT BU BCTPMVUF [FSP XIJDI JT UIF MPXFTU UFNQFSBUVSF QPTTJCMF *U JT UIF UFNQFSBUVSF BU XIJDI FWFSZUIJOH XPVME CF JO JUT MPXFTU FOFSHZ TUBUF Absolute zero DPSSFTQPOET UP , PS ¢ $ VTVBMMZ UBLFO BT ¢ $ BOE JT BMTP UIF UFNQFSBUVSF BU XIJDI UIF WPMVNF PG BO JEFBM HBT XPVME CF [FSP *U JT OPU QPTTJCMF UP BDUVBMMZ SFBDI BCTPMVUF [FSP CVU TDJFOUJTUT IBWF NBOBHFE UP HFU WFSZ DMPTF m BCPVU OBOPLFMWJO

A change of 1 °C is the same as a change of 1 K, and 0 °C is equivalent to 273 K 5P DPOWFSU $ UP , BEE F H $ JT FRVJWBMFOU UP J F , 5P DPOWFSU , UP $ TVCUSBDU F H , JT FRVJWBMFOU UP ¢ J F $ 7PMVNFT PG HBTFT BSF PGUFO HJWFO JO DN BOE TP JU JT JNQPSUBOU UP LOPX IPX UP DPOWFSU CFUXFFO DN BOE EN #FDBVTF EN MJUSF JT FRVJWBMFOU UP DN UP DPOWFSU DN UP EN XF EJWJEF CZ UP HP GSPN DN UP EN

5IF DPOWFSTJPO JT TIPXO JO 'JHVSF 1.11

5IF ,FMWJO TDBMF PG UFNQFSBUVSF JT OBNFE JO IPOPVS PG 8JMMJBN 5IPNQTPO -PSE ,FMWJO m

B 4DPUUJTI NBUIFNBUJDBO BOE QIZTJDJTU XIP ê STU TVHHFTUFE UIF JEFB PG BO BCTPMVUF TDBMF PG UFNQFSBUVSF %FTQJUF NBLJOH NBOZ JNQPSUBOU DPOUSJCVUJPOT UP UIF BEWBODFNFOU PG TDJFODF ,FMWJO IBE EPVCUT BCPVU UIF FYJTUFODF PG BUPNT CFMJFWFE UIBU UIF &BSUI DPVME OPU CF PMEFS UIBO NJMMJPO ZFBST BOE JT PGUFO RVPUFE BT TBZJOH UIBU AIFBWJFS UIBO BJS ë ZJOH NBDIJOFT BSF JNQPTTJCMF divide by 1000 convert cm3 to dm3

1 dm3

*O EJĄ FSFOU DPVOUSJFT BSPVOE UIF XPSME T EJĄ FSFOU TDBMFT PG UFNQFSBUVSF BSF VTFE m F H UIF $FMTJVT BOE 'BISFOIFJU TDBMFT 5IF $FMTJVT BOE 'BISFOIFJU TDBMFT BSF CPUI BSUJê DJBM TDBMFT CVU UIF ,FMWJO TDBMF JT BO BCTPMVUF TDBMF 8IBU JT UIF BEWBOUBHF UP TDJFOUJTUT PG VTJOH BO BCTPMVUF TDBMF 8IZ IBT UIF absolute scale of temperature OPU CFFO BEPQUFE JO FWFSZEBZ MJGF

1000 cm3

convert dm3 to cm3 multiply by 1000

Figure 1.11 Converting between cm3 and dm3.

Worked examples 1.25 a $BMDVMBUF UIF OVNCFS PG NPMFT JO DN PG 0 BU 451 b $BMDVMBUF UIF WPMVNF PG NPM $0 BU 451 a OVNCFS PG NPMFT

WPMVNF JO EN

DN EN EN

OVNCFS PG NPMFT NPM

b WPMVNF OVNCFS PG NPMFT ¤ ¤ EN

1 STOICHIOMETRIC RELATIONSHIPS

29

1.26 $BMDVMBUF UIF WPMVNF PG DBSCPO EJPYJEF DPMMFDUFE BU 451 QSPEVDFE XIFO H PG DBMDJVN DBSCPOBUF EFDPNQPTFT BDDPSEJOH UP UIF FRVBUJPO $B$0 T â&#x2020;&#x2019; $B0 T

$0 H

4UFQ m XPSL PVU UIF OVNCFS PG NPMFT PG $B$0

OVNCFS PG NPMFT PG $B$0 NPM

4UFQ m UIF DIFNJDBM FRVBUJPO UFMMT VT UIBU NPM $B$0 EFDPNQPTFT UP HJWF NPM $0 5IFSFGPSF NPM $B$0 EFDPNQPTFT UP HJWF NPM $0 4UFQ m DPOWFSU UIF OVNCFS PG NPMFT UP WPMVNF

mass of substance

number of moles

molar mass

NPM $0 PDDVQJFT EN BU 451 WPMVNF PG $0 OVNCFS PG NPMFT ¤ WPMVNF PG NPMF EN

WPMVNF PG $0 ¤ EN 5IF WPMVNF PG $0 QSPEVDFE JT EN 1.27 1PUBTTJVN DIMPSBUF 7 EFDPNQPTFT XIFO IFBUFE ,$M0 T â&#x2020;&#x2019; ,$M T

0 H

8IBU NBTT PG QPUBTTJVN DIMPSBUF 7 EFDPNQPTFT UP QSPEVDF DN PG PYZHFO HBT NFBTVSFE BU 451 4UFQ m XPSL PVU UIF OVNCFS PG NPMFT PG 0 5IF WPMVNF PG 0 NVTU ê STU CF DPOWFSUFE UP EN

WPMVNF PG 0 JO EN EN

OVNCFS PG NPMFT PG 0 ¤ ¢ NPM

4UFQ m UIF DIFNJDBM FRVBUJPO UFMMT VT UIBU NPM 0 BSF QSPEVDFE GSPN NPM ,$M0 5IFSFGPSF UIF OVNCFS PG NPMFT PG ,$M0 JT UXP UIJSET PG UIF OVNCFS PG NPMFT PG 0

¢ ¢ ¤ ¤ ¤ NPM

4UFQ m DPOWFSU UIF OVNCFS PG NPMFT PG ,$M0 UP NBTT NPMBS NBTT PG ,$M0 H NPM¢ NBTT PG ,$M0 ¤ ¤ ¢ H 5IF NBTT PG ,$M0 SFRVJSFE JT H

30

Formula for solving moles questions involving volumes of gases "O BMUFSOBUJWF XBZ PG EPJOH UIFTF RVFTUJPOT JT UP VTF B GPSNVMB m1 V2 = n1M1 n2Mv XIFSF m NBTT PG ê STU TVCTUBODF JO H

DJFOU PG ê STU TVCTUBODF n DPFÄ&#x2026; M NPMBS NBTT PG ê STU TVCTUBODF V WPMVNF JO EN PG TFDPOE TVCTUBODF JG JU JT B HBT DJFOU PG TFDPOE TVCTUBODF n DPFÄ&#x2026; Mv NPMBS WPMVNF PG B HBT EN BU 451 5IJT GPSNVMB DBO CF VTFE JG UIF NBTT PG POF TVCTUBODF JT HJWFO BOE UIF WPMVNF PG BOPUIFS TVCTUBODF JT SFRVJSFE PS WJDF WFSTB *G B WPMVNF JT HJWFO BOE B WPMVNF JT SFRVJSFE UIFO BO BMUFSOBUJWF GPSN PG UIJT FRVBUJPO JT V1 V2 = n1 n2

/PUF UIJT JT WFSZ TJNJMBS UP UIF GPSNVMB UIBU XBT VTFE FBSMJFS XJUI NBTTFT

5IFSF JT OP OFFE UP DPOWFSU VOJUT PG WPMVNF UP EN XJUI UIJT FRVBUJPO m CVU V NVTU IBWF UIF TBNF VOJUT BT V

XIFSF V WPMVNF PG ê STU TVCTUBODF JG JU JT B HBT V WPMVNF PG TFDPOE TVCTUBODF )PXFWFS XJUI RVFTUJPOT JOWPMWJOH KVTU HBTFT JU JT VTVBMMZ FBTJFS UP XPSL UIFN PVU VTJOH "WPHBESP T MBX BT EFTDSJCFE FBSMJFS

?

Test yourself Assume that all gases behave as ideal gases and that all measurements are made under the same conditions of temperature and pressure.

26 a $BMDVMBUF UIF WPMVNF PG DBSCPO EJPYJEF QSPEVDFE XIFO DN PG FUIFOF CVSOT JO FYDFTT PYZHFO BDDPSEJOH UP UIF FRVBUJPO $ H H

0 H â&#x2020;&#x2019; $0 H

) 0 M

b $BMDVMBUF UIF WPMVNF PG OJUSJD PYJEF /0

QSPEVDFE XIFO EN PG PYZHFO JT SFBDUFE XJUI FYDFTT BNNPOJB BDDPSEJOH UP UIF FRVBUJPO /) H

0 H â&#x2020;&#x2019; /0 H

) 0 H

27 %FUFSNJOF UIF OVNCFS PG NPMFT QSFTFOU JO FBDI PG UIF GPMMPXJOH BU TUBOEBSE UFNQFSBUVSF BOE QSFTTVSF d DN PG / a EN PG 0 b EN PG $) e DN PG $0

c EN PG 40 28 8PSL PVU UIF WPMVNF PG FBDI PG UIF GPMMPXJOH BU TUBOEBSE UFNQFSBUVSF BOE QSFTTVSF d NPM /) a NPM $ H b NPM 40 e NPM 0 c NPM /

1 STOICHIOMETRIC RELATIONSHIPS

31

29 4PEJVN OJUSBUF 7 EFDPNQPTFT BDDPSEJOH UP UIF FRVBUJPO /B/0 T → /B/0 T

0 H

$BMDVMBUF UIF WPMVNF JO DN PG PYZHFO QSPEVDFE NFBTVSFE BU 451 XIFO H PG TPEJVN OJUSBUF 7 EFDPNQPTF 30 5JO SFBDUT XJUI OJUSJD BDJE BDDPSEJOH UP UIF FRVBUJPO 4O T

)/0 BR

→ 4O0 T

/0 H

) 0 M

*G H PG UJO BSF SFBDUFE XJUI FYDFTT OJUSJD BDJE XIBU WPMVNF PG OJUSPHFO EJPYJEF JO DN JT QSPEVDFE BU 451 31 $BMDVMBUF UIF NBTT PG TPEJVN DBSCPOBUF UIBU NVTU CF SFBDUFE XJUI FYDFTT IZESPDIMPSJD BDJE UP QSPEVDF DN PG DBSCPO EJPYJEF BU 451 /B $0 T

)$M BR

→ /B$M BR

$0 H

) 0 M

A.BDSPTDPQJD NFBOT APO B MBSHF TDBMF 5IF PQQPTJUF JT ANJDSPTDPQJD 5IF NJDSPTDPQJD QSPQFSUJFT PG B HBT BSF UIF QSPQFSUJFT PG UIF QBSUJDMFT UIBU NBLF VQ UIF HBT

32 a 0YZHFO DBO CF DPOWFSUFE UP P[POF 0 CZ QBTTJOH JU UISPVHI B TJMFOU FMFDUSJD EJTDIBSHF 0 H → 0 H

*G DN PG PYZHFO BSF VTFE BOE PG UIF PYZHFO JT DPOWFSUFE UP P[POF DBMDVMBUF UIF UPUBM WPMVNF PG HBT QSFTFOU BU UIF FOE PG UIF FYQFSJNFOU b )ZESPHFO SFBDUT XJUI DIMPSJOF BDDPSEJOH UP UIF FRVBUJPO H H

$M H → )$M H

8IBU JT UIF UPUBM WPMVNF PG HBT QSFTFOU JO UIF DPOUBJOFS BU UIF FOE PG UIF FYQFSJNFOU JG DN PG IZESPHFO BSF SFBDUFE XJUI DN PG DIMPSJOF

Macroscopic properties of ideal gases 4P GBS BMM UIF RVFTUJPOT XF IBWF EFBMU XJUI IBWF JOWPMWFE XPSLJOH PVU WPMVNFT PG HBTFT BU 451 *O PSEFS UP XPSL PVU WPMVNFT PG HBTFT VOEFS PUIFS DPOEJUJPOT XF NVTU VOEFSTUBOE B MJUUMF NPSF BCPVU UIF QSPQFSUJFT PG HBTFT

The relationship between pressure and volume (Boyle’s law) At a constant temperature, the volume of a fixed mass of an ideal gas is inversely proportional to its pressure.

P / Pa

5IJT NFBOT UIBU JG UIF QSFTTVSF PG B HBT JT EPVCMFE BU DPOTUBOU UFNQFSBUVSF UIFO UIF WPMVNF XJMM CF IBMWFE BOE WJDF WFSTB 5IJT SFMBUJPOTIJQ JT JMMVTUSBUFE JO 'JHVSF 1.12 P∝ 0 0

V / cm3

Figure 1.12 The relationship between pressure and volume of a fixed mass of an ideal gas at constant temperature.

1 V

5IF SFMBUJPOTIJQ DBO BMTP CF XSJUUFO BT k P V XIFSF k JT B DPOTUBOU 5IJT DBO CF SFBSSBOHFE UP HJWF PV k

32

5IJT NFBOT UIBU UIF QSPEVDU PG UIF QSFTTVSF BOE WPMVNF PG BO JEFBM HBT BU B QBSUJDVMBS UFNQFSBUVSF JT B DPOTUBOU BOE EPFT OPU DIBOHF BT UIF QSFTTVSF BOE UIF WPMVNF DIBOHF 0UIFS HSBQIT DBO BMTP CF ESBXO UP JMMVTUSBUF UIJT SFMBUJPOTIJQ TFF 'JHVSFT 1.13 BOE 1.14

B HSBQI PG QSFTTVSF BHBJOTU #FDBVTF QSFTTVSF JT QSPQPSUJPOBM UP WPMVNF

XPVME CF B TUSBJHIU MJOF HSBQI UIBU XPVME QBTT UISPVHI UIF PSJHJO WPMVNF BMUIPVHI UIJT HSBQI XJMM OFWFS BDUVBMMZ QBTT UISPVHI UIF PSJHJO m UIF HBT XPVME IBWF UP IBWF JOĂŞ OJUF WPMVNF BU [FSP QSFTTVSF 5IJT JT TIPXO JO 'JHVSF 1.13 #FDBVTF PV k XIFSF k JT B DPOTUBOU B HSBQI PG PV BHBJOTU QSFTTVSF PS WPMVNF XJMM CF B TUSBJHIU IPSJ[POUBM MJOF 5IJT JT TIPXO JO 'JHVSF 1.14 P / Pa

PV / cm3 Pa

0 0

1/ V / cmâ&#x20AC;&#x201C;3

1 Figure 1.13 The relationship between the pressure and volume of a fixed mass of an ideal gas at constant temperature.

0

P/ Pa

0

Figure 1.14 The relationship between PV and P for a fixed mass of an ideal gas at constant temperature.

The relationship between volume and temperature (Charlesâ&#x20AC;&#x2122; law)

V / cm3

*G UIF UFNQFSBUVSF JT JO LFMWJO UIF GPMMPXJOH SFMBUJPOTIJQ FYJTUT CFUXFFO UIF WPMVNF BOE UIF UFNQFSBUVSF The volume of a fixed mass of an ideal gas at constant pressure is directly proportional to its kelvin temperature.

Vâ&#x2C6;?T 5IFSFGPSF JG UIF LFMWJO UFNQFSBUVSF JT EPVCMFE BOE UIF QSFTTVSF SFNBJOT DPOTUBOU UIF WPMVNF PG UIF HBT JT EPVCMFE BOE WJDF WFSTB 5IJT NFBOT UIBU JG BO JEFBM HBT IBT B WPMVNF PG DN BU , JU XJMM IBWF B WPMVNF PG DN BU , JG UIF QSFTTVSF SFNBJOT DPOTUBOU 5IJT JT JMMVTUSBUFE JO 'JHVSF 1.15 5IJT SFMBUJPOTIJQ EPFT OPU XPSL GPS UFNQFSBUVSFT JO Â&#x17E;$ 'JHVSF 1.16

'PS JOTUBODF JG UIF WPMVNF PG BO JEFBM HBT BU Â&#x17E;$ JT DN UIF WPMVNF JU XJMM PDDVQZ BU Â&#x17E;$ XJMM CF BCPVU DN

0 0

T/ K

Figure 1.15 The relationship between the volume and temperature (in kelvin) of a fixed mass of an ideal gas at constant pressure.

"O JEFBM HBT DBO OFWFS MJRVFGZ CFDBVTF UIFSF BSF OP GPSDFT CFUXFFO UIF NPMFDVMFT

1 STOICHIOMETRIC RELATIONSHIPS

33

V / cm3

5IJT JT B MJOFBS SFMBUJPOTIJQ CVU OPU B QSPQPSUJPOBM POF CFDBVTF UIF HSBQI EPFT OPU QBTT UISPVHI UIF PSJHJO

â&#x20AC;&#x201C;273

0

T/ °C

Figure 1.16 The relationship between the volume and temperature (in °C) of a fixed mass of an ideal gas at constant pressure. As can be seen, the temperature at which the volume of an ideal gas is zero will be â&#x2C6;&#x2019;273 °C. This temperature is absolute zero.

The relationship between pressure and temperature

P / Pa

For a fixed mass of an ideal gas at constant volume, the pressure is directly proportional to its absolute temperature: P â&#x2C6;? T *G UIF UFNQFSBUVSF JO kelvin PG B ĂŞ YFE WPMVNF PG BO JEFBM HBT JT EPVCMFE UIF QSFTTVSF XJMM BMTP EPVCMF 'JHVSF 1.17

The overall gas law equation 0 0

T/ K

Figure 1.17 The relationship between the pressure and temperature (kelvin) of a fixed mass of an ideal gas at constant volume.

An ideal gas is one that obeys all of the above laws exactly.

5IF UISFF SFMBUJPOTIJQT BCPWF DBO CF DPNCJOFE UP QSPEVDF UIF GPMMPXJOH FRVBUJPO P1V1 P2V2 = T1 T2 /PUF BOZ VOJUT NBZ CF VTFE GPS P BOE V TP MPOH BT UIFZ BSF DPOTJTUFOU PO CPUI TJEFT PG UIF FRVBUJPO

The temperature must be kelvin.

34

Worked examples 1.28 *G UIF WPMVNF PG BO JEFBM HBT DPMMFDUFE BU $ BOE L1B J F BU 451 JT DN XIBU XPVME CF UIF WPMVNF BU $ BOE L1B P L1B V DN

P L1B V

T $ ,

T $ , ,

5IF VOJUT PG P BOE P BSF DPOTJTUFOU XJUI FBDI PUIFS 5FNQFSBUVSF NVTU CF JO ,

P V P V

T T ¤ ¤ V

3FBSSBOHJOH UIF FRVBUJPO ¤ ¤ V DN ¤

5IF VOJUT PG V BSF UIF TBNF BT UIPTF PG V

5IFSFGPSF UIF WPMVNF PDDVQJFE CZ UIF HBT BU $ BOE L1B JT DN 1.29 8IBU UFNQFSBUVSF JO $ JT SFRVJSFE UP DBVTF BO JEFBM HBT UP PDDVQZ EN BU B QSFTTVSF PG L1B JG JU PDDVQJFT DN BU 451 P L1B

P L1B

V EN

V DN J F

T

T ,

EN PS EN

¤ ¤

T

5IF VOJUT PG P BSF UIF TBNF BT UIPTF PG P

5IF VOJUT PG V BOE V XFSF NBEF DPOTJTUFOU XJUI FBDI PUIFS 8F DPVME IBWF BMTP DIBOHFE V UP DN

3FBSSBOHJOH UIF FRVBUJPO ¤ ¤ ¤ ¤ T ¤ ¤ , T ¤ 5IJT NVTU OPX CF DPOWFSUFE UP $ CZ TVCUSBDUJOH 5FNQFSBUVSF ¢ $ 5IF UFNQFSBUVSF NVTU CF $ GPS UIF HBT UP PDDVQZ B WPMVNF PG EN

1 STOICHIOMETRIC RELATIONSHIPS

35

The ideal gas equation *G UIF SFMBUJPOTIJQT CFUXFFO P V BOE T BSF DPNCJOFE XJUI "WPHBESP T MBX UIF JEFBM HBT FRVBUJPO JT PCUBJOFE PV = nRT A consistent set of units must be used.

