Issuu

CHAPTER 1 INTRODUCTION 1.1

(a) Matter is anything that occupies space and has mass. (b) Mass is a measure of the quantity of matter in an object. (c) Weight is the force that gravity exerts on an object. (d) A substance is matter that has a definite or constant composition and distinct properties. (e) A mixture is a combination of two or more substances in which the substances retain their distinct identities.

1.2

The scientifically correct statement is, “The mass of the student is 56 kg.”

1.3

Table salt dissolved in water is an example of a homogeneous mixture. Oil mixed with water is an example of a heterogeneous mixture.

1.4

A physical property is any property of a substance that can be observed without transforming the substance into some other substance. A chemical property is any property of a substance that cannot be studied without converting the substance into some other substance.

1.5

Density is an example of an intensive property. Mass is an example of an extensive property.

1.6

(a) (b)

An element is a substance that cannot be separated into simple substances by chemical means. A compound is a substance composed of atoms of two or more elements chemically united in fixed proportions.

1.7

(a)

Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are changed.

(b)

Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter (different composition).

(c)

Physical property. The measurement of the boiling point of water does not change its identity or composition.

(d)

Physical property. The measurement of the densities of lead and aluminum does not change their composition.

(e)

Chemical property. When uranium undergoes nuclear decay, the products are chemically different substances.

(a)

Physical change. Helium is not changed in any way by leaking out of the balloon.

(b)

Chemical change in the battery.

(c)

Physical change. The orange juice concentrate can be regenerated by evaporation of the water.

(d)

Chemical change. Photosynthesis changes water, carbon dioxide, and so on, into complex organic matter.

(e)

Physical change. The salt can be recovered unchanged by evaporation.

1.9

(a)

extensive

(b)

extensive

(c)

intensive

1.10

(a)

extensive

(b)

intensive

(c)

intensive

1.8

(d)

extensive

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CHAPTER 1: INTRODUCTION

1.11

(a)

element

(b)

compound

(c)

element

(d)

compound

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CHAPTER 1: INTRODUCTION

compound

(b)

1.12

(a)

1.13

(a) m

1.14

(a) (h)

1.15

Density is the mass of an object divided by its volume. Chemists commonly use g/cm 3 or equivalently g/mL for density. Density is an intensive property.

1.16

? °C = (°F - 32°F) ´

(b) m2

106 1012

(b)

element

(c)

(d) kg 101

(d)

compound

(d)

(e) s

(f) N

102

(e)

element (g) J

103

(h) K (f)

106

(g)

109

5°C 9°F

9°F ö æ ? °F = ç °C ´ + 32°F 5°C ÷ø è

1.17

The density of the sphere is given by:

d =

1.18

m 1.20 ´ 104 g = = 11.4 g/cm 3 V 1.05 ´ 103 cm3

Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid. Rearrange the density equation, Equation (1.1) of the text, to solve for mass. density =

mass volume

Solution: mass density  volume mass of Hg =

1.19

(a) (b) (c) (d)

1.20

13.6 g ´ 95.8 mL = 1.30 ´ 103 g 1 mL

5°C = 41°C 9°F 9° F ö æ ? °F = ç -11.5 °C ´ + 32°F = 11.3 °F 5 °C ÷ø è 9°F ö æ 4 ? °F = ç 6.3 ´ 103 °C ´ ÷ + 32°F = 1.1 ´ 10 °F 5°C ø è ? °C = (105 - 32)°F ´

? °C = (451 - 32)°F ´

5°C = 233°C 9°F

Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between Celsius and Fahrenheit given in Section 1.5 of the text. Substitute the temperature values given in the problem into the appropriate equation.

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CHAPTER 1: INTRODUCTION

Solution: 1K 1°C

(a)

K = (°C + 273°C)

(i)

K 113C  273C 386 K

(ii)

K 37C  273C 3.10  102 K

(iii) K 357C 273C6.30  102 K

(i)

1K 1°C C K  273  77 K  273 196C

(ii)

C 4.2 K  273 269C

(b)

K = (°C + 273°C)

(iii) C 601 K  273 328C 2.7  108

(b)

3.56  102

(c)

9.6  102

1.21

(a)

1.22

Strategy: Writing scientific notation as N 10n, we determine n by counting the number of places that the decimal point must be moved to give N, a number between 1 and 10. If the decimal point is moved to the left, n is a positive integer, the number you are working with is larger than 10. If the decimal point is moved to the right, n is a negative integer. The number you are working with is smaller than 1. (a)

Express 0.749 in scientific notation.

Solution: The decimal point must be moved one place to give N, a number between 1 and 10. In this case, N  7.49 Since 0.749 is a number less than one, n is a negative integer. In this case, n 1. Combining the above two steps: 0.749 7.49  101 (b)

Express 802.6 in scientific notation.

Solution: The decimal point must be moved two places to give N, a number between 1 and 10. In this case, N  8.026 Since 802.6 is a number greater than one, n is a positive integer. In this case, n  2. Combining the above two steps: 802.6 8.026  102 (c)

Express 0.000000621 in scientific notation.

Solution: The decimal point must be moved seven places to give N, a number between 1 and 10. In this case, N  6.21 4 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 1: INTRODUCTION

Since 0.000000621 is a number less than one, n is a negative integer. In this case, n 7. Combining the above two steps: 0.000000621 6.21  107 1.23

(a)

15,200

(b)

0.0000000778

1.24

(a)

Express 3.256  105 in nonscientific notation.

For the above number expressed in scientific notation, n 5. To convert to nonscientific notation, the decimal point must be moved 5 places to the left. 3.256  105 0.00003256 (b)

Express 6.03  106 in nonscientific notation.

For the above number expressed in scientific notation, n 6. The decimal place must be moved 6 places to the right to convert to nonscientific notation. 6.03 106 6,030,000 1.25

(a) (b)

1.26

145.75  (2.3  101)  145.75  0.23 1.4598  102

79500 2.5 ´ 10

7.95 ´ 10 4 2.5 ´ 10

= 3.2 ´ 102

(c)

(7.0  103)  (8.0  104)  (7.0  103)  (0.80  103) 6.2  103

(d)

(1.0  104)  (9.9  106) 9.9  1010

(a)

Addition using scientific notation.

Strategy: Let us express scientific notation as N  10n. When adding numbers using scientific notation, we must write each quantity with the same exponent, n. We can then add the N parts of the numbers, keeping the exponent, n, the same. Solution: Write each quantity with the same exponent, n. Let us write 0.0095 in such a way that n 3. We have decreased 10n by 103, so we must increase N by 103. Move the decimal point 3 places to the right. 0.0095  9.5  103 Add the N parts of the numbers, keeping the exponent, n, the same. 9.5  103  8.5  103 18.0  103 The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10 to express N between 1 and 10 (1.8), we must increase 10n by a factor of 10. The exponent, n, is increased by 1 from 3 to 2. 5 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 1: INTRODUCTION

18.0  1031.80 102 (b)

Division using scientific notation.

Strategy: Let us express scientific notation as N  10n. When dividing numbers using scientific notation, divide the N parts of the numbers in the usual way. To come up with the correct exponent, n, we subtract the exponents. Solution: Make sure that all numbers are expressed in scientific notation. 653  6.53  102 Divide the N parts of the numbers in the usual way. 6.53 5.75  1.14 Subtract the exponents, n. 1.14  102  (8)  1.14  102  8 1.14  1010 (c)

Subtraction using scientific notation.

Strategy: Let us express scientific notation as N  10n. When subtracting numbers using scientific notation, we must write each quantity with the same exponent, n. We can then subtract the N parts of the numbers, keeping the exponent, n, the same. Solution: Write each quantity with the same exponent, n. Let us write 850,000 in such a way that n  5. This means to move the decimal point five places to the left. 850,000  8.5  105 Subtract the N parts of the numbers, keeping the exponent, n, the same. 8.5  105  9.0  105 0.5  105 The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10 to express N between 1 and 10 (5), we must decrease 10n by a factor of 10. The exponent, n, is decreased by 1 from 5 to 4. 0.5 1055  104 (d)

Multiplication using scientific notation.

Strategy: Let us express scientific notation as N  10n. When multiplying numbers using scientific notation, multiply the N parts of the numbers in the usual way. To come up with the correct exponent, n, we add the exponents. Solution: Multiply the N parts of the numbers in the usual way. 3.6 3.6  13 Add the exponents, n. 13  104  (6)13  102

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CHAPTER 1: INTRODUCTION

The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10 to express N between 1 and 10 (1.3), we must increase 10n by a factor of 10. The exponent, n, is increased by 1 from 2 to 3. 13 102 1.3  103 1.27

(a)

four

(b)

two

(c)

five

1.28

(a)

three

(b)

one

(c)

one or two

1.29

(a)

10.6 m

(b)

0.79 g

(c)

16.5 cm2

1.30

(a)

Division

(d)

two, three, or four (d)

(d)

two

1 × 106 g/cm3

Strategy: The number of significant figures in the answer is determined by the original number having the smallest number of significant figures. Solution: 7.310 km = 1.283 5.70 km The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits. Therefore, the answer has only three significant digits. The correct answer rounded off to the correct number of significant figures is: 1.28 (Why are there no units?) (b)

Subtraction

Strategy: The number of significant figures to the right of the decimal point in the answer is determined by the lowest number of digits to the right of the decimal point in any of the original numbers. Solution: Writing both numbers in decimal notation, we have 0.00326 mg  0.0000788 mg 0.0031812 mg The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the decimal point. Therefore, we carry five digits to the right of the decimal point in our answer. The correct answer rounded off to the correct number of significant figures is: 0.00318 mg 3.18  103 mg (c)

Addition

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CHAPTER 1: INTRODUCTION

Since 7.74  107 has only two digits to the right of the decimal point, two digits are carried to the right of the decimal point in the final answer. (d)

Subtraction, addition, and division

Strategy: For subtraction and addition, the number of significant figures to the right of the decimal point in that part of the calculation is determined by the lowest number of digits to the right of the decimal point in any of the original numbers. For the division part of the calculation, the number of significant figures in the answer is determined by the number having the smallest number of significant figures. First, perform the subtraction and addition parts to the correct number of significant figures, and then perform the division.

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CHAPTER 1: INTRODUCTION

Solution: (7.8 m - 0.34 m) 7.5 m = = 3.8 m / s (1.15 s + 0.82 s) 1.97 s Note: We rounded to the correct number of significant figures at the midpoint of this calculation. In practice, it is best to keep all digits in your calculator, and then round to the correct number of significant figures at the end of the calculation.

1.31

1.32

(a)

? dm = 22.6 m ´

1 dm = 226 dm 0.1 m

(b)

? kg = 25.4 mg ´

0.001 g 1 kg ´ = 2.54 ´ 10-5 kg 1 mg 1000 g

(a) Strategy: The problem may be stated as ? mg  242 lb A relationship between pounds and grams is given on the end sheet of your text (1 lb  453.6 g). This relationship will allow conversion from pounds to grams. A metric conversion is then needed to convert grams to milligrams (1 mg  1  103 g). Arrange the appropriate conversion factors so that pounds and grams cancel, and the unit milligrams is obtained in your answer. Solution: The sequence of conversions is lb grams  mg Using the following conversion factors, 453.6 g 1 lb

1 mg 1 ´ 10-3 g

we obtain the answer in one step: ? mg = 242 lb ´

453.6 g 1 mg ´ = 1.10 ´ 108 mg 1 lb 1 ´ 10-3 g

Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many mg are in 1 lb? There is 453,600 mg in 1 lb. (b) Strategy: The problem may be stated as ? m3 68.3 cm3 Recall that 1 cm  1  102 m. We need to set up a conversion factor to convert from cm3 to m3. Solution: We need the following conversion factor so that centimeters cancel and we end up with meters. 1 ´ 10-2 m 1 cm 9 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 1: INTRODUCTION

Since this conversion factor deals with length and we want volume, it must therefore be cubed to give

æ 1 ´ 10-2 m ö 1 ´ 10-2 m 1 ´ 10-2 m 1 ´ 10-2 m ´ ´ = ç ÷ ç 1 cm ÷ 1 cm 1 cm 1 cm è ø

We can write 3

æ 1 ´ 10-2 m ö ? m = 68.3 cm ´ ç ÷ = 6.83 ´ 10-5 m 3 ç 1 cm ÷ è ø 3

Check: We know that 1 cm3 1  106 m3. We started with 6.83  101 cm3. Multiplying this quantity by 1  106 gives 6.83  105. 1 oz $932 ´ = $30.04 31.03 g 1 oz

1.33

? $ = 1.00 g ´

1.34

Calculating the mean for each set of data, we find: Student A: 47.7 mL Student B: 47.1 mL Student C: 47.8 mL From these calculations, we can conclude that the volume measurements made by student B were the most accurate of the three students. The precision in the measurements made by both students B and C are fairly high, although the measurements by student C were the most precise.

1.35

Calculating the mean for each set of data, we find: Student X: 61.5 g Student Y: 62.6 g Student Z: 62.1 g From these calculations, we can conclude that the mass measurements made by student Z were the most accurate of the three students. The precision in the measurements made by both students X and Z are fairly high, while the measurements made by student Y are the least precise of the measurements made.

1.36

1.37

1 mi 5280 ft 12 in 1 min ´ ´ ´ = 81 in/s 13 min 1 mi 1 ft 60 s

(a)

? in/s =

(b)

? m/min =

(c)

? km/h =

(a)

6.0 ft ´

1m = 1.8 m 3.28 ft

168 lb ´

453.6 g 1 kg ´ = 76.2 kg 1 lb 1000 g

1 mi 1609 m ´ = 1.2 ´ 102 m/min 13 min 1 mi

1 mi 1609 m 1 km 60 min ´ ´ ´ = 7.4 km/h 13 min 1 mi 1000 m 1h

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CHAPTER 1: INTRODUCTION

(b)

? km / h =

70 mi 1.609 km ´ = 1.1×102 km / h 1h 1 mi

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1.38

CHAPTER 1: INTRODUCTION

(c)

3.0 ´ 1010 cm 1 in 1 ft 1 mi 3600 s ´ ´ ´ ´ = 6.7 ´ 108 mph 1s 2.54 cm 12 in 5280 ft 1h

(d)

6.0 ´ 103 g of blood ´

(a)

1.42 yr ´

365 day 24 h 3600 s 3.00 ´ 108 m 1 mi ´ ´ ´ ´ = 8.35 ´ 1012 mi 1 yr 1 day 1h 1s 1609 m

(b)

32.4 yd ´

36 in 2.54 cm ´ = 2.96 ´ 103 cm 1 yd 1 in

(c)

3.0 ´ 1010 cm 1 in 1 ft ´ ´ = 9.8 ´ 108 ft/s 1s 2.54 cm 12 in

(d)

? °C = (47.4 - 32.0)°F ´

(e)

9°F ö æ ? °F = ç °C ´ + 32°F 5 °C ÷ø è

0.62 g Pb 6

1 ´ 10 g blood

= 3.7 ´ 10 -3 g Pb

5°C = 8.6°C 9°F

9° F ö æ ? °F = ç -273.15 °C ´ + 32°F = - 459.67°F 5°C ÷ø è 3

(f)

æ 0.01 m ö -5 3 ? m 3 = 71.2 cm3 ´ ç ÷ = 7.12 ´ 10 m 1 cm è ø

(g)

1L æ 1 cm ö ? L = 7.2 m3 ´ ç = 7.2 ´ 103 L ÷ ´ 3 0.01 m è ø 1000 cm

2.70 g

density =

1.40

density =

1.41

See Section 1.4 of your text for a discussion of these terms.

1 cm3

æ 1 cm ö 1 kg 3 3 ´ç ÷ = 2.70 ´ 10 kg/m 1000 g è 0.01 m ø

1.39

0.625 g 1L 1 mL ´ ´ = 6.25 ´ 10-4 g/cm 3 1L 1000 mL 1 cm3

(a)

Chemical property. Iron has changed its composition and identity by chemically combining with oxygen and water.

(b)

Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids, thus changing the composition and identity of the water.

(c)

Physical property. The color of the hemoglobin can be observed and measured without changing its composition or identity.

(d)

Physical property. The evaporation of water does not change its chemical properties. Evaporation is a change in matter from the liquid state to the gaseous state.

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CHAPTER 1: INTRODUCTION

(e)

Chemical property. The carbon dioxide is chemically converted into other molecules. 1 ton

1.42

(132 ´ 109 lb of sulfuric acid) ´

1.43

There are 78.3  117.3  195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor. æ 195.6o C ö ç ÷ ç 100oS ÷ è ø

2.00 ´ 10 lb

= 6.60 ´ 107 tons of sulfuric acid

Set up the equation like a Celsius to Fahrenheit conversion. We need to subtract 117.3C, because the zero point on the new scale is 117.3C lower than the zero point on the Celsius scale. æ 195.6°C ö ? °C = ç ÷ (? °S ) - 117.3°C è 100°S ø

1.44

Solving for ? °S gives:

æ 100°S ö ? °S = (? °C + 117.3°C) ç ÷ è 195.6°C ø

For 25°C we have:

æ 100°S ö ? °S = (25°C + 117.3°C) ç ÷ = 73°S è 195.6°C ø

Volume of rectangular bar  length  width  height density =

1.45

1.46

m 52.7064 g = = 2.6 g/cm 3 V (8.53 cm)(2.4 cm)(1.0 cm)

mass  density  volume 4 (10.0 cm)3] 8.08  104 g 3

(a)

mass (19.3 g/cm3)  [

(b)

æ 1 cm ö -6 mass = (21.4 g/cm ) ´ ç 0.040 mm ´ ÷ = 1.4 ´ 10 g 10 mm ø è

(c)

mass (0.798 g/mL)(50.0 mL) 39.9 g

You are asked to solve for the inner diameter of the bottle. If we can calculate the volume that the cooking oil occupies, we can calculate the radius of the cylinder. The volume of the cylinder is, Vcylinderr2h (r is the inner radius of the cylinder and h is the height of the cylinder). The cylinder diameter is 2r. volume of oil filling bottle =

mass of oil density of oil

volume of oil filling bottle =

1360 g = 1.43 ´ 103 mL = 1.43 ´ 103 cm3 0.953 g/mL

Next, solve for the radius of the cylinder. Volume of cylinder r2h r =

volume p´h

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CHAPTER 1: INTRODUCTION

r =

1.43 ´ 103 cm3 = 4.60 cm p ´ 21.5 cm

The inner diameter of the bottle equals 2r. Bottle diameter 2r 2(4.60 cm) 9.20 cm 1.47

From the mass of the water and its density, we can calculate the volume that the water occupies. The volume that the water occupies is equal to the volume of the flask. volume =

mass density

Mass of water  87.39 g  56.12 g  31.27 g Volume of the flask =

1.48

mass 31.27 g = = 31.35 cm 3 density 0.9976 g/cm 3

The volume of silver is equal to the volume of water it displaces. Volume of silver  260.5 mL  242.0 mL  18.5 mL  18.5 cm3 density =

1.49

194.3 g 18.5 cm 3

= 10.5 g/cm 3

In order to work this problem, you need to understand the physical principles involved in the experiment in Problem 1.48. The volume of the water displaced must equal the volume of the piece of silver. If the silver did not sink, would you have been able to determine the volume of the piece of silver? The liquid must be less dense than the ice in order for the ice to sink. The temperature of the experiment must be maintained at or below 0°C to prevent the ice from melting.

1.50

The speed of light is 3.00  108 m/s. 1.00 ns ´

1s 3.00 ´ 108 m 100 cm 1 in 1 ft ´ ´ ´ ´ = 0.984 ft 9 10 ns 1s 1m 2.54 cm 12 in

Remember that roughly speaking, light travels 1 ft in 1 ns. 1.51

For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C. 5°C ? °C = 0.1°F ´ = 0.056°C 9°F The percent error is the amount of uncertainty in a measurement divided by the value of the measurement, converted to percent by multiplication by 100. Percent error =

known error in a measurement ´ 100% value of the measurement

For the Fahrenheit thermometer,

percent error =

0.056°C ´ 100% = 0.14% 38.9°C

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CHAPTER 1: INTRODUCTION

percent error =

For the Celsius thermometer,

0.1°C ´ 100% = 0.26% 38.9°C

Which thermometer is more accurate?

