02 chap 2 chemf4 bil 2017(csy5p)

MODULE • Chemistry FORM 4

UNIT

CHEMICAL FORMULAE AND EQUATIONS

2

FORMULA DAN PERSAMAAN KIMIA Concept Map / Peta Konsep

RELATIVE MASS / JISIM RELATIF – Relative Atomic Mass / Jisim Atom Relatif – Relative Molecular Mass / Jisim Molekul Relatif – Relative Formula Mass / Jisim Formula Relatif

Mass (g) Jisim (g)

÷ (Molar Mass) g mol–1 ÷ (Jisim Molar) g mol–1 × (Molar Mass) g mol–1 × (Jisim Molar) g mol–1

× Avogadro constant × pemalar Avogadro

NUMBER OF MOL BILANGAN MOL

÷ Molar Volume (dm3 mol–1) ÷ Isi padu Molar (dm3 mol–1) × Molar Volume (dm3 mol–1) × Isi padu Molar (dm3 mol–1)

Numerical problem involving chemical equation (Interpret quantitatively) Masalah pengiraan melibatkan persamaan kimia (Tafsiran secara kuantitatif) Chemical formula Formula kimia

Empirical Formula Formula empirik

Formula of Ionic Compound Formula sebatian ion

Chemical Equation Persamaan kimia

2

÷ Avogadro constant ÷ pemalar Avogadro

Interpret qualitatively Tafsiran kualitatif

Volume of gas (dm3) Isi padu gas (dm3)

UNIT

Numerically equal Nilai sama

Number of particles Bilangan zarah

– Reactants Bahan tindak balas – Products Hasil – Physical state Keadaan fizikal

Molecular Formula Formula molekul

Calculate Percentage by mass of element in a compound Mengira peratusan jisim unsur dalam suatu sebatian Remark / Catatan: – Molar mass is mass of one mol of a substance. The unit is g mol–1. / Jisim molar ialah jisim bagi satu mol suatu bahan. Unitnya g mol–1.

– Molar volume of gas is volume occupied by one mol of any gas, 24 dm3 mol–1 at room conditions or 22.4 dm3 mol–1 at standard temperature and pressure (s.t.p.) / Isi padu molar gas ialah isi padu yang ditempati oleh satu mol sebarang gas, 24 dm3 mol–1 pada keadaan bilik atau 22.4 dm3 mol–1 pada suhu dan tekanan piawai (s.t.p.)

– Avogadro constant is number of particles in one mol of any substance. The value is 6.02 × 1023.

Pemalar avogadro ialah bilangan zarah dalam satu mol sebarang bahan. Nilainya 6.02 × 1023.

Learning objective / Objektif pembelajaran • •

•

• • • •

Understand and apply the concept of relative atomic mass / Memahami dan mengaplikasi konsep jisim atom relatif Analysing the relationship between the number of mole and the number of particles Menganalisis hubungan antara bilangan mol dengan bilangan zarah Analysing the relationship between the number of mole of material with mass Menganalisis hubungan antara bilangan mol bahan dengan jisim Analysing the number of mole of gas with gas volume / Menganalisis bilangan mol gas dengan isi padu gas Synthesis chemical formula / Mensintensis formula kimia Interpret chemical equations / Mentafsirkan persamaan kimia Adopting scientific attitudes and values in the investigation of matter Mengamalkan sikap saintifik dan nilai murni dalam penyiasatan tentang jirim

21

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 21

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

Relative Mass / Jisim Relatif What is relative mass?

Apakah jisim relatif?

A single atom is too small and light and cannot be weighed directly.

The masses of atoms are not determined directly but comparing them with standard atom.

Satu atom adalah terlalu ringan, kecil dan tidak dapat ditimbang secara langsung.

Jisim satu atom tidak ditentukan secara langsung tetapi dengan membandingkannya dengan atom piawai.

How is Relative Atomic Mass calculated?

Bagaimakah Jisim Atom Relatif dikira?

– Carbon-12 (an isotope of carbon) is chosen as a standard atom because its mass can be determined very accurately using mass spectrometer. Carbon-12 isotope is given a mass of exactly 12.00.

Karbon-12 (isotop karbon) dipilih sebagai atom piawai kerana jisimnya dapat ditentukan dengan tepat menggunakan spektrometer jisim. Isotop karbon-12 mempunyai jisim 12.00.

UNIT

– Relative atomic mass based on the carbon-12 scale is the mass of one atom of the elements compared 1 with —– of the mass of an atom of carbon-12: 12 Jisim atom relatif berdasarkan skala atom karbon-12 adalah jisim satu atom unsur berbanding dengan 1 —– jisim satu atom karbon-12: 12

2

– Relative atomic mass of an element (RAM) Jisim atom relatif suatu unsur (JAR) The average mass of one atom of the element / Jisim purata satu atom unsur = 1 1 —– × The mass of atom of carbon-12 / —– × Jisim satu atom karbon-12 12 12 What can be interpreted?

Apakah tafsirannya?

Mass of an atom of Carbon-12 = 12.00 1 —– of the mass of an atom Carbon-12 = 1 12 Jisim satu atom Karbon-12 = 12.00 1 —– jisim satu atom Karbon-12 = 1 12

C-12

Example / Contoh:

Mg

c

c

Relative atomic mass of magnesium / Jisim atom relatif magnesium = 24 Mass of magnesium atom / Jisim atom magnesium = 1 1 —– × Mass of an atom Carbon-12 / —– × Jisim satu atom Karbon-12 12 12

(

)

1 Mass of magnesium atom = 24 —– × mass of an atom Carbon-12 = 24 12

(

)

1 Jisim atom magnesium = 24 —– x jisim satu atom Karbon-12 = 24 12 What is the unit for relative atomic mass? Give reason.

Apakah unit jisim atom relatif?

Berikan alasan.

It has no unit.

The relative atomic mass of an element can also be considered as the number of times the mass of one 1 atom of that element is heavier than —– of Carbon-12 12 Tiada unit.

Jisim atom relatif suatu unsur juga boleh dianggap sebagai bilangan kali jisim atom suatu unsur itu lebih 1 berat daripada —– Karbon-12 12 Example / Contoh: Relative atomic mass of Helium is 4.

– 3 atoms of Helium have the same mass as one Carbon-12 Jisim atom relatif Helium ialah 4.

– 3 atom Helium mempunyai jisim yang sama dengan satu Karbon-12

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 22

22

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 Define Relative Molecular Mass (RMM).

Takrifkan Jisim Molekul Relatif (JMR).

1 The average mass of one molecule of a substance when compared to —– of the mass of carbon-12.

12 1 Purata jisim satu molekul suatu bahan apabila dibandingkan dengan —– jisim karbon-12.

12 Relative Molecular Mass / Jisim Molekul Relatif The average mass of one molecule / Jisim purata satu molekul = 1 1 —– × The mass of an atom of carbon-12 / —– × Jisim satu atom karbon-12 12 12 Remark: The Relative Molecular Mass is used for covalent molecules.

Catatan: Jisim molekul relatif digunakan untuk molekul kovalen.

How to calculate Relative Molecular Mass (RMM)?

Bagaimanakah cara mengira Jisim Molekul Relatif (JMR)?

RMM is obtained by adding up the RAM of all the atoms that are present in the formula.

JMR diperoleh dengan menambah JAR bagi semua atom yang hadir dalam formula.

Calculate Relative Molecular Mass (RMM) for the following molecular substances: Hitung Jisim Molekul Relatif (JMR) bagi molekul bahan berikut: Molecular substance Bahan molekul

Molecular formula Formula molekul

Relative molecular mass Jisim molekul relatif

O2

2 × 16 = 32

Water / Air

H 2O

2 × 1 + 16 = 18

Carbon dioxide / Karbon dioksida

CO2

12 + 2 × 16 = 44

Ammonia / Ammonia

NH3

14 + 3 × 1 = 17

Oxygen / Oksigen

[Relative atomic mass / Jisim atom relatif : O = 16, H = 1, C = 12, N = 14]

2

Calculate Relative Formula Mass (RFM) for the following ionic substances: Hitung Jisim Formula Relatif (JFR) bagi bahan ion berikut: Substance Bahan

UNIT

1

2

Exercise / Latihan

Chemical formula Formula kimia

Relative formula mass Jisim formula relatif

Sodium chloride / Natrium klorida

NaCl

23 + 35.5 = 58.5

Potassium oxide / Kalium oksida

K2 O

2 × 39 + 16 = 94

CuSO4

64 + 32 + 4 × 16 = 160

(NH4)2CO3

2 [14 + 4 × 1] + 12 + 3 × 16 = 96

Aluminium nitrate / Aluminium nitrat

Al(NO3)3

27 + 3 [14 + 3 × 16] = 213

Calcium hydroxide / Kalsium hidroksida

Ca(OH)2

40 + 2 [16 + 1] = 74

Lead(II) hydroxide / Plumbum(II) hidroksida

Pb(OH)2

207 + 2 [16 + 1] = 241

CuSO4•5H2O

64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250

Copper(II) sulphate / Kuprum(II) sulfat Ammonium carbonate / Ammonium karbonat

Hydrated copper(II) sulphate / Kuprum(II) sulfat terhidrat

[Relative atomic mass / Jisim atom relatif : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12, Al = 27, Ca = 40, Pb = 207]

3

The formula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of metal M? Oksida logam M mempunyai formula M2O3. Jisim formula relatif ialah 152. Apakah jisim atom relatif logam M? M = Relative atomic mass for M / Jisim atom relatif untuk M 2M + 3 × 16 = 152 M = 52

4 Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x.

Fosforus membentuk sebatian klorida dengan formula PClx. Jisim molekul relatifnya adalah 208.5. Hitungkan nilai x.

[Relative atomic mass / Jisim atom relatif : P = 31, Cl = 35.5]

31 + x × 35.5 = 208.5 35.5x = 208.5 – 31 35.5x = 177.5

x = 5

23

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 23

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

5

Relative atomic mass of calcium is 40 based on the carbon-12 scale.

Jisim atom relatif kalsium berdasarkan skala karbon-12 ialah 40. (a) State the meaning of the statement above. / Nyatakan maksud pernyataan di atas.

1

Mass of one calcium atom is 40 times greater than mass of one carbon-12 atom.

12 1

Jisim satu atom kalsium adalah 40 kali ganda lebih besar daripada jisim satu atom karbon-12.

12 (b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16]

Berapa kalikah satu atom kalsium lebih berat daripada satu atom oksigen? [Jisim atom relatif: O = 16] Relative atomic mass of calcium Jisim atom relatif kalsium 40

/ = = 2.5 times / kali Relative atomic mass of oxygen Jisim atom relatif oksigen 16

UNIT

(c) How many calcium atoms have the same mass as two atoms of bromine? [Relative atomic mass: Br = 80]

Berapakah bilangan atom kalsium yang mempunyai jisim yang sama dengan dua atom bromin? [Jisim atom relatif: Br = 80]

Number of calcium atom / Bilangan atom kalsium × 40 = 2 × 80 2 × 80

Number of calcium atom / Bilangan atom kalsium = =4 40

2 Mole Concept / Konsep Mol Mole and the Number of Particles Bilangan Mol dan Bilangan Zarah What is Avogadro constant?

Apakah pemalar Avogadro?