Exam tip " TFU PG VOJUT UIBU JT FRVJWBMFOU UP UIJT VTFT WPMVNF JO EN BOE QSFTTVSF JO L1B m JG ZPV VTF UIFTF VOJUT ZPV DBO BWPJE UIF QSPCMFN PG DPOWFSUJOH WPMVNFT JOUP N

8IFSF R JT UIF gas constant BOE n JT UIF OVNCFS PG NPMFT "MUIPVHI R JT B VOJWFSTBM DPOTUBOU JU DBO CF RVPUFE XJUI WBSJPVT VOJUT BOE JUT WBMVF EFQFOET PO UIFTF VOJUT 5IF 4* VOJUT GPS UIF HBT DPOTUBOU BSF + ,¢ NPM¢ BOE UIJT SFRVJSFT UIF GPMMPXJOH TFU PG VOJUT

4* TUBOET GPS 4ZTUÍNF *OUFSOBUJPOBM E 6OJUÊT BOE SFGFST UP UIF JOUFSOBUJPOBMMZ BDDFQUFE TZTUFN PG VOJUT VTFE JO TDJFODF

R = 8.31 J K−1 mol−1 Pressure: N m−2 or Pa Volume: m3 Temperature: K 1 000 000 cm3 ⇔ 1 m3 1000 dm3 ⇔ 1 m3 5P DPOWFSU N UP DN NVMUJQMZ CZ 5P DPOWFSU DN UP N EJWJEF CZ 5P DPOWFSU N UP EN NVMUJQMZ CZ 5P DPOWFSU EN UP N EJWJEF CZ

Worked examples 1.30 "O JEFBM HBT PDDVQJFT DN BU $ BOE L1B 8IBU BNPVOU PG HBT JO NPMFT JT QSFTFOU *G XF VTF UIF WBMVF PG + ,¢ NPM¢ GPS UIF HBT DPOTUBOU BMM WBMVFT NVTU CF DPOWFSUFE UP UIF BQQSPQSJBUF TFU PG VOJUT P L1B ¤ 1B

V DN N ¤ ¢ N

n R + ,¢ NPM¢ T $ , , PV nRT ¤ ¤ ¤ ¢ n ¤ ¤ 3FBSSBOHJOH UIF FRVBUJPO n

¤ ¤ ¤ ¢ NPM ¤

5IF OVNCFS PG NPMFT JT NPM

36

1.31 " HBT IBT B EFOTJUZ PG H EN¢ BU $ BOE ¤ 1B $BMDVMBUF JUT NPMBS NBTT NBTT EFOTJUZ WPMVNF 8F LOPX UIF EFOTJUZ TP XF LOPX UIF NBTT PG EN PG UIF HBT *G XF DBO ê OE UIF OVNCFS PG NPMFT JO EN XF DBO XPSL PVU UIF NPMBS NBTT P ¤ 1B

V EN N ¤ ¢ N

n R + ,¢ NPM¢ T $ , 6TJOH PV nRT n

¤ ¤ ¤ ¢ NPM ¤

5IJT OVNCFS PG NPMFT IBT B NBTT PG H NBTT NPMBS NBTT OVNCFS PG NPMFT

NPMBS NBTT H NPM¢

1.32 8IBU JT UIF NPMBS WPMVNF PG BO JEFBM HBT BU $ BOE ¤ 1B (JWF ZPVS BOTXFS JO N NPM¢ BOE EN NPM¢

5IF NPMBS WPMVNF PG B HBT JT UIF WPMVNF PDDVQJFE CZ POF NPMF PG UIF HBT 8F BSF GBNJMJBS XJUI UIF WBMVF GPS UIF NPMBS WPMVNF PG B HBT BU 451 XIJDI JT EN NPM¢ P ¤ 1B n NPM T $ , ,

V R + ,¢ NPM¢

6TJOH PV nRT V

¤ ¤ N ¤

5IF NPMBS WPMVNF JT N NPM¢ BU $ BOE ¤ 1B 5IJT NVTU CF NVMUJQMJFE CZ UP DPOWFSU UP dm J F EN NPM¢

1 STOICHIOMETRIC RELATIONSHIPS

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1.33 8IFO TPEJVN OJUSBUF 7 PGUFO KVTU DBMMFE TPEJVN OJUSBUF JT IFBUFE JU EFDPNQPTFT UP HJWF TPEJVN OJUSBUF ***

BMTP DBMMFE TPEJVN OJUSJUF BOE PYZHFO HBT 8IFO B DFSUBJO NBTT PG TPEJVN OJUSBUF 7 JT IFBUFE DN PG PYZHFO JT PCUBJOFE NFBTVSFE BU L1B BOE $ $BMDVMBUF UIF NBTT PG TPEJVN OJUSBUF *** GPSNFE /B/0 T → /B/0 T

0 H

8F DBO VTF PV nRT UP XPSL PVU UIF OVNCFS PG NPMFT PG 0 P L1B ¤ 1B

V DN N ¤ ¢ N

n R + ,¢ NPM¢ T $ , ¤ ¤ ¤ ¢ ¤ ¢ NPM 6TJOH PV nRT n ¤ 5IJT HJWFT UIF OVNCFS PG NPMFT PG 0 'SPN UIF DIFNJDBM FRVBUJPO UIF OVNCFS PG NPMFT PG 0 JT IBMG UIF OVNCFS PG NPMFT PG /B/0 5IFSFGPSF UIF OVNCFS PG NPMFT PG /B/0 JT ¤ ¢ ¤ ¤ ¢ NPM 5IF NPMBS NBTT PG /B/0 JT H NPM¢ TP UIF NBTT PG /B/0 JT ¤ ¤ ¢ H

:PV XPVME QSPCBCMZ TBZ UIBU UIF SPPN ZPV BSF TJUUJOH JO BU UIF NPNFOU JT GVMM PG BJS *G IPXFWFS ZPV EP B RVJDL DBMDVMBUJPO NBLJOH B DPVQMF PG BQQSPYJNBUJPOT ZPV TIPVME CF BCMF UP XPSL PVU UIBU UIF WPMVNF PG UIF NPMFDVMFT PG HBT JO UIF SPPN JT POMZ BCPVU PG UIF WPMVNF PG

UIF SPPN m TDJFOUJê D SFBMJUZ JT WFSZ EJĄ FSFOU GSPN PVS FWFSZEBZ SFBMJUZ 5IFSF JT BDUVBMMZ B WFSZ TNBMM QSPCBCJMJUZ UIBU BMM UIFTF NPMFDVMFT DPVME BU BOZ POF UJNF BMM FOE VQ JO UIF TBNF DPSOFS PG UIF SPPN m PVS TVSWJWBM EFQFOET PO UIF GBDU UIBU UIJT QSPCBCJMJUZ JT WFSZ TNBMM

Nature of science " TDJFOUJê D MBX JT B HFOFSBM TUBUFNFOU PGUFO JO NBUIFNBUJDBM GPSN CBTFE PO PCTFSWBUJPO FYQFSJNFOU PG TPNF BTQFDU PG UIF QIZTJDBM XPSME *U XJMM PGUFO JOWPMWF UIF SFMBUJPOTIJQ CFUXFFO WBSJPVT RVBOUJUJFT VOEFS TQFDJê FE DPOEJUJPOT 'PS FYBNQMF #PZMF T MBX EFTDSJCFT UIF SFMBUJPOTIJQ CFUXFFO UIF WPMVNF BOE QSFTTVSF PG B ê YFE NBTT PG BO JEFBM HBT BU DPOTUBOU UFNQFSBUVSF " MBX EPFT OPU FYQMBJO BOZUIJOH m JU JT KVTU B EFTDSJQUJPO PG XIBU IBQQFOT " UIFPSZ JT B XBZ PG FYQMBJOJOH TDJFOUJê D PCTFSWBUJPOT PS MBXT 5P CF BDDFQUFE B UIFPSZ XJMM IBWF CFFO SJHPSPVTMZ UFTUFE CZ FYQFSJNFOUT BOE PCTFSWBUJPOT m GPS FYBNQMF UIF QBSUJDMF UIFPSJFT BOE LJOFUJD UIFPSZ DBO CF VTFE UP FYQMBJO #PZMF T MBX 5IFSF JT OP QSPHSFTTJPO GSPN B UIFPSZ UP B MBX m UIFZ BSF EJĄ FSFOU UIJOHT "WPHBESP T PSJHJOBM IZQPUIFTJT XBT UIBU FRVBM WPMVNFT PG EJĄ FSFOU HBTFT DPOUBJO UIF TBNF OVNCFS PG NPMFDVMFT 5IJT XBT CBTFE PO EFEVDUJPOT GSPN DBSFGVM NFBTVSFNFOUT BOE PCTFSWBUJPOT NBEF CZ PUIFS TDJFOUJTUT TVDI BT (BZ -VTTBD 38

?

Test yourself In all questions, take the value of the ideal gas constant as 8.31 J Kâ&#x2C6;&#x2019;1 molâ&#x2C6;&#x2019;1 33 *G B DFSUBJO NBTT PG BO JEFBM HBT PDDVQJFT 37 8IBU JT UIF NPMBS WPMVNF PG BO JEFBM HBT BU

DN BU Â&#x17E;$ BOE ¤ 1B XIBU WPMVNF ¤ 1B BOE Â&#x17E;$

XPVME JU PDDVQZ BU Â&#x17E;$ BOE ¤ 1B 38 $PQQFS OJUSBUF EFDPNQPTFT XIFO IFBUFE 34 " DFSUBJO NBTT PG BO JEFBM HBT PDDVQJFT DN BU Â&#x17E;$ BOE ¤ 1B "U XIBU UFNQFSBUVSF JO Â&#x17E;$ XJMM JU PDDVQZ DN JG UIF QSFTTVSF SFNBJOT UIF TBNF 35 )PX NBOZ NPMFT PG BO JEFBM HBT BSF QSFTFOU JO B DPOUBJOFS JG JU PDDVQJFT B WPMVNF PG EN BU B QSFTTVSF PG ¤ 1B BOE B UFNQFSBUVSF PG Â&#x17E;$ 36 $BMDVMBUF UIF NPMBS NBTT PG BO JEFBM HBT JG H PG UIF HBT PDDVQJFT B WPMVNF PG DN BU B QSFTTVSF PG ¤ 1B BOE B UFNQFSBUVSF PG ¢ Â&#x17E;$

BDDPSEJOH UP UIF FRVBUJPO $V /0 T â&#x2020;&#x2019; $V0 T

/0 H

0 H

*G H PG DPQQFS OJUSBUF JT IFBUFE BOE UIF HBTFT DPMMFDUFE BU B UFNQFSBUVSF PG Â&#x17E;$ BOE L1B a XIBU WPMVNF JO EN PG PYZHFO JT DPMMFDUFE b XIBU JT UIF UPUBM WPMVNF PG HBT DPMMFDUFE JO cm 39 8IFO B DFSUBJO NBTT PG NBOHBOFTF IFQUPYJEF .O 0 EFDPNQPTFE JU QSPEVDFE DN PG PYZHFO NFBTVSFE BU Â&#x17E;$ BOE ¤ 1B 8IBU NBTT PG NBOHBOFTF IFQUPYJEF EFDPNQPTFE .O 0 BR â&#x2020;&#x2019; .O0 T

0 H

1.3.3 Calculations involving solutions

Learning objective

Solutions

r 4PMWF QSPCMFNT JOWPMWJOH TPMVUJPOT

Solute: a substance that is dissolved in another substance.

Solvent: a substance that dissolves another substance (the solute).

The solvent should be present in excess of the solute.

Solution: the substance that is formed when a solute dissolves in a solvent.

8IFO B TPEJVN DIMPSJEF /B$M solution JT QSFQBSFE /B$M TPMJE UIF solute JT EJTTPMWFE JO XBUFS UIF solvent

4PMVUJPOT JO XBUFS BSF HJWFO UIF TZNCPM BR JO DIFNJDBM FRVBUJPOT aq TUBOET GPS aqueous

/PUF XIFO B TPMVUF JT EJTTPMWFE JO B DFSUBJO WPMVNF PG XBUFS TBZ DN UIF UPUBM WPMVNF PG UIF TPMVUJPO JT OPU TJNQMZ DN PS UIF TVN PG UIF WPMVNFT PDDVQJFE CZ UIF TPMVUF BOE UIF WPMVNF PG UIF TPMWFOU 5IF UPUBM WPMVNF PG TPMVUJPO QSPEVDFE EFQFOET PO UIF GPSDFT PG BUUSBDUJPO CFUXFFO UIF TPMVUF QBSUJDMFT BOE UIF TPMWFOU QBSUJDMFT DPNQBSFE XJUI UIF GPSDFT PG BUUSBDUJPO JO UIF PSJHJOBM TPMWFOU 5IJT JT XIZ DPODFOUSBUJPO JT EFĂŞ OFE JO UFSNT PG UIF WPMVNF PG UIF TPMVUJPO SBUIFS UIBO UIF WPMVNF PG UIF TPMWFOU

1 STOICHIOMETRIC RELATIONSHIPS

39

3FQPSUFE WBMVFT GPS UIF DPODFOUSBUJPO PG HPME JO TFBXBUFS WBSZ HSFBUMZ " WBMVF PG BCPVU ¤ ¢ H EN¢ PS ¤ ¢ NPM EN¢ JT QSPCBCMZ B SFBTPOBCMF FTUJNBUF 5IF WPMVNF PG XBUFS JO UIF PDFBOT JT FTUJNBUFE UP CF BCPVU ¤ EN TP UIFSF JT BO BXGVM MPU PG HPME JO UIF PDFBOT .BOZ QFPQMF JODMVEJOH /PCFM 1SJ[F XJOOJOH TDJFOUJTU 'SJU[ )BCFS IBWF USJFE UP DPNF VQ XJUI XBZT UP FYUSBDU UIF HPME 5IF QSPCMFN JT UIBU UIF DPODFOUSBUJPOT BSF TP MPX

The concentration of a solution is the amount of solute dissolved in a unit volume of solution. The volume that is usually taken is 1 dm3. The amount of solute may be expressed in g or mol therefore the units of concentration are g dmâ&#x2C6;&#x2019;3 or mol dmâ&#x2C6;&#x2019;3.

Concentrations BSF TPNFUJNFT XSJUUFO XJUI UIF VOJU . XIJDI NFBOT NPM EN¢ CVU JT EFTDSJCFE BT ANPMBS 5IVT . XPVME SFGFS UP B A NPMBS TPMVUJPO J F B TPMVUJPO PG DPODFOUSBUJPO NPM EN¢ 5IF SFMBUJPOTIJQ CFUXFFO DPODFOUSBUJPO OVNCFS PG NPMFT BOE WPMVNF PG TPMVUJPO JT concentration (mol dmâ&#x2C6;&#x2019;3) =

number of moles (mol) volume (dm3)

5IJT JT TVNNBSJTFE JO 'JHVSF 1.18 *G UIF DPODFOUSBUJPO JT FYQSFTTFE JO H EN¢ UIF SFMBUJPOTIJQ JT number of moles

concentration in mol dmâ&#x20AC;&#x201C;3

concentration (g dmâ&#x2C6;&#x2019;3) =

mass (g) volume (dm3)

volume in dm3

Figure 1.18 The relationship between concentration, number of moles and volume of solution.

Worked examples 1.34 *G H PG TPEJVN IZESPYJEF /B0) JT EJTTPMWFE JO XBUFS BOE UIF WPMVNF JT NBEF VQ UP DN DBMDVMBUF UIF DPODFOUSBUJPO JO NPM EN¢ BOE H EN¢ Concentration (g dmâ&#x2C6;&#x2019;3) NBTT DPODFOUSBUJPO JO H EN¢ WPMVNF JO EN

WPMVNF JO EN EN

DPODFOUSBUJPO H EN¢

40

Concentration (mol dmâ&#x2C6;&#x2019;3) NPMBS NBTT PG /B0) H NPM¢

OVNCFS PG NPMFT NPM

DPODFOUSBUJPO

OVNCFS PG NPMFT WPMVNF JO EN

DPODFOUSBUJPO NPM EN¢

"MUFSOBUJWFMZ PODF XF IBWF UIF DPODFOUSBUJPO JO H EN¢ XF DBO TJNQMZ EJWJEF CZ UIF NPMBS NBTT UP HFU UIF DPODFOUSBUJPO JO NPM EN¢

DPODFOUSBUJPO NPM EN¢

1.35 $BMDVMBUF UIF OVNCFS PG NPMFT PG IZESPDIMPSJD BDJE )$M QSFTFOU JO DN PG NPM EN¢ IZESPDIMPSJD BDJE OVNCFS PG NPMFT DPODFOUSBUJPO ¤ WPMVNF JO EN

OVNCFS PG NPMFT ¤ NPM

5IFSFGPSF UIF OVNCFS PG NPMFT JT NPM

4RVBSF CSBDLFUT BSF PGUFO VTFE UP EFOPUF DPODFOUSBUJPOT JO NPM dmm 4P <)$M> JOEJDBUFT UIF NPMBS DPODFOUSBUJPO PG IZESPDIMPSJD BDJE BOE XF DPVME XSJUF <)$M> NPM dmm JO UIJT XPSLFE FYBNQMF

Concentrations of very dilute solutions 8IFO EFBMJOH XJUI WFSZ TNBMM DPODFOUSBUJPOT ZPV XJMM PDDBTJPOBMMZ DPNF BDSPTT UIF VOJU parts per million ppm 'PS JOTUBODF JG H PG B TPMVUF JT QSFTFOU JO NJMMJPO HSBNT PG B TPMVUJPO UIFO UIF DPODFOUSBUJPO JT QQN 4P JO HFOFSBM UIF DPODFOUSBUJPO JO QQN JT HJWFO CZ DPODFOUSBUJPO JO QQN

NBTT PG TPMVUF ¤

NBTT PG TPMVUJPO

5IF VOJUT PG NBTT PG TPMVUF BOE NBTT PG TPMVUJPO NVTU CF UIF TBNF TP UIBU UIFZ DBODFM

5IF QQN OPUBUJPO JT NPTU PGUFO VTFE XIFO XSJUJOH BCPVU QPMMVUJPO m F H the concentration of arsenic in drinking water in the US should not exceed 0.010 ppm

Worked examples 1.36 *G B TBNQMF PG H PG XBUFS JT GPVOE UP DPOUBJO NH PG DZBOJEF XIBU JT UIF DZBOJEF DPODFOUSBUJPO JO QQN 5IF NBTT JO NH NVTU ê STU CF DPOWFSUFE UP B NBTT JO H CZ EJWJEJOH CZ NBTT PG DZBOJEF ¤ ¢ H

¤ ¢ ¤ DPODFOUSBUJPO PG DZBOJEF JO QQN QQN

"MUIPVHI UIF DPODFOUSBUJPO JO QQN JT QSPQFSMZ EFê OFE BT BCPWF JU JT PGUFO VTFE JO OFXTQBQFS BSUJDMFT FUD JO B TMJHIUMZ NPSF DPOWFOJFOU CVU OPU DPNQMFUFMZ DPSSFDU XBZ BT UIF NBTT PG TPMVUF JO NH QFS dm MJUSF PG TPMVUJPO 5IJT JT B SFBTPOBCMF BQQSPYJNBUJPO CFDBVTF UIF NBTT PG dm PG XBUFS XIJDI XJMM NBLF VQ NPTU PG UIF TPMVUJPO JT H m J F NH 4P JG UIF DPQQFS DPODFOUSBUJPO JO B TBNQMF PG UBQ XBUFS JT QQN UIJT JT SPVHIMZ FRVJWBMFOU UP NH PG DPQQFS QFS EN PG XBUFS 1 STOICHIOMETRIC RELATIONSHIPS

41

5IF DPNQBOJPO VOJUT PG QBSUT QFS CJMMJPO QQC BOE QBSUT QFS USJMMJPO QQU BSF PGUFO VTFE XIFO EJTDVTTJOH FYUSFNFMZ MPX DPODFOUSBUJPOT m UIFTF VOJUT DBO DBVTF TPNF DPOGVTJPO CFDBVTF UIF EFê OJUJPOT PG CJMMJPO BOE USJMMJPO BSF EJÄ&#x201E; FSFOU JO EJÄ&#x201E; FSFOU DPVOUSJFT *U VTFE UP CF UIBU B CJMMJPO JO #SJUJTI &OHMJTI SFGFSSFE UP ¤ CVU OPX UIF 64 EFê OJUJPO PG ¤ JT UIF DPNNPOMZ BDDFQUFE WBMVF

5IF QQN OPUBUJPO JT BMTP VTFE XIFO EJTDVTTJOH UIF DPODFOUSBUJPOT PG WBSJPVT QPMMVUBOU HBTFT JO BJS *O UIJT DBTF JU JT EFê OFE BT DPODFOUSBUJPO JO QQN

WPMVNF PG HBT ¤

WPMVNF PG BJS

'PS JOTUBODF UIF DBSCPO NPOPYJEF DPODFOUSBUJPO JO B TBNQMF PG BJS NJHIU CF QQN 5IJT JOEJDBUFT UIBU UIFSF XPVME CF dm PG $0 QFS dm PG BJS "U 451 UIJT XPVME SPVHIMZ DPOWFSU UP H $0 QFS NJMMJPO dm PG BJS 4P JO B DMBTTSPPN XJUI EJNFOTJPOT PG BCPVU N ¤ N ¤ N UIFSF XPVME CF BQQSPYJNBUFMZ H PG DBSCPO NPOPYJEF

Working out the concentration of ions 8IFO JPOJD TVCTUBODFT TFF QBHF 119 EJTTPMWF JO XBUFS UIF TVCTUBODF CSFBLT BQBSU JOUP JUT DPOTUJUVFOU JPOT 4P GPS JOTUBODF XIFO DPQQFS **

DIMPSJEF $V$M EJTTPMWFT JO XBUFS JU TQMJUT BQBSU JOUP $V BOE $Mâ&#x2C6;&#x2019; JPOT $V$M BR â&#x2020;&#x2019; $V BR

$Mâ&#x2C6;&#x2019; BR

5IFSFGPSF XIFO NPM $V$M EJTTPMWFT JO XBUFS ¤ NPM J F NPM $Mâ&#x2C6;&#x2019; JPOT BSF QSPEVDFE 5IF DPODFOUSBUJPO PG UIF DIMPSJEF JPOT JT UIFSFGPSF UXJDF UIF DPODFOUSBUJPO PG UIF $V$M

Worked example 1.37 $BMDVMBUF UIF OVNCFS PG NPMFT PG DIMPSJEF JPOT QSFTFOU JO DN PG B NPM EN¢ TPMVUJPO PG JSPO *** DIMPSJEF 'F$M BOE UIF UPUBM DPODFOUSBUJPO PG BMM UIF JPOT QSFTFOU OVNCFS PG NPMFT DPODFOUSBUJPO ¤ WPMVNF JO EN

OVNCFS PG NPMFT PG 'F$M ¤ ¤ ¢ NPM 'F$M

'F$M BR â&#x2020;&#x2019; 'F BR

$Mâ&#x2C6;&#x2019; BR

4P EJTTPMWJOH ¤ ¢ NPM 'F$M QSPEVDFT ¤ ¤ ¢ NPM $Mâ&#x2C6;&#x2019; BR J F ¤ ¢ NPM $Mâ&#x2C6;&#x2019; BR

5IF OVNCFS PG NPMFT PG DIMPSJEF JPOT QSFTFOU JT ¤ ¢ NPM 8IFO POF 'F$M VOJU EJTTPMWFT JO XBUFS GPVS JPOT BSF QSPEVDFE 'F

$Mâ&#x2C6;&#x2019;

4P UIF UPUBM DPODFOUSBUJPO PG UIF JPOT QSFTFOU JT GPVS UJNFT UIF DPODFOUSBUJPO PG UIF 'F$M J F ¤ NPM EN¢ 5IF UPUBM DPODFOUSBUJPO PG JPOT QSFTFOU JT NPM EN¢

42

Titrations Titration JT B UFDIOJRVF GPS ê OEJOH UIF WPMVNFT PG TPMVUJPOT UIBU SFBDU FYBDUMZ XJUI FBDI PUIFS 0OF TPMVUJPO JT BEEFE GSPN B CVSFUUF UP UIF PUIFS TPMVUJPO JO B DPOJDBM ë BTL 'JHVSF 1.19 "O JOEJDBUPS JT PGUFO SFRVJSFE UP EFUFSNJOF UIF FOE QPJOU PG UIF UJUSBUJPO *O PSEFS GPS UIF UFDIOJRVF UP CF VTFE UP EFUFSNJOF UIF DPODFOUSBUJPO PG B QBSUJDVMBS TPMVUJPO UIF DPODFOUSBUJPO PG POF PG UIF TPMVUJPOT JU SFBDUT XJUI NVTU CF LOPXO BDDVSBUFMZ m UIJT JT B TUBOEBSE TPMVUJPO

burette

Worked example 1.38 4VMGVSJD BDJE ) 40 JT UJUSBUFE BHBJOTU DN PG NPM EN¢ TPEJVN IZESPYJEF TPMVUJPO /B0) *U JT GPVOE UIBU DN PG TVMGVSJD BDJE JT SFRVJSFE GPS OFVUSBMJTBUJPO $BMDVMBUF UIF DPODFOUSBUJPO PG UIF TVMGVSJD BDJE /B0) BR

) 40 BR â&#x2020;&#x2019; /B 40 BR

) 0 M

4UFQ m XPSL PVU UIF OVNCFS PG NPMFT PG /B0) OVNCFS PG NPMFT DPODFOUSBUJPO ¤ WPMVNF JO EN

OVNCFS PG NPMFT ¤ ¤ ¢ NPM

4UFQ m UIF CBMBODFE FRVBUJPO UFMMT VT UIBU NPM /B0) SFBDU XJUI NPM ) 40 5IFSFGPSF ¤ ¢ NPM /B0) SFBDU XJUI ¤ ¢ NPM ) 40 J F ¤ ¢ NPM ) 40 5IJT JT UIF

OVNCFS PG NPMFT PG ) 40 JO DN PG ) 40 4UFQ m DPOWFSU OVNCFS PG NPMFT UP DPODFOUSBUJPO OVNCFS PG NPMFT DPODFOUSBUJPO WPMVNF JO EN

DN EN EN

¤ ¢ NPM EN¢ <) 40 >

5IF DPODFOUSBUJPO PG UIF ) 40 JT NPM EN¢

Figure 1.19 Titration set-up.