1.52

Temperature:

9°F ö æ ? °F = ç 24.2°C ´ + 32°F = 75.6° F 5°C ÷ø è

Uncertainty:

? °F = 0.1°C ´

9°F = 0.2°F 5°C

? F  75.6F  0.2F 1.53

To work this problem, we need to convert from cubic feet to L. Some tables will have the conversion factor of 28.3 L  1 ft3, but we can also calculate it using the dimensional analysis method described in Section 1.7. First, converting from cubic feet to liters 3

æ 12 in ö æ 2.54 cm ö 1 mL 1 ´ 10-3 L (5.0 ´ 107 ft 3 ) ´ ç ´ = 1.4 ´ 109 L ÷ ´ç ÷ ´ 3 1 ft 1 in 1 mL è ø è ø 1 cm The mass of vanillin (in g) is: 2.0 ´ 10-11 g vanillin ´ (1.4 ´ 109 L) = 2.8 ´ 10-2 g vanillin 1L The cost is: (2.8 ´ 10-2 g vanillin) ´

1.54

$112 = $0.063 = 6.3¢ 50 g vanillin

The key to solving this problem is to realize that all the oxygen needed must come from the 4% difference (20% - 16%) between inhaled and exhaled air. The 240 mL of pure oxygen/min requirement comes from the 4% of inhaled air that is oxygen. 240 mL of pure oxygen/min  (0.04)(volume of inhaled air/min) Volume of inhaled air/min =

240 mL of oxygen/min = 6000 mL of inhaled air/min 0.04

Since there are 12 breaths per min, volume of air/breath =

1.55

6000 mL of inhaled air 1 min ´ = 5 ´ 10 2 mL/breath 1 min 12 breaths

The mass of the seawater is: (1.5 ´ 1021 L) ´

1 mL 1.03 g ´ = 1.5 ´ 1024 g = 1.5 ´ 10 21 kg seawater 0.001 L 1 mL

Seawater is 3.1% NaCl by mass. The total mass of NaCl in kilograms is: 15 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 1: INTRODUCTION

mass NaCl (kg) = (1.5 ´ 1021 kg seawater) ´ mass NaCl (tons) = (4.7 ´ 1019 kg) ´

1.56

3.1% NaCl = 4.7 ´ 1019 kg NaCl 100% seawater

2.205 lb 1 ton ´ = 5.2 ´ 1016 tons NaCl 1 kg 2000 lb

First, calculate the volume of 1 kg of seawater from the density and the mass. We chose 1 kg of seawater, because the problem gives the amount of Mg in every kg of seawater. The density of seawater is 1.03 g/mL. volume =

mass density

volume of 1 kg of seawater =

1000 g = 971 mL = 0.971 L 1.03 g/mL

In other words, there is 1.3 g of Mg in every 0.971 L of seawater. Next, let us convert tons of Mg to grams of Mg. (8.0 ´ 104 tons Mg) ´

2000 lb 453.6 g ´ = 7.3 ´ 1010 g Mg 1 ton 1 lb

Volume of seawater needed to extract 8.0  104 ton Mg  (7.3 ´ 1010 g Mg) ´

1.57

0.971 L seawater = 5.5 ´ 1010 L of seawater 1.3 g Mg

Assume that the crucible is platinum. Let us calculate the volume of the crucible and then compare that to the volume of water that the crucible displaces. volume =

mass density

Volume of crucible =

860.2 g 21.45 g/cm3

Volume of water displaced =

= 40.10 cm 3

(860.2 - 820.2)g 0.9986 g/cm3

= 40.1 cm 3

The volumes are the same (within experimental error), so the crucible is made of platinum.

1.58

9°F ö æ ? °F = ç °C ´ + 32°F 5 °C ÷ø è

Let temperature t 9 t + 32°F 5 9 t - t = 32°F 5 4 - t = 32°F 5 t =

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CHAPTER 1: INTRODUCTION

t40F 40C 1.59

Volume  surface area  depth Recall that 1 L  1 dm3. Let us convert the surface area to units of dm2 and the depth to units of dm. 2

æ 1000 m ö æ 1 dm ö 16 2 surface area = (1.8 ´ 108 km 2 ) ´ ç ÷ ´ç ÷ = 1.8 ´ 10 dm 1 km 0.1 m è ø è ø depth = (3.9 ´ 103 m) ´

1 dm = 3.9 ´ 104 dm 0.1 m

Volume surface area  depth  (1.8  1016 dm2)(3.9  104dm)  7.0  1020 dm37.0  1020 L

1.60

1.61

(a)

|0.798 g/mL - 0.802 g/mL| ´ 100% = 0.5% 0.798 g/mL

(b)

|0.864 g - 0.837 g| ´ 100% = 3.1% 0.864 g

The diameter of the basketball can be calculated from its circumference. We can then use the diameter of a ball as a conversion factor to determine the number of basketballs needed to circle the equator. Circumference = 2πr

1.62

d = 2r =

circumference 29.6 in = = 9.42 in p p

6400 km ´

1000 m 1 cm 1 in 1 ball ´ ´ ´ = 2.67 ´ 107 basketballs -2 1 km 2.54 cm 9.42 in 1 ´ 10 m

? g Au =

4.0 ´ 10-12 g Au 1 mL ´ ´ (1.5 ´ 1021 L seawater) = 6.0 ´ 1012 g Au 1 mL seawater 0.001 L

value of gold = (6.0 ´ 1012 g Au) ´

1 lb 16 oz $930 ´ ´ = $2.0 ´ 1014 453.6 g 1 lb 1 oz

No one has become rich mining gold from the ocean, because the cost of recovering the gold would outweigh the price of the gold.

1.63

mass of Earth's crust = (5.9 ´ 1021 tons) ´

0.50% crust = 3.0 ´ 1019 tons 100% Earth

mass of silicon in crust = (3.0 ´ 1019 tons crust) ´

1.64

27.2% Si 2000 lb 1 kg ´ ´ = 7.4 ´ 1021 kg Si 100% crust 1 ton 2.205 lb

10 cm  0.1 m. We need to find the number of times the 0.1 m wire must be cut in half until the piece left is 1.3  1010 m long. Let n be the number of times we can cut the Cu wire in half. We can write:

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CHAPTER 1: INTRODUCTION

æ1ö -10 ç 2 ÷ ´ 0.1 m = 1.3 ´ 10 m è ø n

æ1ö -9 ç 2 ÷ = 1.3 ´ 10 m è ø Taking the log of both sides of the equation: æ1ö n log ç ÷ = log(1.3 ´ 10-9 ) è2ø

n 30 times 5000 mi 1 gal gas 9.5 kg CO2 ´ ´ = 5.9 ´ 1011 kg CO 2 1 car 20 mi 1 gal gas

1.65

(250 ´ 106 cars) ´

1.66

Volume  area  thickness. From the density, we can calculate the volume of the Al foil. Volume =

mass 3.636 g = = 1.347 cm3 density 2.699 g / cm3

Convert the unit of area from ft2 to cm2. 2

æ 12 in ö æ 2.54 cm ö 2 1.000 ft ´ ç ÷ ´ç ÷ = 929.0 cm è 1 ft ø è 1 in ø 2

thickness =

1.67

volume 1.347 cm3 = = 1.450 ´ 10-3 cm = 1.450 ´ 10-2 mm 2 area 929.0 cm

First, let us calculate the mass (in g) of water in the pool. We perform this conversion because we know there is 1 g of chlorine needed per million grams of water. (2.0 ´ 104 gallons H 2 O) ´

3.79 L 1 mL 1g ´ ´ = 7.6 ´ 107 g H 2 O 1 gallon 0.001 L 1 mL

Next, let us calculate the mass of chlorine that needs to be added to the pool. (7.6 ´ 107 g H 2 O) ´

1 g chlorine 1 ´ 106 g H 2 O

= 76 g chlorine

The chlorine solution is only 6% chlorine by mass. We can now calculate the volume of chlorine solution that must be added to the pool. 76 g chlorine ´

100% soln 1 mL soln ´ = 1.3 ´ 103 mL of chlorine solution 6% chlorine 1 g soln

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CHAPTER 1: INTRODUCTION

1.68

The mass of water used by 50,000 people in 1 year is: 50, 000 people ´

1 g H2O 150 gal water 3.79 L 1000 mL 365 days ´ ´ ´ ´ = 1.04 ´ 1013 g H 2 O/yr 1 person each day 1 gal 1L 1 mL H 2 O 1 yr

A concentration of 1 ppm of fluorine is needed. In other words, 1 g of fluorine is needed per million grams of water. NaF is 45.0% fluorine by mass. The amount of NaF needed per year in kg is: (1.04 ´ 1013 g H 2 O) ´

1g F 6

1 ´ 10 g H 2 O

100% NaF 1 kg ´ = 2.3 ´ 104 kg NaF 45% F 1000 g

An average person uses 150 gallons of water per day. This is equal to 569 L of water. If only 6.0 L of water is used for drinking and cooking, 563 L of water is used for purposes in which NaF is not necessary. Therefore, the amount of NaF wasted is: 563 L ´ 100% = 99% 569 L 1.69

We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can calculate the thickness of the oil layer from the volume and surface area. 2

æ 1 cm ö 5 2 40 m ´ ç ÷ = 4.0 ´ 10 cm è 0.01 m ø 0.10 mL  0.10 cm3 2

Volume  surface area  thickness

thickness =

volume 0.10 cm3 = = 2.5 ´ 10 -7 cm surface area 4.0 ´ 105 cm 2

Converting to nm: (2.5 ´ 10 -7 cm) ´

1.70

0.01 m 1 nm ´ = 2.5 nm 1 cm 1 ´ 10-9 m

Let the fraction of gold = x and the fraction of sand = (1 – x). We set up an equation to solve for x. (x)(19.3 g/cm3) + (1 – x)(2.95 g/cm3) = 4.17 g/cm3 19.3x – 2.95x + 2.95 = 4.17 x = 0.0746 Converting to a percentage, the mixture contains 7.46% gold.

1.71

Twenty-five grams of the least dense metal (solid A) will occupy the greatest volume of the three metals, and 25.0 g of the most dense metal (solid B) will occupy the least volume. We can calculate the volume occupied by each metal and then add the volume of water (20 mL) to find the total volume occupied by the metal and water. Solid A:

1 mL = 8.6 mL 2.9 g A Total volume = 8.6 mL + 20.0 mL = 28.6 mL 25.0 g A ´

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CHAPTER 1: INTRODUCTION

Solid B:

1 mL = 3.0 mL 8.3 g B Total volume = 3.0 mL + 20.0 mL = 23.0 mL

Solid C:

1 mL = 7.6 mL 3.3 g C Total volume = 7.6 mL + 20.0 mL = 27.6 mL

25.0 g B ´

25.0 g C ´

Therefore, we have: (a) solid C, (b) solid B, and (c) solid A. 1.72

(a)

Data were collected to indicate a high iridium content in clay deposited above sediments formed during the Cretaceous period. A hypothesis was formulated that the iridium came from a large asteroid. When the asteroid impacted Earth, large amounts of dust and debris blocked the sunlight. Plants eventually died, then many plant eating animals gradually perished, then, in turn, meat eating animals starved.

(b)

If the hypothesis is correct, we should find a similarly high iridium content in corresponding rock layers at different locations on Earth. Also, we should expect the simultaneous extinction of other large species in addition to dinosaurs.

(c)

Yes. Hypotheses that survive many experimental tests of their validity may evolve into theories. A theory is a unifying principle that explains a body of facts and/or those laws that are based on them.

(d)

Twenty percent of the asteroid’s mass turned into dust. First, let us calculate the mass of the dust.

? g dust =

0.02 g cm 2

æ 1 cm ö 14 2 17 ´ç ÷ ´ (5.1 ´ 10 m ) = 1.0 ´ 10 g dust 0.01 m è ø

This mass is 20% of the asteroid’s mass. mass of asteroid =

1.0 ´ 1017 g = 5.0 ´ 1017 g = 5.0 ´ 1014 kg 0.20

Converting to tons: (5.0 ´ 1014 kg) ´

1000 g 1 lb 1 ton ´ ´ = 5.5 ´ 1011 tons 1 kg 453.6 g 2000 lb

From the mass and density of the asteroid, we can calculate its volume. Then, from its volume, we can calculate the radius.

Volume =

mass 5.0 ´ 1017 g = = 2.5 ´ 1017 cm3 density 2 g/cm3

Volumesphere =

4 3 pr 3

2.5 ´ 1017 cm3 =

4 3 pr 3

r 4  105 cm 4  103 m The diameter of the asteroid is approximately 5 miles! 20 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 1: INTRODUCTION

1.73

Gently heat the liquid to see if any solid remains after the liquid evaporates. Also, collect the vapor and then compare the densities of the condensed liquid with the original liquid. The composition of a mixed liquid would change with evaporation along with its density.

1.74

This problem is similar in concept to a limiting reagent problem. We need sets of coins with 3 quarters, 1 nickel, and 2 dimes. First, we need to find the total number of each type of coin. Number of quarters = (33.871 ´ 103 g) ´ Number of nickels = (10.432 ´ 103 g) ´ Number of dimes = (7.990 ´ 103 g) ´

1 quarter = 6000 quarters 5.645 g

1 nickel = 2100 nickels 4.967 g

1 dime = 3450 dimes 2.316 g

Next, we need to find which coin limits the number of sets that can be assembled. For each set of coins, we need 2 dimes for every 1 nickel. 2100 nickels ´

2 dimes = 4200 dimes 1 nickel

We do not have enough dimes. For each set of coins, we need 2 dimes for every 3 quarters. 6000 quarters ´

2 dimes = 4000 dimes 3 quarters

Again, we do not have enough dimes, and therefore the number of dimes is our “limiting reagent.” If we need 2 dimes per set, the number of sets that can be assembled is: 3450 dimes ´

1 set = 1725 sets 2 dimes

The mass of each set is: æ 5.645 g ö æ 4.967 g ö æ 2.316 g ö ç 3 quarters ´ ÷ + ç1 nickel ´ ÷ + ç 2 dimes ´ ÷ = 26.53 g/set 1 quarter ø è 1 nickel ø è 1 dime ø è

Finally, the total mass of 1725 sets of coins is: 1725 sets ´

1.75

26.53 g = 4.576 ´ 104 g 1 set

We wish to calculate the density and radius of the ball bearing. For both calculations, we need the volume of the ball bearing. The data from the first experiment can be used to calculate the density of the mineral oil. In the second experiment, the density of the mineral oil can then be used to determine what part of the 40.00 mL volume is due to the mineral oil and what part is due to the ball bearing. Once the volume of the ball bearing is determined, we can calculate its density and radius. From experiment one: Mass of oil  159.446 g  124.966 g  34.480 g

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CHAPTER 1: INTRODUCTION

Density of oil =

34.480 g = 0.8620 g/mL 40.00 mL

From the second experiment: Mass of oil  50.952 g  18.713 g  32.239 g Volume of oil = 32.239 g ´

1 mL = 37.40 mL 0.8620 g

The volume of the ball bearing is obtained by difference. Volume of ball bearing  40.00 mL  37.40 mL  2.60 mL  2.60 cm3

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CHAPTER 1: INTRODUCTION

Now that we have the volume of the ball bearing, we can calculate its density and radius. 18.713 g

= 7.20 g/cm 3 2.60 cm3 Using the formula for the volume of a sphere, we can solve for the radius of the ball bearing. Density of ball bearing =

V =

4 3 pr 3

2.60 cm3 =

4 3 pr 3

r3 0.621 cm3 r0.853 cm 1.76

We want to calculate the mass of the cylinder, which can be calculated from its volume and density. The volume of a cylinder is r2l. The density of the alloy can be calculated using the mass percentages of each element and the given densities of each element. The volume of the cylinder is: Vr2l V(6.44 cm)2(44.37 cm) V 5.78 × 103 cm3 The density of the cylinder is: density  (0.7942)(8.94 g/cm3)  (0.2058)(7.31 g/cm3)  8.60 g/cm3 Now, we can calculate the mass of the cylinder. mass  density × volume mass (8.60 g/cm3)(5.78 × 103 cm3) 4.97 × 104 g The assumption made in the calculation is that the alloy must be homogeneous in composition.

1.77

(a)

The volume of the pycnometer can be calculated by determining the mass of water that the pycnometer holds and then using the density to convert to volume. (43.1195 - 32.0764) g ´

(b)

1 mL = 11.063 mL 0.99820 g

Using the volume of the pycnometer from part (a), we can calculate the density of ethanol. (40.8051 - 32.0764) g = 0.78900 g/mL 11.063 mL

(c)

From the volume of water added and the volume of the pycnometer, we can calculate the volume of the zinc granules by difference. Then, we can calculate the density of zinc. volume of water = (62.7728 - 32.0764 - 22.8476) g ´

1 mL = 7.8630 mL 0.99820 g

volume of zinc granules = 11.063 mL - 7.8630 mL = 3.200 mL

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CHAPTER 1: INTRODUCTION

density of zinc =

1.78

22.8476 g = 7.140 g/mL 3.200 mL

For solids, density is more commonly expressed as 7.140 g/cm 3. When the carbon dioxide gas is released, the mass of the solution will decrease. If we know the starting mass of the solution and the mass of solution after the reaction is complete (given in the problem), we can calculate the mass of carbon dioxide produced. Then, using the density of carbon dioxide, we can calculate the volume of carbon dioxide released. 1.140 g Mass of hydrochloric acid = 40.00 mL ´ = 45.60 g 1 mL Mass of solution before reaction  45.60 g  1.328 g  46.93 g We can now calculate the mass of carbon dioxide by difference. Mass of CO2 released  46.93 g  46.699 g  0.23 g Finally, we use the density of carbon dioxide to convert to liters of CO2 released. Volume of CO2 released = 0.23 g ´

1.79

1L = 0.13 L 1.81 g

As water freezes, it expands. First, calculate the mass of the water at 20C. Then, determine the volume that this mass of water would occupy at 5C. Mass of water = 242 mL ´

0.998 g = 241.5 g 1 mL

Volume of ice at - 5°C = 241.5 g ´

1 mL = 264 mL 0.916 g

The volume occupied by the ice is larger than the volume of the glass bottle. The glass bottle would crack! 1.80

(a)

A concentration of CO of 800 ppm in air would mean that there are 800 parts by volume of CO per 1 million parts by volume of air. Using a volume unit of liters, 800 ppm CO means that there are 800 L of CO per 1 million liters of air. The volume in liters occupied by CO in the room is: 3

æ 1 cm ö 1L 17.6 m ´ 8.80 m ´ 2.64 m = 409 m ´ ç = 4.09 ´ 105 L air ÷ ´ ç 1 ´ 10-2 m ÷ 1000 cm3 è ø 3

4.09 ´ 105 L air ´

(b)

1 ´ 106 L air

= 327 L CO

1 mg  1 × 103 g and 1 L  1000 cm3. We convert mg/m3 to g/L:

0.050 mg 1 m3 (c)

8.00 ´ 102 L CO

1 ´ 10-3 g æ 1 ´ 10-2 m ö 1000 cm3 ´ ´ç = 5.0 ´ 10-8 g / L ÷ ´ ç 1 cm ÷ 1 mg 1 L è ø

1 g  1 × 103 mg and 1 mL  1 × 102dL. We convert mg/dL to g/mL:

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CHAPTER 1: INTRODUCTION

120 mg 1 mg 1 ´ 10-2 dL 3 ´ ´ = 1.20 ´ 10μg / mL 3 1 dL 1 mL 1 ´ 10 mg

1.81

The volume of the bowling ball is given by 3

4 3 4 æ 8.6 in ö æ 2.54 cm ö 3 3 pr = p ç ÷ ´ç ÷ = 5.5 ´10 cm 3 3 è 2 ø è 1 in ø

Starting with an 8 lb bowling ball and assuming two significant figures in the mass, converting pounds to grams gives 453.6 g 8 lb ´ = 3.6 ´103 g 1 lb So the density of an 8 lb bowling ball would be

density =

mass 3.6 ´103 g = = 0.65 g/cm3 3 3 volume 5.5 ´ 10 cm

Carrying out analogous calculations for the higher weight bowling balls gives ball weight (lb) 9 10 11 12 13

mass (g) 3

4.1  10 4.5  103 5.0  103 5.4  103 5.9  103

density (g/cm3) 0.75 0.82 0.91 0.98 1.1

Therefore, we would expect bowling balls that are 11 lb or lighter to float since they are less dense than water. Bowling balls that are 13 lb or heavier would be expected to sink since they are denser than water. The 12 lb bowling ball is borderline but it would probably float. Note that the above calculations were carried out by rounding off the intermediate answers, as discussed in Section 1.6 (p. 17). If we carry an additional digit past the number of significant figures to minimize errors from rounding, the following densities are obtained: ball weight (lb) 10 11 12

density (g/cm3) 0.83 0.91 1.01

The differences in the densities obtained are slight, but the value obtained for the 12 lb ball now suggests that it might sink. This problem illustrates the difference rounding off intermediate answers can make in the final answers for some calculations.

ANSWERS TO REVIEW OF CONCEPTS Chapter 1 Section 1.2 (p. 4)

(c)

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CHAPTER 1: INTRODUCTION

Section 1.3 (p. 7) Section 1.4 (p. 8) Section 1.5 (p. 13) Section 1.6 (p. 18)

Elements: (b) and (d). Compounds: (a) and (c). Chemical change: (b) and (c). Physical change: (d). (a) Top ruler, 4.6 in. Bottom ruler, 4.57 in.

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CHAPTER 2 ATOMS, MOLECULES, AND IONS 2.1

(a) Alpha () particles are helium ions with a charge of +2. (b) Beta () particles are electrons emitted during the decay of certain radioactive substances. (c) Gamma () rays are high-energy radiation. (d) X-rays are also highenergy radiation but with a lower energy than  rays.

2.2

The most common types of radiation known to be emitted by radioactive elements are alpha () radiation, beta () radiation, and gamma () radiation.

2.3

Alpha () particles composed of two protons and two neutrons. Cathode rays are a stream of electrons. Protons, neutrons, and electrons are fundamental particles. Fundamental particles are particles that were once thought to be the indivisible components of all matter.

2.4

Please see Section 2.2 of the text where early experiments on atomic structure are discussed in detail.