A fixed quantity of 6.02 × 1023 of particles.

Bilangan zarah yang tetap iaitu 6.02 × 10 23.

Remark / Catatan: The particles can be atom, ion and molecules Sama ada zarah atom, molekul, dan ion

Why is Avogardro constant useful?

Mengapakah pemalar Avogadro berguna?

It is a way for counting the particles (atoms, ions, or molecules). This is because the size of particles is too small, so it is not possible to count physically.

Suatu cara untuk mengira bilangan zarah (atom, ion, molekul). Ini kerana saiz zarah adalah terlalu kecil, adalah mustahil untuk dikira secara fizikal.

What is mole?

Apakah itu mol?

1 A mole is an amount of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12.

Satu mol ialah jumlah bahan yang mengandungi bilangan zarah seperti mana bilangan atom yang terdapat dalam 12 g karbon-12.

2 A mole of a substance is the amount of substance which contains a constant number of particles (atoms, ions, molecules), which is 6.02 × 1023.

Satu mol bahan adalah jumlah bahan yang mengandungi bilangan zarah (atom, ion, molekul) yang tetap iaitu 6.02 × 10 23.

Example: Contoh:

The concept of mole is the same as the concept of a dozen in our everday life. Dozen is used to represent a quantity: Konsep mol adalah sama dengan konsep sedozen dalam kehidupan harian kita. Dozen digunakan untuk mewakil satu kuantiti:

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 24

1 dozen of pencil Sedozen pensel 1 × 12 pencils / pensel

2 dozens of pencil 2 dozen pensel 2 × 12 pencils / pensel

3 dozens of pencil 3 dozen pensel 3 × 12 pencils / pensel

1 mol of atom / 1 mol atom 1 × 6.02 × 1023 atoms / atom

2 mol of atom / 2 mol atom 2 × 6.02 × 1023 atoms / atom

3 mol of atom / 3 mol atom 3 × 6.02 × 1023 atoms / atom

24

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 Why is mole concept useful?

Mengapakah konsep mol berguna?

When a compound is made up of two or more atoms, for example a covalent molecule or an ionic compound, the mole concept is useful to determine the number of respective particles.

Apabila suatu sebatian dihasilkan dari dua atau lebih atom, sebagai contoh satu molekul kovalen atau satu sebatian ion, konsep mol adalah berguna untuk menentukan bilangan zarah masing-masing.

Example / Contoh: Methane has a formula CH4. 1 methane molecule CH4 is made up of 1 C atom and 4 H atoms which are covalently bonded.

Metana mempunyai formula CH4. 1 molekul metana CH4 terdiri daripada satu atom C dan 4 atom H yang mana terikat secara kovalen.

Is made up of Terdiri daripada

H C

H

H

H 1 CH4 molecule 1 molekul CH4

C

H H

H H

1 carbon atom and 4 hydrogen atoms 1 atom karbon dan 4 atom hidrogen

× Avogadro Constant / Pemalar Avogadro

Number of moles Bilangan mol

÷ Avogadro Constant / Pemalar Avogadro

UNIT

What is the relationship between number of moles and number of particles (atoms/ions/molecules)?

Apakah hubungan antara bilangan mol dengan bilangan zarah (atom/ ion/molekul)?

2

– Hence, 6.02 × 1023 CH4 contains 1 × 6.02 × 1023 C atoms and 4 × 6.02 × 1023 H atoms.

Oleh itu, 6.02 × 10 23 CH4 mengandungi 1 × 6.02 × 10 23 atom C dan 4 × 6.02 ×10 23 atom H. – Applying the mole concept / Mengaplikasi konsep mol: 1 mol of CH4 molecules consists of 1 mol C atoms and 4 mol of H atoms.

1 mol molekul CH4 terdiri daripada 1 mol atom C dan 4 mol atom H.

Number of particles Bilangan zarah

Exercise / Latihan 1

Complete the following table: Lengkapkan jadual berikut: Substance Bahan Chlorine Klorin

Water Air

Ammonia Ammonia

Sulphur dioxide Sulfur dioksida

Formula Formula

Number of atom per molecule/ Number of positive and negative ion Bilangan atom per molekul/ Bilangan ion positif dan negatif

Cl2

Cl : 2

H 2O

NH3

SO2

Magnesium chloride Magnesium klorida

MgCl2

Aluminium oxide Aluminium oksida

Al2O3

6.02 × 1023 Cl2 molecules / molekul 2

× 6.02 × 1023 Cl atoms / atom

6.02 × 1023 H2O molecules / molekul

H : 2

2

O : 1

1

× 6.02 × 1023 H atoms / atom × 6.02 × 1023 O atoms / atom

6.02 × 1023 NH3 molecules / molekul

N : 1

1

H : 3

3

× 6.02 × 1023 N atoms / atom × 6.02 × 1023 H atoms / atom

6.02 × 1023 SO2 molecules / molekul

S : 1

1

O : 2

2

Mg2+ : 1

1

Cl : 2

2

Al3+ : 2

2

O : 3

3

–

2–

25

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 25

Number of particles in 1 mol of substance Bilangan zarah dalam 1 mol bahan

× 6.02 × 1023 S atoms / atom × 6.02 × 1023 O atoms / atom × 6.02 × 1023 Mg2+ ions / ion × 6.02 × 1023 Cl– ions / ion × 6.02 × 1023 Al3+ ions / ion × 6.02 × 1023 O2– ions / ion © Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

2

Complete the following: [Differentiate between “mole” dan “molecule”] Lengkapkan yang berikut: [Bezakan antara “mol” dan “molekul”] (a) 1 mol of Cl2

[Chlorine gas]

1 mol Cl2 [Gas klorin]

(b)

(c) UNIT

2

1 mol of NH3 [Ammonia gas] 1 mol NH3 [Gas ammonia]

1 mol of NH3 4 [Ammonia gas] 1 mol NH 3 4 [Gas ammonia]

(d) 2 mol of MgCl2

[Magnesium chloride]

2 mol MgCl2 [Magnesium klorida]

(e) 2 mol of SO2

[Sulphur dioxide]

2 mol SO2 [Sulfur dioksida]

6.02 × 1023 2 × 6.02 × 1023

molecules of chlorine, Cl2 / molekul klorin, Cl2 atoms of chlorine, Cl / atom klorin, Cl

6.02 × 1023 4

molecules of ammonia, NH3 / molekul ammonia, NH3 1 mol of nitrogen atom, N / mol atom nitrogen, N mol atoms / mol atom 3 mol of hydrogen atoms, H / mol atom hidrogen, H

1 23 4 × 6.02 × 10

1

mol of atom

1

mol atom

molecules of ammonia, NH3 / molekul ammonia, NH3 1 or/atau 0.25 4 mol of N atoms / mol atom N, 0.25 × 6.02 × 1023 number of N atoms / bilangan atom N = 3 or/atau 0.75 4 mol of H atoms / mol atom H, 0.75 × 6.02 × 1023 number of H atoms / bilangan atom H =

2

mol of Mg2+ ions /mol ion Mg2+, number of Mg2+ ions / bilangan ion Mg2+ =

2 × 6.02 × 1023

4

mol of Cl– ions /mol ion Cl–, number of Cl– ions /bilangan ion Cl– = 2 × 6.02 × 1023

4 × 6.02 × 1023

molecules of SO2 / molekul SO2 2 mol of S atoms / mol atom S, number of S atoms / bilangan atom S =

3 × 2 = 6 mol of atoms 3 × 2 = 6 mol atom

2 × 6.02 × 1023

4

mol of O atoms / mol atom O, number of O atoms / bilangan atom O =

4 × 6.02 × 1023

Number of Moles and Mass of Substance / Bilangan Mol dan Jisim Bahan Define of molar mass.

Takrifkan jisim molar.

Molar mass is the mass of one mole of any substance.

Jisim molar adalah jisim satu mol sebarang bahan.

State how to obtain the molar mass for any substance.

Nyatakan bagaimana untuk memperoleh jisim molar untuk sebarang bahan.

Molar mass of any substance is numerically equal to its relative mass (Relative atomic mass / relative formula mass / relative molecular mass).

Jisim molar sebarang bahan mempunyai nilai yang sama dengan jisim relatif (Jisim atom relatif / jisim formula relatif / jisim molekul relatif).

What is the unit of molar mass?

Apakah unit jisim molar?

Molar mass is the relative atomic mass (RAM), relative molecular mass (RMM) and relative formula mass (RFM) of a substance in g mol–1.

Jisim molar adalah jisim atom relatif (JAR), jisim molekul relatif (JMR) dan jisim formula relatif (JFR) suatu bahan dalam g mol–1.

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 26

26

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 Example Contoh

Substance Bahan

Carbon, C Karbon, C

Water, H2O Air, H2O

Aluminium, Al Aluminium, Al

Sodium chloride, NaCl Natrium klorida, NaCl

Relative mass Jisim relatif

12

2(1) + 16 = 18

27

35.5 + 23 = 58.5

O

AI

12

C

H

H

Na+

Cl –

1 mol substance 1 mol bahan

12.01 g

Molar mass Jisim molar What is the relationship between number of moles and any given mass of a substance?

Apakah hubungan antara bilangan mol dengan jisim sebarang bahan?

Example Contoh

27.00 g

18.00 g

58.50 g

58.00 g

27.00 g

12 g

18 g

27 g

58.5 g

12 g mol–1

18 g mol–1

27 g mol–1

58.5 g mol–1

× (RAM/RFM/RMM) g mol–1 × (JAR/JFR/JMR) g mol–1

Number of moles Bilangan mol

÷ (RAM/RFM/RMM) g mol–1 ÷ (JAR/JFR/JMR) g mol–1

2

Mass for 1 mol Jisim 1 mol

18.00 g

UNIT

12.00 g

Mass in gram Jisim dalam gram

(i) Calculate mass of 2 mol of water / Hitungkan jisim 2 mol air.