" TUBOEBSE TPMVUJPO DBO CF NBEF VQ GSPN B TPMJE QSJNBSZ TUBOEBSE

" DFSUBJO NBTT PG UIF TPMVUF JT XFJHIFE PVU BDDVSBUFMZ BOE UIFO EJTTPMWFE JO B TNBMM BNPVOU PG EJTUJMMFE XBUFS JO B CFBLFS 5IJT TPMVUJPO JT UIFO USBOTGFSSFE UP B WPMVNFUSJD ë BTL XBTIJOH PVU UIF CFBLFS XJUI TFWFSBM MPUT PG EJTUJMMFE XBUFS UP FOTVSF UIBU BMM UIF TPMVUF JT USBOTGFSSFE 'JOBMMZ XBUFS JT BEEFE UP NBLF UIF TPMVUJPO VQ UP UIF NBSL TP UIBU UIF UPUBM WPMVNF PG UIF TPMVUJPO JT LOPXO "MUFSOBUJWFMZ UIF DPODFOUSBUJPO PG B TUBOEBSE TPMVUJPO NBZ CF LOPXO CFDBVTF JU IBT CFFO UJUSBUFE BHBJOTU BOPUIFS TUBOEBSE TPMVUJPO

1 STOICHIOMETRIC RELATIONSHIPS

43

Equation for solving moles questions involving solutions 5IF GPMMPXJOH FRVBUJPO NBZ CF VTFE BT BO BMUFSOBUJWF NFUIPE GPS TPMWJOH QSPCMFNT Exam tip 5IJT FRVBUJPO JT VTFGVM GPS TPMWJOH UJUSBUJPO QSPCMFNT JO UIF NVMUJQMF DIPJDF QBQFS

c1v1 c2v2 = n1 n2 XIFSF c DPODFOUSBUJPO PG ê STU TVCTUBODF v WPMVNF PG ê STU TVCTUBODF DJFOU PG ê STU TVCTUBODF n DPFÄ&#x2026; c DPODFOUSBUJPO PG TFDPOE TVCTUBODF v WPMVNF PG TFDPOE TVCTUBODF DJFOU PG TFDPOE TVCTUBODF n DPFÄ&#x2026;

Worked example 1.39 'PS OFVUSBMJTBUJPO DN PG QIPTQIPSJD 7 BDJE ) 10 SFRVJSFT DN PG TPEJVN IZESPYJEF /B0) PG DPODFOUSBUJPO NPM EN¢ 8IBU JT UIF DPODFOUSBUJPO PG UIF QIPTQIPSJD 7 BDJE H 10 BR

/B0) BR â&#x2020;&#x2019; /B 10 BR

) 0 M

-FU ) 10 CF TVCTUBODF BOE /B0) CF TVCTUBODF c v DN n

c NPM EN¢ v DN n

c v c v

n n 5IFSF JT OP OFFE UP DPOWFSU UIF WPMVNF UP EN XIFO UIJT FRVBUJPO JT VTFE TP XF DBO VTF UIF WPMVNF JO DN EJSFDUMZ c ¤ ¤

¤ ¤ NPM EN¢ 3FBSSBOHJOH UIF FRVBUJPO c ¤ 5IF DPODFOUSBUJPO PG ) 10 JT NPM EN¢

44

?

Test yourself 40 a 8IBU NBTT PG TPEJVN TVMGBUF /B 40 NVTU CF VTFE UP NBLF VQ DN PG B NPM EN¢ TPMVUJPO b 8IBU JT UIF DPODFOUSBUJPO PG TPEJVN JPOT JO UIF TPMVUJPO JO a 41 8PSL PVU UIF OVNCFST PG NPMFT PG TPMVUF QSFTFOU JO UIF GPMMPXJOH TPMVUJPOT a DN PG NPM EN¢ /B0) BR

b DN PG NPM EN¢ )$M BR

c DN PG NPM EN¢ ,.O0 BR

42 *G DN PG TVMGVSJD BDJE PG DPODFOUSBUJPO NPM EN¢ JT SFRVJSFE GPS OFVUSBMJTBUJPO PG DN PG QPUBTTJVN IZESPYJEF TPMVUJPO DBMDVMBUF UIF DPODFOUSBUJPO PG UIF QPUBTTJVN IZESPYJEF TPMVUJPO ,0) BR

) 40 BR → , 40 BR

) 0 M

43 $BMDJVN DBSCPOBUF JT SFBDUFE XJUI DN PG NPM EN¢ IZESPDIMPSJD BDJE $B$0 T

)$M BR

→ $B$M BR

$0 H

) 0 M

a 8IBU NBTT PG DBMDJVN DBSCPOBUF JT SFRVJSFE GPS BO FYBDU SFBDUJPO b 8IBU WPMVNF PG DBSCPO EJPYJEF NFBTVSFE BU 451 XJMM CF QSPEVDFE 44 8IBU WPMVNF JO DN PG NPM EN¢ CBSJVN DIMPSJEF NVTU CF SFBDUFE XJUI FYDFTT TPEJVN TVMGBUF UP QSPEVDF H PG CBSJVN TVMGBUF #B$M BR

/B 40 BR

→ #B40 T

/B$M BR

45 *G H PG NBHOFTJVN JT SFBDUFE XJUI DN PG NPM EN¢ IZESPDIMPSJD BDJE DBMDVMBUF UIF WPMVNF PG IZESPHFO HBT QSPEVDFE BU 451 .H T

)$M BR → .H$M BR

) H

Water of crystallisation 4PNF TVCTUBODFT DSZTUBMMJTF XJUI XBUFS BT BO JOUFHSBM QBSU PG UIF DSZTUBM MBUUJDF &YBNQMFT BSF IZESBUFE DPQQFS TVMGBUF $V40 p ) 0 BOE IZESBUFE NBHOFTJVN DIMPSJEF .H$M p ) 0 5IF XBUFS JT OFDFTTBSZ GPS UIF GPSNBUJPO PG UIF DSZTUBMT BOE JT DBMMFE water of crystallisation 4VCTUBODFT UIBU DPOUBJO XBUFS PG DSZTUBMMJTBUJPO BSF EFTDSJCFE BT IZESBUFE XIFSFBT UIPTF UIBU IBWF MPTU UIFJS XBUFS PG DSZTUBMMJTBUJPO BSF EFTDSJCFE BT BOIZESPVT 4P XF UBML BCPVU AIZESBUFE DPQQFS TVMGBUF $V40 p ) 0

BOE ABOIZESPVT DPQQFS TVMGBUF $V40 )ZESBUFE DPQQFS TVMGBUF DBO CF PCUBJOFE BT MBSHF CMVF DSZTUBMT CVU BOIZESPVT DPQQFS TVMGBUF JT XIJUF BOE QPXEFSZ *O UIF DBTF PG $V40 p ) 0 UIF XBUFS DBO CF SFNPWFE CZ IFBUJOH

IFBU

$V40 p ) 0 ⎯→ $V40 ) 0 )PXFWFS UIJT JT OPU BMXBZT UIF DBTF 8IFO .H$M p ) 0 JT IFBUFE NBHOFTJVN PYJEF .H0 JT GPSNFE

IFBU

.H$M p ) 0 ⎯→ .H0 )$M ) 0

1 STOICHIOMETRIC RELATIONSHIPS

45

Worked examples 1.40 8IFO H PG IZESBUFE NBHOFTJVN TVMGBUF .H40 ·xH 0 JT IFBUFE H PG BOIZESPVT NBHOFTJVN TVMGBUF .H40 JT GPSNFE %FUFSNJOF UIF WBMVF PG x JO UIF GPSNVMB NBTT PG XBUFS HJWFO PÄ&#x201E; ¢ J F H NBTT PG .H40 H 5IJT JT OPX CBTJDBMMZ KVTU BO FNQJSJDBM GPSNVMB RVFTUJPO BOE XF OFFE UP ê OE UIF SBUJP CFUXFFO UIF OVNCFST PG NPMFT PG .H40 BOE ) 0 NPMBS NBTT PG ) 0 H NPM¢

NPMBS NBTT PG .H40 H NPM¢

OVNCFS PG NPMFT PG ) 0 NPM

OVNCFS PG NPMFT PG .H40 NPM

%JWJEF CZ UIF TNBMMFS OVNCFS UP HFU UIF SBUJP

5IF WBMVF PG x JT BOE UIF GPSNVMB PG IZESBUFE NBHOFTJVN TVMGBUF JT .H40 p ) 0 1.41 a *G H PG IZESBUFE DPQQFS TVMGBUF $V40 p ) 0 BSF EJTTPMWFE JO XBUFS BOE NBEF VQ UP B WPMVNF PG DN XIBU JT UIF DPODFOUSBUJPO PG UIF TPMVUJPO b 8IBU NBTT PG BOIZESPVT DPQQFS TVMGBUF XPVME CF SFRVJSFE UP NBLF DN PG TPMVUJPO XJUI UIF TBNF DPODFOUSBUJPO BT JO a a NPMBS NBTT PG $V40 p ) 0 H NPM¢

OVNCFS PG NPMFT $V40 p ) 0 NPM

OVNCFS PG NPMFT DPODFOUSBUJPO

NPM EN¢ WPMVNF JO EN

8IFO B IZESBUFE TBMU JT EJTTPMWFE JO XBUFS UIF XBUFS PG DSZTUBMMJTBUJPO KVTU CFDPNFT QBSU PG UIF TPMWFOU BOE UIF TPMVUJPO JT UIF TBNF BT JG UIF BOIZESPVT TBMU XFSF EJTTPMWFE JO XBUFS

4P EJTTPMWJOH H PG $V40 p ) 0 JO XBUFS BOE NBLJOH VQ UIF TPMVUJPO UP DN QSPEVDFT B $V40 TPMVUJPO PG DPODFOUSBUJPO NPM EN¢ b 5IF OVNCFS PG NPMFT PG $V40 QSFTFOU JO DN TPMVUJPO XJMM CF FYBDUMZ UIF TBNF BT BCPWF J F NPM CFDBVTF UIF DPODFOUSBUJPO JT UIF TBNF

NPM $V40 p ) 0 DPOUBJOT NPM $V40

NPMBS NBTT PG $V40 H NPM¢ NBTT PG $V40 NPMBS NBTT ¤ OVNCFS PG NPMFT ¤ H 5IF NBTT PG $V40 SFRVJSFE UP NBLF DN PG B TPMVUJPO PG DPODFOUSBUJPO NPM EN¢ JT H BT PQQPTFE UP H PG $V40 p ) 0 5IF UXP TPMVUJPOT XJMM CF JEFOUJDBM

46

1.42 " H TBNQMF PG IZESBUFE TPEJVN DBSCPOBUF /B $0 ·xH 0 XBT EJTTPMWFE JO XBUFS BOE NBEF VQ UP B UPUBM WPMVNF PG DN 0G UIJT TPMVUJPO DN XBT UJUSBUFE BHBJOTU NPM EN¢ IZESPDIMPSJD BDJE BOE DN PG UIF BDJE XBT SFRVJSFE GPS OFVUSBMJTBUJPO $BMDVMBUF UIF WBMVF PG x JO /B $0 ·xH 0 /B $0 BR

)$M BR â&#x2020;&#x2019; /B$M BR

$0 H

) 0 M

4UFQ m XPSL PVU UIF OVNCFS PG NPMFT PG )$M OVNCFS PG NPMFT DPODFOUSBUJPO ¤ WPMVNF JO EN

OVNCFS PG NPMFT ¤ ¤ ¢ NPM

4UFQ m UIF CBMBODFE FRVBUJPO UFMMT VT UIBU NPM )$M SFBDU XJUI NPM /B $0 5IFSFGPSF ¤ ¢ NPM )$M ¤ ¢ ¤ ¢ NPM /B $0 5IJT JT UIF OVNCFS PG NPMFT PG /B $0 JO DN SFBDU XJUI

5IF PSJHJOBM NBTT PG /B $0 ·xH 0 XBT EJTTPMWFE JO B UPUBM WPMVNF PG DN 5IFSFGPSF UIF OVNCFS PG NPMFT PG /B $0 JO DN PG TPMVUJPO JT ¤ ¢ ¤ J F ¤ ¢ NPM 4UFQ m DPOWFSU OVNCFS PG NPMFT UP NBTT NPMBS NBTT PG /B $0 H NPM¢ NBTT PG ¤ ¢ NPM /B $0 OVNCFS PG NPMFT ¤ NPMBS NBTT NBTT PG /B $0 ¤ ¢ ¤ H 5IF UPUBM NBTT PG /B $0 ·xH 0 H 5IF NBTT PG UIJT UIBU JT EVF UP UIF XBUFS PG DSZTUBMMJTBUJPO ¢ H NBTT

OVNCFS PG NPMFT PG XBUFS PG DSZTUBMMJTBUJPO

NPM NPMBS NBTT 5IF SBUJP NPMFT PG XBUFS PG DSZTUBMMJTBUJPO NPMFT PG TPEJVN DBSCPOBUF DBO CF XPSLFE PVU CZ EJWJEJOH UIF OVNCFS PG NPMFT PG XBUFS CZ UIF OVNCFS PG NPMFT PG TPEJVN DBSCPOBUF

SBUJP ¤ ¢ 5IF WBMVF PG x JT BOE UIF GPSNVMB GPS UIF IZESBUFE TPEJVN DBSCPOBUF JT /B $0 p ) 0

Back titration 5IJT JT B UFDIOJRVF CZ XIJDI B LOPXO FYDFTT PG B QBSUJDVMBS SFBHFOU " JT BEEFE UP BOPUIFS TVCTUBODF 9 TP UIBU UIFZ SFBDU 5IFO UIF FYDFTT " JT UJUSBUFE BHBJOTU BOPUIFS SFBHFOU UP XPSL PVU IPX NVDI " SFBDUFE XJUI UIF TVCTUBODF m BOE UIFSFGPSF IPX NBOZ NPMFT PG 9 XFSF QSFTFOU 5IJT JT VTFGVM XIFO 9 JT BO JNQVSF TVCTUBODF

1 STOICHIOMETRIC RELATIONSHIPS

47

Worked example 1.43 -JNFTUPOF JT JNQVSF DBMDJVN DBSCPOBUF $B$0 H PG MJNFTUPOF JT QVU JOUP B CFBLFS BOE DN PG NPM EN¢ IZESPDIMPSJD BDJE )$M JT BEEFE 5IFZ BSF MFGU UP SFBDU BOE UIFO UIF JNQVSJUJFT BSF ê MUFSFE PÄ&#x201E; BOE UIF TPMVUJPO JT NBEF VQ UP B UPUBM WPMVNF PG DN 0G UIJT TPMVUJPO DN SFRVJSF DN PG NPM EN¢ TPEJVN IZESPYJEF /B0) GPS OFVUSBMJTBUJPO 8PSL PVU UIF QFSDFOUBHF DBMDJVN DBSCPOBUF JO UIF MJNFTUPOF BTTVNF UIBU OPOF PG UIF JNQVSJUJFT SFBDUT XJUI IZESPDIMPSJD BDJE

-FU VT DPOTJEFS UIF ê STU QBSU PG UIF RVFTUJPO A H PG MJNFTUPOF JT QVU JOUP B CFBLFS BOE DN PG NPM EN¢ IZESPDIMPSJD BDJE JT BEEFE $B$0 )$M â&#x2020;&#x2019; $B$M $0 ) 0 5IF MJNFTUPOF JT JNQVSF TP XF DBOOPU XPSL PVU UIF OVNCFS PG NPMFT PG $B$0 QSFTFOU CVU XF EP IBWF FOPVHI JOGPSNBUJPO UP XPSL PVU UIF OVNCFS PG NPMFT PG )$M *G UIF MJNFTUPOF XFSF QVSF $B$0 UIF OVNCFS OVNCFS PG NPMFT PG )$M DPODFOUSBUJPO ¤ WPMVNF JO EN PG NPMFT QSFTFOU JO H XPVME CF NPM

XIJDI XPVME SFBDU XJUI NPM )$M OVNCFS PG NPMFT PG )$M ¤ NPM

5IJT JT FYDFTT )$M BOE XIFO UIF MJNFTUPOF JT SFBDUFE XJUI JU UIFSF XJMM CF TPNF )$M MFGU PWFS 5IF TFDPOE QBSU PG UIF RVFTUJPO JT A5IFZ BSF MFGU UP SFBDU BOE UIFO UIF TPMVUJPO JT NBEF VQ UP B UPUBM WPMVNF PG DN 5IJT DN PG TPMVUJPO OPX DPOUBJOT UIF )$M MFGU PWFS BGUFS UIF SFBDUJPO XJUI UIF $B$0 *O PSEFS UP XPSL PVU UIF OVNCFS PG NPMFT PG )$M UIBU EJE OPU SFBDU XF NVTU DPOTJEFS UIF UIJSE QBSU PG UIF RVFTUJPO A0G UIJT TPMVUJPO DN SFRVJSF DN PG NPM EN¢ TPEJVN IZESPYJEF GPS OFVUSBMJTBUJPO OVNCFS PG NPMFT PG /B0) DPODFOUSBUJPO ¤ WPMVNF JO EN

OVNCFS PG NPMFT PG /B0) ¤ NPM

5IJT SFBDUT XJUI )$M BDDPSEJOH UP UIF FRVBUJPO /B0) )$M â&#x2020;&#x2019; /B$M ) 0 5IFSFGPSF NPM /B0) SFBDU XJUI NPM )$M 5IJT NFBOT UIBU DN PG UIF )$M TPMVUJPO DPOUBJOFE NPM )$M 5IFSFGPSF JO DN PG UIJT TPMVUJPO UIFSF XFSF ¤ J F NPM )$M 5IJT JT UIF OVNCFS PG NPMFT PG )$M MFGU PWFS BGUFS JU IBT SFBDUFE XJUI UIF $B$0 #FDBVTF NPM )$M XBT PSJHJOBMMZ BEEFE UP UIF MJNFTUPOF UIF BNPVOU UIBU SFBDUFE XJUI UIF $B$0 XBT ¢ J F NPM $B$0 )$M â&#x2020;&#x2019; $B$M $0 ) 0 NPM )$M SFBDUT XJUI

J F NPM $B$0

NPMBS NBTT PG $B$0 H NPM¢ NBTT PG $B$0 OVNCFS PG NPMFT ¤ NPMBS NBTT ¤ H

¤ $B$0 JO UIF MJNFTUPOF

48

Linked reactions 4PNFUJNFT UIF QSPEVDU PG POF SFBDUJPO CFDPNFT UIF SFBDUBOU JO B TFDPOE SFBDUJPO " DPNNPO FYBNQMF PG UIJT JT UIF EFUFSNJOBUJPO PG UIF DPODFOUSBUJPO PG DPQQFS JPOT JO TPMVUJPO VTJOH TPEJVN UIJPTVMGBUF

Worked example 1.44 " DN TBNQMF PG B TPMVUJPO PG DPQQFS ** OJUSBUF JT BEEFE UP DN PG NPM EN¢ QPUBTTJVN JPEJEF 5IF JPEJOF QSPEVDFE JT UJUSBUFE BHBJOTU NPM EN¢ TPEJVN UIJPTVMGBUF TPMVUJPO VTJOH TUBSDI JOEJDBUPS OFBS UIF FOE QPJOU DN PG UIF TPEJVN UIJPTVMGBUF TPMVUJPO XBT SFRVJSFE GPS UIF UJUSBUJPO $BMDVMBUF UIF DPODFOUSBUJPO PG UIF DPQQFS ** OJUSBUF TPMVUJPO 5IF JOJUJBM SFBDUJPO PG DPQQFS ** JPOT XJUI JPEJEF JPOT JT

â&#x2C6;&#x2019;

$V BR

* BR â&#x2020;&#x2019; $V* T

* BR reaction 1

5IJT JT B SFEPY UJUSBUJPO BOE UIFTF XJMM CF DPOTJEFSFE BHBJO JO 5PQJD 9

" MBSHF FYDFTT PG JPEJEF JPOT JT BEEFE UP NBLF TVSF UIBU BMM UIF DPQQFS JPOT SFBDU " QSFDJQJUBUF PG $V* JT GPSNFE BT XFMM BT UIF JPEJOF *G XF DBO EFUFSNJOF UIF OVNCFS PG NPMFT PG JPEJOF QSPEVDFE JO UIF TPMVUJPO XF DBO BMTP ê OE UIF OVNCFS PG NPMFT PG DPQQFS JPOT 5IF BNPVOU PG JPEJOF JT EFUFSNJOFE CZ UJUSBUJPO XJUI TPEJVN UIJPTVMGBUF TPMVUJPO 4 0 ¢ BR

* BR â&#x2020;&#x2019; *â&#x2C6;&#x2019; BR

4 0 ¢ BR reaction 2

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1 STOICHIOMETRIC RELATIONSHIPS

49

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¤ ¤ ¢ NPM

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b OVNCFS PG NPMFT PG .$Mx ¤ ¤ ¢ NPM

c 5IF HFOFSBM FRVBUJPO GPS UIF SFBDUJPO JT .$Mx BR

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50

NBTT

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B ¤

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C NPM

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2 LH PG DBSCPO EJPYJEF DPOUBJOT A NPM

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B NPM

C NPM

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0 H â&#x2020;&#x2019; $0 H

) 0 M

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B DN

C DN

D DN

6 8IBU NBTT PG /B S 0 p ) 0 NVTU CF VTFE UP NBLF VQ DN PG B NPM EN¢ TPMVUJPO A H

B H

C H

D H

1 STOICHIOMETRIC RELATIONSHIPS

51

7 DN PG QPUBTTJVN IZESPYJEF ,0) JT FYBDUMZ OFVUSBMJTFE CZ DN PG NPM EN¢ TVMGVSJD BDJE ) 40 5IF DPODFOUSBUJPO PG UIF QPUBTTJVN IZESPYJEF JT A NPM EN¢ B NPM EN¢

C NPM EN¢ D NPM EN¢

8 #BSJVN DIMPSJEF TPMVUJPO SFBDUT XJUI TPEJVN TVMGBUF TPMVUJPO BDDPSEJOH UP UIF FRVBUJPO

#B$M BR

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C NPM EN¢ D NPM EN¢

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52

a LH PG .O 0 BSF IFBUFE XJUI LH PG BMVNJOJVN 8PSL PVU UIF NBYJNVN NBTT PG NBOHBOFTF UIBU DBO CF PCUBJOFE GSPN UIJT SFBDUJPO

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[3]

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[3]

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[3]

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[3]

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e %FUFSNJOF UIF WBMVF PG x

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1 STOICHIOMETRIC RELATIONSHIPS

53

Summary sublimation deposition

melting

Solid

boiling / evaporating

Liquid freezing

Gas condensing

States of matter

The physical and chemical properties of a compound are very different to those of the elements from which it is formed.