2.5

The sample is emitting particles from its nucleus (, , etc.).

2.6

Rutherford used  particles to probe the structure of the atom. Most  particles aimed at thin foils of gold passed through the foil with little or no deflection. A few  particles were deflected at large angles and occasionally an  particle bounced back in the direction from which it had come. Rutherford concluded that most of the atom was empty space with a small, dense, positively charged core (the nucleus).

2.7

First, convert 1 cm to picometers. 1 cm ´

0.01 m 1 pm ´ = 1 ´ 1010 pm 1 cm 1 ´ 10-12 m

? He atoms = (1 ´ 1010 pm) ´

2.8

1 He atom 2

1 ´ 10 pm

= 1 ´ 108 He atoms

Note that you are given information to set up the unit factor relating meters and miles. ratom = 104 rnucleus = 104 ´ 10 cm ´

1m 1 mi ´ = 0.62 mi 100 cm 1609 m

2.9

(a) The atomic number is the number of protons in a nucleus. (b) The mass number is the sum of the number of protons and neutrons in the nucleus of an atom. In an atom, the numbers of protons and electrons are equal. Therefore, if the atomic number of an atom is known, both the number of protons and the electrons are also known.

2.10

The chemical identity of an atom is determined by its number of protons (atomic number). Isotopes of an element contain differing numbers of neutrons, hence the mass numbers of isotopes of an element will differ. Z is the atomic number, A is the mass number, and X represents the symbol of the element.

2.11

For iron, the atomic number Z is 26. Therefore, the mass number A is: A 26  28 54

CHAPTER 2; ATOMS, MOLECULES, AND IONS

2.12

Strategy: The 239 in Pu-239 is the mass number. The mass number (A) is the total number of neutrons and protons present in the nucleus of an atom of an element. You can look up the atomic number (number of protons) on the periodic table. Solution: mass number  number of protons  number of neutrons number of neutrons mass number  number of protons  239  94 145

2.13

2.14

Isotope No. Protons No. Neutrons

3 2 He

4 2 He

24 12 Mg

25 12 Mg

48 22Ti

79 35 Br

195 78 Pt

2 1

2 2

12 12

12 13

22 26

35 44

78 117

Isotope No. Protons No. Neutrons No. Electrons

15 7N

33 16 S

63 29 Cu

84 38 Sr

130 56 Ba

186 74W

202 80 Hg

7 8 7

16 17 16

29 34 29

38 46 38

56 74 56

74 112 74

80 122 80

23 11 Na

(b)

64 28 Ni

2.15

(a)

2.16

The accepted way to denote the atomic number and mass number of an element X is as follows:

A ZX where, A  mass number Z  atomic number (a)

186 74W

(b)

201 80 Hg

2.17

Elements can be grouped together according to their chemical and physical properties in a chart called the periodic table. The periodic table enables us to classify elements (as metals, metalloids, and nonmetals) and correlate their properties in a systematic way. Groups are the vertical columns of the periodic table, and periods are the horizontal rows of the table.

2.18

Metals are good conductors of heat and electricity, while nonmetals are usually poor conductors of heat and electricity. Metals, excluding mercury, are solids, whereas many nonmetals are gases.

2.19

(a) Hydrogen (H2), carbon (C), oxygen (O2), argon (Ar). (b) Sodium (Na), titanium (Ti), tungsten (W), lead (Pb). (c) Silicon (Si), germanium (Ge), arsenic (As), astatine (At).

2.20

Column A is the alkali metals. Two examples are sodium (Na) and potassium (K). Column B is the alkaline earth metals. Two examples are calcium (Ca) and barium (Ba). Column C is the halogens. Two examples are fluorine (F) and iodine (I). Column D is the noble gases. Two examples are argon (Ar) and xenon (Xe).

2.21

Helium and Selenium are nonmetals whose name ends with ium. (Tellerium is a metalloid whose name ends in ium.)

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.22

2.23

(a)

Metallic character increases as you progress down a group of the periodic table. For example, moving down Group 4A, the nonmetal carbon is at the top and the metal lead is at the bottom of the group.

(b)

Metallic character decreases from the left side of the table (where the metals are located) to the right side of the table (where the nonmetals are located).

The following data were measured at 20C. (a)

Li (0.53 g/cm3)

K (0.86 g/cm3)

H2O (0.98 g/cm3)

(b)

Au (19.3 g/cm3)

Pt (21.4 g/cm3)

Hg (13.6 g/cm3)

(c)

Os (22.6 g/cm3)

(d)

Te (6.24 g/cm3)

2.24

F and Cl are Group 7A elements; they should have similar chemical properties. Na and K are both Group 1A elements; they should have similar chemical properties. P and N are both Group 5A elements; they should have similar chemical properties.

2.25

An atom is the basic unit of an element that can enter into chemical combination. A molecule is an aggregate of at least two atoms in a definite arrangement held together by chemical forces (also called chemical bonds).

2.26

Allotropes are two or more forms of the same element that differ significantly in chemical and physical properties. Diamond and graphite are allotropes of carbon. Allotropes of an element differ in structure and properties, whereas isotopes of a given element contain different numbers of neutrons but have similar chemistries.

2.27

Two commonly used molecular models are the ball-and-stick model and the space-filling model.

2.28

(a) Na+

2.29

(a) (b) (c)

This is a polyatomic molecule that is an elemental form of the substance. It is not a compound. This is a polyatomic molecule that is a compound. This is a diatomic molecule that is a compound.

2.30

(a) (b) (c)

This is a diatomic molecule that is a compound. This is a polyatomic molecule that is a compound. This is a polyatomic molecule that is the elemental form of the substance. It is not a compound.

2.31

Elements: Compounds:

2.32

There are more than two correct answers for each part of the problem. (a) (d)

2.33

(b) I

N2, S8, H2 NH3, NO, CO, CO2, SO2

H2 and F2 (b) H2O and C12H22O11 (sucrose)

Ion No. protons No. electrons

(d) SO42

Na 11 10

Ca2 20 18

HCl and CO

Al3 13 10

Fe2 26 24

(c)

I 53 54

F 9 10

S8 and P4

S2 16 18

O2 8 10

N3 7 10

CHAPTER 2; ATOMS, MOLECULES, AND IONS

2.34

The atomic number (Z) is the number of protons in the nucleus of each atom of an element. You can find this on a periodic table. The number of electrons in an ion is equal to the number of protons minus the charge on the ion. number of electrons (ion)  number of protons  charge on the ion K 19 18

Ion No. protons No. electrons

Mg2 12 10

Fe3 26 23

Br 35 36

Mn2 25 23

C4 6 10

Cu2 29 27

2.35

Chemical formulas express the composition of molecules and ionic compounds in terms of chemical symbols. (a) 1:1. (b) 1:3. (c) 1:2. (d) 2:3.

2.36

A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance. An empirical formula shows the elements present and the simplest whole number ratio of the atoms but not necessarily the actual number of atoms in a given molecule.

2.37

Cyclobutane (C4H8) and cyclohexane (C6H12) have the same empirical formula (CH2) but different molecular formulas.

2.38

P4 signifies one molecule that is composed of four P atoms. 4P represents four atoms of P (phosphorus).

2.39

An ionic compound contains cations and anions. Electrical neutrality is maintained because the positive charge of the cations is balanced by the negative charge of the anions.

2.40

Ionic compounds do not consist of discrete molecular units but are three-dimensional networks of ions. The formula of ionic compounds represents the simplest ratio (empirical formula) in which the cation and anion combine.

2.41

(a)

2.42

Strategy: An empirical formula tells us which elements are present and the simplest whole number ratio of their atoms. Can you divide the subscripts in the formula by some factor to end up with smaller whole number subscripts?

(b)

(c)

C9H20

(d)

P2O5

(e)

BH3

Solution: (a) (b) (c) (d)

Dividing both subscripts by 2, the simplest whole number ratio of the atoms in Al2Br6 is AlBr3. Dividing all subscripts by 2, the simplest whole number ratio of the atoms in Na2S2O4 is NaSO2. The molecular formula as written, N2O5, contains the simplest whole number ratio of the atoms present. In this case, the molecular formula and the empirical formula are the same. The molecular formula as written, K2Cr2O7, contains the simplest whole number ratio of the atoms present. In this case, the molecular formula and the empirical formula are the same.

2.43

The molecular formula of glycine is C2H5NO2.

2.44 2.45

The molecular formula of ethanol is C2H6O. Compounds of metals with nonmetals are usually ionic. Nonmetal–nonmetal compounds are usually molecular. Ionic: Molecular:

LiF, BaCl2, KCl SiCl4, B2H6, C2H4

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CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.46

Compounds of metals with nonmetals are usually ionic. Nonmetal–nonmetal compounds are usually molecular. Ionic: Molecular:

NaBr, BaF2, CsCl. CH4, CCl4, ICl, NF3

2.47

(a) (b) (c) (d) (e) (f) (g) (h)

sodium chromate potassium hydrogen phosphate hydrogen bromide (molecular compound) hydrobromic acid lithium carbonate potassium dichromate ammonium nitrite phosphorus trifluoride

(i) (j) (k) (l) (m) (n) (o) (p)

phosphorus pentafluoride tetraphosphorus hexoxide cadmium iodide strontium sulfate aluminum hydroxide sodium carbonate decahydrate sulfite ion hydrogen arsenate ion

2.48

Strategy: When naming ionic compounds, our reference for the names of cations and anions is Table 2.3 of the text. Ions not listed in Table 2.3 can often be named by extension of the name of analogous ions in the same family; for example, because ClO3- is chlorate, BrO3- is named bromate. Keep in mind that if a metal can form cations of different charges, we need to use the Stock system. In the Stock system, Roman numerals are used to specify the charge of the cation. The metals that have only one charge in ionic compounds are the alkali metals (1), the alkaline earth metals (2), Ag, Zn2, Cd2, and Al3. When naming acids, binary acids are named differently than oxoacids. For binary acids, the name is based on the nonmetal. For oxoacids, the name is based on the polyatomic anion. For more detail, see Section 2.7 of the text. Solution: (a)

This is an ionic compound in which the metal cation (K) has only one charge. The correct name is potassium hypochlorite. Hypochlorite is a polyatomic ion with one less O atom than the chlorite ion, ClO2

(b)

silver carbonate

(c)

This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Fe ion. Since each chloride ion has a 1 charge, the Fe ion has a 2 charge. The correct name is iron(II) chloride.

(d)

potassium permanganate

(g)

This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Fe ion. Since the oxide ion has a 2 charge, the Fe ion has a 2 charge. The correct name is iron(II) oxide.

(h)

iron(III) oxide

(i)

This is an ionic compound in which the metal can form more than one cation. Use a Roman numeral to specify the charge of the Ti ion. Since each of the four chloride ions has a 1 charge (total of 4), the Ti ion has a 4 charge. The correct name is titanium(IV) chloride.

(j)

sodium hydride

(e)

(k)

cesium chlorate

lithium nitride

(f)

(l)

hypoiodous acid

sodium oxide

(m) This is an ionic compound in which the metal cation (Na) has only one charge. The O22 ion is called the peroxide ion. Each oxygen has a 1 charge. You can determine that each oxygen only has a 1 charge, because each of the two Na ions has a 1 charge. Compare this to sodium oxide in part (l). The correct name is sodium peroxide. (n)

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CHAPTER 2; ATOMS, MOLECULES, AND IONS

end of the name, we add hexahydrate for the six waters of hydration. The correct name is iron(III) chloride hexahydrate. (o)

The phosphate ion is PO34- , and arsenic is in the same family as phosphorus, so AsO34- is the arsenate ion.

(p)

HSO3- is formed by adding a hydrogen ion ( H + ) to the sulfite ion ( SO32- ), so the ion is named hydrogen sulfite.

2.49

(a) (f) (k)

RbNO2 BCl3 CaSO42H2O

2.50

Strategy: When writing formulas of molecular compounds, the prefixes specify the number of each type of atom in the compound.

(b) (g)

K2S IF7

HBrO4 (NH4)2SO4

(d) (i)

Mg3(PO4)2 AgClO4

(e) (j)

CaHPO4 Fe2(CrO4)3

When writing formulas of ionic compounds, the subscript of the cation is numerically equal to the charge of the anion, and the subscript of the anion is numerically equal to the charge on the cation. If the charges of the cation and anion are numerically equal, then no subscripts are necessary. Charges of common cations and anions are listed in Table 2.3 of the text. Keep in mind that Roman numerals specify the charge of the cation not the number of metal atoms. Remember that a Roman numeral is not needed for some metal cations, because the charge is known. These metals are the alkali metals (1), the alkaline earth metals (2), Ag, Zn2, Cd2, and Al3. When writing formulas of oxoacids, you must know the names and formulas of polyatomic anions (see Table 2.3 of the text). Solution: (a) (b)

(f)

(g)

(h) (i)

The Roman numeral I tells you that the Cu cation has a 1 charge. Cyanide has a 1 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is CuCN. Strontium is an alkaline earth metal. It only forms a 2 cation. The polyatomic ion chlorite, ClO2, has a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. The correct formula is Sr(ClO2)2. Perchloric tells you that the anion of this oxoacid is perchlorate, ClO4. The correct formula is HClO4(aq). Remember that (aq) means that the substance is dissolved in water. Hydroiodic tells you that the anion of this binary acid is iodide, I. The correct formula is HI(aq). Na is an alkali metal. It only forms a 1 cation. The polyatomic ion ammonium, NH4, has a 1 charge and the polyatomic ion phosphate, PO43, has a 3 charge. To balance the charge, you need 2 Na cations. The correct formula is Na2(NH4)PO4. The Roman numeral II tells you that the Pb cation has a 2 charge. The polyatomic ion carbonate, CO32, has a 2 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is PbCO3. The Roman numeral II tells you that the Sn cation has a 2 charge. Fluoride has a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. The correct formula is SnF2. This is a molecular compound. The Greek prefixes tell you the number of each type of atom in the molecule. The correct formula is P4S10. The Roman numeral II tells you that the Hg cation has a 2 charge. Oxide has a -2 charge. Since, the charges are numerically equal, no subscripts are necessary in the formula. The correct formula is HgO.

CHAPTER 2: ATOMS, MOLECULES, AND IONS

(j)

(k)

The Roman numeral I tells you that the Hg cation has a 1 charge. However, this cation exists as Hg22. Iodide has a 1 charge. You need two iodide ions to balance the 2 charge of Hg22. The correct formula is Hg2I2. The Roman numeral II tells you that the Co cation has a 2 charge. Chloride has a 1 charge. Since the charges on the cation and anion are numerically different, the subscript of the cation is numerically equal to the charge on the anion, and the subscript of the anion is numerically equal to the charge on the cation. We add 6H2O at the end of the formula for the six waters of hydration. The correct formula is CoCl26H2O.

2.51

The number of protons  65  35  30. The element that contains 30 protons is zinc, Zn. There are two fewer electrons than protons, so the charge of the cation is 2. The symbol for this cation is Zn2.

2.52

Changing the electrical charge of an atom usually has a major effect on its chemical properties. The two electrically neutral carbon isotopes should have nearly identical chemical properties; that is (c).

2.53

(a) (b) (c) (d)

Species with the same number of protons and electrons will be neutral. A, F, G. Species with more electrons than protons will have a negative charge. B, E. Species with more protons than electrons will have a positive charge. C, D. + 2+ 81  A: 105 B B: 147 N 3 C: 39 D: 66 E: 35 F: 115 B Br 19 K 30 Zn

2.54

(a) (b)

Does this refer to hydrogen atoms or hydrogen molecules? One cannot be sure. NaCl is an ionic compound; it does not form molecules.

2.55

Yes. The law of multiple proportions requires that the masses of sulfur combining with phosphorus must be in the ratios of small whole numbers. For the three compounds shown, four phosphorus atoms combine with three, seven, and ten sulfur atoms, respectively. If the atom ratios are in small whole number ratios, then the mass ratios must also be in small whole number ratios.

2.56

The species and their identification are as follows: (a) (b) (c) (d) (e) (f)

SO2 S8 Cs N2O5 O O2

molecule and compound element and molecule element molecule and compound element element and molecule

2.57

(a)

molecular, C3H8 empirical, C3H8

2.58

(a) (b) (c) (d)

CO2 (s), solid carbon dioxide NaCl, sodium chloride N2O, nitrous oxide CaCO3, calcium carbonate

2.59

Symbol Protons Neutrons Electrons

(b)

molecular, C2H2 empirical, CH

(g) (h) (i) (j) (k) (l)

O3 CH4 KBr S P4 LiF

(c)

molecular, C2H6 empirical, CH3

(e) (f) (g) (h)

G: 199 F

element and molecule molecule and compound compound element element and molecule compound (d)

molecular, C6H6 empirical, CH

CaO, calcium oxide Ca(OH)2, calcium hydroxide NaHCO3, sodium bicarbonate Mg(OH)2, magnesium hydroxide

11 5B

54 2+ 26 Fe

31 3 15 P

196 79 Au

222 86 Rn

5 6 5

26 28 24

15 16 18

79 117 79

86 136 86

CHAPTER 2; ATOMS, MOLECULES, AND IONS

Net Charge 2.60

(a) (b)

2.61

2

3

Ionic compounds are typically formed between metallic (especially Groups 1A, 2A, and aluminum) and nonmetallic elements. In general the transition metals, the actinides and lanthanides have variable charges.

Group 1A metals form M ions. Group 2A metals form Y2 ions. Aluminum forms an Al3 ion. Oxygen forms an O2 ion (oxide). Nitrogen forms an N3 ion (nitride), and the halogens form X ions. Making a table: Nonmetals

1A Metals

2A Metals

Aluminum

Halogens Oxygen Nitrogen

MX M2O M3N

YX2 YO Y3N2

AlX3 Al2O3 AlN

2.62

The symbol 23Na provides more information than 11Na. The mass number plus the chemical symbol identifies a specific isotope of Na (sodium), while combining the atomic number with the chemical symbol tells you nothing new. Can other isotopes of sodium have different atomic numbers?

2.63

The binary Group 7A element acids are HF, hydrofluoric acid; HCl, hydrochloric acid; HBr, hydrobromic acid; and HI, hydroiodic acid. Oxoacids containing Group 7A elements (using the specific examples for chlorine) are HClO4, perchloric acid; HClO3, chloric acid; HClO2, chlorous acid; and HClO, hypochlorous acid. Examples of oxoacids containing other Group A-block elements are H3BO3, boric acid (Group 3A); H2CO3, carbonic acid (Group 4A); HNO3, nitric acid; H3PO4, phosphoric acid (Group 5A); and H2SO4, sulfuric acid (Group 6A). Hydrosulfuric acid, H2S, is an example of a binary Group 6A acid while HCN, hydrocyanic acid, contains both a Group 4A and 5A element.

2.64

Isotope: No. Neutrons:

2.65

2.66

H2, N2, O2, F2, Cl2, He, Ne, Ar, Kr, Xe, Rn

2.67

Cu, Ag, and Au are fairly chemically unreactive. This makes them specifically suitable for making coins and jewelry, which you want to last a very long time.

2.68

They do not have a strong tendency to form compounds. Helium and neon are chemically inert.

2.69

(a) (b) (c)

2.70

The empirical and molecular formulas of acetaminophen are C8H9NO2.

dinitrogen pentoxide (N2O5) boron trifluoride (BF3) dialuminum hexabromide (Al2Br6)

CHAPTER 2: ATOMS, MOLECULES, AND IONS

2.71

CH4, C2H6, and C3H8 each only have one structural formula.

H H

C4H10 has two structural formulas.

H HH H C H

H H

C5H12 has three structural formulas.

H H

2.72

(a)

HH H C

HH H

HH H C H H

H H H

The following strategy can be used to convert from the volume of the Pt cube to the number of Pt atoms. cm3 grams  atoms 1.0 cm3 ´

(b)

21.45 g Pt 1 cm

1 atom Pt 3.240 ´ 10-22 g Pt

= 6.6 ´ 10 22 Pt atoms

Since 74% of the available space is taken up by Pt atoms, 6.6  1022 atoms occupy the following volume: 0.74  1.0 cm3 0.74 cm3 We are trying to calculate the radius of a single Pt atom, so we need the volume occupied by a single Pt atom.

volume Pt atom =

0.74 cm3 6.6 ´ 1022 Pt atoms

= 1.12 ´ 10-23 cm3 /Pt atom

4 3 pr . Solving for the radius: 3 4 V = 1.12 ´ 10-23 cm3 = pr 3 3

The volume of a sphere is

r3 2.67  1024 cm3 r 1.4  108 cm Converting to picometers: radius Pt atom = (1.4 ´ 10-8 cm) ´

0.01 m 1 pm ´ = 1.4 ´ 102 pm 1 cm 1 ´ 10-12 m

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CHAPTER 2; ATOMS, MOLECULES, AND IONS

2.73 Cation Mg2

Anion HCO3

Sr2

Cl

3

Formula Mg(HCO3)2 SrCl2

Name Magnesium bicarbonate



Fe(NO2)3

Iron(III) nitrite



Manganese(II) chlorate

NO2

2

Strontium chloride

Sn4

Br

Mn(ClO3)2 SnBr4

Co2

PO43

Co3(PO4)2

Cobalt(II) phosphate



Hg2I2

Mercury(I) iodide

2

Hg2

ClO3

Tin(IV) bromide



CO3

Cu2CO3

Copper(I) carbonate



3

Li3N

Lithium nitride

2

Al2S3

Aluminum sulfide

2

3

2.74

(a)

(b)

(c)

(d)

(e)

2.75

From left to right: NF3, nitrogen trifluoride; PBr5, phosphorus pentabromide; and SCl2, sulfur dichloride.