Relative molecular mass of H2O / Jisim molekul relatif H2O = 18 Molar mass of 1 mol of H2O / Jisim molar 1 mol H2O = 18 g mol–1 Mass of 2 mol of H2O / Jisim molar 2 mol H2O = Number of moles × Molar mass / Bilangan mol × Jisim molar 18 g mol–1 = 2 mol × 36

=

g

(ii) Calculate number of moles of 45 g of water, H2O / Hitungkan bilangan mol bagi 45 g air, H2O. Number of moles of 45 g of H2O / Bilangan mol 45 g H2O

=

Mass of H2O / Jisim H2O Molar mass / Jisim molar

=

45 g = 2.5 mol 18 g mol–1

Exercise / Latihan Calculate / Hitung: 1 Mass of 3 mol of sodium hydroxide, NaOH Jisim bagi 3 mol natrium hidroksida, NaOH

2 Number of moles in 20 g of sodium hydroxide, NaOH Bilangan mol dalam 20 g natrium hidroksida, NaOH

Molar mass of NaOH / Jisim molar NaOH = (23 + 16 + 1) g mol–1 = 40 g mol–1

Number of moles of sodium hydroxide, NaOH Bilangan mol natrium hidroksida, NaOH 20 g = 40 g mol–1 = 0.5 mol

Mass of 3 mol of sodium hydroxide, NaOH Jisim 3 mol natrium hidroksida, NaOH = 3 mol × 40 g mol–1 = 120 g Answer / Jawapan: 120 g

27

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 27

Answer / Jawapan: 0.5 mol © Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

3 Mass of 2.5 mol of oxygen gas, O2 Jisim 2.5 mol gas oksigen, O2 Molar mass of oxygen gas, O2 Jisim molar gas oksigen, O2 = (16 + 16) g mol–1 = 32 g mol–1 Mass of 2.5 mol of oxygen gas, O2 Jisim 2.5 mol gas oksigen, O2 = 2.5 mol × 32 g mol–1 = 80 g

4 Mass of 0.5 mol of sodium chloride, NaCl Jisim 0.5 mol natrium klorida, NaCl Molar mass of NaCl / Jisim molar NaCl = (23 + 35.5) g mol–1 = 58.5 g mol–1 Mass of 0.5 mol of sodium chloride, NaCl Jisim 0.5 mol natrium klorida, NaCl = 0.5 mol × 58.5 g mol–1 = 29.25 g Answer / Jawapan: 80 g

UNIT

2

5 Number of moles in 37.8 g of zinc nitrate, Zn(NO3)2 Bilangan mol dalam 37.8 g zink nitrat, Zn(NO3)2 Molar mass of zinc nitrate, Zn(NO3)2 Jisim molar zink nitrat, Zn(NO3)2 = [65 + 2 (14 + 3 × 16)] g mol–1 = 189 g mol–1 Number of moles of zinc nitrate, Zn(NO3)2 Bilangan mol zink nitrat, Zn(NO3)2 37.8 g = 189 g mol–1 = 0.2 mol

Answer / Jawapan: 29.25 g 6 Mass of 3.01 × 1023 copper atoms, Cu Jisim 3.01 × 1023 atom kuprum, Cu Number of moles of Cu / Bilangan mol Cu Number of copper atom / Bilangan atom kuprum = Avogadro constant / Pemalar Avogadro 3.01 × 1023 = 0.5 mol 6.02 × 1023 Mass of Cu / Jisim Cu = Number of moles × Molar mass / Bilangan mol × Jisim molar = 0.5 mol × 64 g mol–1 = 32 g =

Answer / Jawapan: 0.2 mol

Answer / Jawapan: 32 g

Number of Moles and Volume of Gas / Bilangan Mol dan Isi Padu Gas Define molar volume of gas.

Nyatakan maksud isi padu molar gas.

Volume occupied by 1 mol of a gas.

Isi padu yang dipenuhi oleh 1 mol gas.

State the molar volume of any gases at room conditions and at standard temperature and pressure (s.t.p).

Nyatakan isi padu molar sebarang gas pada keadaan bilik dan pada suhu dan tekanan piawai (s.t.p).

Molar volume of any gases at room conditions is 24 dm3 mol–1.

Isi padu molar sebarang gas adalah 24 dm3 mol–1 pada keadaan bilik.

Molar volume of any gases at standard temperature and pressure (s.t.p) is 22.4 dm3 mol–1.

Isi padu molar sebarang gas adalah 22.4 dm3 mol–1 pada suhu dan tekanan piawai (s.t.p).

Remark / Catatan: The volume of gas is affected by temperature and pressure.

Isi padu gas dipengaruhi oleh suhu dan tekanan.

Example / Contoh: The diagram shows the molar volume of three gases at room conditions.

Rajah menunjukkan isi padu molar bagi tiga gas pada keadaan bilik.

24 dm3

24 dm3

1 mol oxygen gas, O2 1 mol gas oksigen, O2 (32 g) 6.02 × 1023 O2 molecules / molekul

1 mol ammonia gas, NH3 1 mol gas ammonia, NH3 (17 g) 6.02 × 1023 NH3 molecules / molekul

24 dm3 1 mol carbon dioxide gas, CO2 1 mol gas karbon dioksida, CO2 (44 g) 6.02 × 1023 CO2 molecules / molekul

Remark / Catatan: 1 dm3 = 1 000 cm3

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 28

28

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 Relationship between number of moles and any given volume of gas.

Hubungan antara bilangan mol dan isi padu sebarang gas yang diberi.

Example Contoh

× 24 dm3 mol–1 / 22.4 dm3 mol–1

Number of moles of gas Bilangan mol

÷ 24 dm3 mol–1 / 22.4 dm3 mol–1

(i) 2 mol of carbon dioxide gas occupies

2 mol gas karbon dioksida menempati 0.5

(ii) 16 g of oxygen gas = volume of

44.8

12

dm3 at STP. dm3 pada STP.

mol of oxygen gas. Therefore, 16 g of oxygen gas occupies a

dm at room conditions. [Relative atomic mass: O =16] 3

16 g gas oksigen = 12

44.8

Volume of gas in dm3 Isi padu gas dalam dm3

0.5

mol gas oksigen. Oleh itu, 16 g gas oksigen menempati isi padu

dm pada keadaan bilik. [Jisim atom relatif; O = 16] 3

UNIT

2

Formula for conversion of unit: Formula untuk penukaran unit: Volume of gas in dm3 Isi padu gas dalam dm3

÷ 24 dm3 mol–1/ 22.4 dm3 mol–1

Mass in gram (g) Jisim dalam gram (g)

÷ (RAM/RFM/RMM) g mol–1 ÷ (JAR/JFR/JMR) g mol–1 × (RAM/RFM/RMM) g mol–1 × (JAR/JFR/JMR) g mol–1

× 24 dm3 mol–1/ 22.4 dm3 mol–1

Number of moles Bilangan mol

÷ (6.02 × 1023)

Number of particles Bilangan zarah

× (6.02 × 1023)

Exercise / Latihan 1

A sampel of chlorine gas weighs 14.2 g. Calculate Suatu sampel gas klorin berjisim 14.2 g. Hitungkan [Relative atomic mass / Jisim atom relatif : Cl = 35.5] (a) Number of moles of chlorine atoms / Bilangan mol atom klorin.

Number of moles of chlorine atoms / Bilangan mol atom klorin, Cl =

14.2 g = 0.4 mol 35.5 g mol–1

(b) Number of moles of chlorine molecules (Cl2) / Bilangan mol molekul klorin (Cl2 ).

14.2 g

Number of moles of chlorine molecules / Bilangan mol molekul klorin, Cl2 = = 0.2 mol 71 g mol–1

(c)

Volume of chlorine gas at room conditions. / Isi padu gas klorin pada keadaan bilik.

[Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure] [Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan bilik] Volume of chlorine gas / Isi padu gas klorin = 0.2 mol × 24 dm3 mol–1 = 4.8 dm3

29

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 29

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

2

(a) Calculate the number of atoms in the following substances:

Hitungkan bilangan atom yang terdapat dalam bahan berikut: [Relative atomic mass: N = 14; Zn = 65; Avogadro Constant = 6.02 × 1023] [Jisim atom relatif: N = 14; Zn = 65; Pemalar Avogadro = 6.02 × 1023] (i) 13 g of zinc / 13 g zink 13 g = 0.2 mol 65 g mol–1

Number of zinc atom / Bilangan atom zink = 0.2 × 6.02 × 1023

= 1.204 × 1023

Number of mol of zinc atom / Bilangan mol atom zink =

(ii) 5.6 g of nitrogen gas / 5.6 g gas nitrogen 6.5 g = 0.4 mol 14 g mol–1

Number of N atom / Bilangan atom N = 0.4 × 6.02 × 1023

= 2.408 × 1023

UNIT

2

(b)

Number of moles of N atom / Bilangan mol atom N =

Calculate the number of molecules in the following substances: Hitungkan bilangan molekul dalam bahan berikut: [Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant = 6.02 × 1023] [Jisim atom relatif: N = 14, H = 1, Cl = 35.5, Pemalar Avogadro = 6.02 × 1023] (i) 8.5 g of ammonia gas, NH3 / 8.5 g gas ammonia, NH3

Molar mass of ammonia gas / Jisim molar gas ammonia, NH3 = (14 + 3) g mol–1

= 17 g mol–1

8.5 g 17 g mol–1 = 0.5 mol

Number of moles of ammonia gas / Bilangan mol gas ammonia, NH3 =

Number of molecules in ammonia gas / Bilangan molekul dalam gas ammonia, NH3 = 0.5 mol × 6.02 1023

= 3.01 × 1023 (ii) 14.2 g of chlorine gas, Cl2 / 14.2 g gas klorin, Cl2 Molar mass of chlorine gas, Cl2 / Jisim molas gas klorin, Cl2 = 35.5 × 2 g mol–1 = 71 g mol–1

Jisim klorin

Number of moles of chlorine gas / Bilangan mol gas klorin, Cl2 = Mass of chlorine / Jisim molar Molar mass 14.2 g

= = 0.2 mol 71 g mol–1

Number of chlorine molecules / Bilangan molekul klorin = 0.2 mol × 6.02 × 1023

= 1.204 × 1023 3

A gas jar contains 240 cm3 of carbon dioxide gas. Calculate: Suatu balang gas berisi 240 cm3 gas karbon dioksida. Hitungkan: [Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions] [Jisim atom relatif: C = 12, O = 16; Isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik] (a) Number of moles of carbon dioxide gas / Bilangan mol gas karbon dioksida: 240 cm3

Number of moles of CO2 / Bilangan mol CO2 = = 0.01 mol 24 000 cm3 mol–1 (b) Number of molecules of carbon dioxide gas / Bilangan molekul gas karbon dioksida:

Number of molecules of CO2 / Bilangan molekul CO2 = 0.01 × 6.02 × 1023

= 6.02 × 1021

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 30

30

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

(c) Mass of carbon dioxide gas / Jisim gas karbon dioksida:

Mass of CO2 / Jisim CO2 = 0.01 mol × [12 + 2 × 16] g mol–1

= 0.44 g

4

What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as found in 3.6 g of water?

Berapakah jisim molekul klorin (Cl2) yang mengandungi dua kali ganda bilangan molekul yang terdapat dalam 3.6 g air?

[Relative atomic mass / Jisim atom relatif : H = 1, O = 16, Cl = 35.5] Number of moles of chlorine molecule / Bilangan mol molekul klorin, Cl2 = 2 × Number of moles of water / Bilangan mol air, H2O 3.6 g Number of moles of H2O / Bilangan mol H2O = 18 g mol–1 = 0.2 mol

UNIT

2

Number of moles of chlorine molecule / Bilangan mol molekul klorin = 2 × 0.2 mol = 0.4 mol

Mass of Cl2 / Jisim Cl2 = 0.4 mol × 71 g mol–1

= 28.4 g

5

Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium.

Hitungkan jisim karbon yang mempunyai bilangan atom yang sama seperti yang terdapat dalam 4 g magnesium.