Elements

Compounds When elements combine to form compounds they always combine in fixed ratios depending on the number of atoms.

An element is a pure substance that contains only one type of atom.

A compound is a pure substance formed when two or more elements chemically combine.

components can be separated by physical means

Mixtures

Heterogeneous

components can be separated by mechanical means

54

does not have uniform composition and consists of separate phases

contain two or more substances mixed together

Homogeneous

has the same (uniform) composition throughout the mixture and consists of only one phase

P1V1 P2V2 = T1 T2 Real gases deviate most from ideal behaviour at low temperature and high pressure.

number of moles =

for ideal gases: pV = nRT (T in kelvin)

volume molar volume

Avogadro’s law: equal volumes of ideal gases at the same temperature and pressure contain the same number of molecules.

c1v1 c2v2 = n1 n2

concentration (mol dm–3) =

concentration (ppm) =

number of moles (mol) volume (dm3)

Use the limiting reactant to determine the amount of products in a reaction.

mass of solute x 106 mass of solution

actual yield percentage yield = × 100 theoretical yield

Moles calculations 1 Work out the number of moles of anything you can.

2 Use the chemical equation to work out the number of moles of the quantity you require.

3 Convert moles to the required quantity – volume, mass etc.

number of moles =

molar volume of an ideal gas at STP = 22.7 dm3 mol–1

MOLES

1 mol is the amount of substance which contains the same number of particles as there are carbon atoms in 12 g of 12C.

mass of substance molar mass

molar mass = Mr in g mol–1

Relative molecular mass (Mr): sum of all Ar values in a molecule.

To find the limiting reactant in a reaction, divide the number of moles of each reactant by its coefficient.

The lowest number indicates the limiting reactant.

Avogadro’s constant (L) = 6.02 × 1023 mol−1

mass of 1 molecule =

Relative atomic mass (Ar): average of the masses of the isotopes in a naturally occurring sample of the element relative 1 to 12 mass of a 12C atom.

% by mass of an number of atoms of the element × Ar = element in a compound Mr

molar mass Avogadro’s constant

Empirical formula: simplest whole number ratio of the elements present in a compound.

Molecular formula: total number of atoms of each element present in a molecule of the compound.

1 STOICHIOMETRIC RELATIONSHIPS

55

2 Atomic structure 2.1 The nuclear atom

Learning objectives

r Understand that an atom is

made up of protons, neutrons and electrons r Define mass number, atomic number and isotope r Work out the numbers of protons, neutrons and electrons in atoms and ions r Discuss the properties of isotopes r Calculate relative atomic masses and abundances of isotopes r Understand that a mass spectrometer can be used to determine the isotopic composition of a sample electron

neutron

Figure 2.1 A simple representation of a lithium atom (not to scale).

Particle

Relative mass

Relative charge

proton

1

+1

neutron

1

0 −4

5 × 10

–1

Table 2.1 The properties of protons, neutrons and electrons. The mass of an electron is often regarded as negligible.

Protons and neutrons are made up of other particles called quarks.

56

2.1.1 Atoms In the simplest picture of the atom the negatively charged electrons orbit around the central, positively charged nucleus (Figure 2.1).

The nucleus is made up of protons and neutrons (except for a hydrogen atom, which has no neutrons).

proton

electron

There are approximately 92 naturally occurring elements, plus several more that have been made artificially in nuclear reactions, and probably a few more that have yet to be discovered. As far as we know, there are no more naturally occurring elements – these are the only elements that make up our universe.

Chemistry is the study of how atoms of the various elements are joined together to make everything we see around us. It is amazing when one imagines that the entire Universe can be constructed through combinations of these different elements. With just 92 different building blocks (and in most cases many fewer than this), objects as different as a table, a fish and a piece of rock can be made. It is even more amazing when one realises that these atoms are made up of three subatomic (‘smaller than an atom’) particles, and so the whole Universe is made up of combinations of just three things – protons, neutrons and electrons.

Protons and neutrons, the particles that make up the nucleus, are sometimes called nucleons.

The actual mass of a proton is 1.67 × 10−27 kg and the charge on a proton is +1.6 × 10−19 C. Relative masses and charges, shown in Table 2.1, are used to compare the masses of particles more easily. Because the values are relative, there are no units.

From these values it can be seen that virtually all the mass of the atom is concentrated in the nucleus. However, most of the volume of the atom is due to the electrons – the nucleus is very small compared with the total size of the atom.

The diameter of an atom is approximately 1 × 10−10 m and that of a nucleus between about 1 × 10−14 and 1 × 10−15 m, meaning that a nucleus is about 10 000 to 100 000 times smaller than an atom. So, if the nucleus were the size of the full stop at the end of this sentence, the atom would be between 3 and 30 m across.

None of these particles can be observed directly. These particles were originally ‘discovered’ by the interpretation of experimental data. Do we know or believe in the existence of these particles? If we looked at a science textbook of 200 years ago there would be no mention of protons, electrons and neutrons. If we could look at a chemistry textbook of 200 years in the future will there be any mention of them? Are these particles a true representation of reality, or a device invented by scientists to make sense of experimental data and provide an explanation of the world around them?

Democritus and his teacher Leucippus, fifth-century BC Greek philosophers, are often credited with first suggesting the idea of the atom as the smallest indivisible particle of which all matter is made.

The atomic number (Z) defines an element – it is unique to that particular element. For example, the element with atomic number 12 is carbon and that with atomic number 79 is gold. This means that we could use the atomic number of an element instead of its name. However, the name is usually simpler and more commonly used in everyday speech.

The overall charge on an atom is zero and therefore:

Atomic number (Z) is the number of protons in the nucleus of an atom.

number of protons in an atom = number of electrons

John Dalton (1766–1844) is generally regarded as the founder of modern atomic theory.

The electron was discovered in 1897 by J. J. Thompson at the University of Cambridge, UK.

Atomic number is, however, defined in terms of protons, because electrons are lost or gained when ions are formed in chemical reactions.

element symbol

mass number

A

atomic number

Z

Mass number (A) is the number of protons plus neutrons in the nucleus of an atom.

Therefore: number of neutrons in an atom = mass number − atomic number The full symbol of an element includes the atomic number and the mass number (see Figure 2.2). For example, sodium has an atomic number of 11 and a mass number of 23. The nucleus of a sodium atom contains 11 protons and 12 neutrons (23 − 11). Surrounding the nucleus are 11 electrons. The symbol for sodium is 2131Na.

Ions Ions are charged particles that are formed when an atom loses or gains (an) electron(s).

A positive ion (cation) is formed when an atom loses (an) electron(s) so that the ion has more protons(+) than electrons(−) (Figure 2.3). A negative ion (anion) is formed when an atom gains (an) electron(s) so that the ion has more electrons(−) than protons(+).

X

Figure 2.2 Where to place the mass number (A) and atomic number (Z) in the full symbol of an element. positive charge  the atom has lost 11 protons + 12 neutrons an electron therefore there are 10 electrons

23 11

+

Na

11 protons 8 protons + 8 neutrons

16 8

negative charge  the atom has gained two electrons therefore there are 10 electrons

O

2–

8 protons

Figure 2.3 The number of subatomic particles in the Na+ and O2− ions.

2 ATOMIC STRUCTURE

57

2.1.2 Isotopes The mass number of chlorine is given in many periodic tables as 35.5. It is not possible to have half a neutron – the mass number that is given is an average, taking into account the presence of isotopes.

Isotopes are different atoms of the same element with different mass numbers: i.e. different numbers of neutrons in the nucleus.

Isotope Protons Neutrons Electrons 1 1H 2 1H 3 1H 12 6C 13 6C 14 6C 35 17Cl 37 17Cl

1

0

1

1

1

1

1

2

1

6

6

6

6

7

6

6

8

6

17

18

17

17

20

17

Table 2.2 The numbers of subatomic particles in some common isotopes.

The isotopes of hydrogen are sometimes given different names and symbols: hydrogen-1 is called protium; hydrogen-2 is deuterium (D); and hydrogen-3 is tritium (T).

The two isotopes of chlorine are 35Cl (chlorine-35) and 37Cl (chlorine-37). Most naturally occurring samples of elements are composed of a mixture of isotopes, but usually one isotope is far more abundant than the others and the mass number of the most common isotope is quoted.

The numbers of protons, neutrons and electrons in some isotopes are shown in Table 2.2.

Isotopes have the same chemical properties (they react in exactly the same way) but different physical properties (e.g. different melting points and boiling points).

Isotopes react in the same way because they have the same numbers of electrons, and chemical reactions depend only on the number and arrangement of electrons and not on the composition of the nucleus. For example, both protium and deuterium would react in the same way with nitrogen: N2 + 3H2

2NH3

N2 + 3D2

2ND3

Isotopes have different physical properties because, for example, the different masses mean that their atoms move at different speeds. The boiling point of 1H2 is −253 °C, whereas that of 2H2 is −250 °C. Heavy water (D2O) has a melting point of 3.8 °C and a boiling point of 101.4 °C.

Nature of science In science, things change! Science is an ever changing and increasing body of knowledge. This is what Richard Feynman meant when he talked about science as ‘… the belief in the ignorance of experts’.

An important example of how knowledge and understanding have changed is the development of atomic theory. Since the days of John Dalton (1766-1844) our view of the world around us has changed dramatically. These developments have happened through careful observation and experiment and have gone hand-in-hand with improvements in equipment and the development of new technology.

Radioactivity was discovered towards the end of the 19th century and this gave scientists a new tool to probe the atom. In a famous experiment Geiger and Marsden subjected a thin film of metal foil to a beam of alpha particles (helium nuclei, 4He2+) and found that some of the particles were reflected back at large angles. Their experimental data allowed Rutherford to develop the theory of the nuclear atom.

58

The discovery of subatomic particles (protons, neutrons and electrons) in the late nineteenth and early twentieth centuries necessitated a paradigm shift in science and the development of much more sophisticated theories of the structure of matter.

?

Test yourself 1 Give the number of protons, neutrons and electrons in the following atoms: 238 75 81 92 U 33As 35 Br 2 Give the number of protons, neutrons and electrons in the following ions: 40 2+ 127 − 14 0 3+ 20Ca 53 I 58Ce 3 If you consider the most common isotopes of elements as given in a basic periodic table, how many elements have more protons than neutrons in an atom?

4 The following table shows the number of protons, electrons and neutrons in a series of atoms and ions.

Symbol

Protons

Neutrons

Electrons

D

27

30

25

X

43

54

42

Q

35

44

35

L

27

32

26

M

35

46

36

Z

54

78

54

a Which symbols represent isotopes? b Which symbols represent positive ions?

Relative atomic masses Because of the different isotopes present, it is most convenient to quote an average mass for an atom – this is the relative atomic mass (Ar).

The relative atomic mass (Ar) of an element is the average of the masses of the isotopes in a naturally occurring sample of the 1 of an atom of carbon-12. element relative to the mass of 12

How to calculate relative atomic mass

Worked examples 2.1 Lithium has two naturally occurring isotopes: 6

Li: natural abundance 7%

7

Li: natural abundance 93%

Calculate the relative atomic mass of lithium.

Imagine we have 100 Li atoms: 7 will have mass 6 and 93 will have mass 7.

The average mass of these atoms is: (7 × 6) + (93 × 7) = 6.93 100 Therefore the Ar of Li is 6.93.

2 ATOMIC STRUCTURE

59

2.2 Iridium has a relative atomic mass of 192.22 and consists of Ir-191 and Ir-193 isotopes. Calculate the percentage composition of a naturally occurring sample of iridium.

We will assume that we have 100 atoms and that x of these will have a mass of 191. This means that there will be (100 − x) atoms that have a mass of 193.

The total mass of these 100 atoms will be: 191x + 193(100 − x) The average mass of the 100 atoms will be: Therefore we can write the equation:

191x + 193(100 − x) 100

191x + 193(100 − x) = 192.22 100 191x + 193(100 − x) = 19 222

191x + 19 300 − 193x = 19 222 −2x = 19 222 − 19 300 −2x = −78 Therefore x = 39.

This means that the naturally occurring sample of iridium contains 39% Ir-191 and 61% Ir-193.

Alternatively: Ar − mass number of lighter isotope × 100 = % of heavier isotope difference in mass number of two isotopes In the example here:

(192.22 − 191) × 100 = 61% (193 − 191)

The mass spectrum of an element and relative atomic mass The proportion of each isotope present in a sample of an element can be measured using an instrument called a mass spectrometer (Figure 2.4).

The readout from a mass spectrometer is called a mass spectrum. In a mass spectrum of an element, we get one peak for each individual isotope.

The height of each peak (more properly, the area under each peak) is proportional to the number of atoms of this isotope in the sample tested.

The mass spectrum of magnesium is shown in Figure 2.5.

Figure 2.4 Setting up a mass spectrometer.

60

area under peak

Abundance / %

78.6

Mg

11.3

10.1

0

24 25 Mass : charge ratio (m / z)

26

Figure 2.5 The mass spectrum of magnesium showing the amounts of the different isotopes present.

The scale on the x-axis in a mass spectrum is the mass : charge ratio (m/z or m/e). In order to pass through a mass spectrometer, atoms are bombarded with high-energy electrons to produce positive ions. Sometimes more than one electron is knocked out of the atom, which means that the ions behave differently, as if they have smaller masses – hence the use of mass : charge ratio. We can generally ignore this and assume that the horizontal scale refers to the mass of the isotope (the mass of the electron removed is negligible).

The relative atomic mass can be calculated using: Ar =

?

(78.6 × 24) + (10.1 × 25) + (11.3 × 26) = 24.3 100

Test yourself 5 Chromium has four naturally occurring isotopes, and their masses and natural abundances are shown in the table below.

6 Silicon has three naturally occurring isotopes and their details are given in the table below.

Isotope

Isotope

Natural abundance (%)

Natural abundance (%)

28

92.2

4.35

29

4.7

52

83.79

30

3.1

53

9.50

54

2.36

50

Cr Cr Cr Cr

Calculate the relative atomic mass of chromium to two decimal places.

Si Si Si

Calculate the relative atomic mass of silicon to two decimal places.

7 a Indium has two naturally occurring isotopes: indium-113 and indium-115. The relative atomic mass of indium is 114.82. Calculate the natural abundance of each isotope. b Gallium has two naturally occurring isotopes: gallium-69 and gallium-71. The relative atomic mass of gallium is 69.723. Calculate the natural abundance of each isotope.

2 ATOMIC STRUCTURE

61

2.2 Electron configuration

Learning objectives

r Describe the electromagnetic

spectrum r Describe the emission spectrum of hydrogen r Explain how emission spectra arise increasing energy energy levels

electron

K

outer shell/ outer energy level/ valence shell

Figure 2.6 The electron arrangement of potassium.

2.2.1 The arrangement of electrons in atoms At the simplest level of explanation, the electrons in an atom are arranged in energy levels (shells) around the nucleus. For example, the electron arrangement of potassium can be represented as shown in Figure 2.6 and is written as 2,8,8,1 (or 2.8.8.1).

The lowest energy level, called the first energy level or first shell (sometimes also called the K shell), is the one closest to the nucleus. The shells increase in energy as they get further from the nucleus. The maximum number of electrons in each main energy level is shown in Table 2.3.

The main energy level number is sometimes called the principal quantum number and is given the symbol n. The maximum number of electrons in each shell is given by 2n2. main energy level number

1

2

3

4

5

maximum number of electrons

2

8

18

32

50

Table 2.3 Distribution of electrons in energy levels.

The general rule for filling these energy levels is that the electrons fill them from the lowest energy to the highest (from the nucleus out). The first two energy levels must be completely filled before an electron goes into the next energy level. The third main energy level is, however, only filled to 8 before electrons are put into the fourth main energy level. This scheme works for elements with atomic number up to 20.

The electromagnetic spectrum frequency â&#x2C6;?

1 wavelength

Light is a form of energy. Visible light is just one part of the electromagnetic spectrum (Figure 2.7). increasing frequency

frequency â&#x2C6;? energy

increasing energy

radio waves

microwaves

infrared

visible light

ultraviolet

increasing wavelength

Figure 2.7 The electromagnetic spectrum.

62

X-rays

Îł-rays

The various forms of electromagnetic radiation are usually regarded as waves that travel at the speed of light in a vacuum (3.0 × 108 m s−1) but vary in their frequency/energy/wavelength.

Although electromagnetic radiation is usually described as a wave, it can also display the properties of a particle, and we sometimes talk about particles of electromagnetic radiation called photons.

White light is visible light made up of all the colours of the spectrum.

In order of increasing energy, the colours of the spectrum are: red < orange < yellow < green < blue < indigo < violet.

Would our interpretation of the world around us be different if our eyes could detect light in other regions of the electromagnetic spectrum?

Are there really seven colours in the visible spectrum?

Evidence for energy levels in atoms The hydrogen atom spectrum When hydrogen gas at low pressure is subjected to a very high voltage, the gas glows pink (Figure 2.8). The glowing gas can be looked at through a spectroscope, which contains a diffraction grating and separates the various wavelengths of light emitted from the gas. Because light is emitted by the gas, this is called an emission spectrum.

In the visible region, the spectrum consists of a series of sharp, bright lines on a dark background (Figure 2.9). This is a line spectrum, as opposed to a continuous spectrum, which consists of all the colours merging into each other (Figure 2.10).

hydrogen at low pressure

spectroscope

increasing energy/frequency

Figure 2.8 Observing the emission spectrum of hydrogen.

discharge tube

The lines get closer together at higher frequency/energy.

Figure 2.9 A representation of the atomic emission spectrum of hydrogen.

Each element has its own unique emission spectrum, and this can be used to identify the element.

increasing energy/frequency

Figure 2.10 A continuous spectrum.

Line spectrum – only certain frequencies/wavelengths of light present.

Continuous spectrum – all frequencies/wavelengths of light present.

2 ATOMIC STRUCTURE

63

How an emission spectrum is formed Passing an electric discharge through a gas causes an electron to be promoted to a higher energy level (shell) (Figure 2.11).

The electron is in the lowest energy level. This is called the ground state.

electron in higher energy level

+ ENERGY the electron gains energy and moves to a higher energy level

The electron is in a higher energy level than the ground state.

This is called an excited state.

electron in low energy level

electron energy is E2 (high energy) photon of light emitted energy of photon = (E2 â&#x20AC;&#x201C; E1)

Figure 2.11 An electron can be promoted to a higher energy level in a discharge tube.

The electron is unstable in this higher level and will fall to a lower energy level (Figure 2.12). As it returns from a level at energy E2 to E1, the extra energy (E2 â&#x2C6;&#x2019; E1) is given out in the form of a photon of light.This contributes to a line in the spectrum.

The energy levels can also be shown as in Figure 2.13.

electron energy is E1 (lower energy)

Because light is given out, this type of spectrum is an emission spectrum.

Each line in the spectrum comes from the transition of an electron from a high energy level to a lower one.

64

4 Increasing energy

Figure 2.12 When an electron falls from a higher to a lower energy level in an atom, a photon of light is emitted.

6 5 3 2

1

Figure 2.13 How the lines arise in the emission spectrum of hydrogen.

The fact that a line spectrum is produced provides evidence for electrons being in energy levels (shells): i.e. electrons in an atom are allowed to have only certain amounts of energy (Figure 2.14).

energy level 5

5

5

4

4

increasing energy

3 3 2

these transitions would all produce lines in the emission spectrum

2

(there are others)

energy level 1

1

a

b

Figure 2.14 a Electrons in energy levels: only transitions between two discrete energy levels are possible, and a line spectrum is produced. b If the electrons in an atom could have any energy, all transitions would be possible. This would result in a continuous spectrum.

Different series of lines Figure 2.15 shows a representation of the emission spectrum of hydrogen across the infrared, visible and ultraviolet regions. The series in each region consists of a set of lines that get closer together at higher frequency. Each series is named after its discoverer.

The different series of lines occur when electrons fall back down to different energy levels. infrared

visible

Infrared and ultraviolet radiation can be detected only with the aid of technology â&#x20AC;&#x201C; we cannot interact with them directly. Does this have implications as to how we view the knowledge gained from atomic spectra in these regions?

ultraviolet

Exam tip The names of the series do not have to be learnt for the examination.

3 Paschen series

2 Balmer series

1 Lyman series

Figure 2.15 A representation of the emission spectrum of hydrogen. The colours and lines in the spectrum in the infrared and ultraviolet regions are just for illustrative purposes.

All the transitions that occur in the visible region of the spectrum (those we can see) involve electrons falling down to level 2 (creating the Balmer series).

All transitions down to level 1 occur in the ultraviolet region.Therefore we can deduce that the energy difference between level 1 and any other level is bigger than that between level 2 and any other higher level.

The atomic emission spectrum of hydrogen is relatively simple because hydrogen atoms contain only one electron. Ions such as He+ and Li2+, which also contain one electron, would have spectra similar to hydrogen â&#x20AC;&#x201C; but not exactly the same because the number of protons in the nucleus also influences the electron energy levels.

2 ATOMIC STRUCTURE

65

electron can have any energy outside the atom inside the atom

electron can only have certain amounts of energy

electrons falling from outside the atom â&#x2C6;&#x17E; 6 5 4

Convergence The lines in the emission spectrum get closer together at higher frequency/energy (Figure 2.16). increasing energy/frequency

3 2

convergence limit

Figure 2.16 A representation of the Lyman series of hydrogen in the ultraviolet region of the electromagnetic spectrum.

Figure 2.17 The purple arrow represents the transition giving rise to the convergence limit in the Lyman series for hydrogen.

Eventually, at the convergence limit, the lines merge to form a continuum. Beyond this point the electron can have any energy and so must be free from the influence of the nucleus, i.e. the electron is no longer in the atom (Figure 2.17).

Extension The convergence limit is not usually observed, but the frequency can be worked out by plotting a graph of the difference in frequency of successive lines against their frequency and extrapolating the graph to give the frequency when the difference in frequency between successive lines is zero.

Nature of science Advances in technology are often accompanied by advances in science â&#x20AC;&#x201C; Bunsenâ&#x20AC;&#x2122;s development of his burner with a high-temperature flame in the 1850s enabled spectroscopic analysis of substances.

Scientific theories often develop from a need to explain natural phenomena. For instance, Niels Bohr, building on the work of Rutherford, proposed a model for the atom in which electrons orbit the nucleus and only exist in certain allowed energy levels. He then used this model to explain the line spectra of hydrogen and other elements.

?

Test yourself 8 Arrange the following in order of: a increasing energy b decreasing wavelength ultraviolet radiation infrared radiation microwaves orange light green light 9 Describe how a line in the Lyman series of the hydrogen atom spectrum arises.

66

10 Draw an energy level diagram showing the first four energy levels in a hydrogen atom and mark with an arrow on this diagram one electron transition that would give rise to: a a line in the ultraviolet region of the spectrum b a line in the visible region of the spectrum c a line in the infrared region of the spectrum.