2.76

The change in energy is equal to the energy released. We call this E. Similarly, m is the change in mass. E Because m = 2 , we have c 1000 J (1.715 ´ 103 kJ) ´ DE 1 kJ Dm = = = 1.91 ´ 10-11 kg = 1.91 ´ 10-8 g 2 8 2 c (3.00 ´ 10 m/s)

æ 1 kg × m 2 ö ÷ Note that we need to convert kJ to J so that we end up with units of kg for the mass. çç1 J = ÷ s2 è ø We can add together the masses of hydrogen and oxygen to calculate the mass of water that should be formed. 12.096 g  96.000  108.096 g The predicted change (loss) in mass is only 1.91 × 108 g, which is too small a quantity to measure accurately. Therefore, for all practical purposes, the law of conservation of mass is assumed to hold for ordinary chemical processes. 2.77

(a)

Rutherford’s experiment is described in detail in Section 2.2 of the text. From the average magnitude of scattering, Rutherford estimated the number of protons (based on electrostatic interactions) in the nucleus.

(b)

Assuming that the nucleus is spherical, the volume of the nucleus is: V =

4 3 4 pr = p(3.04 ´ 10-13 cm)3 = 1.18 ´ 10-37 cm3 3 3

The density of the nucleus can now be calculated. d =

m 3.82 ´ 10-23 g = = 3.24 ´ 1014 g/cm 3 37 3 V 1.18 ´ 10 cm

To calculate the density of the space occupied by the electrons, we need both the mass of 11 electrons and the volume occupied by these electrons. 10 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part .

CHAPTER 2: ATOMS, MOLECULES, AND IONS

The mass of 11 electrons is: 11 electrons ´

9.1095 ´ 10-28 g = 1.0020 ´ 10-26 g 1 electron

The volume occupied by the electrons will be the difference between the volume of the atom and the volume of the nucleus. The volume of the nucleus was calculated above. The volume of the atom is calculated as follows:

186 pm ´ Vatom =

1 ´ 10-12 m 1 cm ´ = 1.86 ´ 10-8 cm 1 pm 1 ´ 10-2 m

4 3 4 pr = p(1.86 ´ 10-8 cm)3 = 2.70 ´ 10-23 cm3 3 3

VelectronsVatomVnucleus (2.70  1023 cm3)  (1.18  1037 cm3)  2.70  1023 cm3 As you can see, the volume occupied by the nucleus is insignificant compared to the space occupied by the electrons. The density of the space occupied by the electrons can now be calculated. d =

m 1.0020 ´ 10-26 g = = 3.71 ´ 10 -4 g/cm 3 -23 3 V 2.70 ´ 10 cm

The above results do support Rutherford’s model. Comparing the space occupied by the electrons to the volume of the nucleus, it is clear that most of the atom is empty space. Rutherford also proposed that the nucleus was a dense central core with most of the mass of the atom concentrated in it. Comparing the density of the nucleus with the density of the space occupied by the electrons also supports Rutherford’s model. 2.78

The formula of the ionic compound is XY2. Element X is most likely in Group 4B and element Y is most likely in Group 6A. A possible compound is TiO2, titanium(IV) oxide. Other choices are elements in Group 4A: SnO2 [tin(IV) oxide] and PbO2 [lead(IV) oxide].

2.79

Two different structural formulas for the molecular formula C2H6O are:

H O

In the second hypothesis of Dalton’s Atomic Theory, he states that in any compound, the ratio of the number of atoms of any two of the elements present is either an integer or a simple fraction. In the above two compounds, the ratio of atoms is the same. This does not necessarily contradict Dalton’s hypothesis, but Dalton was not aware of chemical bond formation and structural formulas. 2.80

The diameter of a pea is about 0.5 cm, and the radius of the nucleus is about 5  10–13 cm. Therefore, the expansion factor for the radius of the nucleus is 0.5 cm / 2 5 ´ 10

–13

= 5 ´ 1011 times

CHAPTER 2; ATOMS, MOLECULES, AND IONS

A typical atom has a radius of about 100 pm (10–8 cm). Using that value as the distance of the electron from the nucleus in an atom, then if the atom was scaled to the size of a pea, the electron would be (1  10–8 cm)  (5  1011) = 5  103 cm or 50 m (roughly half a football field) from the nucleus! 2.81

Over long periods of time (on a geological scale), minerals containing potassium and sodium are slowly decomposed by wind and rain, and their K+ and Na+ ions are converted to more soluble compounds. Eventually, rain leaches these compounds out of the soil and carries them to the sea. Plants take up many of the K+ ions along the way, while the Na+ ions are free to move on to the sea because they are not needed for biological functions by the plants.

2.82

The acids, from left to right, are chloric acid, nitrous acid, hydrocyanic acid, and sulfuric acid.

ANSWERS TO REVIEW OF CONCEPTS Section 2.1 (p. 31) Section 2.2 (p. 36) Section 2.3 (p. 37) Section 2.4 (p. 39) Section 2.5 (p. 40)

Section 2.6 (p. 45) Section 2.7 (p. 52) Section 2.8 (p. 53)

Yes, the ratio of atoms represented by B that combine with A in these two compounds is (2/1): (5/2) or 4:5. The proton and neutron have approximately the same mass. (a) 78. (b)17O. Chemical properties change more markedly across a period. (a) S8 signifies one molecule of sulfur that is composed of 8 sulfur atoms. 8S represents 8 individual atoms of sulfur. (b) (a) 15 protons, 18 electrons. (b) 22 protons, 18 electrons. (a) Mg(NO3)2 (b) Al2O3 (c) LiH (d) Na2S. (a) Aluminum sulfide (b) IF7 Two.

CHAPTER 3 STOICHIOMETRY 3.1

One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atoms. We cannot weigh a single atom, but it is possible to determine the mass of one atom relative to another experimentally. The first step is to assign a value to the mass of one atom of a given element so that it can be used as a standard.

3.2

12.00 amu. On the periodic table, the mass is listed as 12.01 amu because this is an average mass of the naturally occurring mixture of isotopes of carbon.

3.3

The value 197.0 amu is an average value (an average atomic mass). If we could examine gold atoms individually, we would not find an atom with a mass of 197.0 amu. However, the average mass of a gold atom in a typical sample of gold is 197.0 amu.

3.4

You need the mass of each isotope of the element and each isotope’s relative abundance.

3.5

(34.968 amu)(0.7553)  (36.956 amu)(0.2447) 35.45 amu

3.6

Strategy: Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope. It would seem that there are two unknowns in this problem, the fractional abundance of 6Li and the fractional abundance of 7Li. However, these two quantities are not independent of each other; they are related by the fact that they must sum to 1. Start by letting x be the fractional abundance of 6Li. Since the sum of the two abundances must be 1, we can write Abundance 7Li  (1 x) Solution: Average atomic mass of Li

6.941 amu x(6.0151 amu)  (1 x)(7.0160 amu) 6.941 1.0009x 7.0160 1.0009x 0.075 x 0.075

x 0.075 corresponds to a natural abundance of 6Li of 7.5%. The natural abundance of 7Li is (1 x)  0.925 or 92.5%.

3.7

æ 6.022 ´ 1023 amu ö The conversion factor required is ç ÷ ç ÷ 1g è ø

? g = 13.2 amu ´ 3.8

1g 6.022 ´ 1023 amu

= 2.19 ´ 10-23 g

æ 6.022 ´ 1023 amu ö The unit factor required is ç ÷ ç ÷ 1g è ø

CHAPTER 3: STOICHIOMETRY

? amu = 8.4 g ´

6.022 ´ 1023 amu = 5.1 ´ 1024 amu 1g

3.9

The mole is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 grams of the carbon-12 isotopes. The unit for mole used in calculations is mol. A mole is a unit like a pair, dozen, or gross. A mole is the amount of substance that contains 6.022  1023 particles. Avogadro’s number (6.022  1023) is the number of atoms in exactly 12 g of the carbon-12 isotopes.

3.10

The molar mass of an atom is the mass of one mole, 6.022 × 1023 atoms, of that element. Units are g/mol.

3.11

In one year: (7.0 ´ 109 people) ´

Total time =

3.12

365 days 24 h 3600 s 2 particles ´ ´ ´ = 4.4 ´ 1017 particles/yr 1 yr 1 day 1h 1 person

6.022 ´ 1023 particles 4.4 ´ 1017 particles/yr

= 1.4 ´ 106 yr

The thickness of the book in miles would be: 0.0036 in 1 ft 1 mi ´ ´ ´ (6.022 ´ 1023 pages) = 3.4 ´ 1016 mi 1 page 12 in 5280 ft

The distance, in miles, traveled by light in one year is: 1.00 yr ´

365 day 24 h 3600 s 3.00 ´ 108 m 1 mi ´ ´ ´ ´ = 5.88 ´ 1012 mi 1 yr 1 day 1h 1s 1609 m

The thickness of the book in light-years is:

(3.4 ´ 1016 mi) ´

1 light-yr 5.88 ´ 1012 mi

= 5.8 ´ 103 light - yr

It will take light 5.8  103 years to travel from the first page to the last one! 6.022 ´ 1023 S atoms = 3.07 ´ 10 24 S atoms 1 mol S

3.13

5.10 mol S ´

3.14

(6.00 ´ 109 Co atoms) ´

3.15 3.16

1 mol Co 6.022 ´ 1023 Co atoms

= 9.96 ´ 10-15 mol Co

1 mol Ca = 1.93 mol Ca 40.08 g Ca Strategy: We are given moles of gold and asked to solve for grams of gold. What conversion factor do we need to convert between moles and grams? Arrange the appropriate conversion factor so moles cancel and the unit grams is obtained for the answer. 77.4 g of Ca ´

CHAPTER 3: STOICHIOMETRY

Solution: The conversion factor needed to covert between moles and grams is the molar mass. In the periodic table (see inside front cover of the text), we see that the molar mass of Au is 197.0 g. This can be expressed as 1 mol Au  197.0 g Au From this equality, we can write two conversion factors. 1 mol Au 197.0 g Au

and

197.0 g Au 1 mol Au

The conversion factor on the right is the correct one. Moles will cancel, leaving the unit grams for the answer. We write ? g Au = 15.3 mol Au ´

197.0 g Au = 3.01 ´ 103 g Au 1 mol Au

Check: Does a mass of 3010 g for 15.3 mole of Au seem reasonable? What is the mass of 1 mole of Au?

3.17

3.18

(a)

200.6 g Hg 1 mol Hg ´ = 3.331 ´ 10-22 g/Hg atom 1 mol Hg 6.022 ´ 1023 Hg atoms

(b)

20.18 g Ne 1 mol Ne ´ = 3.351 ´ 10-23 g/Ne atom 1 mol Ne 6.022 ´ 1023 Ne atoms

(a) Strategy: We can look up the molar mass of arsenic (As) on the periodic table (74.92 g/mol). We want to find the mass of a single atom of arsenic (unit of g/atom). Therefore, we need to convert from the unit mole in the denominator to the unit atom in the denominator. What conversion factor is needed to convert between moles and atoms? Arrange the appropriate conversion factor so mole in the denominator cancels, and the unit atom is obtained in the denominator. Solution: The conversion factor needed is Avogadro’s number. We have 1 mol  6.022  1023 particles (atoms) From this equality, we can write two conversion factors. 1 mol As 6.022 ´ 1023 As atoms

and

6.022 ´ 10 23 As atoms 1 mol As

The conversion factor on the left is the correct one. Moles will cancel, leaving the unit atoms in the denominator of the answer. We write

? g/As atom = (b)

74.92 g As 1 mol As ´ = 1.244 ´ 10-22 g/As atom 1 mol As 6.022 ´ 1023 As atoms

Follow same method as part (a). ? g/Ni atom =

58.69 g Ni 1 mol Ni ´ = 9.746 ´ 10-23 g/Ni atom 23 1 mol Ni 6.022 ´ 10 Ni atoms

Check: Should the mass of a single atom of As or Ni be a very small mass?

CHAPTER 3: STOICHIOMETRY

3.19

1.00 ´ 1012 Pb atoms ´

3.20

Strategy: The question asks for atoms of Cu. We cannot convert directly from grams to atoms of copper. What unit do we need to convert grams of Cu to in order to convert to atoms? What does Avogadro’s number represent?

1 mol Pb 6.022 ´ 10

Pb atoms

207.2 g Pb = 3.44 ´ 10-10 g Pb 1 mol Pb

Solution: To calculate the number of Cu atoms, we first must convert grams of Cu to moles of Cu. We use the molar mass of copper as a conversion factor. Once moles of Cu are obtained, we can use Avogadro’s number to convert from moles of copper to atoms of copper. 1 mol Cu  63.55 g Cu The conversion factor needed is 1 mol Cu 63.55 g Cu

Avogadro’s number is the key to the second conversion. We have 1 mol  6.022  1023 particles (atoms) From this equality, we can write two conversion factors. 1 mol Cu

and

6.022 ´ 1023 Cu atoms

6.022 ´ 1023 Cu atoms 1 mol Cu

The conversion factor on the right is the one we need because it has the number of Cu atoms in the numerator, which is the unit we want for the answer. Let us complete the two conversions in one step. grams of Cu  moles of Cu  number of Cu atoms ? atoms of Cu = 0.063 g Cu ´

1 mol Cu 6.022 ´ 1023 Cu atoms ´ = 6.0 ´ 1020 Cu atoms 63.55 g Cu 1 mol Cu

Check: Should 0.063 g of Cu contain fewer than Avogadro’s number of atoms? What mass of Cu would contain Avogadro’s number of atoms?

3.21

1 mol H 6.022 ´ 1023 H atoms ´ = 6.57 ´ 10 23 H atoms 1.008 g H 1 mol H

For hydrogen:

1.10 g H ´

For chromium:

14.7 g Cr ´

1 mol Cr 6.022 ´ 1023 Cr atoms ´ = 1.70 ´ 1023 Cr atoms 52.00 g Cr 1 mol Cr

There are more hydrogen atoms than chromium atoms.

3.22

2 Pb atoms ´

1 mol Pb 6.022 ´ 10

(5.1 ´ 10-23 mol He) ´

Pb atoms

207.2 g Pb = 6.881 ´ 10-22 g Pb 1 mol Pb

4.003 g He = 2.0 ´ 10-22 g He 1 mol He

CHAPTER 3: STOICHIOMETRY

2 atoms of lead have a greater mass than 5.1  1023 mol of helium. 3.23

Using the appropriate atomic masses, (a) (b) (c) (d) (e) (f) (g)

3.24

CH4 NO2 SO3 C6H6 NaI K2SO4 Ca3(PO4)2

12.01 amu  4(1.008 amu) 16.04 amu 14.01 amu  2(16.00 amu) 46.01 amu 32.07 amu  3(16.00 amu) 80.07 amu 6(12.01 amu)  6(1.008 amu) 78.11 amu 22.99 amu  126.9 amu 149.9 amu 2(39.10 amu)  32.07 amu  4(16.00 amu) 174.27 amu 3(40.08 amu)  2(30.97 amu)  8(16.00 amu) 310.2 amu

Strategy: How do molar masses of different elements combine to give the molar mass of a compound? Solution: To calculate the molar mass of a compound, we need to sum all the molar masses of the elements in the molecule. For each element, we multiply its molar mass by the number of moles of that element in one mole of the compound. We find molar masses for the elements in the periodic table (inside front cover of the text).

3.25

(a)

molar mass Li2CO3  2(6.941 g)  12.01 g  3(16.00 g) 73.89 g

(b)

molar mass CS2  12.01 g  2(32.07 g) 76.15 g

(c)

molar mass CHCl3 12.01 g  1.008 g  3(35.45 g) 119.4 g

(d)

molar mass C6H8O6  6(12.01 g)  8(1.008 g)  6(16.00 g) 176.12 g

(e)

molar mass KNO3  39.10 g  14.01 g  3(16.00 g) 101.11 g

(f)

molar mass Mg3N2  3(24.31 g)  2(14.01 g) 100.95 g

To find the molar mass (g/mol), we simply divide the mass (in g) by the number of moles. 152 g = 409 g/mol 0.372 mol

3.26

The molar mass of acetone, C3H6O, is 58.08 g. We use the molar mass and Avogadro’s number as conversion factors to convert from grams to moles to molecules of acetone. 0.435 g C3H 6 O ´

3.27

1 mol C3H 6 O 6.022 ´ 1023 molecules C3 H 6 O ´ = 4.51 ´ 1021 molecules C3 H 6O 58.08 g C3 H 6 O 1 mol C3H 6 O

We use the molar mass of squaric acid (114.06 g), Avogadro’s number, and the subscripts in the formula of squaric acid, C4H2O4, to convert from grams of squaric acid to moles of squaric acid to molecules of squaric acid, and finally to atoms of C, H, or O. We first convert to molecules of squaric acid. 1.75 g C 4 H 2 O 4 ´

1 mol C 4 H 2O 4 6.022 ´ 10 23 molecules C 4 H 2 O 4 ´ = 9.24 ´ 10 21 molecules C 4 H 2 O 4 114.06 g C4 H 2 O 4 1 mol C 4 H 2 O 4

Next, we convert to atoms of C, H, and O using the subscripts in the formula as conversion factors. 9.24 ´ 1021 molecules C4 H 2 O 4 ´

4 atoms C = 3.70 ´ 1022 C atoms 1 molecule C4 H 2O 4

3.28

CHAPTER 3: STOICHIOMETRY

9.24 ´ 1021 molecules C4 H 2O 4 ´

2 atoms H = 1.85 ´ 1022 H atoms 1 molecule C4 H 2O4

9.24 ´ 1021 molecules C4 H 2O4 ´

4 atoms O = 3.70 ´ 1022 O atoms 1 molecule C4 H 2O 4

Strategy: We are asked to solve for the number of C, S, H, and O atoms in 7.14  103 g of dimethyl sulfoxide (DMSO). We cannot convert directly from grams DMSO to atoms. What unit do we need to obtain first before we can convert to atoms? How should Avogadro’s number be used here? How many atoms of C, S, H, or O are in 1 molecule of DMSO? Solution: Let us first calculate the number of C atoms in 7.14  103 g of dimethyl sulfoxide. First, we must convert grams of DMSO to number of molecules of DMSO. This calculation is similar to Problem 3.26. The molecular formula of DMSO shows there are two C atoms in one DMSO molecule, which will allow us to convert to atoms of C. We need to perform three conversions: grams of DMSO  moles of DMSO  molecules of DMSO  atoms of C The conversion factors needed for each step are: 1) the molar mass of DMSO, 2) Avogadro’s number, and 3) the number of C atoms in 1 molecule of DMSO. We complete the three conversions in one calculation. 7.14 ´ 103 g DMSO ´

1 mol DMSO 6.022 ´ 10 23 DMSO molecules 2 C atoms ´ ´ 78.14 g DMSO 1 mol DMSO 1 molecule DMSO

1.10  1026 C atoms The above method utilizes the ratio of molecules (DMSO) to atoms (carbon). We can also solve the problem by reading the formula as the ratio of moles of DMSO to moles of carbon using the following conversions: grams of DMSO  moles of DMSO  moles of C  atoms of C Try it. Check: Does the answer seem reasonable? We have 7.14  103 g DMSO. How many atoms of C would 78.14 g of DMSO contain? We could calculate the number of atoms of the remaining elements in the same manner or we can use the atom ratios from the molecular formula. The sulfur atom to carbon atom ratio in a DMSO molecule is 1:2, the hydrogen atom to carbon atom ratio is 6:2 or 3:1, and the oxygen atom to carbon atom ratio is 1:2.

3.29

? atoms of S = (1.10 ´ 1026 C atoms) ´

1 S atom = 5.50 ´ 1025 S atoms 2 C atoms

? atoms of H = (1.10 ´ 1026 C atoms) ´

3 H atoms = 3.30 ´ 1026 H atoms 1 C atom

? atoms of O = (1.10 ´ 1026 C atoms) ´

1 O atom = 5.50 ´ 1025 O atoms 2 C atoms

The molar mass of C19H38O is 282.5 g.

CHAPTER 3: STOICHIOMETRY

1.0 ´ 10-12 g ´

1 mol 6.022 ´ 1023 molecules ´ = 2.1 ´ 109 molecules 282.5 g 1 mol

Notice that even though 1.0  1012 g is an extremely small mass, it still comprised over a billion pheromone molecules!

3.30

Mass of water = 2.56 mL ´

1.00 g = 2.56 g 1.00 mL

Molar mass of H2O  (16.00 g)  2(1.008 g)  18.02 g/mol ? H 2O molecules = 2.56 g H 2 O ´

1 mol H 2 O 6.022 ´ 1023 molecules H 2O ´ 18.02 g H 2O 1 mol H 2 O

8.56  1022 molecules 3.31

Please see Section 3.4 of the text.

3.32

The relative abundance of each isotope can be determined from the area of the peak in the mass spectrum for that isotope.