[Relative atomic mass / Jisim atom relatif : C = 12, Mg = 24] 4g 1 Number of moles of magnesium/ Bilangan mol magnesium = 24 g mol–1 = mol 6 Number of moles of carbon / Bilangan mol karbon = Number of moles of magnesium / Bilangan mol magnesium 1 = mol 6 1 Mass of carbon / Jisim karbon = mol × 12 g mol–1 = 2 g 6

6

Compare the number of molecules in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your answer.

Bandingkan bilangan molekul dalam 32 g sulfur dioksida (SO2 ) dengan 7 g gas nitrogen (N2 ). Terangkan jawapan anda.

[Relative atomic mass / Jisim atom relatif : S = 32, O = 16, N = 14] 32 g Number of moles of molecules in 32 g SO2 / Bilangan mol molekul dalam 32 g SO2 = 64 g mol–1 = 0.5 mol 7g Number of moles of molecules in 7 g N2 / Bilangan mol molekul dalam 7 g N2 = 28 g mol–1 = 0.25 mol Number of molecules in 32 g SO2 is two times more than 7 g N2.

Bilangan molekul dalam 32 g SO2 adalah dua kali lebih banyak daripada 7 g N2.

Number of mole in sulphur dioxide molecules is two times more than number of mole of nitrogen molecules.

Bilangan mol molekul SO2 adalah dua kali lebih banyak daripada bilangan mol molekul nitrogen.

7

Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer.

Bandingkan bilangan atom dalam 1.28 g oksigen dengan bilangan atom dalam 1.3 g zink. Terangkan jawapan anda.

[Relative atomic mass / Jisim atom relatif : O = 16, Zn = 65] 1.28 g Number of moles of O atoms in 1.28 g O2 / Bilangan mol atom O dalam 1.28 g O2 = 16 g mol–1 = 0.08 mol 1.30 g Number of moles of Zn atoms in 1.3 g Zn / Bilangan mol atom Zn dalam 1.3 g Zn = 65 g mol–1 = 0.02 mol Number of oxygen atoms in 1.28 g oxygen is 4 times more than number of zinc atoms in 1.3 g zinc.

Bilangan atom oksigen dalam 1.28 g oksigen adalah 4 kali lebih banyak daripada bilangan atom zink dalam 1.3 g zink.

Number of mol of oxygen atom is 4 times more than zinc atom.

Bilangan mol atom oksigen adalah 4 kali lebih banyak daripada atom zink.

31

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 31

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

Chemical Formulae / Formula Kimia Define chemical formula.

Takrifkan formula kimia.

A set of chemical symbols for atoms of elements in whole numbers subscript representing chemical substances.

Satu set simbol kimia bagi atom-atom unsur dengan subskrip nombor bulat yang mewakili bahan kimia.

Example / Contoh: Substance / Bahan

What information can be obtained from the chemical formula?

Apakah maklumat yang boleh diperoleh dari formula kimia?

Chemical formula / Formula kimia

Water / Air

H 2O

Ammonia / Ammonia

NH3

Propane / Propana

C 3H 8

Example / Contoh: Substance Bahan

Chemical formula Formula kimia

Information Maklumat

UNIT

(i)

2 Ammonia Ammonia

Elements present in the substance Unsur yang terdapat dalam sebatian Ammonia is made up of nitrogen and hydrogen Ammonia terdiri daripada nitrogen dan hidrogen

(ii) Number of atoms of each element in the compound

Bilangan atom setiap unsur yang terdapat dalam sebatian

Ammonia molecule consists of 1 nitrogen atom and 3 hyrogen atoms

Molekul ammonia terdiri daripada 1 atom nitrogen dan 3 atom hidrogen

NH3

(iii) Relative formula mass / relatif molecular mass Jisim formula relatif / jisim molekul relatif Relative molecular mass = 14 + (3 × 1) = 17 Jisim molekul relatif = 14 + (3 × 1) = 17 Define empirical formula.

Takrifkan formula empirik.

A formula that shows the simplest whole number ratio of atoms of each element in a compound.

Formula yang menunjukkan nisbah nombor bulat teringkas bagi bilangan atom setiap unsur yang terdapat dalam sebatian.

Define molecular formula.

Takrifkan formula molekul.

Molecular formula of a compound is a formula that shows the actual number of atoms of each element that are present in a molecule of the compound.

Formula molekul suatu sebatian menunjukkan bilangan sebenar atom bagi setiap unsur yang terdapat dalam satu molekul sebatian.

What is the relationship between molecular formula and empirical formula?

Apakah hubungan antara formula molekul dan formula empirik?

Molecular formula = (Empirical formula)n , where n is an integer.

Formula molekul = (Formula empirik)n, di mana n ialah integer.

Determine the empirical formula and the value of n. / Nyatakan formula empirik dan nilai n.

Compound Sebatian

Molecular formula Formula molekul

Empirical formula Formula empirik

Value of n Nilai n

Water Air

H 2O

H 2O

1

Carbon dioxide Karbon dioksida

CO2

CO2

1

H2SO4

H2SO4

1

Ethene Etena

C 2H 4

CH2

2

Benzene Benzena

C 6H 6

CH

6

Glucose Glukosa

C6H12O6

CH2O

6

Sulphuric acid Asid sulfurik

Remark / Catatan: The molecular formula and the empirical formula of a compound will be the same if the value of n = 1 but different if the value of n > 1. / Formula molekul dan formula empirik suatu sebatian akan sama sekiranya nilai n = 1 tetapi akan berbeza sekiranya nilai n > 1.

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 32

32

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 Example Contoh

1

The empirical formula for chlorinated hydrocarbon is CHCl2. The relative formula mass of this compound is 168. Find the molecular formula of this compound. / Formula empirik bagi hidrokarbon berklorin ialah CHCl2. Jisim formula relatif sebatian ini ialah 168. Cari formula molekul sebatian ini.

(CHCl2)n = 168 Molecular formula / Formula molekul (12 + 1 + [2 × 35.5])n = 168 = (Empirical formula / Formula empirik)n (84)n = 168 = (CHCl2)2

n = 2 = C2H2Cl4

Experiments to determine empirical formula of metal oxide: / Eksperimen untuk menentukan formula empirik oksida logam: Empirical formula of magnesium oxide Formula empirik magnesium oksida

Empirical formula of copper(II) oxide Formula empirik kuprum(II) oksida

Set-up of apparatus / Susunan radas:

Set-up of apparatus / Susunan radas: Copper(II) oxide Kuprum(II) oksida Magnesium Magnesium

Heat Panaskan

2

UNIT

2

Hydrogen gas Gas hidrogen

Heat Panaskan

Reaction occurs / Tindak balas yang berlaku: Magnesium is burnt in a crucible to react with oxygen to form magnesium oxide.

Magnesium dipanaskan dengan kuat di dalam mangkuk pijar untuk bertindak balas dengan oksigen membentuk magnesium oksida.

Reaction occurs / Tindak balas yang berlaku: Hydrogen gas is passed through the heated copper(II) oxide.

Hydrogen reduces copper(II) oxide to form copper and water.

Gas hidrogen dilalukan melalui kuprum(II) oksida yang dipanaskan.

Hidrogen menurunkan kuprum(II) oksida kepada kuprum dan air.

Balanced equation / Persamaan kimia seimbang: 2Mg + O2 → 2MgO

Balanced equation / Persamaan kimia seimbang: CuO + H2 → Cu + H2O

This method can also be used to determine the empirical formulae of reactive metal oxide such as aluminium oxide and zinc oxide.

Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida logam reaktif seperti aluminium oksida dan zink oksida.

This method can also be used to determine the empirical formulae of less reactive metal oxide such as lead oxide and tin oxide.

Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida logam kurang reaktif seperti plumbum oksida dan stanum oksida

Experiment to Determine Empirical Formula of Magnesium Oxide Eksperimen untuk Menentukan Formula Empirik Magnesium Oksida In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide: Semasa eksperimen ini, magnesium bertindak balas dengan oksigen dalam udara untuk membentuk asap putih, magnesium oksida: Magnesium + Oxygen → Magnesium oxide Magnesium + Oksigen → Magnesium oksida

Material / Bahan : Magnesium ribbon, sand paper / Pita magnesium, kertas pasir Apparatus /Radas : Crucible with lid, tongs, Bunsen burner, pipe-clay triangle, balance and tripod stand / Mangkuk pijar dengan penutup, penyepit, penunu Bunsen, segi tiga tanah liat, penimbang dan tungku kaki tiga

Set-up of apparatus / Susunan radas: Magnesium ribbon / Pita magnesium

Heat / Panaskan

Procedure / Langkah: (a) A crucible and its lid are weighed and the mass recorded.

Mangkuk pijar dengan penutup ditimbang dan jisimnya dicatatkan.

33

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 33

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

(b)

(c)

(d)

(e) (f)

(g)

(h)

UNIT

(i)

(j)

(k)

2

10 cm of magnesium ribbon is cleaned with sand paper.

10 cm pita magnesium dibersihkan dengan menggunakan kertas pasir.

The magnesium ribbon is coiled loosely and placed in the crucible.

Pita magnesium digulung dan diletakkan dalam mangkur pijar.

The crucible together with the lid and magnesium ribbon are weighed again.

Mangkuk pijar bersama dengan penutup dan pita magnesium ditimbang.

The apparatus is set up as shown in the diagram. / Radas disusun seperti ditunjukkan dalam rajah.

The crucible is heated strongly without its lid. When the magnesium starts to burn, the crucible is covered with its lid.

Mangkuk pijar dipanaskan dengan kuat tanpa penutup. Apabila pita magnesium mula terbakar, mangkuk pijar ditutup dengan penutup.

The lid of the crucible is lifted from time to time using a pair of tongs.

Penutup dibuka sekali sekala dengan menggunakan penyepit.

When the magnesium ribbon stops burning, the lid is removed and the crucible is heated strongly for another 2 minutes.

Apabila pita magnesium berhenti terbakar, penutup dibuka dan mangkuk pijar dipanaskan dengan kuat selama 2 minit lagi.

The crucible, the lid and its content are allowed to cool down to room temperature.

Mangkuk pijar, penutup dan kandungannya dibiarkan sejuk ke suhu bilik.

The crucible, lid and its content are weighed again and the mass recorded.

Mangkuk pijar, penutup dan kandungannya ditimbang sekali lagi dan jisimnya dicatatkan.

The process of heating, cooling and weighing are repeated until a constant mass is obtained.

Proses pemanasan, penyejukan dan penimbangan diulang beberapa kali sehingga jisim tetap diperoleh.

Precaution steps / Langkah berjaga-jaga: Step taken / Langkah yang diambil sand paper

Magnesium ribbon is cleaned with

kertas pasir

Pita magnesium perlu digosok dengan crucible lid The Penutup mangkuk pijar

Purpose / Tujuan .

oxide layer

To remove the .

is lifted from time to time.

dibuka sekali sekala. crucible lid replaced The is then quickly.

Penutup mangkuk pijar ditutup semula kemudian dengan cepat.

on the surface of the magnesium ribbon. lapisan oksida pada permukaan magnesium oksida.