2.2.2 Full electron configurations

Learning objectives

The emission spectra of atoms with more than one electron, along with other evidence such as ionisation energy data (see later), suggest that the simple treatment of considering that electrons in atoms occupy only main energy levels is a useful first approximation but it can be expanded.

r Determine the full electron

configuration of atoms with up to 36 electrons r Understand what is meant by an orbital and a subshell (subenergy level)

Sub-energy levels and orbitals Each main energy level in an atom is made up of sub-energy levels (subshells). The first main energy level consists solely of the 1s sub-level, the second main energy level is split into the 2s sub-level and the 2p sub-level.

The sub-levels in each main energy level up to 4 are shown in Table 2.4.

Sub-levels

s 1

1s

2

2s

2p

3

3s

3p

3d

4

4s

4p

4d

p

d

f

2

4f

2

6

2

6

10

2

6

10

4s

3p 3s

sub-level

2p 2s

main energy level

Number of electrons in each sub-level

3d

increasing energy

Main energy level

4p

14

Table 2.4 The sub-levels in each main energy level up to level 4.

Within any main energy level (shell) the ordering of the sub-levels (subshells) is always s < p < d < f, but there are sometimes reversals of orders between sub-levels in different energy levels. The relative energies of the subshells are shown in Figure 2.18.

1s

Figure 2.18 The ordering of the energy levels and sub-levels within an atom. The sub-levels within a main energy level are shown in the same colour.

The Aufbau (building-up) principle (part 1) The Aufbau principle is simply the name given to the process of working out the electron configuration of an atom.

So the full electron configuration of sodium (11 electrons) can be built up as follows: r The first two electrons go into the 1s sub-level → 1s2: this sub-level is now full. r The next two electrons go into the 2s sub-level → 2s2: this sub-level is now full. r The next six electrons go into the 2p sub-level → 2p6: this sub-level is now full. r The last electron goes into the 3s sub level → 3s1.

The full electron configuration of sodium is therefore 1s2 2s2 2p6 3s1 (Figure 2.19). This can also be abbreviated to [Ne]3s1, where the electron configuration of the previous noble gas is assumed and everything after that is given in full.

increasing energy

Electrons fill sub-levels from the lowest energy level upwards – this gives the lowest possible (potential) energy.

3s

2p 2s

sodium 1s

Figure 2.19 The arrangement of electrons in energy levels for a sodium atom.

[1s2 2s2 2p6] 3s1 from neon

2 ATOMIC STRUCTURE

67

The full electron configuration of iron (26 electrons) is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 (Figure 2.20). Note that because the 4s sub-level is lower in energy than the 3d sub-level, it is filled first. In other words, two electrons go into the fourth main energy level before the third main energy level is filled. This can be written as [Ar]4s2 3d6.

3d 4s 3p

increasing energy

3s

2p

This is sometimes written as [Ar]3d6 4s2 to keep the sub-levels in order of the main energy levels.

2s

The full electronic configuration of germanium (32 electrons) is: 1s 2s2 2p6 3s2 3p6 4s2 3d10 4p2. Or, in abbreviated form: [Ar]4s2 3d10 4p2.

The order in which the sub-levels are filled can be remembered most easily from the periodic table. For example, Selenium (Se) is in period 4 and 4 along in the p block â&#x20AC;&#x201C; therefore the last part of the electron configuration is 4p4. The full electron configuration can be worked out by following the arrows in Figure 2.21: Hâ&#x2020;&#x2019;He 1s2 Liâ&#x2020;&#x2019;Be 2s2 Bâ&#x2020;&#x2019;Ne 2p6 Naâ&#x2020;&#x2019;Mg 3s2 Alâ&#x2020;&#x2019;Ar 3p6 2 10 Kâ&#x2020;&#x2019;Ca 4s Scâ&#x2020;&#x2019;Zn 3d Gaâ&#x2020;&#x2019;Se 4p4 (remember to go down 1 in the d block) Therefore the electron configuration of selenium is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4.

Figure 2.22 shows an alternative way of remembering the order in which sub-levels are filled.

iron

2

1s

Figure 2.20 The arrangements of electrons in energy levels for an iron atom. start here

1s 2s

2p

3s

3p

3d

4s

4p

4d

4f

Note: all the atoms in the same group (vertical column) of the periodic table have the same outer shell electron configuration. For example, all the elements in group 16 (like Se) have the outer shell electron configuration ns2 np4, where n is the period number.

Figure 2.22 Draw out the sub-levels in each main energy level. Starting at 1s, follow the arrows to give the ordering of the sub-levels.

1

2

1 H

number of electrons in outer shell

group number â&#x2C6;&#x2019;10 = number of electrons in outer shell

13 14 15 16 17 18 He 1s2

period number gives number of highest main energy level occupied

1

2

2 Li

Be

3 Na Mg

number of electrons in s sub-level

2s2 3s2

number of electrons in p sub-level

1

2

3

4

5

6

B

C

N

O

F

Ne

10

AI

Si

P

S

CI

Ar

Ga Ge As Se

Br

Kr

I

Xe

number of electrons in d sub-level

1

2

3

4

Sc

Ti

V

Cr Mn Fe Co Ni Cu Zn

5 Rb Sr

Y

Zr Nb Mo Tc

6 Cs Ba

La

Hf

4

K

Ca

s block

4s2

Ta

W

5

6

7

8

9

Ru Rh Pd Ag Cd

Re Os d block

Ir

Pt Au Hg

3d10

In

Sn Sb Te

TI

Pb Bi

Po At Rn

p block

Figure 2.21 Electron configurations can be worked out from the periodic table. The â&#x20AC;&#x2DC;p blockâ&#x20AC;&#x2122; is so named because the highest occupied TVC MFWFM JT B Q TVC MFWFM 5IF QFSJPE OVNCFS JOEJDBUFT UIF IJHIFTU PDDVQJFE NBJO FOFSHZ MFWFM 4PNF FYDFQUJPOT UP UIF HFOFSBM SVMFT GPS filling sub-levels are highlighted in pink. Helium has the configuration 1s2 and has no p electrons, despite the fact that it is usually put in the p block.

68

2p6 3p6 4p6

?

Test yourself 11 Give the full electron configurations of the following atoms: a N c Ar e V b Si d As

Orbitals

An orbital can contain a maximum of two electrons.

Electrons occupy atomic orbitals in atoms.

An orbital is a region of space in which there is a high probability of finding an electron. It represents a discrete energy level.

There are four different types of atomic orbital: s p d f The first shell (maximum number of electrons 2) consists of a 1s orbital and this makes up the entire 1s sub-level. This is spherical in shape (Figure 2.23a).

The 1s orbital is centred on the nucleus (Figure 2.23b). The electron is moving all the time and the intensity of the colour here represents the probability of finding the electron at a certain distance from the nucleus.

The darker the colour the greater the probability of the electron being at that point. This represents the electron density.

The electron can be found anywhere in this region of space (except the nucleus – at the centre of the orbital) but it is most likely to be found at a certain distance from the nucleus.

The second main energy level (maximum number of electrons 8) is made up of the 2s sub-level and the 2p sub-level. The 2s sub-level just consists of a 2s orbital, whereas the 2p sub-level is made up of three 2p orbitals. The 2s orbital (like all other s orbitals) is spherical in shape and bigger than the 1s orbital (Figure 2.24). p orbitals have a ‘dumb-bell’ shape (Figure 2.25). Three p orbitals make up the 2p sub-level. These lie at 90° to each other and are named appropriately as px, py, pz (Figure 2.26). The px orbital points along the x-axis. The three 2p orbitals all have the same energy – they are described as degenerate.

a

b

Figure 2.23 a The shape of a 1s orbital; b the electron density in a 1s orbital.

2s orbital

1s orbital

Figure 2.24 A cross section of the electron density of the 1s and 2s orbitals together.

z y x

pz a

b

Figure 2.25 a The shape of a 2p orbital; b the electron density in a 2p orbital.

px

py

Figure 2.26 The three p orbitals that make up a p sub-level point at 90° to each other.

2 ATOMIC STRUCTURE

69

2s

Figure 2.28 One of the five d orbitals in the 3d sub-level.

Figure 2.27 The 2s and 2p sub-levels in the second main energy level.

Figure 2.27 shows the orbitals that make up the 2s and 2p sub-levels in the second main energy level.

The third shell (maximum 18 electrons) consists of the 3s, 3p and 3d sub-levels. The 3s sub-level is just the 3s orbital; the 3p sub-level consists of three 3p orbitals; and the 3d sub-level is made up of five 3d orbitals.

One of the five 3d orbitals is shown in Figure 2.28.

The fourth shell (maximum 32 electrons) consists of one 4s, three 4p, five 4d and seven 4f orbitals. The seven 4f orbitals make up the 4f sublevel. One of the f orbitals is shown in Figure 2.29.

Figure 2.29 One of the f orbitals in the 4f sub-level.

Main energy level (shell)

s

p

d

1

1

2

1

3

3

1

3

5

4

1

3

5

f

Table 2.5 The number of orbitals in each energy level.

7

Within any subshell, all the orbitals have the same energy (they are degenerate) – e.g. the three 2p orbitals are degenerate and the five 3d orbitals are degenerate.

The number of orbitals in each energy level is shown in Table 2.5.

The diagrams of atomic orbitals that we have seen here are derived from mathematical functions that are solutions to the Schrödinger equation. Exact solutions of the Schrödinger equation are only possible for a system involving one electron, i.e. the hydrogen atom. It is not possible to derive exact mathematical solutions

Electrons can be regarded as either spinning in one direction (clockwise);

or in the opposite direction (anticlockwise).

Putting electrons into orbitals – the Aufbau principle (part 2) As well as moving around in space within an orbital, electrons also have another property called spin.

There are two rules that must be considered before electrons are put into orbitals.

1 The Pauli exclusion principle: the maximum number of electrons in an orbital is two. If there are two electrons in an orbital, they must have opposite spin. orbital correct

70

for more complex atoms. What implications does this have for the limit of scientific knowledge? When we describe more complex atoms in terms of orbitals, we are actually just extending the results from the hydrogen atom and gaining an approximate view of the properties of electrons in atoms.

2 Hund’s rule: electrons fill orbitals of the same energy (degenerate orbitals) so as to give the maximum number of electrons with the same spin.

Here we can see how three electrons occupy the orbitals of the 2p sub-level:

Exam tip These diagrams are sometimes described as ‘orbital diagrams’, ‘arrows in boxes’ or ‘electrons in boxes’.

2p sub-level py

px

pz

By contrast, these higher energy situations do not occur:

px

py

pz

2p sub-level px

py

pz

Figures 2.30a and b show the full electron configuration of oxygen and silicon atoms, respectively.

There are a small number of exceptions to the rules for filling sublevels – i.e. electron configurations that are not quite as expected. Two of these exceptions are chromium and copper, which, instead of having electron configurations of the form [Ar]3dn4s2 have only one electron in the 4s sub-level: 5 1 24Cr: [Ar]3d 4s

10 1 29Cu: [Ar]3d 4s

The reasons for this are complex and beyond the level of the syllabus – but in general, having the maximum number of electron spins the same within a set of degenerate orbitals gives a lower energy (more stable) situation.

increasing energy

2p sub-level

2p4 2s2

O

1s2

8

3p2 3s2 2p6 2s2 1s2

14

Si

Figure 2.30 Electron configuration: a PYZHFO b silicon.

Nature of science Scientific theories are constantly being modified, improved or replaced as more data become available and the understanding of natural phenomena improves. The most up-to-date theory of the structure of the atom involves quantum mechanics and this has replaced previous theories.

Developments in apparatus and techniques have been essential in the advancement of science. For instance, J.J. Thomson used electric and magnetic fields to investigate the properties of cathode rays, which led to the discovery of the electron.

?

Test yourself 12 Draw out the full electron configurations of the following atoms, showing electrons in boxes: a C b P c Cr

2 ATOMIC STRUCTURE

71

2.3 Electrons in atoms (HL)

Learning objectives

2.3.1 Ionisation energy and the convergence limit

r Solve problems using E = hv r Explain successive ionisation

data for elements r Explain the variation in first ionisation energy across a period and down a group

6

∞

5

As discussed on page 66, the lines in the emission spectrum of an atom get closer together at higher frequency/energy.

Eventually, at the convergence limit, the lines merge to form a continuum. Beyond this point the electron can have any energy and so must be free from the influence of the nucleus, i.e. the electron is no longer in the atom (Figure 2.17 on page 66). Knowing the frequency of the light emitted at the convergence limit enables us us to work out the ionisation energy of an atom – the energy for the process: M(g) → M+(g) + e−

4

3

2

The ionisation energy is the minimum amount of energy required to remove an electron from a gaseous atom.

The ionisation energy for hydrogen represents the minimum energy for the removal of an electron (from level 1 to ∞) (Figure 2.31), and the frequency of the convergence limit in the Lyman series represents the amount of energy given out when an electron falls from outside the atom to level 1 (∞ to 1). These are therefore the same amount of energy.

The relationship between the energy of a photon and the frequency of electromagnetic radiation

1

Figure 2.31 The ionisation process in a hydrogen atom.

Light, and other forms of electromagnetic radiation, exhibit the properties of both waves and particles – this is known as wave–particle duality.

The energy (E) of a photon is related to the frequency of the electromagnetic radiation: E = hv where v is the frequency of the light (Hz or s−1) h is Planck’s constant (6.63 × 10−34 J s) This equation can be used to work out the differences in energy between various levels in the hydrogen atom.

Worked example 2.3 The frequency of a line in the visible emission spectrum of hydrogen is 4.57 × 1014 Hz. Calculate the energy of the photon emitted.

E=h Therefore E = 6.63 ×10−34 × 4.57 × 1014 = 3.03 × 10−19 J This line in the spectrum represents an electron falling from level 3 to level 2 and so the energy difference between these two levels is 3.03 × 10−19 J. 72

The wavelength of the light can be worked out from the frequency using the equation: c = vλ where λ is the wavelength of the light (m) c is the speed of light (3.0 × 108 ms–1) The two equations E = hv and c = vλ can be combined: E=

hc λ

This relates the energy of a photon to its wavelength.

Worked example 2.4 If the frequency of the convergence limit in the Lyman series for hydrogen is 3.28 × 1015 Hz, calculate the ionisation energy of hydrogen in kJ mol−1.

E=h Therefore E = 6.63 × 10−34 × 3.28 × 1015 = 2.17 × 10−18 J This represents the minimum amount of energy required to remove an electron from just one atom of hydrogen, but we are required to calculate the total energy required to remove one electron from each atom in 1 mole of hydrogen atoms – therefore we must multiply by Avogadro’s constant.

The energy required is 2.17 × 10−18 × 6.02 × 1023, or 1.31 × 106 J mol−1.

Dividing by 1000 gives the answer in kJ mol−1, so the ionisation energy of hydrogen is 1.31 × 103 kJ mol−1.

The ionisation energy of hydrogen can be obtained only from a study of the series of lines when the electron falls back to its ground state (normal) energy level – in other words, only the Lyman series, where the electron falls back down to level 1.

Ionisation energy and evidence for energy levels and sub-levels The first ionisation energy for an element is the energy for the process: M(g) → M+(g) + e− The full definition is the energy required to remove one electron from each atom in one mole of gaseous atoms under standard conditions, but see later.

2 ATOMIC STRUCTURE

73

The second ionisation energy is: M+(g) → M2+(g) + e− The nth ionisation energy is: M(n−1)+(g) → Mn+(g) + e− The highest energy electrons are removed first. Figure 2.32 shows this for potassium, in which the highest energy electron is the 4s1, and this is the first to be removed: 4s1

electron removed

4s0

3p6 3s2

3p6 3s2

2p6

2p6

2s2

2s2

1s2

1s2

K

K+

Figure 2.32 The first ionisation of potassium. repulsion between electrons

nucleus

only repulsions between outer electrons shown

removed

electron

attraction between electron and nucleus

less repulsion between electrons - electrons pulled in closer to nucleus

Figure 2.33 When an electron is removed from an atom, the remaining electrons are drawn closer to the nucleus due to reduced repulsion.

74

The second ionisation energy is always higher than the first, and this can be explained in two ways: 1 Once an electron has been removed from an atom, a positive ion is created. A positive ion attracts a negatively charged electron more strongly than a neutral atom does. More energy is therefore required to remove the second electron from a positive ion.

2 Once an electron has been removed from an atom, there is less repulsion between the remaining electrons. They are therefore pulled in closer to the nucleus (Figure 2.33). If they are closer to the nucleus, they are more strongly attracted and more difficult to remove.

Successive ionisation energies of potassium The graph in Figure 2.34 shows the energy required to remove each electron in turn from a gaseous potassium atom.

The simple electron arrangement of potassium is 2,8,8,1 and this can be deduced directly from Figure 2.34. The large jumps in the graph occur between the main energy levels (shells).

The outermost electron in potassium is furthest from the nucleus and therefore least strongly attracted by the nucleus – so this electron is easiest to remove. It is also shielded (screened) from the full attractive force of the nucleus by the other 18 electrons in the atom (Figure 2.35).

6.0

2

Plotting log10 of these numbers reduces the range. The 1st ionisation energy of potassium is 418 kJ mol−1, whereas the 19th is 475 000 kJ mol−1. It would be very difficult to plot these values on a single graph.

5.5 8 electrons

log10(IE / kJ mol –1)

5.0 4.5 8 electrons

4.0 3.5 3.0 2.5 2.0

1

1

2

3 4

5

6

7 8 9 10 11 12 13 14 15 16 17 18 19 Number of ionisation energy

Figure 2.34 Successive ionisation energies (IE) for potassium.

A log scale is used here to allow all the data to be plotted on one graph, but although on one level this has made the data easier to interpret and support the explanations that have been given, it has also distorted the data. The difference between the first and second ionisation energies of potassium is about 2600 kJ mol−1, but the difference between the 18th and 19th ionisations energies is over 30 000 kJ mol−1! How can the way data are presented

be used by scientists to support their theories? Can you find examples where the scale on a graph has been chosen to exaggerate a particular trend – is scientific knowledge objective or is it a matter of interpretation and presentation? The arguments for and against human-made climate change are a classic example of where the interpretation and presentation of data are key in influencing public opinion.

Complete shells of electrons between the nucleus and a particular electron reduce the attractive force of the nucleus for that electron. There are three full shells of electrons between the outermost electron and the nucleus, and if this shielding were perfect the effective nuclear charge felt by the outer electron would be 1+ (19+ in nucleus – 18 shielding electrons). This shielding is not perfect, however, and the effective nuclear charge felt by the outermost electron is higher than +1.

K 19+

An alternative view of shielding is that the outer electron is attracted by the nucleus but repelled by the inner electrons.

Figure 2.35 The outer electron in a potassium atom is shielded from the full attractive force of the nucleus by the inner shells of electrons (shaded in blue).

2 ATOMIC STRUCTURE

75

Extension

K 19+

a

K 19+

b

9th electron

K 19+

c

10th electron

Once the first electron has been removed from a potassium atom, the next electron is considerably more difficult to remove (there is a large jump between first and second ionisation energies). This is consistent with the electron being removed from a new main energy level (shell). This electron is closer to the nucleus and therefore more strongly attracted (Figure 2.36a). It is also shielded by fewer electrons (the ten electrons in the inner main energy levels), because electrons in the same shell do not shield each other very well (they do not get between the electron and the nucleus).

The ionisation energy now rises steadily as the electrons are removed successively from the same main energy level. There is no significant change in shielding, but as the positive charge on the ion increases it becomes more difficult to remove a negatively charged electron (less electron–electron repulsion, so the electrons are pulled in closer to the nucleus).

There is another large jump in ionisation energies between the ninth and the tenth (Figure 2.36b and c) because the ninth electron is the last to be removed from the third main energy level but the tenth is the first to be removed from the second level. The tenth electron is significantly closer to the nucleus and is less shielded than the ninth electron to be removed.

Graphs of successive ionisation energy give us information about how many electrons are in a particular energy level. Consider the graph log10(ionisation energy / kJ mol–1)

Figure 2.36 The ionisation energy depends on which main energy level the electron is removed from.

In electrostatics, a sphere of charge behaves like a point charge at its centre; therefore relative to the outer electron, spheres of charge inside (the electron shells) behave as if their charge is at the nucleus. The charge felt by the outer electron is (19+) + (18−) = 1+ acting at the nucleus.

The electrons do not form perfect spheres of charge, and the movement of the outer electron is not simply in an orbit around the nucleus as shown, and this is why the effective nuclear charge felt by the outer electron in potassium is greater than 1. There are various ways of estimating or calculating the effective nuclear charge for a particular electron in an atom (e.g. Slater’s rules). Calculations suggest that the effective nuclear charge felt by the outer electron in potassium is about 3.5+.

5.5 5.0 4.5 4.0 3.5 3.0 2.5

0

1

2

3

4

5 6 7 8 9 10 Number of ionisation energy

Figure 2.37 The successive ionisation energies of silicon.

76

11

12

13

14

4000 3500

3226

3000 2500 2000

1574

1500 785

1000 500 0

0

1 2 3 4 Number of ionisation energy

5

Figure 2.38 The first four ionisation energies of silicon.

Test yourself 13 a The frequency of a line in the emission spectrum of hydrogen is 7.31 × 1014 Hz. Calculate the energy of the photon emitted. b The energy of a photon is 1.53 × 10−18 J. Calculate the frequency of the electromagnetic radiation.

14 The table shows the successive ionisation of some elements.

Deduce which group in the periodic table each element is in.

outside the atom

∞

3p 3s

Ionisation energy / kJ mol−1 Number of ionisation energy

4348

4500

less energy required

?

5000

Ionisation energy / kJ mol–1

for silicon shown in Figure 2.37. There is a large jump in the ionisation energy between the fourth and the fifth ionisation energies, which suggests that these electrons are removed from different main energy levels. It can therefore be deduced that silicon has four electrons in its outer main energy level (shell) and is in group 14 of the periodic table.

If a graph of ionisation energy (rather than log10 ionisation energy) is plotted for the removal of the first few electrons from a silicon atom, more features can be seen (Figure 2.38). For example there is a larger jump in the ionisation energy between the second and third ionisation energies.

The full electron configuration of silicon is 1s22s22p63s23p2. The first two electrons are removed from the 3p sub-level (subshell), whereas the third electron is removed from the 3s sub-level (Figure 2.39). The 3p sublevel is higher in energy than the 3s sub-level, and therefore less energy is required to remove the electron. This provides evidence for the existence of sub energy levels (subshells) in an atom.

Element X Element Z Element Q

1

1 085

736

1 400

2

2 349

1 448

2 851

3

4 612

7 719

4 570

4

6 212

10 522

7 462

5

37 765

13 606

9 429

6

47 195

17 964

53 174

2p 2s

1s

Figure 2.39 .PSF FOFSHZ JT SFRVJSFE UP remove an electron from the 3s sub-level of silicon than from the 3p sub-level.

We are using reasoning to deduce the existence of energy levels in an atom. Do we know that energy levels exist?

2 ATOMIC STRUCTURE

77

Variation in ionisation energy across a period

The general trend is that ionisation energy increases from left to right across a period.

The first ionisation energies for the elements in period 2, from lithium to neon, are plotted in Figure 2.40.

Ne 2000

Ionisation energy / kJ molâ&#x20AC;&#x201C;1

F 1500

N

O

C 1000

500

Be

B

Li

Li 3+

0 Period 2

Figure 2.40 The first ionisation energies for the period 2 elements.