3.33

Since there are only two isotopes of carbon, there are only two possibilities for CF4. 12 19 + 13 19 + 6 C 9 F4 (molecular mass 88 amu) and 6 C 9 F4 (molecular mass 89 amu)

There would be two peaks in the mass spectrum. 3.34

Since there are two hydrogen isotopes, they can be paired in three ways: 1H-1H, 1H-2H, and 2H-2H. There will then be three choices for each sulfur isotope. We can make a table showing all the possibilities (masses in amu): 32 33 34 36 S S S S 1

H2 H2H 2 H2 1

34 35 36

35 36 37

36 37 38

38 39 40

There will be seven peaks of the following mass numbers: 34, 35, 36, 37, 38, 39, and 40. Very accurate (and expensive!) mass spectrometers can detect the mass difference between two 1H and one 2 H. How many peaks would be detected in such a “high-resolution” mass spectrum? 3.35

The percent composition is the percent by mass of each element in a compound. For NH3, we would speak of the mass % of nitrogen (N) and the mass % of hydrogen (H) in the compound. What percentage of the mass of a sample of ammonia is due to nitrogen and what percentage of the mass is due to hydrogen?

3.36

If you know the percent composition by mass of an unknown compound, you can determine its empirical formula.

3.37

An empirical formula tells us which elements are present and the simplest whole number ratio of their atoms. A definition of empirical is something that is derived from experiment and observation rather than from theory. When chemists analyze an unknown compound, the first step is usually the determination of the compound’s empirical formula. With additional information, it is possible to deduce the molecular formula.

CHAPTER 3: STOICHIOMETRY

3.38

The approximate molar mass.

3.39

Molar mass of SnO2 (118.7 g)  2(16.00 g)  150.7 g

3.40

%Sn =

118.7 g/mol ´ 100% = 78.77% 150.7 g/mol

%O =

(2)(16.00 g/mol) ´ 100% = 21.23% 150.7 g/mol

Strategy: Recall the procedure for calculating a percentage. Assume that we have 1 mole of CHCl3. The percent by mass of each element (C, H, and Cl) is given by the mass of that element in 1 mole of CHCl3 divided by the molar mass of CHCl3, then multiplied by 100 to convert from a fractional number to a percentage. Solution: The molar mass of CHCl3 12.01 g/mol  1.008 g/mol  3(35.45 g/mol)  119.4 g/mol. The percent by mass of each of the elements in CHCl3 is calculated as follows: %C =

12.01 g/mol ´ 100% = 10.06% 119.4 g/mol

%H =

1.008 g/mol ´ 100% = 0.8442% 119.4 g/mol

%Cl =

3(35.45) g/mol ´ 100% = 89.07% 119.4 g/mol

Check: Do the percentages add to 100%? The sum of the percentages is (10.06%  0.8442%  89.07%)  99.97%. The small discrepancy from 100% is due to the way we rounded off. 3.41

The molar mass of cinnamic alcohol, C9H10O, is 134.17 g/mol. (a)

(b)

%C =

(9)(12.01 g/mol) ´ 100% = 80.56% 134.17 g/mol

%H =

(10)(1.008 g/mol) ´ 100% = 7.513% 134.17 g/mol

%O =

16.00 g/mol ´ 100% = 11.93% 134.17 g/mol

0.469 g C9 H10O ´

1 mol C9 H10 O 6.022 ´ 1023 molecules C9 H10 O ´ 134.17 g C9 H10 O 1 mol C9 H10 O

2.11  1021 molecules C9H10O 3.42

Compound Molar mass (g) N% by mass (a) (NH2)2CO 2(14.01 g) ´ 100% = 46.65% 60.06 60.06 g

CHAPTER 3: STOICHIOMETRY

(b) 80.05 (c) 59.08 (d)

NH4NO3 2(14.01 g) ´ 100% = 35.00% 80.05 g HNC(NH2)2 3(14.01 g) ´ 100% = 71.14% 59.08 g 14.01 g ´ 100% = 82.27% 17.03 g

NH317.03

Ammonia, NH3, is the richest source of nitrogen on a mass percentage basis. 3.43

Assume you have exactly 100 g of substance. nC = 44.4 g C ´

1 mol C = 3.70 mol C 12.01 g C

nH = 6.21 g H ´

1 mol H = 6.16 mol H 1.008 g H

nS = 39.5 g S ´

1 mol S = 1.23 mol S 32.07 g S

nO = 9.86 g O ´

1 mol O = 0.616 mol O 16.00 g O

Thus, we arrive at the formula C3.70H6.16S1.23O0.616. Dividing by the smallest number of moles (0.616 mol) gives the empirical formula, C6H10S2O. To determine the molecular formula, divide the molar mass by the empirical mass. molar mass 162 g = » 1 empirical molar mass 162.3 g

Hence, the molecular formula and the empirical formula are the same, C6H10S2O. 3.44

METHOD 1: Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. The percentage of oxygen is found by difference: 100%  (19.8%  2.50%  11.6%)  66.1% In 100 g of PAN, there will be 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O. Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest whole number ratio of their atoms. This ratio is also a mole ratio. Use the molar masses of these elements as conversion factors to convert to moles. nC = 19.8 g C ´

1 mol C = 1.65 mol C 12.01 g C

nH = 2.50 g H ´

1 mol H = 2.48 mol H 1.008 g H

nN = 11.6 g N ´

1 mol N = 0.828 mol N 14.01 g N

CHAPTER 3: STOICHIOMETRY

nO = 66.1 g O ´

1 mol O = 4.13 mol O 16.00 g O

Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript. The formula is C1.65H2.48N0.828O4.13. Dividing the subscripts by 0.828 gives the empirical formula, C2H3NO5. To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one. molar mass ³ 1 (integer values) empirical molar mass In this case, molar mass 120 g = » 1 empirical molar mass 121.05 g Hence, the molecular formula and the empirical formula are the same, C2H3NO5. METHOD 2: Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar mass to convert to moles of each element. nC = (0.198) ´ (120 g) ´

1 mol C = 1.98 mol C » 2 mol C 12.01 g C

nH = (0.0250) ´ (120 g) ´

nN = (0.116) ´ (120 g) ´ nO = (0.661) ´ (120 g) ´

1 mol H = 2.98 mol H » 3 mol H 1.008 g H

1 mol N = 0.994 mol N » 1 mol N 14.01 g N

1 mol O = 4.96 mol O » 5 mol O 16.00 g O

Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular formula. The formula is C2H3NO5. Step 3: Try to reduce the molecular formula to a simpler whole number ratio to determine the empirical formula. The formula is already in its simplest whole number ratio. The molecular and empirical formulas are the same. The empirical formula is C2H3NO5. 1 mol Fe 2 O3 2 mol Fe ´ = 0.308 mol Fe 159.7 g Fe2 O3 1 mol Fe2 O3

3.45

24.6 g Fe2 O3 ´

3.46

Using unit factors we convert: g of Hg  mol Hg  mol S  g S ? g S = 246 g Hg ´

3.47

1 mol Hg 1 mol S 32.07 g S ´ ´ = 39.3 g S 200.6 g Hg 1 mol Hg 1 mol S

The balanced equation is: 2Al(s)  3I2(s) ¾¾ ® 2AlI3(s) Using unit factors, we convert: g of Al  mol of Al  mol of I2 g of I2

CHAPTER 3: STOICHIOMETRY

20.4 g Al ´

3.48

3 mol I 2 253.8 g I2 1 mol Al ´ ´ = 288 g I 2 26.98 g Al 2 mol Al 1 mol I 2

Strategy: Tin(II) fluoride is composed of Sn and F. The mass due to F is based on its percentage by mass in the compound. How do we calculate mass percent of an element? Solution: First, we must find the mass % of fluorine in SnF2. Then, we convert this percentage to a fraction and multiply by the mass of the compound (24.6 g) to find the mass of fluorine in 24.6 g of SnF2. The percent by mass of fluorine in tin(II) fluoride, is calculated as follows: mass % F =

mass of F in 1 mol SnF2 ´ 100% molar mass of SnF2 =

2(19.00 g) ´ 100% = 24.25% F 156.7 g

Converting this percentage to a fraction, we obtain 24.25/100  0.2425. Next, multiply the fraction by the total mass of the compound. ? g F in 24.6 g SnF2 (0.2425)(24.6 g) 5.97 g F Check: As a ball-park estimate, note that the mass percent of F is roughly 25 percent, so that a quarter of the mass should be F. One quarter of approximately 24 g is 6 g, which is close to the answer. Note: This problem could have been worked in a manner similar to Problem 3.46. You could complete the following conversions: g of SnF2 mol of SnF2 mol of F  g of F 3.49

In each case, assume 100 g of compound. (a)

2.1 g H ´

1 mol H = 2.1 mol H 1.008 g H

65.3 g O ´

1 mol O = 4.08 mol O 16.00 g O

32.6 g S ´

1 mol S = 1.02 mol S 32.07 g S

This gives the formula H2.1S1.02O4.08. Dividing by 1.02 gives the empirical formula, H2SO4. (b)

20.2 g Al ´

1 mol Al = 0.749 mol Al 26.98 g Al

79.8 g Cl ´

1 mol Cl = 2.25 mol Cl 35.45 g Cl

This gives the formula, Al0.749Cl2.25. Dividing by 0.749 gives the empirical formula, AlCl3. 3.50

(a) Strategy: In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form the compound. Therefore, we need to convert from mass percent to moles in order to

CHAPTER 3: STOICHIOMETRY

determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the mass of each element in the compound? How do we then convert from grams to moles? Solution: If we have 100 g of the compound, then each percentage can be converted directly to grams. In this sample, there will be 40.1 g of C, 6.6 g of H, and 53.3 g of O. Because the subscripts in the formula represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of moles of each element so that nC = 40.1 g C ´

1 mol C = 3.34 mol C 12.01 g C

nH = 6.6 g H ´

1 mol H = 6.5 mol H 1.008 g H

nO = 53.3 g O ´

1 mol O = 3.33 mol O 16.00 g O

Thus, we arrive at the formula C3.34H6.5O3.33, which gives the identity and the mole ratios of atoms present. However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.33).

3.34 » 1 3.33

6.5 » 2 3.33

3.33 = 1 3.33

This gives the empirical formula, CH2O. Check: Are the subscripts in CH2O reduced to the smallest whole numbers? (b)

Following the same procedure as part (a), we find: nC = 18.4 g C ´

1 mol C = 1.53 mol C 12.01 g C

nN = 21.5 g N ´

1 mol N = 1.53 mol N 14.01 g N

nK = 60.1 g K ´

1 mol K = 1.54 mol K 39.10 g K

Dividing by the smallest number of moles (1.53 mol) gives the empirical formula, KCN. 3.51

The molar mass of CaSiO3 is 116.17 g/mol. %Ca =

40.08 g ´ 100% = 34.50% 116.17 g

%O =

(3)(16.00 g) ´ 100% = 41.32% 116.17 g

%Si =

28.09 g ´ 100% = 24.18% 116.17 g

Check to see that the percentages sum to 100%. (34.50%  24.18%  41.32%)  100.00%. 3.52

The empirical molar mass of CH is approximately 13.02 g. Let us compare this to the molar mass to determine the molecular formula. Recall that the molar mass divided by the empirical mass will be an integer greater than or equal to one.

CHAPTER 3: STOICHIOMETRY

molar mass ³ 1 (integer values) empirical molar mass

In this case, molar mass 78 g = » 6 empirical molar mass 13.02 g

Thus, there are six CH units in each molecule of the compound, so the molecular formula is (CH)6 or C6H6. 3.53

Find the molar mass corresponding to each formula. For C4H5N2O:

4(12.01 g)  5(1.008 g)  2(14.01 g)  (16.00 g)  97.10 g

For C8H10N4O2:

8(12.01 g)  10(1.008 g)  4(14.01 g)  2(16.00 g)  194.20 g

The molecular formula is C8H10N4O2. 3.54

METHOD 1: Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. In 100 g of MSG there will be 35.51 g C, 4.77 g H, 37.85 g O, 8.29 g N, and 13.60 g Na. Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest whole number ratio of their atoms. This ratio is also a mole ratio. Let nC, nH, nO, nN, and nNa be the number of moles of elements present. Use the molar masses of these elements as conversion factors to convert to moles. nC = 35.51 g C ´

nH = 4.77 g H ´

1 mol C = 2.957 mol C 12.01 g C

1 mol H = 4.73 mol H 1.008 g H

nO = 37.85 g O ´

1 mol O = 2.366 mol O 16.00 g O

nN = 8.29 g N ´

1 mol N = 0.592 mol N 14.01 g N

nNa = 13.60 g Na ´

1 mol Na = 0.5916 mol Na 22.99 g Na

Thus, we arrive at the formula C2.957H4.73O2.366N0.592Na0.5916 that gives the identity and the ratios of atoms present. However, chemical formulas are written with whole numbers. Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript.

2.957 = 4.998 » 5 0.5916 0.592 N: = 1.00 0.5916

4.73 = 8.00 0.5916 0.5916 Na : = 1 0.5916 H:

2.366 = 3.999 » 4 0.5916

This gives us the empirical formula for MSG, C5H8O4NNa. To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one. 13 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 3: STOICHIOMETRY

molar mass ³ 1 (integer values) empirical molar mass

In this case, molar mass 169 g = » 1 empirical molar mass 169.11 g

Hence, the molecular formula and the empirical formula are the same, C5H8O4NNa. It should come as no surprise that the empirical and molecular formulas are the same since MSG stands for monosodium glutamate. METHOD 2: Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar mass to convert to moles of each element. nC = (0.3551) ´ (169 g) ´

1 mol C = 5.00 mol C 12.01 g C

nH = (0.0477) ´ (169 g) ´

1 mol H = 8.00 mol H 1.008 g H

nO = (0.3785) ´ (169 g) ´

1 mol O = 4.00 mol O 16.00 g O

nN = (0.0829) ´ (169 g) ´

1 mol N = 1.00 mol N 14.01 g N

nNa = (0.1360) ´ (169 g) ´

1 mol Na = 1.00 mol Na 22.99 g Na

Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular formula. The formula is C5H8O4NNa. 3.55

A chemical reaction is a process in which a substance (or substances) is changed into one or more new substances. In this reaction, hydrogen and oxygen, the reactants, are changed into the product, water.

3.56

A chemical equation uses chemical symbols to show what happens during a chemical reaction. A chemical reaction is a process in which a substance (or substances) is changed into one or more new substances.

3.57

To accurately show what happens during a chemical reaction, a chemical equation must be balanced. The law of conservation of mass is obeyed by a balanced chemical equation.

3.58

(g), (l), (s), (aq).

3.59

The balanced equations are as follows: (a)

2C  O2 2CO

(f)

2O3 3O2

(b)

2CO  O2 2CO2

(g)

2H2O2 2H2O  O2

(c)

H2 Br2 2HBr

(h)

N2 3H2 2NH3

(d)

2K  2H2O  2KOH  H2

(i)

Zn  2AgCl  ZnCl2 2Ag

(e)

2Mg  O2 2MgO

(j)

S8 8O2 8SO2

(k)

2NaOH  H2SO4 Na2SO4 2H2O

(m) 3KOH  H3PO4 K3PO4 3H2O

CHAPTER 3: STOICHIOMETRY

(l) 3.60

Cl2 2NaI  2NaCl  I2

(n)

CH4 4Br2 CBr4 4HBr

The balanced equations are as follows: (a)

2N2O5 2N2O4 O2

(b)

2KNO3 2KNO2 O2

(c)

NH4NO3 N2O  2H2O

(d)

NH4NO2  N2 2H2O

(e)

2NaHCO3 Na2CO3 H2O  CO2

(f)

P4O10 6H2O  4H3PO4

(g)

2HCl  CaCO3 CaCl2 H2O  CO2

(h)

2Al  3H2SO4 Al2(SO4)3 3H2

(i)

CO2 2KOH  K2CO3 H2O(j)

CH4 2O2 CO2 2H2O

(k)

Be2C  4H2O  2Be(OH)2 CH4

(l)

3Cu  8HNO3 3Cu(NO3)2 2NO  4H2O

(m) S  6HNO3 H2SO4 6NO2 2H2O

(n)

2NH3 3CuO  3Cu  N2 3H2O

3.61

Stoichiometry is the quantitative study of reactants and products in a chemical reaction; therefore, it is based on the Law of Conservation of Mass. A balanced chemical equation is essential to solving stoichiometric problems so that the “mole method” can be applied correctly.

3.62

The steps of the mole method are shown in Figure 3.8 of the text.

3.63

On the reactants side, there are 8 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 4 D atoms. Writing an equation, 8A  4B  4C  4D Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the equation by four gives, 2A  B  C  D The correct answer is choice (c).

3.64

On the reactants side, there are 6 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 2 D atoms. Writing an equation, 6A  4B  4C  2D Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the equation by two gives, 3A  2B  2C  D The correct answer is choice (d).

3.65

The mole ratio from the balanced equation is 2 mole CO2: 2 mole CO. 3.60 mol CO ´

3.66

2 mol CO2 = 3.60 mol CO2 2 mol CO

Si(s)  2Cl2(g)   SiCl4(l)

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CHAPTER 3: STOICHIOMETRY

Strategy: Looking at the balanced equation, how do we compare the amounts of Cl2 and SiCl4? We can compare them based on the mole ratio from the balanced equation. Solution: Because the balanced equation is given in the problem, the mole ratio between Cl2 and SiCl4 is known: 2 mole Cl2 1 mole SiCl4. From this relationship, we have two conversion factors. 2 mol Cl2 1 mol SiCl 4

1 mol SiCl4 2 mol Cl 2

and

Which conversion factor is needed to convert from moles of SiCl4 to moles of Cl2? The conversion factor on the left is the correct one. Moles of SiCl4 will cancel, leaving units of “mol Cl2” for the answer. We calculate moles of Cl2 reacted as follows: ? mol Cl 2 reacted = 0.507 mol SiCl4 ´

2 mol Cl2 = 1.01 mol Cl 2 1 mol SiCl4

Check: Does the answer seem reasonable? Should the moles of Cl2 reacted be double the moles of SiCl4 produced? 3.67

Starting with the amount of ammonia produced (6.0 mole), we can use the mole ratio from the balanced equation to calculate the mole of H2 and N2 that reacted to produce 6.0 mole of NH3. 3H2(g)  N2(g)  2NH3(g)

3.68

? mol H 2 = 6.0 mol NH3 ´

3 mol H 2 = 9.0 mol H 2 2 mol NH3

? mol N 2 = 6.0 mol NH3 ´

1 mol N 2 = 3.0 mol N 2 2 mol NH3

Starting with the 9.8 mole of CH3OH, we can use the mole ratio from the balanced equation to calculate the moles of H2O formed. 2CH3OH(l)  3O2(g)  2CO2(g)  4H2O(l) ? mol H 2O = 9.8 mol CH3OH ´

3.69

It is convenient to use the unit ton-mol in this problem. We normally use a g-mol. 1 g-mol SO2 has a mass of 64.07 g. In a similar manner, 1 ton-mol of SO2 has a mass of 64.07 tons. We need to complete the following conversions: tons SO2 ton-mol SO2 ton-mol S  ton S. (1.3 ´ 108 tons SO 2 ) ´

3.70

4 mol H 2O = 20 mol H 2 O = 2.0 ´ 101 mol H 2 O 2 mol CH3OH

1 ton-mol SO2 1 ton-mol S 32.07 ton S ´ ´ = 6.5 ´ 107 tons S 64.07 ton SO2 1 ton-mol SO2 1 ton-mol S

(a)

2NaHCO3   Na2CO3 H2O  CO2

(b)

Molar mass NaHCO3 22.99 g  1.008 g  12.01 g  3(16.00 g)  84.01 g Molar mass CO2 12.01 g  2(16.00 g)  44.01 g

CHAPTER 3: STOICHIOMETRY

The balanced equation shows one mole of CO2 formed from two molesof NaHCO3. mass NaHCO 3 = 20.5 g CO 2 ´

2 mol NaHCO3 84.01 g NaHCO3 1 mol CO 2 ´ ´ 44.01 g CO 2 1 mol CO 2 1 mol NaHCO3

78.3 g NaHCO3 3.71

The balanced equation shows a mole ratio of 1 mole NH3 1 mole NCl3. 2.94 g NH3 ´

3.72

1 mol NH3 1 mol NCl3 120.4 g NCl3 ´ ´ = 20.8 g NCl 3 17.03 g NH3 1 mol NH 3 1 mol NCl3

C6H12O6   2C2H5OH  2CO2 glucose ethanol Strategy: We compare glucose and ethanol based on the mole ratio in the balanced equation. Before we can determine moles of ethanol produced, we need to convert to moles of glucose. What conversion factor is needed to convert from grams of glucose to moles of glucose? Once moles of ethanol are obtained, another conversion factor is needed to convert from moles of ethanol to grams of ethanol. Solution: The molar mass of glucose will allow us to convert from grams of glucose to moles of glucose. The molar mass of glucose  6(12.01 g)  12(1.008 g)  6(16.00 g)  180.16 g. The balanced equation is given, so the mole ratio between glucose and ethanol is known; that is 1 mole glucose  2 mole ethanol. Finally, the molar mass of ethanol will convert moles of ethanol to grams of ethanol. This sequence of three conversions is summarized as follows: grams of glucose  moles of glucose  moles of ethanol  grams of ethanol ? g C2 H 5OH = 500.4 g C6 H12 O6 ´

1 mol C6 H12 O6 2 mol C2 H5 OH 46.07 g C2 H 5OH ´ ´ 180.16 g C6 H12O6 1 mol C 6 H12O6 1 mol C 2 H 5OH

255.9 g C2H5OH Check: Does the answer seem reasonable? Should the mass of ethanol produced be approximately half the mass of glucose reacted? Twice as many moles of ethanol are produced compared to the moles of glucose reacted, but the molar mass of ethanol is about one-fourth that of glucose.