Untuk membuang

oxygen To allow magnesium .

Untuk membenarkan magnesium dengan To prevent fumes of

from the air to react with oksigen

. magnesium oxide

from escaping. magnesium oksida dari terbebas.

Untuk mengelakkan wasap The process of weighing

heating

,

and

penyejukan

dan penimbangan jisim tetap diperoleh.

diulang beberapa kali sehingga

constant mass

is repeated until a

is obtained.

Proses pemanasan

cooling

,

Observation / Pemerhatian: brightly Magnesium burns Magnesium terbakar dengan pepejal putih membentuk

masuk dan bertindak balas

To ensure magnesium react oxygen to form

completely

with magnesium oxide

membentuk

magnesium oksida

white fumes

to release terang

membebaskan

.

white solid and is formed. wasap putih dan kemudiannya

.

Inference / Inferens:

Magnesium is a

Magnesium adalah logam yang

Magnesium reacts with

Magnesium bertindak balas dengan

Result / Keputusan:

reactive oxygen

metal. reaktif

.

magnesium oxide in the air to form . oksigen magnesium oksida dalam udara membentuk

Description / Penerangan

x

Mass of crucible + lid + magnesium / Jisim mangkuk pijar + penutup + magnesium

y

Mass of crucible + lid + magnesium oxide / Jisim mangkuk pijar + penutup + magnesium oksida

z

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 34

.

Mass (g) / Jisim (g)

Mass of crucible + lid / Jisim mangkuk pijar + penutup

Š Nilam Publication Sdn. Bhd.

.

Untuk memastikan semua magnesium telah bertindak balas lengkap oksigen dengan untuk

34

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

Calculation / Pengiraan: Element / Unsur

O

Mass (g) Jisim (g)

y–x

z–y

Number of mole of atoms Bilangan mol atom

y–x 24

z–y 16

p

q

Simplest ratio of moles Nisbah mol teringkas

Empirical formula of magnesium oxide is

MgpOq

.

Formula empirik magnesium oksida ialah

MgpOq

.

3

Experiment to Determine Empirical Formula of Copper(II) oxide Eksperimen untuk Menentukan Formula Empirik Kuprum(II) oksida

2

Mg

UNIT

Copper(II) oxide + Hydrogen → Copper + Water / Kuprum(II) oksida + Hidrogen → Kuprum + Air

Set-up of apparatus / Susunan radas: Copper(II) oxide / Kuprum(II) oksida Burning of hydrogen gas Nyalaan gas hidrogen

Hydrogen gas / Gas hidrogen

Combustion tube Heat / Panaskan

Tabung pembakaran

Anhydrous calcium chloride, CaCl2 / Kalsium klorida kontang, CaCl2

Precaution steps / Langkah berjaga-jaga: Step taken / Langkah yang ambil

Purpose / Tujuan

Hydrogen gas is passed through anhydrous calcium chloride.

Gas hidrogen dialirkan melalui kalsium klorida kontang.

dry Anhydrous calcium chloride absorb water vapour to the hydrogen gas. mengeringkan Kalsium klorida kontang menyerap wap air untuk gas hidrogen.

Dry

hydrogen gas is passed through the

combustion tube for 5 to 10 minutes.

Gas hidrogen

kering

To

remove

air

all the

(The mixture of hydrogen gas and dialirkan melalui tabung

pembakaran selama 5 hingga 10 minit.

lighted).

Untuk

mengeluarkan

(Campuran hidrogen dan

in the combustion tube. air will cause explosion when

semua udara dalam tabung pembakaran. udara menghasilkan letupan apabila

dinyalakan) The gas that comes out from the small hole is collected in the test tube. Then, a lighted wooden splinter is placed

If the gas burns quietly without ‘pop’ sound , all the removed has been from the combustion tube.

at the mouth of the test tube.

Jika gas terbakar tanpa bunyi ‘pop’ , semua gas telah

Gas yang keluar daripada lubang kecil dikumpul dalam sebuah menyala letakkan tabung uji. Kayu uji di di

daripada tabung pembakaran.

air dikeluarkan

mulut tabung uji.

35

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 35

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 continuous

The flow of hydrogen gas must be throughout the experiment.

berterusan

Gas hidrogen dialirkan secara

copper To prevent hot copper(II) oxide form sepanjang

eksperimen.

The process of

heating

cooling

, constant

and weighing are

repeated until a Proses pemanasan

mass is obtained. , penyejukan dan penimbangan tetap diulangi beberapa kali sehingga jisim diperoleh.

Observation / Pemerhatian:

The

black

Warna

To ensure all

perang

UNIT

2

Inference / Inferens:

Copper(II) oxide reacts with hydrogen to produce the brown

Kuprum(II) oksida bertindak balas dengan hidrogen untuk menghasilkan

Result / Keputusan:

.

copper metal

. logam kuprum

yang berwarna perang.

Mass (g) / Jisim (g)

Mass of combustion tube + porcelain dish Jisim tabung pembakaran + piring tanah liat

x

Mass of combustion tube + porcelain dish + copper(II) oxide Jisim tabung pembakaran + piring tanah liat + kuprum(II) oksida

y

Mass of combustion tube + porcelain dish + copper Jisim tabung pembakaran + piring tanah liat + kuprum

z

Calculation / Pengiraan: Element / Unsur

Cu

O

Mass (g) Jisim (g)

z–x

y–z

Number of mole of atoms Bilangan mol atom

z–x 64

y–z 16

p

q

Simplest ratio of moles Nisbah mol teringkas

copper has changed to . kuprum(II) oksida telah bertukar kepada

.

Description / Penerangan

panas daripada bertindak balas kuprum(II) oksida .

copper(II) oxide

brown

kuprum(II) oksida menjadi

to

dan membentuk

Untuk memastikan semua kuprum .

colour of copper(II) oxide turns

hitam

again. kuprum

Untuk mengelakkan oksigen dengan

oxygen

from reacting with

Empirical formula of copper(II) oxide is

CupOq CupOq

.

Formula empirik kuprum(II) oksida ialah

4

Explain why the set-up of apparatus to determine the empirical formula in both experiments are different.

Terangkan mengapa susunan radas untuk menentukan formula empirik dalam kedua-dua eksperimen itu berbeza. reactive oxidised (a) Magnesium is metal (above hydrogen in reactivity series). Magnesium easily to form magnesium oxide reaktif . / Magnesium adalah logam (terletak di atas hidrogen dalam siri kereaktifan. teroksida

Magnesium mudah (b) Copper is below hydrogen gas

hydrogen

membentuk

.

magnesium oksida

.

in the metal reactivity series. Oxygen in copper(II) oxide can be

to form copper and water. hidrogen Kuprum berada di bawah dalam siri kereaktifan. Kuprum(II) oksida boleh oleh

gas hidrogen

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 36

removed

by

disingkirkan

untuk membentuk kuprum dan air.

36

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

5

To calculate the empirical formula of a compound, the following table can be used as a guide: Untuk menghitung formula empirik suatu sebatian, jadual di bawah boleh digunakan sebagai panduan: Calculation steps / Langkah pengiraan: (a) Calculate the mass of each element in the compound.

Hitungkan jisim setiap unsur dalam sebatian. (b) Convert the mass of each element to number of mole of atom.

Tukar jisim setiap unsur kepada bilangan mol atom. (c) Calculate the simplest ratio of moles of atom of the elements.

Hitungkan nisbah bilangan mol atom teringkas unsur-unsur tersebut.

Element / Unsur Mass of element (g) / Jisim unsur (g) Number of mole of atom / Bilangan mol atom Simplest ratio of moles / Nisbah mol teringkas

When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical formula of metal X oxide. / Apabila 11.95 g oksida logam X diturunkan oleh hidrogen, 10.35 g logam X terhasil. Hitungkan formula empirik bagi oksida logam X. [Relative atomic mass / Jisim atom relatif : X = 207, O = 16] Element / Unsur

X

O

10.35

1.6

10.35 = 0.05 207

1.6 = 0.1 16

Ratio of moles / Nisbah mol

1

2

Simplest ratio of moles / Nisbah mol teringkas

1

2

Mass of element (g) / Jisim unsur (g) Number of moles of atoms / Bilangan mol atom

XO2

Empirical formula / Formula empirik:

2

A certain compound contains the following composition: Satu sebatian mengandungi komposisi unsur seperti berikut: Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass / Jisim atom relatif: O = 16, Na = 23, Br = 80] (Assume that 100 g of substance is used / Anggap 100 g bahan digunakan)

.

Element / Unsur

Na

Br

O

Mass of element (g) / Jisim unsur (g)

15.23

52.98

31.79

Number of moles of atoms / Bilangan mol atom

0.66

0.66

1.99

Ratio of moles / Nisbah mol

1

1

3.01

Simplest ratio of moles / Nisbah mol teringkas

1

1

3

NaBrO3

Empirical formula / Formula empirik:

3

2.08 g of element X combines with 4.26 g of element Y to form a compound with formula XY3. Calculate the relative atomic mass of element X. [RAM: Y = 35.5] / 2.08 g unsur X bergabung dengan 4.26 g unsur Y untuk membentuk sebatian dengan formula XY3 . Hitung jisim atom relatif unsur X. [JAR: Y = 35.5]

.

Element / Unsur

X

Y

Mass of element (g) / Jisim unsur (g)

2.08

4.26

Number of moles of atom Bilangan mol atom

2.08 x

Simplest ratio of moles Nisbah mol teringkas

1

4.26 = 0.12 35.5 3

37

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 37

UNIT

1

2

Exercise / Latihan

x = relative atomic mass of X / jisim atom relatif bagi X Mol X 1

= Mol Y 3 2.08 x 1

= 3 0.12

x = 52 Š Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

4

2.07 g of element Z reacts with bromine to form 3.67 g of a compound with empirical formula ZBr2. Calculate the relative atomic mass of element Z. [RAM: Br = 80] 2.07 g unsur Z bertindak balas dengan bromin membentuk 3.67 g sebatian dengan formula empirik ZBr2. Hitung jisim atom relatif bagi unsur Z. [JAR: Br = 80] Element / Unsur

Z

Br

Mass of element (g) Jisim unsur (g)

2.07

3.67 – 2.07 = 1.6

Number of moles of atoms Bilangan mol atom

2.07 z

1.6 = 0.02 80

1

2

Simplest ratio of moles Nisbah mol teringkas

UNIT

z = relative atomic mass of Z / jisim atom relatif bagi Z Mol Z 1

= Mol Br 2 2.07 z 1

= 2 0.02

2

z = 207

5

The empirical formula of compound X is CH2 and relative molecular mass is 56. Determine the molecular formula of compound X. / Formula empirik sebatian X adalah CH2 dan jisim atom relatif adalah 56. Tentukan formula molekul sebatian X.

[Relative atomic mass / Jisim atom relatif: H = 1; C = 12]

(12 + 2)n = 56 56

n= =4 14

Molecular formula / Formula molekul = (CH2)4 = C4H8

6

2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86.

2.58 g suatu hidrokarbon mengandungi 2.16 g karbon. Jisim molekul relatif bagi hidrokarbon ini ialah 86.