The nuclear charge increases from lithium (3+) to neon (10+) as protons are added to the nucleus (Figure 2.41). The electrons are all removed from the same main energy level and, because electrons in the same energy level do not shield each other very well, there is no big change in shielding.

Therefore the attractive force on the outer electrons increases from left to right across the period, and the outer electron is more difficult to remove for neon. The neon atom is also smaller than the lithium atom, and so the outer electron is closer to the nucleus and more strongly held.

Ne 10+

This can also be explained in terms of the effective nuclear charge felt by the outer electron in neon being higher.

Figure 2.41 Ne has more protons in the nucleus, but the amount of shielding from inner electrons is roughly the same as in lithium. outside the atom

Be

less energy required

2p

â&#x2C6;&#x17E;

Be 1s22s2 2p

2s B

Figure 2.42 The 2p subshell in boron is higher in energy than the 2s subshell in beryllium.

78

There are two exceptions to the general increase in ionisation energy across a period.

Despite the fact that boron has a higher nuclear charge (more protons in the nucleus) than beryllium the ionisation energy is lower. The electron configurations of beryllium and boron are:

2s

B 1s22s22p1

The major difference is that the electron to be removed from the boron atom is in a 2p sub-level, whereas it is in a 2s sub-level in beryllium. The 2p sub-level in B is higher in energy than the 2s sub-level in beryllium (Figure 2.42), and therefore less energy is required to remove an electron from boron.

Extension An alternative, more in-depth, explanation is that the 2p electron in boron is shielded to a certain extent by the 2s electrons, and this increase in shielding from beryllium to boron offsets the effect of the increase in nuclear charge. 2s electrons shield the 2p electrons because there is a significant probability of the 2s electron being closer to the nucleus and therefore getting between the 2p electron and the nucleus.

The second exception is that the first ionisation energy of oxygen is lower than that of nitrogen.

The electron configurations for nitrogen and oxygen are: N 1s22s22p3

O 1s22s22p4

The major difference is that oxygen has two electrons paired up in the same p orbital, but nitrogen does not (Figure 2.43). An electron in the same p orbital as another electron is easier to remove than one in an orbital by itself because of the repulsion from the other electron.

When two electrons are in the same p orbital they are closer together than if there is one in each p orbital. If the electrons are closer together, they repel each other more strongly. If there is greater repulsion, an electron is easier to remove.

Down a group in the periodic table the ionisation energy decreases.

2p3

2p4

N

O

Figure 2.43 Electrons in the 2p sub-level of OJUSPHFO BOE PYZHFO

The variation in first ionisation energy down a group is discussed on page 92.

The transition metals The transition metals will be considered in more detail in a later topic, but they are mentioned here for completeness. These elements represent a slight departure from the â&#x20AC;&#x2DC;last in, first outâ&#x20AC;&#x2122; rule for ionisation energy.

Although the sub-levels are filled in the order 4s and then 3d, the 4s electrons are always removed before the 3d electrons.

The full electron configuration for an iron atom is 1s22s22p63s23p64s23d6.

The electron configuration for Fe2+ is 1s22s22p63s23p63d6.The electron configuration for Fe3+ is 1s22s22p63s23p63d5.

Nature of science Theories in science must be supported by evidence. Experimental evidence from emission spectra and ionisation energy is used to support theories about the electron arrangements of atoms.

?

Test yourself 15 Work out the full electron configurations of the following ions: b Cr3+ c Co2+ d Rb+ a Ca2+

Extension When removing electrons, we should not really think about the order they were put into the atom but consider the stability of the final ion. The electron configuration of the final ion will be that which generates the ion of lowest energy. s electrons are generally better at shielding other electrons than are d electrons; by removing the 4s electrons, the shielding of the remaining 3d electrons is reduced, and these are lowered in energy. If the 3d electrons are removed, there is no real energy advantage in terms of reduced shielding â&#x20AC;&#x201C; therefore it is less favourable to remove the 3d electrons.

Overall, what this all amounts to is that, in the ion, the 3d sub-level is lower in energy than the 4s orbital.

2 ATOMIC STRUCTURE

79

Exam-style questions 1 Which of the following contains 50 neutrons?

A

50 23 V

B

89 + 39 Y

C

91 + 40 Zr

D

86 + 37 Rb

79 − 35 Br

D

40 2+ 20 Ca

2 Which of the following has more electrons than neutrons?

A

9 2+ 4 Be

B

31 3− 15 P

C

3 Rhenium has two naturally occurring isotopes, 185Re and 187Re. The relative atomic mass of rhenium is 186.2.

What are the natural abundances of these isotopes?

A B C D

40% 185Re and 60% 187Re 60% 185Re and 40% 187Re 12% 185Re and 88% 187Re 88% 185Re and 12% 187Re

4 Which of the following electron transitions in the hydrogen atom will be of highest energy?

A n=8 → n=4 B n=7 → n=2

C n=9 → n=3 D n=6 → n=2

HL 5 Within any main energy level the correct sequence, when the sub-energy levels (subshells) are arranged in order of

increasing energy, is: A s B d

p s

f f

d p

C s D s

p d

d p

f f

HL 6 Which of the following electron configurations is not correct?

A Mg: 1s22s22p63s2 B Cu: 1s22s22p63s23p64s23d9

C Ge: 1s22s22p63s23p63d104s24p2 D Br: 1s22s22p63s23p64s23d104p5

−34 14 HL 7 Planck’s constant is 6.63 × 10 J s. The energy of a photon of light with frequency 5.00 × 10 Hz is:

A 7.54 × 1047 J B 1.33 × 10−48 J

C 3.32 × 10−19 J D 1.33 × 10−20 J

HL 8 Which of the following does not have three unpaired electrons?

A P

B V

C Mn3+

D Ni3+

HL 9 In which of the following does the second element have a lower first ionisation energy than the first?

A B C D

80

Si Na Be Ar

C Mg B Ne

HL 10 The first four ionisation energies of a certain element are shown in the table below.

Number of ionisation energy

Ionisation energy / kJ mol−1

1

418

2

3046

3

4403

4

5866

In which group in the periodic table is this element?

A group 1

B group 2

C group 13

D group 14

11 a Define the terms atomic number and isotopes.

[3]

b State the number of protons, neutrons and electrons in an atom of 5276 Fe.

[2]

c A sample of iron from a meteorite is analysed and the following results are obtained.

Isotope

Abundance / %

54

5.80

56

91.16

57

3.04

Fe Fe Fe

i Name an instrument that could be used to obtain this data. ii Calculate the relative atomic mass of this sample, giving your answer to two decimal places.

12 a Describe the difference between a continuous spectrum and a line spectrum.

[1] [2] [2]

b Sketch a diagram of the emission spectrum of hydrogen in the visible region, showing clearly the relative energies of any lines.

[2]

c Explain how a line in the visible emission spectrum of hydrogen arises.

[3]

HL d The frequencies of two lines in the emission spectrum of hydrogen are given in the table.

Calculate the energy difference between levels 5 and 6 in a hydrogen atom.

Higher level

Lower level

Frequency / Hz

5

3

2.34 × 1014

6

3

2.74 × 1014

[2]

13 a Write the full electron configuration of an atom of potassium.

[1]

HL b Write an equation showing the second ionisation energy of potassium.

[2]

HL c Explain why the second ionisation energy of potassium is substantially higher than its first ionisation

energy.

[3]

HL d State and explain how the first ionisation energy of calcium compares with that of potassium.

[3]

2 ATOMIC STRUCTURE

81

14 a Write the full electron configuration of the O2â&#x2C6;&#x2019; ion. b Give the formula of an atom and an ion that have the same number of electrons as an O2â&#x2C6;&#x2019; ion.

HL c Explain why the first ionisation energy of oxygen is lower than that of nitrogen.

[1] [2] [2]

HL d Sketch a graph showing the variation of the second ionisation energy for the elements in period 2 of

the periodic table from lithium to neon.

82

[3]

Summary ATOMS

contain

filled from lowest to highest energy

nucleus electrons

contains

arranged in energy levels (shells)

protons neutrons

To SUBSHELLS on OFYU page

number of protons = atomic number number of neutrons + protons = mass number

Isotopes are atoms of the same element that have different mass numbers.

can be separated by mass spectrometry

Atomic emission spectra are caused by electrons falling from a higher energy level to a lower one.

line spectra – only certain GSFRVFODJFT BSF QSFTFOU

series of lines spectrum becomes continuous at the convergence limit

HL

HL

E = h can be used to work out the ionisation energy of hydrogen from the convergence limit of the Lyman series

To IONISATION ENERGY PO OFYU QBHF Lyman series – electron falls to energy level 1

Balmer series – electron falls to energy level 2

Paschen series – electron falls to energy level 3

emitted radiation is ultraviolet

emitted radiation is visible light

emitted radiation is infrared

all part of the electromagnetic spectrum

2 ATOMIC STRUCTURE

83

Summary – continued SUBSHELLS

contain

orbitals

types

s (spherical) p (dumb-bell shaped)

px T NBY F–) Q NBY F–) E NBY F–) G NBY F–)

py

pz

increasing energy in a shell

The full electron configuration of an atom shows the distribution of its electrons over its subshells.

example

The Pauli exclusion principle: 5IF NBYJNVN OVNCFS PG FMFDUSPOT in an orbital is 2. If there are 2 electrons in an orbital, they must have opposite spins.

Ge: 1s22s22p63s23p64s23d104p2

remember

24Cr:

[Ar]3d54s1

29Cu:

Hund’s rule: Electrons fill degenerate orbitals UP HJWF UIF NBYJNVN OVNCFS PG electrons with the same spin.

[Ar]3d104s1

HL IONISATION ENERGY

The electron with the highest energy is removed first.

The 4s electrons are removed first from transition metals.

first ionisation energy FOFSHZ SFRVJSFE UP remove the first electron)

nth ionisation energy FOFSHZ SFRVJSFE to remove the nth electron)

nth ionisation energy > (n – 1)th ionisation energy because the electron is removed from a more positive ion

(in general) increases across a period as the nuclear charge increases

large jump when an electron is removed from a new main energy level (shell)

The electron in the new energy level is closer to the nucleus and so more strongly attracted.

BUT!

B has a lower 1st ionisation energy than Be because the electrons are removed from different subshells.

84

O has a lower 1st ionisation energy than N because O has 2e– paired in the same p orbital.

The periodic table 3 3.1 The periodic table

Learning objectives

The elements in the periodic table are arranged in order of atomic number starting with hydrogen, which has atomic number 1. The groups are the vertical columns in the periodic table and the periods are the horizontal rows (Figure 3.1).

Most of the elements in the periodic table are metals – these are shown in yellow in Figure 3.1. The elements shown in pink are non-metals.

period number

group number: 1

2

3

4

5

6

7

8

9

10

11

12

r Understand how the elements in the periodic table are arranged r Understand the terms group and period r Understand how the electron configuration of an element relates to its position in the periodic table

13

14

1

H

2

Li Be

B

C

3

Na Mg

AI

Si

4

15

16

17

He

1

3

11

K

19

2

O

F

Ne

S

CI

Ar

Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br

Kr

4

5

12

13

Ca Sc 20

21

22

V

23

24

25

26

Rb Sr

6

Cs Ba La Hf Ta W Re Os

7

Fr Ra Ac

55

87

38

56

88

Y

Ti

5

37

18

39

57

27

28

29

30

31

6

14

32

N 7

P

15

33

8

16

34

Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te 40

72

41

73

42

74

43

75

44

76

81

82

51

83

52

I

84

53

Xe

Pt Au Hg TI Pb Bi Po At Rn 80

50

36

Ir

79

49

35

18

46

78

48

17

10

45

77

47

9

85

54

86

89

Figure 3.1 The periodic table showing the distribution of metals, metalloids and non-metals.

There are some elements, such as Si, Ge and Sb, that have some of the properties of both metals and non-metals – these are called metalloids and are shaded green.

The symbols of the elements that are solid at room temperature and pressure are shown in black in Figure 3.1, whereas those that are gases are in blue and liquids are in red.

Some of the groups in the periodic table are given names. Commonly used names are shown in Figure 3.2. The noble gases are sometimes also called the ‘inert gases’.

In the periodic table shown in Figure 3.1 it can be seen that the atomic numbers jump from 57 at La (lanthanum) to 72 at Hf (hafnium). This is because some elements have been omitted – these are the lanthanoid elements. The actinoid elements, which begin with Ac (actinium), have also been omitted from Figure 3.1. Figure 3.2 shows a long form of the periodic table, showing these elements as an integral part.

There is some disagreement among chemists about just which elements should be classified as metalloids – polonium and astatine are sometimes included in the list.

Hydrogen is the most abundant element in the Universe: about 90% of the atoms in the Universe are hydrogen. Major uses of hydrogen include making ammonia and the hydrogenation of unsaturated vegetable oils.

3 THE PERIODIC TABLE

85

LANTHANOIDS

ELEMENTS

HALOGENS NOBLE GASES

ALKALI METALS

TRANSITION ACTINOIDS

Figure 3.2 The long form of the periodic table, with the names of some of the groups. Hydrogen, though sometimes placed in group 1, does not count as an alkali metal. The lanthanoids and actinoids are also often called ‘lanthanides’ and ‘actinides’.

Mendeleev puzzled over the arrangement of elements in the periodic table until he had a dream in which he claims to have seen the arrangement. Kekulé also came up with the ring structure of benzene after a dream. Does it matter how a scientist comes up with a hypothesis? What is the difference between a scientific and a non-scientific hypothesis? Is it the origins of the hypothesis or the fact that it can be tested experimentally (is falsifiable) that makes it scientific?

Tin has a non-metallic allotrope (grey tin) and prolonged exposure of tin to low temperatures can bring about the transformation to the brittle non-metallic form.

The transformation is sometimes called ‘tin pest’ and there are tales (not necessarily true!) that Napoleon lost the Russian campaign in 1812 because his soldiers had buttons made of tin – in the cold Russian winter the buttons became brittle and their trousers fell down!

86

Metallic properties The metallic and non-metallic properties of elements can be related to ionisation energies (see later in this topic).

A metallic structure consists of a regular lattice of positive ions in a sea of delocalised electrons (page 160). To form a metallic structure, an element must be able to lose electrons fairly readily to form positive ions.

Going across a period the ionisation energy increases and so elements lose electrons less easily. So, metallic structures are formed by elements on the left-hand side of the periodic table, which have lower ionisation energies.

Going down a group in the periodic table, ionisation energy decreases, therefore elements are much more likely to exhibit metallic behaviour lower down a group. This can be seen especially well in group 14 going from non-metallic carbon at the top, through the metalloids (Si and Ge) to the metals tin and lead at the bottom.

In general, metallic elements tend to have large atomic radii, low ionisation energies, less exothermic electron affinity values and low electronegativity.

3.1.1 The periodic table and electron configurations

Four elements are named after the small village of Ytterby in Sweden – yttrium, terbium, erbium and ytterbium.

Electrons in the outer shell (the highest main energy level) of an atom are sometimes called valence electrons. The group number of an element is related to the number of valence electrons. All the elements in group 1 have one valence electron (one electron in their outer shell); all the elements in group 2 have two valence electrons. For elements in groups 13 –18, the number of valence electrons is given by (group number –10); so the elements in group 13 have three valence electrons and so on. The period number indicates the number of shells (main energy levels) in the atom – or which is the outer shell (main energy level).

The periodic table is divided into blocks according to the highest energy subshell (sub-level) occupied by electrons. So in the s block all the elements have atoms in which the outer shell electron configuration is ns1 or ns2 (where n is the shell number) and in the p block it is the p subshell that is being filled (Figure 3.3).

1

2

number of electrons in outer shell

13 14 15 16 17 18

group number −10 = number of electrons in outer shell

period number: gives number of highest main energy level occupied

number of electrons in p sub-level

1 H

He 1s2

1

2

2 Li

Be

3 Na Mg

2s2 3s2

number of electrons in d sub-level

1

2

3

4

Sc

Ti

V

Cr Mn Fe Co Ni Cu Zn

5 Rb Sr

Y

Zr Nb Mo Tc

6 Cs Ba

La

Hf

4

K

Ca

4s2

Ta

W

s block

5

6

7

8

9

10

Ru Rh Pd Ag Cd

Re Os

Ir

Pt Au Hg

d block

3d10

1

2

3

4

5

6

B

C

N

O

F

Ne

AI

Si

P

S

CI

Ar

Ga Ge As Se

Br

Kr

I

Xe

In

Sn Sb Te

TI

Pb Bi

2p6 3p6 4p6

Po At Rn

p block

Figure 3.3 Division of the periodic table into blocks.

Consider sulfur – this element is in period 3 and group 16, and so has three shells (the highest occupied shell is the third) and 16 – 10 = 6 electrons in its outer shell. It is in the p block – therefore its highest energy occupied subshell is a p subshell and the outer shell electron configuration is 3s23p4 (six valence electrons).

The noble gases (group 18) have either two (He) or eight electrons (Ne–Rn) in their outer shell. Helium belongs in the s block because its highest energy occupied subshell is 1s, but it is usually put in group 18 with the other noble gases.

3 THE PERIODIC TABLE

87

Nature of science

The development of science is not without controversy with regard to who has discovered what.

Scientists publish work to make their material available to other scientists and also to establish prior claim on discoveries. For example, the German chemist Julius Lothar Meyer was working on the arrangements of elements at the same time as Mendeleev and came to very similar conclusions – so why is Mendeleev remembered as the father of the modern periodic table rather than Meyer?

Learning objectives

r Understand trends in atomic radius, ionic radius, first ionisation energy, electron affinity and electronegativity across a period r Understand trends in atomic radius, ionic radius, first ionisation energy, electron affinity and electronegativity down a group

88

Scientists look for patterns in data. They gather evidence, not necessarily just from their own work but also from the published work of other scientists, and analyse the data to discover connections and to try to come up with general laws.

The modern periodic table has developed from one originally conceived by Russian chemist Dmitri Mendeleev in 1869. Mendeleev suggested that the elements were arranged in order of atomic weight (what we would now call relative atomic mass) and produced a table in which elements with similar chemical properties were arranged in vertical groups. Mendeleev took several risks when presenting his data – he suggested that some elements had not been discovered and left spaces for them in his table. Not only did he leave spaces but he also predicted the properties of these unknown elements – he made his hypotheses falsifiable, which added great weight to his theory. The predictions he made were later found to be extremely accurate – the mark of a good theory is that it should be able to be used to predict results that can be experimentally confirmed or refuted. He also suggested that the atomic weights of some elements were incorrect – he realised that tellurium (Te) belonged in the same group as O, S and Se but its atomic weight was higher than iodine and so it should be placed after iodine. Instead of abandoning his theory, he questioned the accuracy of the atomic weight of tellurium and placed it before iodine. This is, of course, the correct place, but Mendeleev’s assumption that the atomic weight was lower than that of iodine was not correct.

Henry Moseley, working at the beginning of the 20th century established the connection between atomic number and the periodic table. Like Mendeleev he realised that there were still some elements to be discovered and proposed that three elements between Al and Au were yet to be discovered.

3.2 Physical properties 3.2.1 Variation of properties down a group and across a period In the next few sections, we will consider how various physical properties vary down a group and across a period in the periodic table.

Atomic radius The atomic radius is basically used to describe the size of an atom. The larger the atomic radius, the larger the atom.

The atomic radius is usually taken to be half the internuclear distance in a molecule of the element. For example, in a diatomic molecule such as

chlorine, where two identical atoms are joined together, the atomic radius K would be defined as shown in Figure 3.4.

atomic radius

Atomic radius increases down a group.

This is because, as we go down a group in the periodic table the atoms have increasingly more electron shells. For example, potassium has four shells of electrons but lithium has only two (Figure 3.5).

Figure 3.4 The atomic radius of chlorine atoms in a molecule.

Li

Extension

K

It is possible to define two different atomic radii: the covalent radius and the van der Waalsâ&#x20AC;&#x2122; radius.

Figure 3.5 Potassium and lithium atoms.

Na

Although the nuclear charge is higher for K, the number of electrons and hence the repulsion between electrons is also greater, and this counteracts any effects due to a greater number of protons in the nucleus.

11+

Li Atomic radius decreases across a period.

Figure 3.6 shows the variation in atomic radius across period 3 in the periodic table. The reason that atomic radius decreases across a period is that nuclear charge increases across the period with no significant change in shielding. The shielding remains approximately constant because atoms in the same period have the same number of inner shells.

Sodium and chlorine (Figure 3.7) have the same number of inner shells of electrons (and hence the amount of shielding is similar). However, chlorine has a nuclear charge of 17+ whereas sodium has a nuclear charge of only 11+. This means that the outer electrons are pulled in more strongly in chlorine than in sodium and the atomic radius is smaller.

CI

17+

Atomic radius / pm

200 180 160 140

Figure 3.7 Sodium and chlorine atoms.

Inner shells, which shield the outer electrons, are highlighted in blue.

Na Mg AI

120 100

Si

80

Extension P

S

CI

Period 3

Figure 3.6 The variation in atomic radius across period 3. No atomic radius is shown for argon because it does not form covalent bonds and the internuclear distance between atoms bonded together therefore cannot be measured.

Although it is not possible to measure an atomic radius for Ar, it is possible to measure a value for the van der Waalsâ&#x20AC;&#x2122; radius of this element.

3 THE PERIODIC TABLE

89

Ionic radius The ionic radius is a measure of the size of an ion.

In general, the ionic radii of positive ions are smaller than their atomic radii, and the ionic radii of negative ions are greater than their atomic radii.

Figure 3.8 shows a comparison of the atomic and ionic radii (1+ ion) for the alkali metals. Each ion is smaller than the atom from which it is formed (by loss of an electron).

300 Cs Rb

Radius / pm

250

K Na

200 150

Cs+

Li K+

100 50

Rb+

Na+ Li+

0 Group 1

Figure 3.8 Atomic and ionic radii for the alkali metals.

250 I– Br–

Radius / pm

200

150

CI–

F– I Br

100

50

CI F

0 Group 17

Figure 3.9 A comparison of size between halogen atoms and their ions.

90

Na is larger than Na+ because the former has one extra shell of electrons – the electron configuration of Na is 2,8,1, whereas that of Na+ is 2,8. Also, they both have the same nuclear charge pulling in the electrons (11+), but there is a greater amount of electron–electron repulsion in Na because there are 11 electrons compared with only 10 in Na+. The electron cloud is therefore larger in Na than in Na+ because there are more electrons repelling for the same nuclear charge pulling the electrons in.

The fact that negative ions are larger than their parent atoms can be seen by comparing the sizes of halogen atoms with their ions (1−) in Figure 3.9. Cl− is larger than Cl because it has more electrons for the same nuclear charge and, therefore, greater repulsion between electrons. Cl has 17 electrons and 17 protons in the nucleus. Cl− also has 17 protons in the nucleus, but it has 18 electrons. The repulsion between 18 electrons is greater than between 17 electrons, so the electron cloud expands as an extra electron is added to a Cl atom to make Cl−.

The variation of ionic radius across a period is not a clear-cut trend, because the type of ion changes going from one side to the other – positive ions are formed on the left-hand side of the period and negative ions on the right-hand side.