CHAPTER 3: STOICHIOMETRY

The liters of ethanol can be calculated from the density and the mass of ethanol. volume =

mass density

Volume of ethanol obtained =

3.73

255.9 g = 324 mL = 0.324 L 0.789 g/mL

The mass of water lost is just the difference between the initial and the final masses. Mass H2O lost  15.01 g  9.60 g  5.41 g moles of H 2 O = 5.41 g H 2O ´

3.74

The balanced equation shows that eight mole of KCN are needed to combine with four mole of Au. ? mol KCN = 29.0 g Au ´

3.75

The balanced equation is: 1.0 kg CaCO3 ´

3.76

1 mol H 2 O = 0.300 mol H 2O 18.02 g H 2 O

1 mol Au 8 mol KCN ´ = 0.294 mol KCN 197.0 g Au 4 mol Au

CaCO3(s) ¾¾ ® CaO(s)  CO2(g)

1 mol CaCO3 1000 g 1 mol CaO 56.08 g CaO ´ ´ ´ = 5.6 ´ 102 g CaO 1 kg 100.09 g CaCO3 1 mol CaCO3 1 mol CaO

(a)

NH4NO3(s)   N2O(g)  2H2O(g)

(b)

Starting with moles of NH4NO3, we can use the mole ratio from the balanced equation to find moles of N2O. Once we have moles of N2O, we can use the molar mass of N2O to convert to grams of N2O. Combining the two conversions into one calculation, we have: mol NH4NO3 mol N2O  g N2O ? g N 2O = 0.46 mol NH 4 NO3 ´

3.77

1 mol N 2 O 44.02 g N 2O ´ = 2.0 ´ 101 g N 2 O 1 mol NH 4 NO3 1 mol N 2 O

(a)

2NH3(g) + H2SO4(aq) → (NH4)2SO4(aq)

(b)

Sulfuric acid is in excess. First, let us calculate the moles of ammonia reacted to produce 20.3 g of ammonium sulfate. 20.3 g (NH 4 )2 SO 4 ´

2 mol NH3 1 mol (NH 4 )2 SO4 ´ = 0.307 mol NH3 reacted 132.15 g (NH 4 )2 SO4 1 mol (NH 4 )2 SO4

The number of moles of sulfuric acid reacted will be half the moles of ammonia reacted (see mole ratio from the balance equation). The starting mass of NH3 is: 0.307 mol NH3 ´

17.03 g NH3 = 5.23 g NH 3 1 mol NH3

The starting mass of H2SO4 is the amount reacted plus the amount unreacted: 18 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 3: STOICHIOMETRY

0.154 mol H 2SO 4 ´

98.09 g H 2SO 4 = 15.1 g H 2SO 4 reacted 1 mol H 2SO4

15.1 g H2SO4reacted + 5.89 g H2SO4 unreacted = 21.0 g H2SO4 3.78

The balanced equation for the decomposition is: 2KClO3(s)   2KCl(s)  3O2(g) ? g O2 = 46.0 g KClO3 ´

1 mol KClO3 3 mol O 2 32.00 g O2 ´ ´ = 18.0 g O2 122.55 g KClO3 2 mol KClO3 1 mol O 2

3.79

The reactant used up first in a reaction is called the limiting reagent. Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent. The maximum amount of product formed depends on how much of the limiting reagent is present. When this reactant is used up, no more product can be formed. In a reaction with one reactant, the one reactant is by definition the limiting reagent.

3.80

If you are making sandwiches and have four pieces of bread, you can only make two sandwiches, no matter how much mayonnaise, mustard, sandwich meat, and so on that you have. The bread limits the number of sandwiches that you can make.

3.81

2A  B  C (a)

The number of B atoms shown in the diagram is 5. The balanced equation shows 2 mole A  1 mole B. Therefore, we need 10 atoms of A to react completely with 5 atoms of B. There are only 8 atoms of A present in the diagram. There are not enough atoms of A to react completely with B. A is the limiting reagent.

(b)

3.82

There are 8 atoms of A. Since the mole ratio between A and B is 2:1, 4 atoms of B will react with 8 atoms of A, leaving 1 atom of B in excess. The mole ratio between A and C is also 2:1. When 8 atoms of A react, 4 molecules of C will be produced.

B C

N2 3H2 2NH3 9 mole of H2 will react with 3 mole of N2, leaving 1 mole of H2 in excess. The mole ratio between N2 and NH3 is 1:2. When 3 mole of N2 react, 6 mole of NH3 will be produced.

3.83

This is a limiting reagent problem. Let us calculate the moles of NO2 produced assuming complete reaction for each reactant. 2NO(g)  O2(g)  2NO2(g)

CHAPTER 3: STOICHIOMETRY

0.886 mol NO ´ 0.503 mol O2 ´

2 mol NO2 = 0.886 mol NO 2 2 mol NO

2 mol NO 2 = 1.01 mol NO 2 1 mol O2

NO is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.886 mole NO2. 3.84

Strategy: Note that this reaction gives the amounts of both reactants, so it is likely to be a limiting reagent problem. The reactant that produces fewer moles of product is the limiting reagent because it limits the amount of product that can be produced. How do we convert from the amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, NO2, formed by the given amounts of O3 and NO to determine which reactant is the limiting reagent. Solution: We carry out two separate calculations. First, starting with 0.740 g O3, we calculate the number of moles of NO2 that could be produced if all the O3 reacted. We complete the following conversions. grams of O3 moles of O3 moles of NO2 Combining these two conversions into one calculation, we write ? mol NO 2 = 0.740 g O3 ´

1 mol O3 1 mol NO2 ´ = 0.0154 mol NO2 48.00 g O3 1 mol O3

Second, starting with 0.670 g of NO, we complete similar conversions. grams of NO  moles of NO  moles of NO2 Combining these two conversions into one calculation, we write ? mol NO2 = 0.670 g NO ´

1 mol NO2 1 mol NO ´ = 0.0223 mol NO 2 30.01 g NO 1 mol NO

The initial amount of O3 limits the amount of product that can be formed; therefore, it is the limiting reagent. The problem asks for grams of NO2 produced. We already know the moles of NO2 produced, 0.0154 mole. Use the molar mass of NO2 as a conversion factor to convert to grams (molar mass NO2 46.01 g). ? g NO 2 = 0.0154 mol NO 2 ´

46.01 g NO 2 = 0.709 g NO 2 1 mol NO 2

Check: Does your answer seem reasonable? 0.0154 mole of product is formed. What is the mass of 1 mole of NO2? Strategy: Working backwards, we can determine the amount of NO that reacted to produce 0.0154 mole of NO2. The amount of NO left over is the difference between the initial amount and the amount reacted. Solution: Starting with 0.0154 mole of NO2, we can determine the moles of NO that reacted using the mole ratio from the balanced equation. We can calculate the initial moles of NO starting with 0.670 g and using molar mass of NO as a conversion factor. mol NO reacted = 0.0154 mol NO2 ´

1 mol NO = 0.0154 mol NO 1 mol NO2

20 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 3: STOICHIOMETRY

mol NO initial = 0.670 g NO ´

1 mol NO = 0.0223 mol NO 30.01 g NO

mol NO remaining  mol NO initial  mol NO reacted. mol NO remaining 0.0223 mol NO  0.0154 mol NO 0.0069 mol NO 3.85

3.86

C3H8(g)  5O2(g) ¾¾ ® 3CO2(g)  4H2O(l)

(a)

The balanced equation is:

(b)

The balanced equation shows a mole ratio of 3 mole CO2:1 mole C3H8. The mass of CO2 produced is: 3 mol CO 2 44.01 g CO 2 3.65 mol C3H8 ´ ´ = 482 g CO 2 1 mol C3H8 1 mol CO2

This is a limiting reagent problem. Let us calculate the moles of Cl2 produced assuming complete reaction for each reactant. 0.86 mol MnO2 ´

48.2 g HCl ´

1 mol Cl 2 = 0.86 mol Cl 2 1 mol MnO 2

1 mol Cl2 1 mol HCl ´ = 0.330 mol Cl2 36.46 g HCl 4 mol HCl

HCl is the limiting reagent; it limits the amount of product produced. It will be used up first. The amount of product produced is 0.330 mole Cl2. Let us convert this to grams. ? g Cl 2 = 0.330 mol Cl 2 ´

70.90 g Cl2 = 23.4 g Cl 2 1 mol Cl 2

3.87

The theoretical yield of a reaction is the amount of product that would result if all the limiting reagents reacted. When the limiting reactant is used up, no more product can be formed.

3.88

There are many reasons why the actual yield is less than the theoretical yield. Some reactions are reversible, so they do not proceed 100% from reactants to product. There could be impurities in the starting materials. Sometimes it is difficult to recover all the products. There may be side reactions that lead to additional products.

3.89

The balanced equation is given:

CaF2 H2SO4 ¾¾ ® CaSO4 2HF

The balanced equation shows a mole ratio of 2 mole HF:1 mole CaF2. The theoretical yield of HF is: (6.00 ´ 103 g CaF2 ) ´

1 mol CaF2 2 mol HF 20.01 g HF 1 kg ´ ´ ´ = 3.08 kg HF 78.08 g CaF2 1 mol CaF2 1 mol HF 1000 g

The actual yield is given in the problem (2.86 kg HF). % yield =

actual yield ´ 100% theoretical yield

% yield =

2.86 kg ´ 100% = 92.9% 3.08 kg

CHAPTER 3: STOICHIOMETRY

3.90

(a)

Start with a balanced chemical equation. It is given in the problem. We use NG as an abbreviation for nitroglycerin. The molar mass of NG  227.1 g/mol. 4C3H5N3O9   6N2 12CO2 10H2O  O2

Map out the following strategy to solve this problem. g NG  mol NG  mol O2 g O2 Calculate the grams of O2 using the strategy above. ? g O 2 = 2.00 ´ 102 g NG ´

(b)

3.91

1 mol O2 32.00 g O2 1 mol NG ´ ´ = 7.05 g O2 227.1 g NG 4 mol NG 1 mol O 2

The theoretical yield was calculated in part (a), and the actual yield is given in the problem (6.55 g). The percent yield is: % yield =

actual yield ´ 100% theoretical yield

% yield =

6.55 g O 2 ´ 100% = 92.9% 7.05 g O 2

The balanced equation shows a mole ratio of 1 mole TiO2:1 mole FeTiO3. The molar mass of FeTiO3 is 151.73 g/mol, and the molar mass of TiO2 is 79.88 g/mol. The theoretical yield of TiO2 is: 8.00 ´ 106 g FeTiO3 ´

1 mol FeTiO3 1 mol TiO 2 79.88 g TiO2 1 kg ´ ´ ´ 151.73 g FeTiO3 1 mol FeTiO3 1 mol TiO 2 1000 g

4.21  103kg TiO2 The actual yield is given in the problem (3.67  103 kg TiO2). % yield =

3.92

actual yield 3.67 ´ 103 kg ´ 100% = ´ 100% = 87.2% theoretical yield 4.21 ´ 103 kg

This is a limiting reagent problem. Let us calculate the moles of Li 3N produced assuming complete reaction for each reactant. 6Li(s)  N2(g)  2Li3N(s) 12.3 g Li ´

2 mol Li3 N 1 mol Li ´ = 0.591 mol Li3 N 6.941 g Li 6 mol Li

33.6 g N 2 ´

2 mol Li3 N 1 mol N 2 ´ = 2.40 mol Li3 N 28.02 g N 2 1 mol N 2

Li is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.591 mole Li3N. Let us convert this to grams. ? g Li3 N = 0.591 mol Li3 N ´

34.83 g Li3 N = 20.6 g Li 3 N 1 mol Li3 N

This is the theoretical yield of Li3N. The actual yield is given in the problem (5.89 g). The percent yield is:

CHAPTER 3: STOICHIOMETRY

% yield =

3.93

actual yield 5.89 g ´ 100% = ´ 100% = 28.6% theoretical yield 20.6 g

Start by letting x be the fractional abundance of 69Ga. Since the sum of the two abundances must be 1, we can write: Abundance 71Ga  (1 x) Average atomic mass of Ga 

69.72 amu x(68.9256 amu)  (1 x)(70.9247 amu) 69.72 1.9991x 70.9247 x 0.603

x 0.603 corresponds to a natural abundance of 69Ga of 60.3%. The natural abundance of 71Ga is (1 x)  0.397 or 39.7%. 3.94

Start by letting x be the fractional abundance of 85Rb. Since the sum of the two abundances must be 1, we can write: Abundance 87Rb  (1 x) Average atomic mass of Rb 85.47 amu x(84.912 amu)  (1 x)(86.909 amu) 85.47 1.997x 86.909 1.997x 1.44 x 0.721 x 0.721 corresponds to a natural abundance of 85Rb of 72.1%. The natural abundance of 87Rb is (1 x)  0.279 or 27.9%.

3.95

All the carbon from the hydrocarbon reactant ends up in CO2, and all the hydrogen from the hydrocarbon reactant ends up in water. In the diagram, we find 4 CO2 molecules and 6 H2O molecules. This gives a ratio between carbon and hydrogen of 4:12. We write the formula C4H12, which reduces to the empirical formula CH3. The empirical molar mass equals approximately 15 g, which is half the molar mass of the hydrocarbon. Thus, the molecular formula is double the empirical formula or C2H6. Since this is a combustion reaction, the other reactant is O2. We write: C2H6 O2 CO2 H2O Balancing the equation, 2C2H6 7O2 4CO2 6H2O

3.96

2H2(g)  O2(g)  2H2O(g) We start with 8 molecules of H2 and 3 molecules of O2. The balanced equation shows 2 mole H2 1 mole O2. If 3 molecules of O2 react, 6 molecules of H2 will react, leaving 2 molecules of H2 in excess. The balanced equation also shows 1 mole O2 2 mole H2O. If 3 molecules of O2 react, 6 molecules of H2O will be produced. After complete reaction, there will be 2 molecules of H2 and 6 molecules of H2O. The correct diagram is choice (b).

3.97

First, let us convert to moles of HNO3 produced.

CHAPTER 3: STOICHIOMETRY

1.00 ton HNO3 ´

1 mol HNO3 2000 lb 453.6 g ´ ´ = 1.44 ´ 104 mol HNO3 1 ton 1 1b 63.02 g HNO3

Now, we will work in the reverse direction to calculate the amount of reactant needed to produce 1.44  103 mol of HNO3. Realize that since the problem says to assume an 80% yield for each step, the amount of 100% reactant needed in each step will be larger by a factor of , compared to a standard stoichiometry 80% calculation where a 100% yield is assumed. Referring to the balanced equation in the last step, we calculate the moles of NO2. (1.44 ´ 104 mol HNO3 ) ´

2 mol NO2 100% ´ = 3.60 ´ 10 4 mol NO 2 1 mol HNO3 80%

Now, let us calculate the amount of NO needed to produce 3.60  104 mol NO2. Following the same procedure as above and referring to the balanced equation in the middle step, we calculate the moles of NO. (3.60 ´ 104 mol NO 2 ) ´

1 mol NO 100% ´ = 4.50 ´ 104 mol NO 1 mol NO2 80%

Now, let us calculate the amount of NH3 needed to produce 4.5  104 mol NO. Referring to the balanced equation in the first step, the moles of NH3 is: (4.50 ´ 104 mol NO) ´

4 mol NH3 100% ´ = 5.63 ´ 10 4 mol NH3 4 mol NO 80%

Finally, converting to grams of NH3: 5.63 ´ 104 mol NH3 ´

3.98

17.03 g NH3 = 9.59 ´ 105 g NH 3 1 mol NH3

We assume that all the Cl in the compound ends up as HCl and all the O ends up as H2O. Therefore, we need to find the number of moles of Cl in HCl and the number of moles of O in H2O. mol Cl = 0.233 g HCl ´

1 mol HCl 1 mol Cl ´ = 0.00639 mol Cl 36.46 g HCl 1 mol HCl

mol O = 0.403 g H 2 O ´

1 mol H 2 O 1 mol O ´ = 0.0224 mol O 18.02 g H 2 O 1 mol H 2 O

Dividing by the smallest number of moles (0.00639 mol) gives the formula, ClO3.5. Multiplying both subscripts by two gives the empirical formula, Cl2O7. 3.99

(a) (b) (c) (d) (e)

C5H12(l)+ 8O2(g) → 5CO2(g) + 6H2O(l) NaHCO3(s)+ HCl(aq) → CO2(g) + NaCl(aq) + H2O(l) 6Li(s)+ N2(g) → 2Li3N(s) PCl3(l)+ 3H2O(l) → H3PO3(aq) + 3HCl(g) 3CuO(s)+ 2NH3(g) → 3Cu(s) + N2(g) + 3H2O(l)

3.100

This is a calculation involving percent composition. Remember,

CHAPTER 3: STOICHIOMETRY

percent by mass of each element =

mass of element in 1 mol of compound ´ 100% molar mass of compound

The molar masses are Al, 26.98 g/mol; Al2(SO4)3, 342.2 g/mol; H2O, 18.02 g/mol. Thus, using x as the number of H2O molecules, æ ö 2(molar mass of Al) mass % Al = ç ÷ ´ 100% è molar mass of Al 2 (SO4 )3 + x(molar mass of H 2 O) ø æ ö 2(26.98 g) 8.20% = ç ÷ ´ 100% 342.2 g + x (18.02 g) è ø

x 17.53 Rounding off to a whole number of water molecules, x 18. Therefore, the formula is Al2(SO4)318 H2O.

3.101

The amount of Fe that reacted is

1 ´ 664 g = 83.0 g reacted 8 .

The amount of Fe remaining is 664 g  83.0 g  581 g remaining. Thus, 83.0 g of Fe reacts to form the compound Fe2O3, which has two mole of Fe atoms per 1 mole of compound. The mass of Fe2O3 produced is: 83.0 g Fe ´

1 mol Fe 2O3 159.7 g Fe2 O3 1 mol Fe ´ ´ = 119 g Fe 2O3 55.85 g Fe 2 mol Fe 1 mol Fe 2O3

The final mass of the iron bar and rust is 581 g Fe  119 g Fe2O37.00 × 102 g 3.102

The mass of oxygen in MO is 39.46 g  31.70 g  7.76 g O. Therefore, for every 31.70 g of M, there is 7.76 g of O in the compound MO. The molecular formula shows a mole ratio of 1 mole M:1 mole O. First, calculate moles of M that react with 7.76 g O. mol M = 7.76 g O ´

molar mass M =

1 mol O 1 mol M ´ = 0.485 mol M 16.00 g O 1 mol O .

31.70 g M = 65.4 g/mol 0.485 mol M .

Thus, the atomic mass of M is 65.4 amu. The metal is most likely Zn. Zn(s)  H2SO4(aq) ¾¾ ® ZnSO4(aq)  H2(g)

3.103

(a)

(b)

We assume that a pure sample would produce the theoretical yield of H2. The balanced equation shows a mole ratio of 1 mole H2:1 mole Zn. The theoretical yield of H2 is: 3.86 g Zn ´

1 mol H 2 2.016 g H 2 1 mol Zn ´ ´ = 0.119 g H 2 65.39 g Zn 1 mol Zn 1 mol H 2

percent purity =

0.0764 g H 2 ´ 100% = 64.2% 0.119 g H 2

CHAPTER 3: STOICHIOMETRY

We assume that the impurities are inert and do not react with the sulfuric acid to produce hydrogen.