[Relative atomic mass / Jisim atom relatif : H = 1; C = 12] (i) Calculate the empirical formula of the hydrocarbon. / Hitungkan formula empirik bagi hidrokarbon ini.

Element Unsur

C

H

Mass of element (g) / Jisim (g)

2.16

2.58 – 2.16 = 0.42

Number of moles of atoms / Bilangan mol atom

0.18

0.42

Ratio of moles / Nisbah mol

1

Simplest ratio of moles / Nisbah mol teringkas

3

2

1 7 = 3 3 7

Empirical formula / Formula empirik = C3H7

(ii) Determine the molecular formula of the hydrocarbon. / Tentukan formula molekul hidrokarbon tersebut.

(3 × 12 + 7 × 1)n = 86 86

n= =2 43

Molecular formula / Formula molekul = (C3H7)2 = C6H14

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 38

38

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

7

The diagram below shows the structural formula for benzene molecule.

Rajah di bawah menunjukkan formula struktur bagi benzena.

H H C C C H H C C C H H (a) Name the elements that make up benzene. / Namakan unsur yang membentuk benzena.

Carbon and hydrogen / Karbon dan hidrogen (b) What are the molecular formula and empirical formula for benzene?

Apakah formula molekul dan formula empirik bagi benzena?

Molecular formula / Formula molekul: C6H6

2

Empirical formula / Formula empirik: CH

UNIT

(c) Compare and contrast the molecular formula and empirical formula for benzene.

Banding dan bezakan formula molekul dan formula empirik bagi benzena.

• Both empirical formula and molecular formula shows benzene is made up of

carbon

and

hydrogen

Kedua-dua fomula molekul dan formula empirik menunjukkan benzena terdiri daripada unsur hidrogen .

elements. karbon dan

• Molecular formula shows the actual Each benzene molecule consists of

number of carbon atoms and hydrogen atoms in benzene molecule.

6 carbon atoms and 6 hydrogen atoms. karbon Formula molekul menunjukkan bilangan sebenar bagi atom dan atom hidrogen dalam molekul molekul 6 karbon 6 benzena. Setiap benzena terdiri daripada atom dan atom hidrogen .

• Empirical formula shows the simplest ratio of number of carbon atoms to hydrogen atoms, the simplest carbon hydrogen 1:1 ratio of number of atoms to atoms in benzene is . nisbah paling ringkas karbon Formula empirik benzena menunjukkan bilangan atom kepada atom hidrogen .

Nisbah paling ringkas bilangan atom

karbon

kepada hidrogen dalam benzena adalah

1:1 .

Percentage Composition by Mass of an Element in a Compound Peratus Komposisi Unsur Mengikut Jisim dalam Sebatian Formula Formula

% composition by mass of an element = % komposisi unsur mengikut jisim =

Example Contoh

Jumlah JAR unsur dalam suatu sebatian × 100% JMR/JFR sebatian

Calculate the percentage composition by mass of nitrogen in the following compounds: Hitungkan peratusan nitrogen mengikut jisim dalam sebatian berikut: [Relative atomic mass / Jisim atom relatif : N = 14, H = 1, O = 16, S = 32, K = 39] (i) (NH4)2SO4 (ii) KNO3 14 2 × 14 %N = × 100% %N = × 100% 101 132

= 21.2%

39

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 39

Total RAM of the element in the compound × 100% RMM/RFM of compound

= 13.9%

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4 Exercise Latihan

The diagram below shows two different fertilisers used by the farmers onto the crops so that the crops can grow faster and bigger and thus, their crop yields can be increased.

Rajah di bawah menunjukkan dua baja berbeza yang digunakan oleh petani ke atas tanaman mereka supaya tanaman mereka dapat tumbuh dengan lebih pantas dan lebih besar dan seterusnya hasil tanaman mereka dapat ditingkatkan.

FERTILISER BAJA B Ammonium nitrate / Ammonia nitrat, NH4NO3

FERTILISER BAJA A Urea / Urea, (NH2)2CO

By using your knowledge in chemistry, determine the best fertiliser to be used by the farmers.

Dengan menggunakan pengetahuan kimia anda, tentukan baja terbaik untuk digunakan oleh petani.

UNIT

Answer / Jawapan: Urea, (NH2)2CO / Urea, (NH2)2CO Percentage of N / Peratusan N =

28 × 100% = 46.67% 60

2 Ammonium nitrate, NH4NO3 / Ammonia nitrat, NH4NO3 28 Percentage of N / Peratusan N = × 100% = 35% 80 Urea , (NH2)2CO is the best fertiliser because it contains higher percentage of nitrogen by mass.

Urea, (NH2)2CO adalah baja terbaik kerana it mengandungi peratusan jisim yang lebih tinggi.

Chemical Formula for Ionic Compounds / Formula Kimia bagi Sebatian Ion 1

2

Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is obtained by exchanging the charges on each ion.

The formula obtained will be XnYm. Formula kimia sebatian ion yang mengandungi ion X m + dan Y n– boleh diperoleh melalui pertukaran bilangan cas setiap ion.

Formula yang diperoleh ialah XnYm. Example / Contoh: (i) Sodium oxide / Natrium oksida Ion / Ion

Na+

O2–

Charges / Bilangan cas

+1

–2

Exchange of charges / Pertukaran bilangan cas

2

1

Smallest ratio / Nisbah teringkas

2

1

2 Na+

O2–

Number of combining ions / Bilangan ion yang bergabung Formula / Formula

Na2O

(ii) Copper(II) nitrate / Kuprum(II) nitrat Cu2+ NO3– +2 –1

1

2

➾ Cu(NO3)2

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 40

(iii) Zinc oxide / Zink oksida Zn2+ O2– +2 –2

2

(Simplest Ratio / Nisbah teringkas)

1 ➾ ZnO

2 1 (Simplest Ratio / Nisbah teringkas)

40

11/10/2017 10:23 AM

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 41

41 CaCO3 Calcium carbonate Kalsium karbonat

CaO Calcium oxide Kalsium oksida

CuO Copper(II) oxide Kuprum(II) oksida

MgO Magnesium oxide Magnesium oksida

ZnO Zinc oxide Zink oksida

PbO Lead(II) oxide Plumbum(II) oksida

Al2O3 Aluminium oxide Aluminium oksida

Ca2+ Calcium ion Ion kalsium

Cu2+ Copper(II) ion Ion kuprum(II)

Mg2+ Magnesium ion Ion magnesium

Zn2+ Zinc ion Ion zink

Pb2+ Lead(II) ion Ion plumbum(II)

Al 3+ Aluminium ion Ion aluminium

Al2(CO3)3 Aluminium carbonate Aluminium karbonat

PbCO3 Lead(II) carbonate Plumbum(II) karbonat

ZnCO3 Zinc carbonate Zink karbonat

MgCO3 Magnesium carbonate Magnesium karbonat

CuCO3 Copper(II) carbonate Kuprum(II) karbonat

(NH4)2CO3 Ammonium carbonate Ammonium karbonat

NH4 + Ammonium ion Ion ammonium

Ag2CO3 Silver carbonate Argentum karbonat

Ag2O Silver oxide Argentum oksida

Ag+ Silver ion Ion argentum

Na2CO3 Sodium carbonate Natrium karbonat H2CO3 Carbonic acid Asid karbonik

Na2O Sodium oxide Natrium oksida

Na+ Sodium ion Ion natrium

K2CO3 Potassium carbonate Kalium karbonat

CO32–, Carbonate ion Ion karbonat

H+ Hydrogen ion Ion hidrogen

K2O Potassium oxide Kalium oksida

K+ Potassium ion Ion kalium

O2–, Oxide ion Ion oksida

AgBr Silver bromide Argentum bromida

HBr Hydrobromic acid Asid hidrobromik

NaBr Sodium bromide Natrium bromida

KBr Potassium bromide Kalium bromida

Br–, Bromide ion Ion bromida

Al2(SO4)3 Aluminium sulphate Aluminium sulfat

PbSO4 Lead(II) sulphate Plumbum sulfat

ZnSO4 Zinc sulphate Zink sulfat

AlCl3 Aluminium chloride Aluminium klorida

PbCl2 Lead(II) chloride Plumbum klorida

ZnCl2 Zinc chloride Zink klorida

AlBr3 Aluminium bromide Aluminium bromida

PbBr2 Lead(II) bromide Plumbum bromida

ZnBr2 Zinc bromide Zink bromida

MgBr2 Magnesium bromide Magnesium bromida

MgCl2 MgSO4 Magnesium Magnesium sulphate chloride Magnesium sulfat Magnesium klorida

CaBr2 Calcium bromide Kalsium bromida CuBr2 Copper(II) bromide Kuprum(II) bromida

CaCl2 Calcium chloride Kalsium klorida

NH4Cl NH4Br Ammonium chloride Ammonium bromide Ammonium klorida Ammonium Bromida

AgCl Silver chloride Argentum klorida

HCl Hydrocloric acid Asid hidroklorik

NaCl Sodium chloride Natrium klorida

KCl Potassium chloride Kalium klorida

Cl–, Chloride ion Ion klorida

CuCl2 Copper(II) chloride Kuprum(II) klorida

CuSO4 Copper(II) sulphate Kuprum(II) sulfat

CaSO4 Calcium sulphate Kalsium sulfat

(NH4)2SO4 Ammonium sulphate Ammonium sulfat

Ag2SO4 Silver sulphate Argentum sulfat

H2SO4 Sulphuric acid Asid sulfurik

Na2SO4 Sodium sulphate Natrium sulfat

K2SO4 Potassium sulphate Kalium sulfat

SO42–, Sulphate ion Ion sulfat

UNIT

2

AlI3 Aluminium iodide Aluminium iodida

PbI2 Lead(II) iodide Plumbum iodida

ZnI2 Zinc iodide Zink iodida

MgI2 Magnesium iodide Magnesium iodida

CuI2 Copper(II) iodide Kuprum(II) iodida

CaI2 Calcium iodide Kalsium iodida

NH4I Ammonium iodide Ammonium iodida

AgI Silver iodide Argentum iodida

HI Hydroiodic acid Asid hidroiodik

NaI Sodium iodide Natrium iodida

KI Potassium iodide Kalium iodida

I–, Iodide ion Ion iodida

Aktiviti 1: Tulis Formula Kimia Dan Nama Bagi Bahan Kimia Berikut

Al(NO3)3 Aluminium nirate Aluminium nitrat

Pb(NO3)2 Lead(II) nitrate Plumbum nitrat

Pb(OH)2 Lead(II) hydroxide Plumbum(II) hidroksida Al(OH)3 Aluminium hydroxide Aluminium hidroksida