300

Si 4–

250 P 3– Radius / pm

200

CI –

S 2– 150

Na+

+

100

Na

Mg2+

50

AI

3+

Si4+

11+

0 Period 3

Figure 3.10 Variation of ionic radius of positive and negative ions across period 3.

For positive ions there is a decrease in ionic radius as the charge on the ion increases, but for negative ions the size increases as the charge increases (Figure 3.10).

Let us consider Na+ and Mg2+ – both ions have the same electron configuration, but Mg2+ has one more proton in the nucleus P3–(Figure 3.11). Because there is the same number of electrons in both ions, the amount of electron–electron repulsion is the same; however, the higher nuclear charge in Mg2+ means that the electrons are pulled in more strongly and so the ionic radius is smaller.

15+ 3− 2− Now let us consider P and S . Both ions have the same number of electrons. S2− has the higher nuclear charge and, therefore, because the amount of electron–electron repulsion is the same in both ions, the electrons are pulled in more strongly in S2− (Figure 3.12).

Mg2+

12+

Figure 3.11 Mg2+ is smaller than Na+.

S2–

P3–

16+

15+

Figure 3.12 S2− is smaller than P3−.

S2–

16+

3 THE PERIODIC TABLE

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The full definition of first ionisation energy is: the energy required to remove one electron from each atom in one mole of gaseous atoms under standard conditions.

First ionisation energy The first ionisation energy of an element is the energy required to remove the outermost electron from a gaseous atom – that is, the energy for the process: M(g) → M+(g) + e−

Variation in first ionisation energy down a group Down any group in the periodic table, the first ionisation energy decreases.

electron closer to the nucleus Li 3+

The decrease in first ionisation energy down a group is shown in Figure 3.13. The size of the atom increases down the group so that the outer electron is further from the nucleus and therefore less strongly attracted by the nucleus (Figure 3.14).

Although the nuclear charge also increases down a group, this is largely balanced out by an increase in shielding down the group, as there are more electron energy levels (shells). It is the increase in size that governs the change in first ionisation energy.

1600

K 19+

First ionisation energy / kJ mol

–1

Li

1200 1000

K Rb

800

Cs

600 400

Figure 3.14 Potassium has a lower first ionisation energy than lithium. Electrons that shield the outer electron are highlighted in blue.

Na

1400

0

Group 1

Figure 3.13 First ionisation energy for group 1.

Variation in first ionisation energy across a period The general trend is that first ionisation energy increases from left to right across a period. This is because of an increase in nuclear charge across the period.

The nuclear charge increases from Na (11+) to Ar (18+) as protons are added to the nucleus. The electrons are all removed from the same main energy level (third shell) and electrons in the same energy level do not shield each other very well. Therefore the attractive force on the outer electrons due to the nucleus increases from left to right across the period

92

First ionisation energy / kJ mol

–1

1600 Ar

1400 1200

CI

1000

S

P

800

400

Si

Mg

600

AI

Na 0

Period 3

Figure 3.15 The variation in first ionisation energy across period 3 in the periodic table.

Na

11+

Ar

The increase in first ionisation energy (Figure 3.15) can also be explained in terms of the effective nuclear charge felt by the outer electron in argon being higher. The effective nuclear charge felt by the outer electron in a sodium atom would be 11 (nuclear charge) − 10 (number of inner shell electrons), i.e. 1+ if shielding were perfect.

The effective nuclear charge felt by the outer electrons in an argon atom would be 18 (nuclear charge) − 10 (number of inner shell electrons), i.e. 8+ if shielding were perfect.

18+

Figure 3.16 Sodium and argon atoms.

and the outer electron is more difficult to remove from an argon atom (Figure 3.16). The argon atom is also smaller than the sodium atom and, therefore, the outer electron is closer to the nucleus and more strongly held.

There are two exceptions to the general increase in first ionisation energy across a period, and these are discussed on page 78.

Exam tip The exceptions to the trend are required knowledge for all students - refer to page 78.

Electron affinity The first electron affinity involves the energy change when one electron is added to a gaseous atom: X(g) + e− → X–(g) It is defined more precisely as the enthalpy change when one electron is added to each atom in one mole of gaseous atoms under standard conditions. Electron affinity is difficult to measure experimentally and data are incomplete.

The first electron affinity is exothermic for virtually all elements – it is a favourable process to bring an electron from far away (infinity) to the outer shell of an atom, where it feels the attractive force of the nucleus.

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Variation of electron affinity down group 17 A graph of electron affinity down group 17 is shown in Figure 3.17. The general trend is that electron affinity decreases down a group, but it can be seen that chlorine has the most exothermic value for electron affinity. A similar trend in electron affinity values is seen going down group 16 and group 14.

The electron affinity becomes less exothermic from Cl to I as the size of the atom increases. The electron is brought into the outer shell of the atom and as the atom gets bigger there is a weaker attraction between the added electron and the nucleus as it is brought to a position which is further from the nucleus.

–290

0

10

20

Atomic number 30

40

60

50 l

–300

Electron affinity / kJ mol–1

–310

–320

–330

Br

F

–340

–350

Cl

–360

Figure 3.17 Electron affinity values of group 17 elements.

Extension Electron–electron repulsion also affects the electron affinity and as the atom gets smaller the electrons are, on average, closer together and there is more electron–electron repulsion. This means that the electron affinity should be less exothermic when an electron is added to a smaller atom.

Going from F to Cl the electron affinity becomes more exothermic because the decrease in electron–electron repulsion outweighs the fact that there is less attraction between the electron and the nucleus.

Variation in electron affinity across a period The general trend in electron affinity from group 13 to group 17 is shown in Figure 3.18.

94

0

12

Electron affinity / kJ mol–1

–50

13

14

AI

Atomic number 15

16

17

18

P

–100 –150

Si

–200

S

–250 –300 –350

Cl

–400

Figure 3.18 Electron affinity values across period 3.

The general trend is that the electron affinity becomes more exothermic. This is because of an increase in nuclear charge and a decrease in atomic radius from left to right across the period. For instance, F has a higher nuclear charge and a smaller radius than O and so the electron will be more strongly attracted when it is brought into the outer shell of the F atom.

Extension Phosphorus has a less exothermic electron affinity than silicon because of its electron configuration. P has three unpaired electrons in three separate p orbitals and when one electron is added this electron must be paired up in the same p orbital as another electron – this introduces an extra repulsion term that is not present in Si. The arguments being used here are very similar to those for the variation of first ionisation energy across a period discussed in Topic 2.

Electronegativity Electronegativity is a measure of the attraction of an atom in a molecule for the electron pair in the covalent bond of which it is a part.

In a covalent bond between two different atoms, the atoms do not attract the electron pair in the bond equally. How strongly the electrons are attracted depends on the size of the individual atoms and their nuclear charge.

Electronegativity decreases down a group – this is because the size of the atoms increases down a group. Consider hydrogen bonded to either F or Cl (Figure 3.19). The bonding pair of electrons is closer to the F nucleus in HF than it is to the Cl nucleus in HCl. Therefore the electron pair is more strongly attracted to the F nucleus in HF and F has a higher electronegativity than Cl.

Electronegativity is discussed in more detail on page 128.

3 THE PERIODIC TABLE

95

attraction for these electrons

attraction for H electrons these 1+

a

H 1+

F 9+ shielding F 9+ shielding

a

H 1+

b

Chlorine’s higher nuclear charge does not make it more electronegative than fluorine because the shielding from inner shells (shown with blue shading in Figure 3.19) increases from F to Cl such that the effective nuclear charge felt by the bonding electrons is approximately the same in each case (+7 if shielding were perfect).

H 1+ distance greater between nucleus and bonding pair of electrons

attraction Electronegativity increases across a period – the reason for this is for these the increase in nuclear charge across the period with electrons no significant change in shielding. The shielding remains approximately constant because atoms in the same period have the same number of inner shells.

H N So, if an N–H bond is compared with an 3.20), 1+ F–H bond (Figure 7+ the electrons in the N–H bond are attracted by the seven protons in the nucleus, but the electrons in the F–H bond are attractedshielding by the nine a protons in the F nucleus. In both cases the shielding is approximately the same (because of two inner shell electrons).

CI 17+

CI 17+

attraction for these electrons

attraction for these electrons

Figure 3.19distance Hydrogen bonded to a fluorine greater nucleus and bbetween chlorine.

H 1+

b

and bonding pair of electrons

N 7+

H 1+

shielding

a

higher nuclear charge

F 9+ shielding

b

Figure 3.20 Hydrogen bonded to a nitrogen and b fluorine.

?

Test yourself

attraction for these electrons

1 Give the names of the following elements: a the element in period 3 and group 14 b the element in period 5 and group 16 H 1+ but c the element in the same group as sulfur in period 6 d a halogen in period 5 b e an element in the same period as potassium that has five outer shell electrons

2 State whether the following properties increase or decrease across a period: a electronegativity b atomic radius 3 Arrange the following in order of increasing radius (smallest first): a Ba Mg Sr Ca Na+ F− b O2− c Na Na+ K Al3+ − d S Cl I Cl− S2−

96

higher nuclear charge following

4 Are the true or false? a A germanium atom is smaller than a silicon atom, but silicon has a higher first ionisation F 9+ energy. b Selenium has a higher first ionisation energy shielding and electronegativity than sulfur. c Antimony has a higher first ionisation energy and electronegativity than tin. d Cl− is bigger than Cl, but Se2− is smaller than Se. e Iodine has a higher electronegativity than tellurium but a lower electronegativity than bromine.

5 Based on the following data, which element (X or Y) is more likely to be a metal?

First ionisation Atomic −1 energy / kJ mol radius / nm

Electronegativity

X

736

0.136

1.3

Y

1000

0.104

2.6

3.2.2 Properties of elements in group 1 and group 17 Group 1 elements The elements in group 1 are known as the alkali metals. They are all highly reactive, soft, low melting point metals (Table 3.1). They are placed together in group 1 for two reasons – they all have one electron in their outer shell and they react in very similar ways (similar chemical properties).

Learning objectives

r Understand that elements in

the same group have similar chemical properties and show a gradual variation in physical properties r Describe some reactions of elements in group 1 and group 17

The reactions of an element are determined by the number of electrons in the outer shell (highest main energy level) of their atoms. Because elements in the same group in the periodic table have the same number of electrons in their outer shell, they react in basically the same way.

The bonding in all these elements is metallic. The solid is held together by electrostatic attraction between the positive ions in the lattice and the delocalised electrons (see page 160).

The attraction for the delocalised, negatively-charged, electrons is due to the nucleus of the positive ion. As the ions get larger as we go down the group, the nucleus becomes further from the delocalised electrons and the attraction becomes weaker (Figure 3.22). This means that less energy is required to break apart the lattice going down group 1.

Element lithium sodium

caesium

Liquid sodium is used as a coolant in some nuclear reactors.

Symbol Atomic number Electron configuration Density / g cm−3 Melting point / °C Li

3

Na

potassium K rubidium

Melting point decreases down group 1 (Figure 3.21).

Rb

0.53

180

1330

1

11

[Ne]3s

0.97

98

890

19

1

0.86

64

774

1

1.53

39

688

1

1.87

29

679

37

Cs

[He]2s

Boiling point / °C

1

55

[Ar]4s [Kr]5s

[Xe]6s

Table 3.1 Similarities in alkali metal properties.

200

Li

smaller distance between nucleus and delocalised electrons

Melting point / °C

150

100

Na

Li +

K

Li +

Rb

Rb

+

+

– Rb

50

Cs

+ Li

– + Li

+

+

Rb

Rb

0 Group 1

Figure 3.21 Variation in melting point in group 1.

Figure 3.22 The delocalised electrons are attracted more strongly in lithium than in rubidium.

3 THE PERIODIC TABLE

97

Reactions of the elements in group 1 The elements in group 1 are all reactive metals that react readily with, among other things, oxygen, water and halogens. The atoms all have one electron in their outer shell, and virtually all reactions involve the loss of this outer shell electron to form a positive ion, M+. The reactions become more vigorous going down the group because the ionisation energy decreases as the size of the atom increases. This means that, for example, caesium loses its outer electron to form a positive ion much more easily than sodium and will react more vigorously.

Reaction with oxygen The alkali metals react vigorously with oxygen and all tarnish rapidly in air. The general equation for the reaction is:

M2O is a basic oxide that will dissolve in water to form an alkaline solution, containing M+ and OH− ions.

4M(s) + O2(g) → 2M2O(s) Reaction with water The alkali metals react rapidly with water. The general equation for the reaction is:

2M(s) + 2H2O(l) → 2MOH(aq) + H2(g)

Lithium, sodium and potassium are all less dense than water.

An alkaline solution is formed. The alkali metal hydroxides are strong bases and ionise completely in aqueous solution (page 321).

The reaction with water becomes more vigorous going down the group – sodium melts into a ball, fizzes rapidly and moves around on the surface of the water; potassium bursts into flames (lilac); and caesium explodes as soon as it comes into contact with water.

Group 17 elements The elements in group 17 are known as the halogens. They are all nonmetals consisting of diatomic molecules (X2). Some properties are given in Table 3.2.

Element

Symbol

fluorine

F

chlorine

Cl

Atomic number 9 17

Colour

[He]2s22p5 2

5

10

2

[Ne]3s 3p

5

Melting point / °C

Boiling point / °C

pale yellow

−220

−188

gas

yellow–green

−101

−35

gas

−7

59

MJRVJE

114

184

solid

bromine

Br

35

[Ar]3d 4s 4p

deep red MJRVJE PSBOHF vapour

iodine

I

53

[Kr]4d105s25p5

grey shiny solid, purple vapour

Table 3.2 Properties of halogens.

98

Electron configuration

Physical state at room temperature and pressure

Variation of melting point in group 17 The melting points of the halogens increase going down the group (Figure 3.23).

150 l2

100

Melting point / °C

50 0 Br2

Group 7 halogens

–50 Cl2

–100 –150 –200

F2

–250

Figure 3.23 Variation in melting point in group 17.

As the relative molecular masses of the X2 halogen molecules increase, the London forces (page 148) between molecules get stronger. This means that more energy must be supplied to separate the molecules from each other.

Reactions of the elements in group 17 All the atoms of the elements in group 17 have seven electrons in their outer shell and react either by gaining an electron to form X− ions or by forming covalent compounds. Reactivity decreases down the group, and fluorine is the most reactive element known, reacting directly with virtually every other element in the periodic table. The variation in reactivity of the halogens cannot be as easily explained as for the alkali metals. The very high reactivity of fluorine can be explained in terms of an exceptionally weak F–F bond and the strength of the bonds it forms with other atoms. The reactivity in terms of the formation of X− ions can be related to a decrease in electron affinity (energy released when an electron is added to a neutral atom) going down the group as the electron is added to a shell further away from the nucleus, but this is only part of the story and several factors must be considered when explaining the reactivity of the halogens.

The halogens all react with the alkali metals to form salts. The general equation is: 2M(s) + X2(g) → 2MX(s) The salts formed are all white/colourless, fairly typical ionic compounds.

They contain M+ and X− ions. All alkali metal chlorides, bromides and iodides are soluble in water and form colourless, neutral solutions.

Chlorine is produced by the electrolysis of brine. Worldwide annual production is about 60 million tonnes. Chlorine and its compounds are involved in the production of about 90% of the most important pharmaceuticals.

Its biggest single use is in the production of PVC.

How vigorous the reaction is depends on the particular halogen and alkali metal used – the most vigorous reaction occurs between fluorine and caesium, and the least vigorous between lithium and iodine.

3 THE PERIODIC TABLE

99

Displacement reactions of halogens These are reactions between a solution of a halogen and a solution containing halide ions – they are discussed in more detail on page 404.

A small amount of a solution of a halogen is added to a small amount of a solution containing a halide ion, and any colour changes are observed (see Table 3.3). Potassium chloride, bromide and iodide solutions are all colourless. The colours of chlorine, bromine and iodine solutions are shown in Figure 3.24.

Orange colour, due to the production of bromine.

Red-brown colour, due to the production of iodine.

KCl(aq)

KBr(aq)

KI(aq)

Cl2(aq)

no reaction

orange solution

dark red–brown solution

Br2(aq)

no reaction

no reaction

dark red–brown solution

I2(aq)

no reaction

no reaction

no reaction

Table 3.3 Results of reactions between halogen solutions and solutions containing halide ions.

The reactions that occur are: Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(aq) Ionic equation: Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq) Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq) Ionic equation: Cl2(aq) + 2I−(aq) → 2Cl−(aq) + I2(aq) CI2(aq)

Br2(aq)

I2(aq)

Figure 3.24 Chlorine solution is pale yellow–green (almost colourless if it is dilute), bromine solution is orange, and iodine solution is red–brown.

100

Br2(aq) + 2KI(aq) → 2KBr(aq) + I2(aq) Ionic equation: Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq) The more reactive halogen displaces the halide ion of the less reactive halogen from solution – chlorine displaces bromide ions and iodide ions from solution, and bromine displaces iodide ions from solution.

These reactions are all redox reactions (Topic 9), in which a more reactive halogen oxidises a less reactive halide ion. Chlorine is a stronger oxidising agent than bromine and iodine; it will oxidise bromide ions to bromine, and iodide ions to iodine. Bromine is a stronger oxidising agent than iodine and will oxidise iodide ions to iodine. In terms of electrons, chlorine has the strongest affinity for electrons and will remove electrons from bromide ions and iodide ions.

3.2.3 Oxides of period 2 and period 3 elements

Learning objectives

r Describe the changes from basic

Oxides of elements may be classified as basic, acidic or amphoteric. The nature of the oxides changes across a period and Table 3.4 shows how the oxides change from basic to amphoteric to acidic across period 3.

In general, metallic oxides are basic and non-metallic oxides are acidic.

A basic oxide is one that will react with an acid to form a salt and, if soluble in water, will produce an alkaline solution. Sodium oxide reacts with water to form sodium hydroxide:

to acidic oxides across a period r Write equations for the reactions of oxides with water and predict the acidity of the resulting solutions

Na2O(s) + H2O(l) â&#x2020;&#x2019; 2NaOH(aq) Sodium oxide reacts with acids such as sulfuric acid to form salts: Na2O(s) + H2SO4(aq) â&#x2020;&#x2019; Na2SO4(aq) + H2O(l) Magnesium oxide, because of the relatively high charges on the ions, is not very soluble in water but it does react to a small extent to form a solution of magnesium hydroxide, which is alkaline:

Exam tip Reactions highlighted like this must be learnt for examinations.

MgO(s) + H2O(l) â&#x2020;&#x2019; Mg(OH)2(aq) Aluminium is on the dividing line between metals and non-metals and forms an amphoteric oxide â&#x20AC;&#x201C; these have some of the properties of a basic oxide and some of an acidic oxide. Aluminium is exhibiting properties between those of a metal (basic) oxide and those of a non-metal (acidic) oxide.

Aluminium oxide does not react with water but it does display amphoteric behaviour in that it reacts with both acids and bases to form salts: reaction with acids: Al2O3 + 6H+â&#x2020;&#x2019; 2Al3+ + 3H2O reaction with alkalis/bases: Al2O3 + 2OHâ&#x2C6;&#x2019; + 3H2O â&#x2020;&#x2019; 2Al(OH)4â&#x2C6;&#x2019; Amphoteric oxides react both with acids and with bases.

Formula of oxide

Sodium

Magnesium

Aluminium

Silicon

Phosphorus

Na2O

MgO

Al2O3

SiO2

P4O10

Nature of element

metal

Nature of oxide Reaction with water Solution formed

basic

Sulfur SO2 SO3

non-metal amphoteric

acidic

soluble, reacts

sparingly soluble, some reaction

insoluble

soluble, reacts

alkaline

slightly alkaline

â&#x20AC;&#x201C;

acidic

Table 3.4 5IF BDJEoCBTF OBUVSF PG TPNF QFSJPE PYJEFT

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101

The remaining oxides in Table 3.4 are all acidic oxides. An acidic oxide is one that reacts with bases/alkalis to form a salt and, if soluble in water, will produce an acidic solution.

P4O6 (phosphorus(III) oxide) and P4O10 (phosphorus(V) oxide) form phosphoric(III) and phosphoric(V) acid, respectively, when they react with water: Phosphoric(V) acid is an ingredient of Coca-Cola®.

P4O6(s) + 6H2O(l) → 4H3PO3(aq) P4O10(s) + 6H2O(l) → 4H3PO4(aq) SO2 (sulfur(IV) oxide) and SO3 (sulfur(VI) oxide) form sulfuric(IV) and sulfuric(VI) acid, respectively, when they react with water: SO2(g) + H2O(l) → H2SO3(aq) SO3(g) + H2O(l) → H2SO4(aq) Nitrogen oxides There are many oxides of nitrogen, ranging in formula from N2O to N2O5. Two of the most environmentally important are nitrogen(II) oxide (NO) and nitrogen(IV) oxide (NO2).

Nitrogen reacts with oxygen at very high temperatures to form NO (nitrogen monoxide, nitric oxide or nitrogen(II) oxide):

N2(g) + O2(g) → 2NO(g) This reaction occurs in the internal combustion engine. NO is virtually insoluble in water and is classified as a neutral oxide.

NO can be oxidised in the atmosphere to NO2, which can react with water to produce nitric(V) acid (HNO3), which is one of the acids responsible for acid deposition (see Subtopic 8.5). NO2 can be classified as an acidic oxide: 2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq) N2O (nitrogen(I) oxide, nitrous oxide) is another neutral oxide. N2O is also known as laughing gas and major uses include as an anaesthetic and as the propellant in ‘squirty cream’.

102

Non-metal oxides such as SO2 are produced in various industrial processes and when coal is burnt. This can be responsible for acid rain, which can, among other things, kill fish in lakes and trees in forests.

Nitrogen oxides (NOx) may be formed in internal combustion engines, and these are involved in the formation of photochemical smog in cities (Figure 3.25).

Figure 3.25 A photochemical smog over Hong Kong.

Nature of science Science and the technology that develops from it have been used to solve many problems â&#x20AC;&#x201C; but it can also cause them. The development of industrial processes that produce acidic gases led to acid rain being a major environmental problem. Acid rain, and its associated problems, is important to people across the world and it is vital that scientists work to improve the public understanding of the issues involved. Scientists also work as advisors to politicians in developing policies to solve these problems.

Advancements in science have often arisen from finding patterns in data. An understanding of the patterns in physical and chemical properties of elements in the periodic table has allowed chemists to make new substances. For instance, the knowledge that sulfur formed a range of compounds with nitrogen probably led scientists to attempt to make selium-nitrogen and tellurium-nitrogen compounds.

?

Test yourself 6 Write balanced equations for the following reactions: a rubidium with water b potassium with bromine c chlorine solution with potassium bromide solution d sodium oxide with water e sulfur(VI) oxide with water

7 State whether trends down the group in each of the following properties are the same or different when group 1 and group 17 are compared: a electronegativity b reactivity c melting point d ionisation energy 8 State whether an acidic or alkaline solution will be formed when each of the following is dissolved in/reacted with water: a SO3 b MgO c Na

3 THE PERIODIC TABLE

103

Learning objectives

3.3 First-row d-block elements (HL)

r Describe the characteristic

3.3.1 The transition elements (d block)

properties of transition metals r Explain why transition metals have variable oxidation numbers r Explain the formation and describe the shape of complex ions r Explain why transition metal complex ions are coloured r Explain the factors that affect the colour of a transition metal complex r Understand the magnetic properties of transition metal atoms and ions r Describe some uses of transition metals and their compounds as catalysts

Exam tip Remember that chromium and copper have slightly different electron configurations.