The wording of the problem suggests that the actual yield is less than the theoretical yield. The percent yield will be equal to the percent purity of the iron(III) oxide. We find the theoretical yield: (2.62 ´ 103 kg Fe 2O3 ) ´

1000 g Fe 2 O3 1 mol Fe2 O3 2 mol Fe 55.85 g Fe 1 kg Fe ´ ´ ´ ´ 1 kg Fe 2 O3 159.7 g Fe 2 O3 1 mol Fe 2O3 1 mol Fe 1000 g Fe

 1.83  103 kg Fe percent yield =

percent yield =

3.105

actual yield ´ 100% theoretical yield 1.64 ´ 103 kg Fe 1.83 ´ 103 kg Fe

´ 100% = 89.6% = purity of Fe 2O 3

The balanced equation is C6H12O6 6O2 ¾¾ ® 6CO2 6H2O 6 mol CO 2 44.01 g CO 2 365 days 5.0 ´ 102 g glucose 1 mol glucose ´ ´ ´ ´ ´ (7.0 ´ 109 people) 1 day 180.16 g glucose 1 mol glucose 1 mol CO 2 1 yr

1.9  1015 g CO2/yr 3.106

The carbohydrate contains 40 percent carbon; therefore, the remaining 60 percent is hydrogen and oxygen. The problem states that the hydrogen to oxygen ratio is 2:1. We can write this 2:1 ratio as H2O. Assume 100 g of compound. 40.0 g C ´

1 mol C = 3.33 mol C 12.01 g C

60.0 g H 2 O ´

1 mol H 2 O = 3.33 mol H 2 O 18.02 g H 2 O

Dividing by 3.33 gives CH2O for the empirical formula. To find the molecular formula, divide the molar mass by the empirical mass. molar mass 178 g = » 6 empirical mass 30.03 g

Thus, there are six CH2O units in each molecule of the compound, so the molecular formula is (CH2O)6 or C6H12O6. 3.107

The mass of the metal (X) in the metal oxide is 1.68 g. The mass of oxygen in the metal oxide is 2.40 g  1.68 g  0.72 g oxygen. Next, find the number of moles of the metal and of the oxygen. moles X = 1.68 g ´

1 mol X = 0.0301 mol X 55.9 g X

moles O = 0.72 g ´

1 mol O = 0.045 mol O 16.00 g O

This gives the formula X0.0301O0.045. Dividing by the smallest number of moles (0.0301 mole) gives the formula X1.00O1.5. Multiplying by two gives the empirical formula, X2O3. 26 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 3: STOICHIOMETRY

The balanced equation is X2O3(s)  3CO(g) ¾¾ ® 2X(s)  3CO2(g) 3.108

Both compounds contain only Mn and O. When the first compound is heated, oxygen gas is evolved. Let us calculate the empirical formulas for the two compounds, then we can write a balanced equation. (a) Compound X: Assume 100 g of compound. 63.3 g Mn ´ 36.7 g O ´

1 mol Mn = 1.15 mol Mn 54.94 g Mn

1 mol O = 2.29 mol O 16.00 g O

Dividing by the smallest number of moles (1.15 mole) gives the empirical formula, MnO2. Compound Y: Assume 100 g of compound. 72.0 g Mn ´

28.0 g O ´

1 mol Mn = 1.31 mol Mn 54.94 g Mn

1 mol O = 1.75 mol O 16.00 g O

Dividing by the smallest number of moles gives MnO1.33. Recall that an empirical formula must have whole number coefficients. Multiplying by a factor of 3 gives the empirical formula Mn3O4. (b) The unbalanced equation is MnO2   Mn3O4 O2 Balancing by inspection gives 3MnO2   Mn3O4 O2 3.109

We carry an additional significant figure throughout this calculation to avoid rounding errors. Assume 100 g of sample. Then, mol Na = 32.08 g Na ´

mol O = 36.01 g O ´

1 mol Na = 1.3954 mol Na 22.99 g Na

1 mol O = 2.2506 mol O 16.00 g O

mol Cl = 19.51 g Cl ´

1 mol Cl = 0.55035 mol Cl 35.45 g Cl

Since Cl is only contained in NaCl, the moles of Cl equal the moles of Na contained in NaCl. mol Na(in NaCl)0.55035 mol The number of moles of Na in the remaining two compounds is 1.3954 mol  0.55035 mol  0.84505 mol Na. To solve for moles of the remaining two compounds, let x  moles of Na2SO4 y  moles of NaNO3 Then, from the mole ratio of Na and O in each compound, we can write 2xy  mol Na  0.84505 mol 4x 3y mol O  2.2506 mol 27 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 3: STOICHIOMETRY

Solving two equations with two unknowns gives x  0.14228  mol Na2SO4 and y 0.56050  mol NaNO3 Finally, we convert to mass of each compound to calculate the mass percent of each compound in the sample. Remember, the sample size is 100 g. mass % NaCl = 0.55035 mol NaCl ´

58.44 g NaCl 1 ´ ´ 100% = 32.16% NaCl 1 mol NaCl 100 g sample

mass % Na 2SO 4 = 0.14228 mol Na 2SO 4 ´

mass % NaNO3 = 0.56050 mol NaNO3 ´

3.110

142.05 g Na 2SO 4 1 ´ ´ 100% = 20.21% Na2 SO 4 1 mol Na 2SO 4 100 g sample

85.00 g NaNO3 1 ´ ´ 100% = 47.64% NaNO3 1 mol NaNO3 100 g sample

When magnesium burns in air, magnesium oxide (MgO) and magnesium nitride (Mg3N2) are produced. Magnesium nitride reacts with water to produce ammonia gas. Mg3N2(s)  6H2O(l)  3Mg(OH)2(s)  2NH3(g) From the amount of ammonia produced, we can calculate the mass of Mg3N2 produced. The mass of Mg in that amount of Mg3N2 can be determined and then the mass of Mg in MgO can be determined by difference. Finally, the mass of MgO can be calculated. 2.813 g NH3 ´

1 mol NH3 1 mol Mg 3 N 2 100.95 g Mg 3 N 2 ´ ´ = 8.335 g Mg 3 N 2 17.034 g NH3 2 mol NH3 1 mol Mg 3 N 2

The mass of Mg in 8.335 g Mg3N2 can be determined from the mass percentage of Mg in Mg3N2. (3)(24.31 g Mg) ´ 8.335 g Mg 3 N 2 = 6.022 g Mg 100.95 g Mg3 N 2

The mass of Mg in the product MgO is obtained by difference 21.496 g Mg  6.022 g Mg  15.474 g Mg. The mass of MgO produced can now be determined from this mass of Mg and the mass percentage of Mg in MgO. 40.31 g MgO ´ 15.474 g Mg = 25.66 g MgO 24.31 g Mg 3.111

The balanced equations are CH4 2O2 ¾¾ ® CO2 2H2O

2C2H6 7O2 ¾¾ ® 4CO2 6H2O

If we let x mass of CH4, then the mass of C2H6 is (13.43 x) g. Next, we need to calculate the mass of CO2 and the mass of H2O produced by both CH4 and C2H6. The sum of the masses of CO2 and H2O will add up to 64.84 g. ? g CO 2 (from CH 4 ) = x g CH 4 ´

1 mol CH 4 1 mol CO 2 44.01 g CO2 ´ ´ = 2.744 x g CO2 16.04 g CH 4 1 mol CH 4 1 mol CO 2

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CHAPTER 3: STOICHIOMETRY

? g H 2 O (from CH4 ) = x g CH 4 ´

1 mol CH 4 2 mol H 2O 18.02 g H 2O ´ ´ = 2.247 x g H 2 O 16.04 g CH 4 1 mol CH 4 1 mol H 2O

? g CO 2 (from C 2 H6 ) = (13.43 - x) g C2 H 6 ´

1 mol C 2 H 6 4 mol CO 2 44.01 g CO 2 ´ ´ 30.07 g C 2 H 6 2 mol C 2 H 6 1 mol CO 2

 2.927(13.43 x) g CO2 ? g H 2O (from C2 H6 ) = (13.43 - x) g C2 H6 ´

1 mol C2 H 6 6 mol H2 O 18.02 g H 2O ´ ´ 30.07 g C2 H 6 2 mol C2 H 6 1 mol H 2 O

 1.798(13.43 x) g H2O Summing the masses of CO2 and H2O: 2.744x g  2.247x g  2.927(13.43 x) g  1.798(13.43 x) g  64.84 g 0.266x 1.383 x 5.20 g The fraction of CH4 in the mixture is

5.20 g = 0.387 13.43 g .

3.112

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

3.113

The molecular formula of isoflurane is C3H2ClF5O. The mass percentage of each element is: %C =

(3)(12.01 g) ´ 100% = 19.53% 184.50 g

%H =

(2)(1.008 g) ´ 100% = 1.093% 184.50 g

%Cl =

35.45 g ´ 100% = 19.21% 184.50 g

%F =

(5)(19.00) g ´ 100% = 51.49% 184.50 g

%O =

16.00 g ´ 100% = 8.672% 184.50 g

Check: 19.53%  1.093%  19.21%  51.49%  8.672%  100.00%. 3.114

For the first step of the synthesis, the yield is 80% or 0.8. For the second step, the yield will be 80% of 0.8 or (0.8  0.8) = 0.64. For the third step, the yield will be 80% of 0.64 or (0.8  0.8  0.8) = 0.512. We see that the percent yield will be: Percent yield = (0.8)n where n = number of steps in the reaction. For 24 steps, Percent yield = (0.8)24 = 0.0047 = 0.47%.

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CHAPTER 3: STOICHIOMETRY

3.115

The mass of water lost upon heating the mixture is (5.020 g  2.988 g)  2.032 g water. Next, if we let x mass of CuSO45H2O, then the mass of MgSO47H2O is (5.020 x)g. We can calculate the amount of water lost by each salt based on the mass % of water in each hydrate. We can write: (mass CuSO45H2O)(% H2O)  (mass MgSO47H2O)(% H2O)  total mass H2O  2.032 g H2O Calculate the % H2O in each hydrate. % H 2 O (CuSO 4 × 5H 2 O) =

(5)(18.02 g) ´ 100% = 36.08% H 2 O 249.7 g

% H 2 O (MgSO4 × 7H 2 O) =

(7)(18.02 g) ´ 100% = 51.17% H 2 O 246.5 g

Substituting into the equation above gives: (x)(0.3608)  (5.020 x)(0.5117)  2.032 g 0.1509x 0.5367 x 3.557 g  mass of CuSO45H2O Finally, the percent by mass of CuSO45H2O in the mixture is: 3.557 g ´ 100% = 70.86% 5.020 g

3.116

The decomposition of KClO3 produces oxygen gas (O2) that reacts with Fe to produce Fe2O3. 4Fe  3O2 2Fe2O3 When the 15.0 g of Fe is heated in the presence of O2 gas, any increase in mass is due to oxygen. The mass of oxygen reacted is: 17.9 g  15.0 g  2.9 g O2 From this mass of O2, we can now calculate the mass of Fe2O3 produced and the mass of KClO3 decomposed. 2.9 g O2 ´

2 mol Fe2 O3 159.7 g Fe2 O3 1 mol O2 ´ ´ = 9.6 g Fe2O 3 32.00 g O2 3 mol O2 1 mol Fe2 O3

The balanced equation for the decomposition of KClO3 is 2KClO3 2KCl  3O2. The mass of KClO3 decomposed is: 2 mol KClO3 122.55 g KClO3 1 mol O 2 2.9 g O 2 ´ ´ ´ = 7.4 g KClO 3 32.00 g O2 3 mol O 2 1 mol KClO3 3.117

(a)

We need to compare the mass % of K in both KCl and K2SO4. %K in KCl =

39.10 g ´ 100% = 52.45% K 74.55 g

%K in K 2SO4 =

2(39.10 g) ´ 100% = 44.87% K 174.27 g

The price is dependent on the %K. 30 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 3: STOICHIOMETRY

Price of K 2SO4 %K in K 2SO 4 = Price of KCl %K in KCl Price of K 2SO 4 = Price of KCl × Price of K 2SO4 =

(b)

%K in K 2SO4 %K in KCl

$0.055 44.87% ´ = $0.047 / kg kg 52.45%

First, calculate the number of moles of K in 1.00 kg of KCl. (1.00 ´ 103 g KCl) ´

1 mol KCl 1 mol K ´ = 13.4 mol K 74.55 g KCl 1 mol KCl

Next, calculate the amount of K2O needed to supply 13.4 mol K. 13.4 mol K ´

3.118

1 mol K 2 O 94.20 g K 2 O 1 kg ´ ´ = 0.631 kg K 2O 2 mol K 1 mol K 2 O 1000 g

Possible formulas for the metal bromide could be MBr, MBr 2, MBr3, and so on. Assuming 100 g of compound, the moles of Br in the compound can be determined. From the mass and moles of the metal for each possible formula, we can calculate a molar mass for the metal. The molar mass that matches a metal on the periodic table would indicate the correct formula. Assuming 100 g of compound, we have 53.79 g Br and 46.21 g of the metal (M). The moles of Br in the compound are: 1 mol Br 53.79 g Br ´ = 0.67322 mol Br 79.90 g Br If the formula is MBr, the mole of M is also 0.67322 mol, so the molar mass of the metal can be calculated as follows: 46.21 g M MBr: = 68.64 g/mol (no such metal) 0.67322 mol M If the formula is MBr2, the mole of M is 0.67322/2  0.33661 mol, and the molar mass of M is: MBr2 :

46.21 g M = 137.3 g/mol (The metal is Ba. The formula is BaBr2 ) 0.33661 mol M

If the formula is MBr3, the mole of M is 0.67322/3  0.22441 mol, and the molar mass of M is: MBr3:

46.21 g M = 205.9 g/mol (no such metal) 0.22441 mol M

We can stop here because MBr4 and higher would give unreasonably large molar masses for M.

3.119

The balanced equations for the combustion of octane are: 2C8H18 25O2 ¾¾ ® 16CO2 18H2O 2C8H18 17O2 ¾¾ ® 16CO  18H2O

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CHAPTER 3: STOICHIOMETRY

The quantity of octane burned is 2650 g (1 gallon with a density of 2.650 kg/gallon). Let x be the mass of octane converted to CO2; therefore, (2650 x) g is the mass of octane converted to CO. The amounts of CO2 and H2O produced by x g of octane are: x g C8 H18 ´

1 mol C8 H18 16 mol CO 2 44.01 g CO 2 ´ ´ = 3.083 x g CO2 114.2 g C8 H18 2 mol C8 H18 1 mol CO 2

x g C8 H18 ´

1 mol C8 H18 18 mol H 2 O 18.02 g H 2 O ´ ´ = 1.420 x g H 2 O 114.2 g C8 H18 2 mol C8H18 1 mol H 2O

The amounts of CO and H2O produced by (2650 x) g of octane are: (2650 - x) g C8 H18 ´

1 mol C8 H18 16 mol CO 28.01 g CO ´ ´ = (5200 - 1.962 x ) g CO 114.2 g C8 H18 2 mol C8 H18 1 mol CO

(2650 - x ) g C8 H18 ´

1 mol C8 H18 18 mol H 2 O 18.02 g H 2 O ´ ´ = (3763 - 1.420 x ) g H 2 O 114.2 g C8 H18 2 mol C8 H18 1 mol H 2 O

The total mass of CO2 CO  H2O produced is 11530 g. We can write: 11530 g  3.083x 1.420x 5200  1.962x 3763  1.420x x 2290 g Since x is the amount of octane converted to CO2, we can now calculate the efficiency of the process. efficiency =

3.120

g octane converted 2290 g ´ 100% = ´ 100% = 86.49% g octane total 2650 g

The surface area of the water can be calculated assuming that the dish is circular. surface area of water = r2 = (10 cm)2 = 3.1 × 102 cm2 The cross-sectional area of one stearic acid molecule in cm2 is: 2

æ 1´ 10-9 m ö æ 1 cm ö2 0.21 nm 2 ´ ç = 2.1´ 10-15 cm2 /molecule ÷ ´ ç 1 nm ÷ çè 0.01 m ÷ø è ø

Assuming that there is no empty space between molecules, we can calculate the number of stearic acid molecules that will fit in an area of 3.1 × 102 cm2.

3.1´ 102 cm 2 ´

1 molecule 2.1´10

-15

= 1.5 ´ 1017 molecules

Next, we can calculate the moles of stearic acid in the 1.4 × 10−4 g sample. Then, we can calculate Avogadro’s number (the number of molecules per mole). 1.4 ´ 10-4 g stearic acid ´

1 mol steric acid = 4.9 ´ 10-7 mol stearic acid 284.5 g stearic

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CHAPTER 3: STOICHIOMETRY

Avogadro's number ( N A ) =

1.5 ´1017 molecules -7

4.9 ´ 10 mol

= 3.1×1023 molecules / mol

The value obtained is only a factor of two lower than the accepted value for Avogadro’s number, a remarkable agreement given the simplicity of the approach.

ANSWERS TO REVIEW OF CONCEPTS Section 3.1 (p. 62) Section 3.2 (p. 66) Section 3.3 (p. 68) Section 3.4 (p. 69)

Section 3.5 (p. 72) Section 3.6 (p. 75) Section 3.7 (p. 79) Section 3.8 (p. 83) Section 3.9 (p. 86) Section 3.10 (p. 87)

193

Ir.The average atomic mass of Ir (192.21 amu) is closer to 192.96 amu than 190.96 amu, so there are more 193Irisotopes than 191Ir. (b) Molecular mass = 192.12 amu, molar mass = 192.12 g When isotopes of the two chlorine ions arrive at the detector of a mass spectrometer, a current is registered for each type of ion. The amount of current generated is directly proportional to the number of ions, so it enables us to determine the relative abundance of each isotope. A weighted average of the masses of the two isotopes based on relative abundance gives the average mass of chlorine. The percent composition by mass of Hg is greater than that of O. You need only to compare the relative masses of one Hg atom and six O atoms. C5H10. Essential part: The correct formulas and the number of formula units in the equation. Helpful part: The physical states of the reactants and products. (a) The equation is 2NO(g)  O2(g)  2NO2(g). Diagram (d) shows that NO is the limiting reagent. No. A percent yield greater than the theoretical yield would violate the law of conservation of mass; however, the calculated theoretical yield can appear to be greater than the theoretical yield due to impurities in the product.

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CHAPTER 4 REACTIONS IN AQUEOUS SOLUTIONS 4.1

In dissolving a solid in a liquid, the solid is typically the solute, the substance present in a smaller amount, and the liquid is the solvent, the substance present in a larger amount. The two substances together form a solution, a homogeneous mixture of two or more substances.

4.2

A nonelectrolyte is a solution that does not conduct electricity. A strong electrolyte solution contains a large number of ions and therefore the solution conducts electricity. A weak electrolyte solution contains a small number of ions, so the solution conducts electricity but not as strongly as a strong electrolyte.

4.3

The process in which an ion is surrounded by water molecules arranged in a specific manner is called hydration. Water has a positive end (the H atoms) and a negative end (the O atom), or positive and negative “poles,” which can interact with ions in solution.

4.4

The symbol “” means that the reactions proceeds to completion. The symbol “” means that the reaction is reversible. The system eventually reaches equilibrium where the rate at which reactants go to products is equal to the rate at which products return to reactants. In some reactions, the equilibrium position favors reactants, in others, the equilibrium position favors products.

4.5

There are dissolved ions in the water; therefore, the solution conducts electricity.

4.6

Li+(aq), F(aq), and H2O(l)

4.7

(a) is a strong electrolyte. The compound dissociates completely into ions in solution. (b) is a nonelectrolyte. The compound dissolves in water, but the molecules remain intact. (c) is a weak electrolyte. A small amount of the compound dissociates into ions in water.

4.8

When NaCl dissolves in water it dissociates into Na and Cl ions. When the ions are hydrated, the water molecules will be oriented so that the negative end of the water dipole interacts with the positive sodium ion and the positive end of the water dipole interacts with the negative chloride ion. The negative end of the water dipole is near the oxygen atom and the positive end of the water dipole is near the hydrogen atoms. The diagram that best represents the hydration of NaCl when dissolved in water is choice (c).

4.9

Ionic compounds, strong acids, and strong bases (metal hydroxides) are strong electrolytes. Weak acids and weak bases are weak electrolytes. Molecular substances other than acids or bases are nonelectrolytes. (a) very weak electrolyte

(b)

strong electrolyte (ionic compound)

(d)

weak electrolyte (weak acid)

(e) nonelectrolyte (molecular compound—neither acid nor base) 4.10

4.11

(b)

nonelectrolyte

(c) weak electrolyte (weak base) (d) strong electrolyte (strong base) Since solutions must be electrically neutral, any flow of positive species (cations) must be balanced by the flow of negative species (anions). Therefore, the correct answer is (d).

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

4.12

(a) Solid NaCl does not conduct. The ions are locked in a rigid lattice structure. (b) Molten NaCl conducts. The ions can move around in the liquid state. (c) Aqueous NaCl conducts. NaCl dissociates completely to Na(aq) and Cl(aq) in water.

4.13

Measure the conductance to see if the solution carries an electrical current. If the solution is conducting, then you can determine whether the solution is a strong or weak electrolyte by comparing its conductance with that of a known strong electrolyte.

4.14

Since HCl dissolved in water conducts electricity, then HCl(aq) must actually exists as H(aq) cations and Cl(aq) anions. Since HCl dissolved in benzene solvent does not conduct electricity, then we must assume that the HCl molecules in benzene solvent do not ionize but rather exist as un-ionized molecules.

4.15

In a molecular equation, the formulas of the compounds are written as though all species existed as molecules or whole units; whereas, an ionic equation shows dissolved species as free ions.

4.16

The advantage of writing a net ionic equation is that the equation only shows the species that actually take part in the reaction.

4.17

Refer to Table 4.2 of the text to solve this problem. AgCl is insoluble in water. It will precipitate from solution. NaNO3 is soluble in water and will remain as Na and NO3 ions in solution. Diagram (c) best represents the mixture.

4.18

Refer to Table 4.2 of the text to solve this problem. Mg(OH)2 is insoluble in water. It will precipitate from solution. KCl is soluble in water and will remain as K and Cl ions in solution. Diagram (b) best represents the mixture.

4.19

Refer to Table 4.2 of the text to solve this problem. (a) Ca3(PO4)2 is insoluble. (b) Mn(OH)2 is insoluble. (c) AgClO3 is soluble. (d) K2S is soluble.