Zn(NO3)2 Zinc nitrate Zink nitrat

Mg(NO3)2 Magnesium nitrate Magnesium nitrat

Cu(NO3)2 Copper(II) nitrate Kuprum(II) nitrat

Ca(NO3)2 Calcium nitrate Kalsium nitrat

NH4NO3 Ammonium nitrate Ammonium nitrat

AgNO3 Silver nitrate Argentum nitrat

HNO3 Nitric acid Asid nitrik

NaNO3 Sodium nitrate Natrium nitrat

KNO3 Potassium nitrate Kalium nitrat

NO3–, Nitrate ion Ion nitrat

Zn(OH)2 Zinc hydroxide Zink hidroksida

Mg(OH)2 Magnesium hydroxide Magnesium hidroksida

Cu(OH)2 Copper(II) hydroxide Kuprum(II) hidroksida

Ca(OH)2 Calcium hydroxide Kalsium hidroksida

AgOH Silver hydroxide Argentum hidroksida

NaOH Sodium hydroxide Natrium hidroksida

KOH Potassium hydroxide Kalium hidroksida

OH–, Hydroxide ion Ion hidroksida

ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS

MODULE • Chemistry FORM 4

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

UNIT

2

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 42

CaO

CuO

MgO

ZnO

PbO

Al2O3

Calcium ion Ion kalsium

Copper(II) ion Ion kuprum(II)

Magnesium ion Ion magnesium

Zinc ion Ion zink

Lead(II) ion Ion plumbum(II)

Aluminium ion Ion aluminium

Ammonium ion Ion ammonium

Al2(CO3)3

PbCO3

ZnCO3

MgCO3

CuCO3

CaCO3

(NH4)2CO3

Ag2CO3

Ag2O

Silver ion Ion argentum

Na2CO3

H2CO3

Na2O

Sodium ion Ion natrium

K2CO3

Carbonate ion Ion karbonat

Hydrogen ion Ion hidrogen

K2O

Potassium ion Ion kalium

Oxide ion Ion oksida

Al2(SO4)3

PbSO4

ZnSO4

MgSO4

CuSO4

CaSO4

(NH4)2SO4

Ag2SO4

H2SO4

Na2SO4

K2SO4

Sulphate ion Ion sulfat

AlCl3

PbCl2

ZnCl2

MgCl2

CuCl2

CaCl2

NH4Cl

AgCl

HCl

NaCl

KCl

Chloride ion Ion klorida

AlBr3

PbBr2

ZnBr2

MgBr2

CuBr2

CaBr2

NH4 Br

AgBr

HBr

NaBr

KBr

Bromide ion Ion bromida

AlI3

PbI2

ZnI2

MgI2

CuI2

CaI2

NH4 I

AgI

HI

NaI

KI

Iodide ion Ion iodida

Al(OH)3

Pb(OH)2

Zn(OH)2

Mg(OH)2

Cu(OH)2

Ca(OH)2

AgOH

NaOH

KOH

Hydroxide ion Ion hidroksida

Al(NO3)3

Pb(NO3)2

Zn(NO3)2

Mg(NO3)2

Cu(NO3)2

Ca(NO3)2

NH4 NO3

AgNO3

HNO3

NaNO3

KNO3

Nitrate ion Ion nitrat

ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULAE OF THE FOLLOWING COMPOUNDS Aktiviti 2: Tanpa Merujuk Kepada Jadual Aktiviti 1, Tuliskan Formula Kimia Bagi Sebatian Berikut

MODULE • Chemistry FORM 4

42

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

ACTIVITY 3: WRITE THE CHEMICAL FORMULAE AND TYPE OF PARTICLES FOR THE FOLLOWING ELEMENT/COMPOUND Aktiviti 3: Tulis Formula Kimia dan Jenis Zarah untuk Unsur/Sebatian Berikut Type of particles Jenis zarah

Formula Formula

Type of particles Jenis zarah

Na2SO4

Ion

Zinc carbonate Zink karbonat

ZnCO3

Ion

Ammonium carbonate Ammonium karbonat

(NH4)2CO3

Ion

Ammonium carbonate Ammonium karbonat

(NH4)2CO3

Ion

Magnesium nitrate Magnesium nitrat

Mg(NO3)2

Ion

Silver chloride Argentum klorida

AgCl

Ion

Hydrochloric acid Asid hidroklorik

HCl

Ion

Sulphuric acid Asid sulfurik

H2SO4

Ion

Potassium oxide Kalium oksida

K2O

Ion

Copper(II) nitrate Kuprum(II) nitrat

Cu(NO3)2

Ion

Magnesium oxide Magnesium oksida

MgO

Ion

Hydrogen gas Gas hidrogen

H2

Molecule Molekul

PbCO3

Ion

Carbon dioxide gas Gas karbon dioksida

CO2

Molecule Molekul

Fe2(SO4)3

Ion

Oxygen gas Gas oksigen

O2

Molecule Molekul

Magnesium chloride Magnesium klorida

MgCl2

Ion

Aluminium sulphate Aluminium sulfat

Al2(SO4)3

Ion

Zinc sulphate Zink sulfat

ZnSO4

Ion

Lead(II) chloride Plumbum(II) klorida

PbCl2

Ion

Silver nitrate Argentum nitrat

AgNO3

Ion

Potassium iodide Kalium iodida

KI

Ion

(NH4)2SO4

Ion

Copper(II) carbonate Kuprum(II) karbonat

CuCO3

Ion

ZnO

Ion

Potasium carbonate Kalium karbonat

K2CO3

Ion

HNO3

Ion

Sodium hydroxide Natrium hidroksida

NaOH

Ion

Ammonia gas Gas ammonia

NH3

Molecule Molekul

Aqueous ammonia Ammonia akueus

NH3(aq)

Ion and Molecule Ion dan Molekul

Magnesium Magnesium

Mg

Atom

Ammonium chloride Ammonium klorida

NH4Cl

Ion

Zinc Zink

Zn

Atom

Nitrogen dioxide gas Gas nitrogen dioksida

NO2

Molecule Molekul

CuSO4

Ion

Sodium chloride Natrium klorida

NaCl

Ion

I2

Molecule Molekul

Silver Argentum

Ag

Atom

Cl2

Molecule Molekul

Bromine Bromin

Br2

Molecule Molekul

Sodium sulphate Natrium sulfat

Lead(II) carbonate Plumbum(II) karbonat Iron(III) sulphate Ferum(III) sulfat

Ammonium sulphate Ammonium sulfat Zinc oxide Zink oksida Nitric acid Asid nitrik

Copper(II) sulphate Kuprum(II) sulfat Iodine Iodin Chlorine Klorin

Compound / Element Sebatian / Unsur

43

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 43

2

Formula Formula

UNIT

Compound / Element Sebatian / Unsur

Š Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

Chemical Equations / Persamaan Kimia 1

A chemical equation summerises what happen during chemical reaction.

Suatu persamaan kimia merumuskan perkara yang berlaku semasa tindak balas kimia.

Reactants (Substances take part in the reaction) Bahan tindak balas (Bahan mengambil bahagian dalam tindak balas)

Produce / Menghasilkan

Products (Substances that are produced) Hasil (Bahan yang dihasilkan)

Example / Contoh: Reaction between zinc powder and hydrochloric acid produce zinc chloride aqueous and hydrogen gas.

Tindak balas antara serbuk zink dan asid hidroklorik menghasilkan zink klorida akueus dan gas hidrogen.

Identify the reactant and products.

Kenal pasti bahan dan hasil.

Reactants: Zinc and hydrochloric acid Bahan tindak balas: Zink dan asid hidroklorik

UNIT

Products: Zinc chloride and hydrogen gas Hasil: Zink klorida dan gas hidrogen

2

Write the word equation.

Tuliskan persamaan perkataan.

Zinc + hyrochloric acid ➝ Zinc chloride + hydrogen gas Zink + asid hidroklorik ➝ Zink klorida + gas hidrogen

Write the chemical formula of the reactants and products. List down the number of atoms of each element on both sides of the equation.

Tuliskan formula kimia bahan dan hasil. Senaraikan bilangan atom bagi setiap unsur pada kedua-dua bahagian persamaan itu.

Zn atom H atom Cl atom

Balance the number of atoms of each element on both sides of the equation by adjusting the coefficients in front of the formulae.

Imbangkan bilangan atom setiap jenis unsur dengan menambahkan pekali di hadapan formula kimia

Zn + HCI Left / Kiri 1 1 1

Zn atom H atom Cl atom

➝

Zn + 2HCI Left / Kiri 1 2 2

ZnCl2 + H2 Right / Kanan 1 (Balanced / Seimbang) 2 (Not balanced / Tidak seimbang) 2 (Not balanced / Tidak seimbang) ➝

ZnCl2 + H2 Right / Kanan 1 (Balanced / Seimbang) 2 (Balanced / Seimbang) 2 (Balanced / Seimbang)

Put the state symbols for every reactant and products: Letakkan simbol keadaan bagi setiap bahan dan hasil: Remark / Catatan: State symbols / Simbol keadaan (s) / (p)

(I) / (c)

(g)

(aq) / (ak)

Solid Pepejal

Liquid Cecair

Gas Gas

Aqueous Akueus

Zn (s/p) + 2HCI (aq/ak) ➝ ZnCl2 (aq/ak) + H2 (g)

Qualitative interpretation of the chemical equation (the reactants and the products) Tafsiran kualitatif persamaan kimia (bahan tindak balas dan hasil tindak balas)

(i) The reactants are solid zinc and hydrochloric acid.

Bahan tindak balas adalah pepejal zink dan asid hidroklorik.

Quantitative interpretation of the chemical equation (the coefficient of each formula shows the number of moles of reactants react and products formed) Tafsiran kuantitatif persamaan kimia (pekali setiap formula menunjukkan bilangan mol bahan dan hasil tindak balas)

Zn + 2HCI ➝ ZnCl2 + H2 Coefficient 1 2 1 1 Pekali

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 44

(ii) The products are zinc chloride solution and hydrogen gas.

Hasil tindak balas adalah larutan zink klorida dan gas hidrogen.

1 mol of zinc reacts with 2 mol of hydrochloric acid to produce 1 mol of zinc chloride and 1 mol of hydrogen.

1 mol zink bertindak balas dengan 2 mol asid hidroklorik untuk menghasilkan 1 mol zink klorida dan 1 mol hidrogen.

44

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

Exercise / Latihan Write a balanced chemical equation for each of the following reactions: Tulis persamaan kimia seimbang bagi setiap tindak balas yang berikut: 1 Zinc carbonate Zinc oxide + Carbon dioxide / Zink karbonat Zink oksida + Karbon dioksida ZnCO3 ZnO + CO2 2

Sulphuric acid + Sodium hydroxide H2SO4 + 2NaOH Na2SO4 + 2H2O

Sodium sulphate + Water / Asid sulfurik + Natrium hidroksida

3

Silver nitrate + Sodium chloride Silver chloride + Sodium nitrate Argentum nitrat + Natrium klorida Argentum klorida + Natrium nitrat AgNO3 + NaCl AgCl + NaNO3

4

Copper(II) oxide + Hydrochloric acid Kuprum(II) oksida + Asid hidroklorik CuO + 2HCl CuCl2 + H2O

5

Sodium + Water Sodium hydroxide + Hydrogen / Natrium + Air 2Na + 2H2O 2NaOH + H2

Natrium sulfat + Air

2

Copper(II) chloride + Water Kuprum(II) klorida + Air

UNIT

Natrium hidroksida + Hidrogen

Numerical Problems involving Chemical Equations Penghitungan Berkaitan Persamaan Kimia Example / Contoh: The equation shows the reaction between zinc and hydrochloric acid.