?

Sc

Transition elements Ti

V

Cr Mn Fe Co Ni Cu Zn

Figure 3.26 Zinc is not a transition element but the classification of scandium is more controversial.

Exam tip Scandium is regarded as a transition element on the syllabus.

The ‘transition elements’ are often called the ‘transition metals’.

104

The first-row d-block elements are:

Sc 21

Ti

V

22

23

Cr Mn Fe Co Ni Cu Zn 24

25

26

27

28

29

30

There are also two other rows of d-block elements.

They are called ‘d-block’ elements because the subshell being filled across this series is the 3d subshell. The electron configurations range from [Ar]4s23d1 for scandium to [Ar]4s23d10 for zinc:

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

[Ar]4s23d1 [Ar]4s23d2 [Ar]4s23d3 [Ar]4s13d5 [Ar]4s23d5 [Ar]4s23d6 [Ar]4s23d7 [Ar]4s23d8 [Ar]4s13d10 [Ar]4s23d10

The transition elements can be defined as different from ‘the d-block elements’, and the definition we will use here is: a transition element is an element that forms at least one stable ion with a partially filled d subshell According to this definition, zinc is not counted as a transition element because the only ion it forms is the 2+ ion, with electron configuration 1s22s22p63s23p63d10 (full d subshell). Zinc (Figure 3.26) does not exhibit some of the typical characteristic properties of transition metals detailed below (e.g. it does not form coloured compounds).The inclusion/exclusion of scandium as a transition element according to this definition is much more controversial.

In virtually every compound scandium has oxidation number +3 (no d electrons) however, it also forms a couple of compounds (ScH2 and CsScCl3) with formal oxidation number +2, but the bonding in these compounds is more complicated and they do not necessarily contain the 2+ ion.

Properties of the transition elements We have already studied the variation in properties of a set of eight elements across the periodic table when we looked at the properties of period 3 elements. The transition elements also form a set of eight elements across the periodic table, but these are much more similar to each other than the elements across period 3. For instance, they are all metals rather than showing a change from metal to non-metal.

The variation in first ionisation energy and atomic radius of the transition elements and period 3 elements are compared in Figures 3.27 and 3.28. It can be seen that the variation of ionisation energy and atomic radius across the series of the transition elements is much smaller than across period 3.

Key Transition elements Ar

1400 P

1000

400

Ti

S

Si

Mg

800 600

180

CI

1200

Cr Mn

V

Fe

Co

Ni

Cu

AI

Na

Period 3

200

Atomic radius / pm

First ionisation energy / kJ mol â&#x20AC;&#x201C;1

Key Period 3

1600

Na Mg

160 140

0

AI

Ti

120 100

200

Transition elements

Mn V

Cr

Si

Fe

P

Co

S

Ni

Cu

CI

80 Period 3 / transition elements

Period 3 / transition elements

Figure 3.27 A comparison of the variation of first ionisation energy across period 3 with that across the transition metal series.

Figure 3.28 A comparison of the variation of atomic radius across period 3 with that across the transition metal series.

Because of their similarity it is possible to draw up a list of characteristic properties of transition elements: r 5SBOTJUJPO FMFNFOUT BSF BMM UZQJDBM NFUBMT m UIFZ IBWF IJHI NFMUJOH points and densities. r 5SBOTJUJPO FMFNFOUT DBO FYIJCJU NPSF UIBO POF PYJEBUJPO OVNCFS JO compounds/complexes. r 5SBOTJUJPO FMFNFOUT GPSN DPNQMFY JPOT r 5SBOTJUJPO FMFNFOUT GPSN DPMPVSFE DPNQPVOET DPNQMFYFT r 5SBOTJUJPO FMFNFOUT BOE UIFJS DPNQPVOET DPNQMFYFT DBO BDU BT DBUBMZTUT in many reactions. r $PNQPVOET PG USBOTJUJPO FMFNFOUT DBO FYIJCJU NBHOFUJD QSPQFSUJFT

Exam tip The last five properties are the most important for examinations.

Ionisation of transition elements Transition elements form positive ions. The electron configurations of some transition metal ions are shown in Table 3.5.

Element

Electron configuration

Cr

[Ar]4s13d5 2

5

Mn

[Ar]4s 3d

Fe

[Ar]4s23d6

Co

[Ar]4s23d7

Cu

[Ar]4s13d10

Ion

Electron configuration

Cr2+

[Ar]3d4

Cr3+

[Ar]3d3

2+

Mn

[Ar]3d5

Fe2+

[Ar]3d6

Fe3+

[Ar]3d5

Co2+

[Ar]3d7

Cu+

[Ar]3d10

Cu2+

[Ar]3d9

The 4s electrons are always removed before the 3d electrons when an ion is formed.

Table 3.5 Electron configurations of transition metals and their ions.

3 THE PERIODIC TABLE

105

3.3.2 Variable oxidation numbers Oxidation numbers are discussed further on page 369.

Oxidation number and oxidation state are often used interchangeably.

The positive oxidation numbers (oxidation states) exhibited by the transition elements are shown in Figure 3.29. The greatest number of different oxidation numbers and the highest oxidation numbers are found in the middle of the series. From titanium to maganese there is an increase in the total number of electrons in the 4s and 3d subshells, so the maximum oxidation number increases. Maganese has the electron configuration [Ar]4s23d5 and therefore a maximum oxidation number of +7. Iron has eight electrons in the 4s and 3d subshells and would be expected to have a maximum oxidation number of +8, but the ionisation energy increases from left to right across the transition elements series and it becomes more difficult to reach the highest oxidation numbers towards the right-hand side of the series. The chemistry of copper and nickel is, for the same reason, dominated by the lower oxidation numbers.

All transition metals show oxidation number +2. In most cases this is because they have two electrons in the 4s subshell, and removal of these generates an oxidation number of +2.

Mn

Ti

Exam tip The oxidation numbers highlighted in Figure 3.29 are mentioned specifically in the data booklet.

Cr

7

Fe

V

6

6

6

5

5

5

5

Co

Ni

Cu

4

4

4

4

4

4

4

4

3

3

3

3

3

3

3

3

2

2

2

2

2

2

2

2

1

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

Figure 3.29 0YJEBUJPO OVNCFST PG USBOTJUJPO NFUBMT JO DPNQPVOET /PU BMM PYJEBUJPO numbers are common.

Why more than one oxidation number?

The 4s and 3d subshells are close in energy, and there are no big jumps in the successive ionisation energies when the 4s and 3d electrons are removed. Therefore the number of electrons lost will depend on a variety of factors such as lattice enthalpy, ionisation energy and hydration enthalpy. Electrons are not removed in order to generate the nearest noble gas electron configuration. The graph in Figure 3.30 shows a comparison of the first seven ionisation energies of magnesium and manganese. It can be seen that there is a very large jump between the second and third ionisation energies of magnesium, but that there are no such jumps for manganese.

106

Ionisation energy / kJ mol–1

25 000 Mg

20 000 15 000

Mn

10 000 5 000 0

1

2

3

4 5 6 Number of ionisation energy

7

8

Figure 3.30 Comparison of successive ionisation energies of magnesium and manganese.

Magnetic properties of transition metal compounds There are two forms of magnetism we need to be concerned with – paramagnetism and diamagnetism.

Paramagnetism is caused by unpaired electrons – paramagnetic substances are attracted by a magnetic field.

Diamagnetism is caused by paired electrons – diamagnetic substances are repelled slightly by a magnetic field.

All substances have some paired electrons and so all substances exhibit diamagnetism. However, the diamagnetic effect is much smaller than the paramagnetic effect and so, if there are any unpaired electrons present, the paramagnetic effect will dominate and the substance will be paramagnetic overall and attracted by a magnetic field. The more unpaired electrons, the greater the paramagnetism (magnetic moment).

Consider the electron configurations of two transition metal ions: 3d

4s

Fe2+ [Ar] Cr

3+

[Ar]

Both contain unpaired electrons and their compounds are paramagnetic – so both FeCl2 and CrCl3 are paramagnetic. Because an Fe2+ ion has four unpaired electrons and a Cr3+ ion has only three, the iron(II) compound is more paramagnetic (higher magnetic moment).

The Cu+ ion has the following electron configuration: 3d

4s

[Ar]

All the electrons are paired so compounds of copper(I), such as CuCl, are diamagnetic only.

Extension The situation is more complicated with complex ions. Depending on the energy difference between the higher and lower set of d orbitals and the amount of energy required to pair up two electrons in the same d orbital (overcoming the repulsions), complexes can be high spin (maximum number of unpaired electrons) or low spin (maximum number of electrons in the lower set of d orbitals). How paramagnetic a substance is then depends on the ligands because they influence the splitting of the d orbitals.

3 THE PERIODIC TABLE

107

A ligand must possess a lone pair CI of electrons.

CI CI

Si

CI

CI

A ligand is a Lewis bbase. a

P

ligand

H H

O

O

H

O

2+

Fe

H H O H

H

O

coordinate covalent bond

O

H H H H

Complex ions A complex ion consists of a central metal ion surrounded by ligands â&#x20AC;&#x201C; transition metal ions form many complexes.

Ligands are negative ions or neutral molecules that have lone pairs of electrons. They use the lone pairs to bond to a metal ion to form a complex ion. Coordinate covalent bonds (dative bonds) are formed between the ligand and the transition metal ion.

The structure of [Fe(H2O)6]2+ is shown in Figure 3.31. H2O is the ligand in this complex ion. The shape of this complex ion is octahedral and it is called the hexaaquairon(II) ion. Other ways of drawing this are shown in Figure 3.32.

H

Figure 3.31 " DPNQMFY JPO JT GPSNFE XIFO ligands bond to a transition metal ion. The ligands donate lone pairs into vacant orbitals (3d, 4s or 4p) on the transition metal ion.

Extension There is a strong case for considering the bonding in a transition metal complex ion as having a significant ionic component. Crystal field theory and ligand field theory consider the bonding from a more ionic point of view.

H H

O

O

H

O

H H

H

O

Fe

H H O H

2+

H

O

H H O H

H H O H

H

O

O

Fe H

O

2+

H

O H

H H H H

Figure 3.32 Alternative representations of the [Fe(H2O)6]2+ DPNQMFY JPO

All transition elements, with the exception of titanium, form an octahedral complex ion with the formula [M(H2O)6]2+ in aqueous solution.

Complex ions can have various shapes depending on the number of ligands. However, shapes cannot be worked out using the valence shell electron-pair repulsion theory (see page 137) because more subtle factors also govern the overall shape. If a complex ion contains 6 ligands it will almost certainly be octahedral, but complexes containing 4 ligands may be tetrahedral or square planar (Figure 3.33).

2â&#x20AC;&#x201C;

Cl Co

Cl Cl

N C

2â&#x20AC;&#x201C;

NC Ni CN

Cl

C N

[CoCl4 ]2â&#x20AC;&#x201C;

[Ni(CN)4 ]2â&#x20AC;&#x201C;

2â&#x2C6;&#x2019;

2â&#x20AC;&#x201C;

Figure 3.33 [CoCl4] is tetrahedral but [Ni(CN)4] JT TRVBSF QMBOBS

Complex ions can undergo substitution reactions in which, for example, H2O ligands are replaced by other ligands. For example, in the addition of concentrated hydrochloric acid to blue copper(II) sulfate solution: [Cu(H2O)6]2+(aq) + 4Clâ&#x2C6;&#x2019;(aq) blue

108

[CuCl4]2â&#x2C6;&#x2019;(aq) + 6H2O(l) yellow

As the acid is added, the yellow [CuCl4]2− complex ion is formed. So, the solution changes colour from blue to green (a mixture of blue and yellow). According to Le Chatelier’s principle (see Topic 7) the position of equilibrium shifts to the right as Cl− is added.

The oxidation number of a transition metal in a complex ion The oxidation number of a transition metal in a complex ion can be worked out from the charges on the ligands. Ligands may be either neutral or negatively charged (see Table 3.6).

In [Fe(H2O)6]2+ all the ligands are neutral water molecules. The overall charge on the ion is just due to the iron ion, so the oxidation number of iron must be +2.

In [Ni(CN)4]2− all the ligands have a 1− charge, so the total charge from all four ligands is 4−. The overall charge on the ion is 2−; so, the oxidation number of nickel must be +2 to cancel out 2− from the 4− charge.

Neutral ligands

1− ligands

H2O

Cl−

NH3

CN−

CO

Br−

Table 3.6 Charges on ligands.

Exam tip Oxidation numbers are discussed in more detail in Topic 9.

Working out the overall charge on a complex ion If the oxidation number (charge) of the central transition metal ion and the charges on the ligands are known, the overall charge on the complex ion can be worked out.

Worked example 3.1 Platinum(II) can form a complex ion with 1 ammonia and 3 chloride ligands. What is the overall charge and formula of the complex ion?

Platinum(II) has a charge of 2+ Ammonia is a neutral ligand (NH3) Chloride has a 1– charge (Cl-) The overall charge is (2+) + (0) + 3(1–) = 1– The formula of the complex ion is: [Pt(NH3)Cl3]–

Catalytic ability Transition elements and their compounds/complexes can act as catalysts.

For example, finely divided iron is the catalyst in the Haber process in the production of ammonia: N2(g) + 3H2(g)

2NH3(g)

Iron in the above reaction is a heterogeneous catalyst (one that is in a different physical state to the reactants) but transition metal compounds often act as homogeneous catalysts (ones that are in the same phase as the reactants). The ability to act as a catalyst relies on a transition metal atom or ion having varying oxidation numbers and also being able to coordinate to other molecules/ions to form complex ions.

Some scientists believe that the bonding between a transition metal and a ligand is purely ionic. All scientists have the same experimental data available to them – to what extent is scientific knowledge objective and to what extent is it a matter of interpretation and belief?

3 THE PERIODIC TABLE

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Nature of science Science is about finding patterns and these patterns allow us to make predictions. However, it is not always the patterns that are most interesting and careful observation is essential to spot anomalies and exceptions to patterns that could lead to new discoveries and theories. Zinc can be regarded as anomalous in the first row of the d block, for instance, it does not generally form coloured compounds; this has resulted in it not being included as a transition element.

3.4 Coloured complexes (HL) Complex ion

Colour

[Cu(H2O)6]2+

blue 2+

[Cu(NH3)4(H2O)2]

deep blue/violet

2+

[Fe(SCN)(H2O)5]

blood red

2+

[Ni(H2O)6]

green

The colours of some complex ions are shown in Table 3.7.

In a gaseous transition metal ion, all the 3d orbitals have the same energy â&#x20AC;&#x201C; that is, they are degenerate. However, when the ion is surrounded by ligands in a complex ion, these d orbitals are split into two groups. In an octahedral complex ion there are two orbitals in the upper group and three orbitals in the lower groups (Figure 3.34).

Table 3.7 5IF DPMPVST PG TPNF DPNQMFY JPOT d orbitals degenerate

ML6

Extension

2+

M2+(g)

All the d orbitals in a complex ion are higher in energy than the d orbitals in an isolated gaseous ion.

d orbitals split into two groups in a complex ion

E N E R G Y

Figure 3.34 5IF TQMJUUJOH PG E PSCJUBMT JO B DPNQMFY JPO

Extension The electrons in dz and dx â&#x20AC;&#x201C; y orbitals are repelled more by the ligand electrons because they point directly at the ligands â&#x20AC;&#x201C; greater repulsion leads to higher energy.

2

2

2

Extension The d orbitals are split in different ways in different-shaped complex ions.

The splitting may be regarded as being caused by the repulsion between the electrons in the metal ion d orbitals and the lone pairs on the ligands. Two of the metal ion d orbitals point directly at the ligands and so are raised in energy, whereas the other three d orbitals point between the ligands and are lowered in energy relative to the other two d orbitals.

Energy in the form of a certain frequency of visible light can be absorbed to promote an electron from the lower set of orbitals to the higher set (Figure 3.35). light energy

absorption of light energy causes an electron to be promoted to the higher set of d orbitals

Figure 3.35 "CTPSQUJPO PG MJHIU CZ B DPNQMFY JPO

110

electron

When white light passes through copper sulfate solution (Figure 3.36), orange light is absorbed, promoting an electron from the lower set of d orbitals to the higher set. This means that the light coming out contains all the colours of the spectrum except orange and so appears blue, the complementary colour to orange.

orange light missing white light CuSO4(aq) orange light absorbed When orange light is removed from white light it appears blue.

Figure 3.36 Colour and absorption.

orange red

yellow

violet

green blue

400

500 600 wavelength / nm

700

White light is a mixture of all colours (frequencies) of visible light.

Figure 3.37 " DPMPVS XIFFM o BMPOH XJUI UIF BQQSPYJNBUF XBWFMFOHUIT PG WJTJCMF light. Complementary colours are opposite each other in the colour wheel, therefore blue is complementary to orange and green is complementary to red.

For a substance to appear coloured, certain frequencies of light in the visible region of the spectrum must be absorbed.

The colour of a substance will appear to an observer as the complementary colour to the light that is absorbed. A colour wheel (Figure 3.37) shows which pairs of colours are complementary (opposite each other in the colour wheel). If we know the colour of the complex ion, the colour of light that is absorbed can be worked out, and vice versa.

For example, because a solution of nickel(II) chloride is green, it must absorb red light – the complementary colour to green.

The formation of coloured substances requires the presence of a partially filled d subshell.

Let us consider the Sc3+ ion or the Ti4+ ion. These both have no electrons in the 3d subshell and so are colourless, as it is not possible to absorb energy to promote a 3d electron.

Extension Cr2O72− (orange), CrO42− (yellow) and MnO−4 (purple) are all very highly coloured, but they have no d electrons. They are coloured because of a different mechanism from the one described here.

3 THE PERIODIC TABLE

111

The Cu+ ion and the Zn2+ ion both have ten 3d electrons (Figure 3.38), and as there is no space in the upper set of orbitals it is not possible to promote an electron to the upper set of orbitals. No light in the visible region of the spectrum is absorbed and these ions are colourless.

+

2+

Figure 3.38 A Cu or Zn ion has ten 3d electrons.

Extension The actual theory behind the spectra of transition metal complexes is significantly more complex than described here, and the simple idea of an electron being promoted from the lower set of d orbitals to the upper set is only really applicable to transition metal ions with one d electron (d9 ions also produce relatively simple spectra). This is evident by the fact that transition metal ions will usually absorb more than one frequency of electromagnetic radiation − not just one as predicted by the simple model.

For transition metals with more than one d electron, the repulsion between d electrons is important in determining the energies of the various energy states. The absorption of electromagnetic radiation by a transition metal ion could be better described as ‘causing a rearrangement of electrons within the d orbitals’.

Most aqueous solutions of Fe3+ are actually yellow, but they do not contain [Fe(H2O)6]3+(aq).

What do we mean when we say that a solution of copper sulfate is blue? Is blueness a property of copper sulfate solution, or is the blueness in our minds? What colour would copper sulfate solution be in orange light? Or in the dark?

Factors that affect the colour of transition metal complexes At the simplest level, the colours of transition metal complexes can be related to the amount of splitting of the d orbitals. For example, if there is a greater difference in energy between the lower and higher set of d orbitals then a higher frequency (shorter wavelength) of light will be absorbed and the complementary colour will be different. The following factors all have a part to play.

Identity of the metal Complexes of different metals in the same oxidation state have different colours. For example, Mn2+(aq) (3d5) is very pale pink/colourless but Fe2+(aq) (3d6) is pale green.

Different metal ions have different electron configurations and, because colours are caused by electron transitions, different arrangements of electrons give rise to different colours due to different amounts of repulsion between electrons.

If isoelectronic (same number of electrons) transition metal ions complexes are considered, such as [Mn(H2O)6]2+ and [Fe(H2O)6]3+ (both metal ions have five 3d electrons) then there will be a greater amount of splitting of d orbitals in [Fe(H2O)6]3+. A higher nuclear charge on the metal ion (26+ for Fe and 25+ for Mn) for the same number of electrons causes the ligands to be pulled in more closely in an Fe3+ complex, so that there is greater repulsion between the ligand electrons and the d electrons of the transition metal ion – and therefore greater splitting of the d orbitals.

Oxidation number The same metal has different colours in different oxidation states. For example:

[Fe(H2O)6]2+(aq) is pale green and [Fe(H2O)6]3+(aq) is pale violet.

In general, for complex ions containing the same metal and the same ligands, the greater the oxidation number of the transition metal, the greater the splitting of the d orbitals.

112

There are two reasons for this: r the electron configurations of the ions are different r a higher charge on the metal ion causes the ligands to be pulled in more closely, so that there is greater repulsion between the ligand electrons and the d electrons of the transition metal ion – and therefore greater splitting of the d orbitals.

Nature of the ligand The same metal ion can exhibit different colours with different ligands.

This is mainly because of the different splitting of the d orbitals caused by different ligands.

Ligands can be arranged into a spectrochemical series according to how much they cause the d orbitals to split:

I− < Br− < Cl− < F – < OH− < H2O < NH3 < CO ≈ CN− So a chloride ion causes greater splitting of the d orbitals than an iodide ion, and an ammonia molecule causes greater splitting of the d orbitals than a water molecule.

[Cu(NH3)4(H2O)2]2+ has a larger energy gap between the two sets of d orbitals than [Cu(H2O)6]2+ and absorbs a shorter wavelength (higher frequency) of light (Figure 3.39). [Cu(NH3)4(H2O)2]2+(aq) is dark blue/ violet and absorbs more in the yellow–green (higher frequency) region of the visible spectrum.

A full explanation of the spectrochemical series is difficult at this level.

The fact that fluoride ions cause greater splitting of d orbitals than iodide ions can be explained in terms of charge density (charge per unit volume) of the ligand – both F− and I− have the same charge but the F− ion is much smaller and therefore causes greater repulsion of the metal ion d electrons and greater splitting of the d orbitals. This explanation cannot, however, be extended to the rest of the spectrochemical series – as can be seen by the fact that CO, a neutral ligand, causes greater splitting of d orbitals than negatively charged ligands that would be expected to have a higher charge density. However, the spectrochemical series can [Cu(NH3)4(H2O6]2+

Ligands that cause greater splitting of d orbitals are called stronger field ligands.

A water molecule is more polar than an ammonia molecule and there would be expected to be a higher charge density on the O in H2O than the N in NH3.

[Cu(H2O6]2+

Absorbance

1.0

0.5

0.0

500

700 Wavelength / nm

900

increasing energy of radiation

Figure 3.39 [Cu(H2O)6]2+ BR JT CMVF BOE BCTPSCT NPTUMZ BU UIF SFEoPSBOHF FOE PG UIF spectrum. [Cu(NH3)4(H2O)2]2+ BR JT EBSL CMVF WJPMFU BOE BCTPSCT NPSF JO UIF ZFMMPXo green (shorter wavelength) region of the visible spectrum.

3 THE PERIODIC TABLE

113