Persamaan menunjukkan tindak balas antara zink dengan asid hidroklorik.

Zn + 2HCl

ZnCl2 + H2

Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at room conditions.

[Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions] Hitungkan jisim zink yang perlu ditindakbalaskan dengan asid hidroklorik berlebihan untuk menghasilkan 6 dm3 gas hidrogen pada keadaan bilik. [Jisim atom relatif: Zn = 65, Cl = 35.5, 1 mol gas menempati 24 dm3 pada suhu bilik] Write a balanced equation Tuliskan persamaan seimbang Write the information from the question above the equation Tulis maklumat daripada soalan di atas persamaan

?g ?g Zn(s) +

Convert the given quantitity into moles by using the method shown in the chart below.

Tukarkan kuantiti yang diberi kepada mol menggunakan teknik dalam carta di bawah.

excess berlebihan 2HCl ➝ ZnCl2 +

6 dm3 6 dm3 H2

Number of moles of H2 / Bilangan mol H2 6 dm3 = 24 dm3 mol–1 = 0.25 mol

Use the mole ratio of the substances involved to find the number of moles of the other substance.

Gunakan nisbah mol bahan yang terlibat untuk mencari bilangan mol bahan lain.

From the equation, Daripada persamaan, 1 mol H2 : 1 mol Zn 0.25 mol H2 : 0.25 mol Zn

Convert the mole into the quantity required in the question by using the method shown in the chart below.

Tukar mol kepada kuantiti yang dikehendaki menggunakan carta di bawah.

Mass of Zn / Jisim Zn = 0.25 mol × 65 g mol–1 = 16.25 g

Remark / Catatan: The coefficient of each formula shows the number of moles of reactants react and products formed. / Pekali setiap formula menunjukkan bilangan mol bahan tindak balas yang bertindak balas dan hasil yang terbentuk.

Mass (g) Jisim (g)

÷ (RAM/RFM/RMM) g mol–1 ÷ (JAR/JFR/JMR) g mol–1 × (RAM/RFM/RMM) g mol–1 × (JAR/JFR/JMR) g mol–1

No. of moles (n) Bilangan mol (n)

45

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 45

× 24 dm3 mol–1 / 22.4 dm3 mol–1 ÷ 24 dm3 mol–1 / 22.4 dm3 mol–1

Volume of gas (dm3) Isi padu gas (dm3)

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

Exercise / Latihan 1

The equation shows the reaction between potassium and oxygen.

Persamaan berikut menunjukkan tindak balas antara kalium dengan oksigen.

4K + O2

2K2O

Calculate the mass of potassium required to produce 23.5 g of potassium oxide. [Relative atomic mass: K = 39, O = 16] Hitungkan jisim kalium yang diperlukan untuk menghasilkan 23.5 g kalium oksida. [Jisim atom relatif: K = 39, O = 16] 23.5 g 23.5 Number of moles of K2O / Bilangan mol K2O = = = 0.25 mol (2 × 39 + 16) g mol–1 94 From the equation / Daripada persamaan,

2 mol K2O : 4 mol K

0.25 mol K2O : 0.5 mol K

UNIT

Mass of K / Jisim K = 0.5 mol × 39 g mol–1 = 19.5 g

2 2

The equation shows the decomposition of hydrogen peroxide.

Persamaan menunjukkan penguraian hidrogen peroksida.

H 2O 2

H 2O + O 2

Balance the equation above. Calculate the number of moles of H2O2 that decomposes if 11.2 dm3 oxygen gas is collected at STP. [Relative atomic mass: H = 1, O = 16, molar volume of gas = 22.4 dm3 mol–1 at STP] Seimbangkan persamaan di atas. Hitung bilangan mol H2O2 yang telah terurai sekiranya 11.2 dm3 gas oksigen dikumpulkan pada STP. [Jisim atom relatif: H = 1, O = 16, isi padu molar gas = 22.4 dm3 mol–1 pada STP] 2H2O2 2H2O + O2 Number of moles of O2 / Bilangan mol O2

11.2 dm3 = = 0.5 mol

3 –1 22.4 dm mol From the equation / Daripada persamaan, 1 mol O2 : 2 mol H2O2 0.5 mol O2 : 1.0 mol H2O2

3

8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate produced. [Relative atomic mass: N = 14, O = 16, Cu = 64] 8.0 g serbuk kuprum(II) oksida dicampurkan kepada asid nitrik cair yang berlebihan dan dihangatkan. Hitungkan jisim kuprum(II) nitrat yang terhasil. [Jisim atom relatif: N = 14, O = 16, Cu = 64] CuO + 2HNO3 Cu(NO3)2 + H2O 8g Number of moles of CuO / Bilangan mol CuO = = 0.1 mol (64 + 16) g mol–1 From the equation / Daripada persamaan,

1 mol CuO : 1 mol Cu(NO3)2

0.1 CuO : 0.1 mol Cu(NO3)2 Mass of Cu(NO3)2 / Jisim Cu(NO3)2 = 0.1 mol × 188 g mol–1 = 18.8 g

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 46

46

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

4

1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1at STP] HOTS

1.3 g zink bertindak balas dengan asid sulfurik cair yang berlebihan. Hasil tindak balas ialah zink sulfat dan hidrogen.

Hitungkan isi padu hidrogen yang terbebas pada STP. [Jisim atom relatif: Zn = 65, isi padu molar gas = 22.4 dm3 mol–1 pada STP]

Zn + H2SO4 ZnSO4 + H2 1.3 g

Number of moles of Zn / Bilangan mol Zn =

65 g mol–1

= 0.02 mol

Volume of H2 / Isi padu H2 = 0.02 mol × 22.4 dm3 mol–1 = 0.448 dm3

= 448 cm3

5

0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate the volume of chlorine gas that has reacted. [Relative atomic mass: Na = 23, Molar volume of gas = 24 dm3 mol–1 at room conditions] 0.46 g natrium terbakar lengkap dalam gas klorin pada keadaan bilik menghasilkan natrium klorida. Hitungkan isi padu klorin yang diperlukan untuk bertindak balas lengkap. [Jisim atom relatif: Na = 23, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

2Na + Cl2

Number of moles of Na / Bilangan mol Na =

2NaCl

UNIT

2

From the equation / Daripada persamaan, 1 mol Zn : 1 mol H2 0.02 mol Zn : 0.02 mol H2

0.46 g

23 g mol–1 = 0.02 mol

Volume of Cl2 / Isi padu Cl2 = 0.01 mol × 24 dm3 mol–1

= 0.24 dm3

= 240 cm3

From the equation / Daripada persamaan, 2 mol Na : 1 mol Cl2 0.02 mol Na : 0.01 mol Cl2

6

The equation shows the combustion of propane gas.

Persamaan menunjukkan pembakaran gas propana.

C3H8 + 5O2

3CO2 + 4H2O

720 cm3 of propane gas (C3H8) at room conditions burns in excess oxygen. Calculate the mass of carbon dioxide formed.

[Relative atomic mass: C = 12, O = 16, Molar volume of gas = 24 dm3 mol–1 at room conditions] 720 cm3 gas propana (C3H8) pada keadaan bilik terbakar dalam oksigen berlebihan. Hitungkan jisim karbon dioksida yang terbentuk. [Jisim atom relatif: C = 12, O = 16, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

C3H8 + 5O2

Number of moles of C3H8 / Bilangan mol C3H8 =

3CO2 + 4H2O

720 cm3

24 000 cm3 mol–1 = 0.03 mol

Mass of CO2 / Jisim CO2 = 0.09 mol × 44 g mol–1 = 3.96 g

From the equation / Daripada persamaan, 1 mol C3H8 : 3 mol CO2 0.03 mol C3H8 : 0.09 mol CO2

47

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 47

© Nilam Publication Sdn. Bhd.

11/10/2017 10:23 AM

MODULE • Chemistry FORM 4

7

UNIT

The diagram below shows a car fitted with air bag which will inflate in an accident.

Rajah di bawah menunjukkan sebuah kereta dilengkapi dengan beg udara yang akan mengembang apabila berlaku kemalangan.

2

The air bag contains solid sodium azide, NaN3 which will decompose rapidly to form sodium and nitrogen gas. The nitrogen gas formed fills the air bag.

[Relative atomic mass: N = 14; H = 1; Na = 23 and 1 mol of gas occupies the volume of 24 dm3 at room temperature and pressure] Beg udara mengandungi pepejal natrium azida, NaN3 yang akan terurai dengan cepat membentuk natrium dan gas nitrogen.

Gas nitrogen yang terbentuk akan mengisi beg udara itu.

[Jisim atomic relatif: N = 14; H = 1; Na = 23 dan 1 mol gas menempati isi padu 24 dm3 pada suhu dan tekanan bilik] (a) Construct equation for the decomposition of sodium azide. / Bina persamaan kimia bagi penguraian natrium azida.

2NaN3 → 2Na + 3N2 (b) In an accident, an air bag fills with 72 dm3 of nitrogen at room temperature and pressure. What is the mass of sodium azide needed to provide 72 dm3 of nitrogen?

Dalam suatu kemalangan, beg udara mengandungi 72 dm3 gas nitrogen pada suhu dan tekanan bilik. Berapakah jisim natrium azida yang diperlukan untuk menghasilkan 72 dm3 gas nitrogen?

Number of moles of nitrogen / Bilangan mol nitrogen =

72 dm3 = 3 mol 24 dm3 mol–1

Number of moles of NaN3 / Bilangan mol NaN3 = 2 mol Mass of NaN3/ Jisim NaN3 = 2 mol × [23 + 3(14)] g mol–1

= 130 g (c) Sodium azide, NaN3, reacts with dilute hydrochloric acid to produce sodium chloride and compound A. Compound A contains 2.33% of hydrogen and 97.7% of nitrogen by mass.

Natrium azida NaN3, bertindak balas dengan asid hidroklorik cair untuk menghasilkan natrium klorida dan sebatian A. Sebatian A mengandungi 2.33% hidrogen dan 97.7% nitrogen berdasarkan jisim.

(i) What is the empirical formula for compound A?

Apakah formula empirik sebatian A?

Element Unsur

H

N

Mass / g Jisim / g

2.33

97.7

Number of mole Bilangan mol

2.33 ——– = 2.33 1

97.7 ——– = 6.98 14

Simplest ratio Nisbah teringkas

2.33 ——– = 1 2.33

6.98 ——– ≈ 3 2.33

Empirical formula / Formula empirik : HN3

(ii) Construct the equation for the reaction between sodium azide and dilute hydrochloric acid.

Bina persamaan bagi tindak balas antara natrium azida dan asid hidroklorik cair.

NaN3 + HCl → NaCl + HN3 Soalan Tambahan Additional Question

© Nilam Publication Sdn. Bhd.

02 Chap 2 ChemF4 Bil 2017(CSY5p).indd 48

48

11/10/2017 10:23 AM