4.20

Strategy: Although it is not necessary to memorize the solubilities of compounds, you should keep in mind the following useful rules: all ionic compounds containing alkali metal cations, the ammonium ion, and the nitrate, bicarbonate, and chlorate ions are soluble. For other compounds, refer to Table 4.2 of the text. Solution: (a) CaCO3 is insoluble. Most carbonate compounds are insoluble. (b) ZnSO4 is soluble. Most sulfate compounds are soluble. (c) Hg(NO3)2 is soluble. All nitrate compounds are soluble. (d) HgSO4 is insoluble. Most sulfate compounds are soluble but those containing Ag, Ca2, Ba2, Hg2, and Pb2 are insoluble. (e) NH4ClO4 is soluble. All ammonium compounds are soluble.

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

4.21

(a)

  Ionic: 2Ag(aq)  2NO3(aq)  2Na(aq)  SO42(aq) ¾¾ ® Ag2SO4(s)  2Na (aq)  2NO3 (aq)

Net ionic: 2Ag(aq)  SO42(aq) ¾¾ ® Ag2SO4(s) (b)

 2 Ionic: Ba2(aq)  2Cl(aq)  Zn2(aq)  SO42(aq) ¾¾ ® BaSO4(s)  Zn (aq)  2Cl (aq)

Net ionic: Ba2(aq)  SO42(aq) ¾¾ ® BaSO4(s) (c)

  Ionic: 2NH4(aq)  CO32(aq)  Ca2(aq)  2Cl(aq) ¾¾ ® CaCO3(s)  2NH4 (aq)  2Cl (aq)

Net ionic: Ca2(aq)  CO32(aq) ¾¾ ® CaCO3(s) 4.22

(a) Strategy: Recall that an ionic equation shows dissolved ionic compounds in terms of their free ions. A net ionic equation shows only the species that actually take part in the reaction. What happens when ionic compounds dissolve in water? What ions are formed from the dissociation of Na2S and ZnCl2? What happens when the cations encounter the anions in solution? Solution: In solution, Na2S dissociates into Na and S2 ions and ZnCl2 dissociates into Zn2 and Cl ions. According to Table 4.2 of the text, zinc ions (Zn2) and sulfide ions (S2) will form an insoluble compound, zinc sulfide (ZnS), while the other product, NaCl, is soluble and remains in solution. This is a precipitation reaction. The balanced molecular equation is: Na2S(aq)  ZnCl2(aq) ¾¾ ® ZnS(s)  2NaCl(aq) The ionic and net ionic equations are:   Ionic: 2Na(aq)  S2(aq)  Zn2(aq)  2Cl(aq) ¾¾ ® ZnS(s)  2Na (aq)  2Cl (aq)

Net ionic: Zn2(aq)  S2(aq) ¾¾ ® ZnS(s) Check: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to the number of atoms on each side, and the number of positive and negative charges on the left-hand side of the equation is the same. (b) Strategy: What happens when ionic compounds dissolve in water? What ions are formed from the dissociation of K3PO4 and Sr(NO3)2? What happens when the cations encounter the anions in solution? Solution: In solution, K3PO4 dissociates into K and PO43 ions and Sr(NO3)2 dissociates into Sr2 and NO3 ions. According to Table 4.2 of the text, strontium ions (Sr2) and phosphate ions (PO43) will form an insoluble compound, strontium phosphate [Sr3(PO4)2], while the other product, KNO3, is soluble and remains in solution. This is a precipitation reaction. The balanced molecular equation is: 2K3PO4(aq)  3Sr(NO3)2(aq) ¾¾ ® Sr3(PO4)2(s)  6KNO3(aq) The ionic and net ionic equations are:   Ionic: 6K(aq)  2PO43(aq)  3Sr2(aq)  6NO3(aq) ¾¾ ® Sr3(PO4)2(s)  6K (aq)  6NO3 (aq)

Net ionic: 3Sr2(aq)  2PO43(aq) ¾¾ ® Sr3(PO4)2(s) Check: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to the number of atoms on each side, and the number of positive and negative charges on the left-hand side of the equation is the same. 3 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

(c) Strategy: What happens when ionic compounds dissolve in water? What ions are formed from the dissociation of Mg(NO3)2 and NaOH? What happens when the cations encounter the anions in solution? Solution: In solution, Mg(NO3)2 dissociates into Mg2 and NO3 ions and NaOH dissociates into Na and OH ions. According to Table 4.2 of the text, magnesium ions (Mg2) and hydroxide ions (OH) will form an insoluble compound, magnesium hydroxide [Mg(OH)2], while the other product, NaNO3, is soluble and remains in solution. This is a precipitation reaction. The balanced molecular equation is: Mg(NO3)2(aq)  2NaOH(aq) ¾¾ ® Mg(OH)2(s)  2NaNO3(aq) The ionic and net ionic equations are:   Ionic: Mg2(aq)  2NO3(aq)  2Na(aq)  2OH(aq) ¾¾ ® Mg(OH)2(s)  2Na (aq)  2NO3 (aq)

Net ionic: Mg2(aq)  2OH(aq) ¾¾ ® Mg(OH)2(s) Check: Note that because we balanced the molecular equation first, the net ionic equation is balanced as to the number of atoms on each side, and the number of positive and negative charges on the left-hand side of the equation is the same. 4.23

(a)

Both reactants are soluble ionic compounds. The other possible ion combinations, Na 2SO4 and Cu(NO3)2, are also soluble.

(b)

Both reactants are soluble. Of the other two possible ion combinations, KCl is soluble, but BaSO 4 is insoluble and will precipitate. Ba2(aq)  SO42(aq)  BaSO4(s)

4.24

(a)

Add chloride ions. KCl is soluble but AgCl is not.

(b)

Add hydroxide ions. Ba(OH)2 is soluble, but Pb(OH)2 is insoluble.

(c)

Add carbonate ions. (NH4)2CO3 is soluble, but CaCO3 is insoluble.

(d)

Add sulfate ions. CuSO4 is soluble, but BaSO4 is insoluble.

4.25

Please see Section 4.3 of the text.

4.26

Arrhenius acids produce H+ in water. Arrhenius bases produce OH in water. A Brønsted acid is a proton (H+) donor. A Brønsted base is a proton (H+) acceptor. The Brønsted definitions do not require acids and bases to be in aqueous solution.

4.27

Hydrochloric acid (HCl) is a monoprotic acid, sulfuric acid (H2SO4) is a diprotic acid, and phosphoric acid (H3PO4) is a triprotic acid.

4.28

In general, an acid–base neutralization reaction produces salt and water. At the neutralization point, the correct amount of base has been added to completely react with the acid. The resulting solution only contains salt and water. A salt is an ionic compound made up of a cation (other than H+) and an anion (other than OH and O2). NaF, BaSO4, and KBr are salts.

4.29

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

4.30 (e)

(a) weak base (b) weak acid strong acid (first stage of ionization)

4.31

(a)

HI dissolves in water to produce H and I, so HI is a Brønsted acid.

(b)

CH3COO can accept a proton to become acetic acid CH3COOH, so it is a Brønsted base.

(c)

H2PO4 can either accept a proton, H, to become H3PO4 and thus behaves as a Brønsted base, or can donate a proton in water to yield H and HPO42, thus behaving as a Brønsted acid.

(d)

HSO4 can either accept a proton, H, to become H2SO4 and thus behaves as a Brønsted base, or can donate a proton in water to yield H and SO42, thus behaving as a Brønsted acid.

4.32

(d) weak acid (g) strong base

Strategy: What are the characteristics of a Brønsted acid? Does it contain at least an H atom? With the exception of ammonia, most Brønsted bases that you will encounter at this stage are anions. Solution: (a) PO43 in water can accept a proton to become HPO42 and is thus a Brønsted base. (b)

ClO2 in water can accept a proton to become HClO2 and is thus a Brønsted base.

(c)

NH4 dissolved in water can donate a proton H thus behaving as a Brønsted acid.

(d)

HCO3 can either accept a proton to become H2CO3, thus behaving as a Brønsted base. Or, HCO3 can donate a proton to yield H and CO32, thus behaving as a Brønsted acid.

Comment: The HCO3 species is said to be amphoteric because it possesses both acidic and basic properties. 4.33

Recall that strong acids and strong bases are strong electrolytes. They are completely ionized in solution. An ionic equation will show strong acids and strong bases in terms of their free ions. A net ionic equation shows only the species that actually take part in the reaction. (a)

  Ionic: H(aq)  Br(aq)  NH3(aq) ¾¾ ® NH4 (aq)  Br (aq)  Net ionic: H(aq)  NH3(aq) ¾¾ ® NH4 (aq)

(b)

Ionic: 3Ba2(aq)  6OH(aq)  2H3PO4(aq) ¾¾ ® Ba3(PO4)2(s)  6H2O(l) Net ionic: 3Ba2(aq)  6OH(aq)  2H3PO4(aq) ¾¾ ® Ba3(PO4)2(s)  6H2O(l)

(c)

 2 Ionic: 2H(aq)  2ClO4(aq)  Mg(OH)2(s) ¾¾ ® Mg (aq)  2ClO4 (aq)  2H2O(l) 2+ Net ionic: 2H(aq)  Mg(OH)2(s) ¾¾ ® Mg (aq) + 2H2O(l)

4.34

Strategy: Recall that strong acids and strong bases are strong electrolytes. They are completely ionized in solution. An ionic equation will show strong acids and strong bases in terms of their free ions. Weak acids and weak bases are weak electrolytes. They only ionize to a small extent in solution. Weak acids and weak bases are shown as molecules in ionic and net ionic equations. A net ionic equation shows only the species that actually take part in the reaction. (a)

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

Solution: CH3COOH is a weak acid. It will be shown as a molecule in the ionic equation. KOH is a strong base. It completely ionizes to K and OH ions. Since CH3COOH is an acid, it donates an H to the base, OH, producing water. The other product is the salt, CH3COOK, which is soluble and remains in solution. The balanced molecular equation is: CH3COOH(aq)  KOH(aq) ¾¾ ® CH3COOK(aq)  H2O(l) The ionic and net ionic equations are:   Ionic: CH3COOH(aq)  K(aq)  OH(aq) ¾¾ ® CH3COO (aq)  K (aq)  H2O(l)  Net ionic: CH3COOH(aq)  OH(aq) ¾¾ ® CH3COO (aq)  H2O(l)

(b) Solution: H2CO3 is a weak acid. It will be shown as a molecule in the ionic equation. NaOH is a strong base. It completely ionizes to Na and OH ions. Since H2CO3 is an acid, it donates an H to the base, OH, producing water. The other product is the salt, Na2CO3, which is soluble and remains in solution. The balanced molecular equation is: H2CO3(aq)  2NaOH(aq) ¾¾ ® Na2CO3(aq)  2H2O(l) The ionic and net ionic equations are:  2 Ionic: H2CO3(aq)  2Na(aq)  2OH(aq) ¾¾ ® 2Na (aq)  CO3 (aq)  2H2O(l) 2 Net ionic: H2CO3(aq)  2OH(aq) ¾¾ ® CO3 (aq)  2H2O(l)

(c) Solution: HNO3 is a strong acid. It completely ionizes to H and NO3 ions. Ba(OH)2 is a strong base. It completely ionizes to Ba2 and OH ions. Since HNO3 is an acid, it donates an H to the base, OH, producing water. The other product is the salt, Ba(NO3)2, which is soluble and remains in solution. The balanced molecular equation is: 2HNO3(aq)  Ba(OH)2(aq) ¾¾ ® Ba(NO3)2(aq)  2H2O(l) The ionic and net ionic equations are:  2 Ionic: 2H(aq)  2NO3(aq)  Ba2(aq)  2OH(aq) ¾¾ ® Ba (aq)  2NO3 (aq)  2H2O(l)   Net ionic: 2H(aq)  2OH(aq) ¾¾ ® 2H2O(l) or H (aq)  OH (aq) ¾¾ ® H2O(l)

4.35

A half-reaction explicitly shows the electrons involved in a redox reaction. An oxidation reaction refers to the half-reaction that involves a loss of electrons. A reduction reaction is a half-reaction that involves a gain of electrons. A reducing agent is a substance that can donate electrons to another substance, thereby reducing this other substance. An oxidizing agent is a substance that can accept electrons from another substance, thereby oxidizing this other substance. Redox reactions are electron-transfer reactions.

4.36

An oxidation number signifies the number of charges that an atom would have in a molecule (or an ionic compound) if electrons were transferred completely. For other than ionic compounds, the oxidation number is used simply to track electrons in redox reactions. The oxidation numbers are not actual charges in the molecule.

4.37

(a) +1 and +2, respectively.

4.38

No. Oxidation and reduction go together. One substance loses electrons and another gains those electrons.

(b) +3, +4, +5, +6, and +7, respectively.

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

4.39

4.40

Even though the problem does not ask you to assign oxidation numbers, you need to be able to do so in order to determine what is being oxidized or reduced. (i) Half Reactions

(ii) Oxidizing Agent

(iii) Reducing Agent

(a)

Sr  Sr2 2e O2 4e 2O2

(b)

Li  Lie H2 2e 2H

(c)

Cs  Cse Br2 2e 2Br

Br2

(d)

Mg  Mg2 2e N2 6e 2N3

Strategy: In order to break a redox reaction down into an oxidation half-reaction and a reduction halfreaction, you should first assign oxidation numbers to all the atoms in the reaction. In this way, you can determine which element is oxidized (loses electrons) and which element is reduced (gains electrons). Solution: In each part, the reducing agent is the reactant in the first half-reaction and the oxidizing agent is the reactant in the second half-reaction. The coefficients in each half-reaction have been reduced to smallest whole numbers. (a)

The product is an ionic compound whose ions are Fe3 and O2.  3 Fe ¾¾ ® Fe  3e 2 O2 4e ¾¾ ® 2O

O2 is the oxidizing agent; Fe is the reducing agent. (b)

Na does not change in this reaction. It is a “spectator ion.”  2Br ¾¾ ® Br2 2e  Cl2 2e ¾¾ ® 2Cl

Cl2 is the oxidizing agent; Br is the reducing agent. (c)

Assume SiF4 is made up of Si4 and F.  4 Si ¾¾ ® Si  4e  F2 2e ¾¾ ® 2F

F2 is the oxidizing agent; Si is the reducing agent. (d)

Assume HCl is made up of H and Cl.   H2 ¾¾ ® 2H  2e  Cl2 2e ¾¾ ® 2Cl

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

Cl2 is the oxidizing agent; H2 is the reducing agent. 4.41

The oxidation number for hydrogen is 1 (rule 4) and for oxygen is 2 (rule 3). The oxidation number for sulfur in S8 is zero (rule 1). Remember that in a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero, and in an ion the sum of oxidation numbers of all elements in the ion must equal the net charge of the ion (rule 6). H2S (2), S2 (2), HS (2) < S8 (0) < SO2 (4) < SO3 (6), H2SO4 (6) The number in parentheses denotes the oxidation number of sulfur.

4.42

Strategy: In general, we follow the rules listed in Section 4.4 of the text for assigning oxidation numbers. Remember that all alkali metals have an oxidation number of 1 in ionic compounds, and in most cases hydrogen has an oxidation number of 1 and oxygen has an oxidation number of 2 in their compounds. Solution: All the compounds listed are neutral compounds, so the oxidation numbers must sum to zero (Rule 6, Section 4.4 of the text). Let the oxidation number of P x. (a) (b) (c)

x 1  (3)(-2)  0, x5 x (3)(1)  (2)(-2)  0, x1 x (3)(1)  (3)(-2)  0, x3

(d) (e) (f)

x (3)(1)  (4)(-2)  0, x5 2x (4)(1)  (7)(-2)  0, 2x 10, x5 3x (5)(1)  (10)(-2)  0, 3x 15, x5

The molecules in part (a), (e), and (f) can be made by strongly heating the compound in part (d). Are these oxidation–reduction reactions? Check: In each case, does the sum of the oxidation numbers of all the atoms equal the net charge on the species, in this case zero? 4.43

4.44

See Section 4.4 of the text. (a)

ClF: F 1 (rule 5), Cl 1 (rule 6)

(b)

IF7: F 1 (rule 5), I 7 (rules 5 and 6)

(c)

CH4: H 1 (rule 4), C 4 (rule 6)

(d)

C2H2: H 1 (rule 4), C 1 (rule 6)

(e)

C2H4: H 1 (rule 4), C 2 (rule 6),

(f)

K2CrO4: K 1 (rule 2), O 2 (rule 3), Cr 6 (rule 6)

(g)

K2Cr2O7: K 1 (rule 2), O 2 (rule 3), Cr 6 (rule 6)

(h)

KMnO4: K 1 (rule 2), O 2 (rule 3), Mn 7 (rule 6)

(i)

NaHCO3: Na 1 (rule 2), H 1 (rule 4), O 2 (rule 3), C 4 (rule 6)

(j)

Li2: Li 0 (rule 1)

(k)

(l)

KO2: K 1 (rule 2), O 1/2 (rule 6)

(m) PF6: F 1 (rule 5), P 5 (rule 6)

(n)

KAuCl4: K 1 (rule 2), Cl 1 (rule 5), Au 3 (rule 6)

(b) (d) (e)

NaIO3: Na 1 (rule 2), O 2 (rule 3), I 5 (rule 6)

VO2 ® VO 2

NO-2 ® NO3+6

Cr3+ ® CrO24-

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

(b) CaI2, 1 (g) PtCl42, 2

4.45

(a) Cs2O, 1 (f) MoO42, 6 (k) SbF6, 5

(d) H3AsO3, 3 (i) SnF2, 2

4.46

(a)

N: 3

(b)

O: 1/2

(c)

C: 1

(d)

C: 4

(e)

C: 3

(f)

O: 2

(g)

B: 3

(h)

W: 6

(e) TiO2, 4 (j) ClF3, 3

4.47

If nitric acid is a strong oxidizing agent and zinc is a strong reducing agent, then zinc metal will probably reduce nitric acid when the two react; that is, N will gain electrons and the oxidation number of N must decrease. Since the oxidation number of nitrogen in nitric acid is 5 (verify!), then the nitrogen-containing product must have a smaller oxidation number for nitrogen. The only compound in the list that does not have a nitrogen oxidation number less than 5 is N2O5 (what is the oxidation number of N in N2O5?). This is never a product of the reduction of nitric acid.

4.48

Strategy: Hydrogen displacement: Any metal above hydrogen in the activity series will displace it from water or from an acid. Metals below hydrogen will not react with either water or an acid. Solution: Only (b) Li and (d) Ca are above hydrogen in the activity series, so they are the only metals in this problem that will react with water.

4.49

In order to work this problem, you need to assign the oxidation numbers to all the elements in the compounds. In each case, oxygen has an oxidation number of 2 (rule 3). These oxidation numbers should then be compared to the range of possible oxidation numbers that each element can have. Molecular oxygen is a powerful oxidizing agent. In SO3 alone, the oxidation number of the element bound to oxygen (S) is at its maximum value (6); the sulfur cannot be oxidized further. The other elements bound to oxygen in this problem have less than their maximum oxidation number and can undergo further oxidation.

4.50

(a)

Cu(s)  HCl(aq)  no reaction, since Cu(s) is less reactive than the hydrogen from acids.

(b)

I2(s)  NaBr(aq)  no reaction, since I2(s) is less reactive than Br2(l).

(c)

Mg(s)  CuSO4(aq)  MgSO4(aq)  Cu(s), since Mg(s) is more reactive than Cu(s). Net ionic equation: Mg(s)  Cu2(aq)  Mg2(aq)  Cu(s).

(d)

Cl2(g)  2KBr(aq)  Br2(l)  2KCl(aq), since Cl2(g) is more reactive than Br2(l). Net ionic equation: Cl2(g)  2Br(aq)  2Cl(aq)  Br2(l).

Molarity =

moles of solute

4.51

. Because moles of solute are in the equation, molarity is a convenient unit for liters of solution performing solution stoichiometry calculations. Also, because the denominator is the volume of solution, it is easy to prepare a solution of a given molarity. See Figure 4.15 of the text.

4.52

The steps are shown in Figure 4.15 of the text.

4.53

The mass of Ba(OH)2∙8H2O required to make 500.0 mL of a solution that is 0.1500 M hydroxide ions is: 0.5000 L ´

0.1500 mol OH - 1 mol Ba(OH)2 × 8H 2O 315.4 g Ba(OH) 2 × 8H 2 O ´ ´ 1 L soln 1 mol Ba(OH) 2 × 8H 2 O 2 mol OH -

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

= 11.83 g Ba(OH)2∙8H2O 4.54

Strategy: How many moles of NaNO3 does 250 mL of a 0.707 M solution contain? How would you convert moles to grams? Solution: From the molarity (0.707 M), we can calculate the moles of NaNO3 needed to prepare 250 mL of solution. 0.707 mol NaNO3 Moles NaNO3 = ´ 250 mL soln = 0.177 mol 1000 mL soln Next, we use the molar mass of NaNO3 as a conversion factor to convert from moles to grams. M (NaNO3)  85.00 g/mol. 0.177 mol NaNO3 ´

85.00 g NaNO3 = 15.0 g NaNO3 1 mol NaNO3

To make the solution, dissolve 15.0 g of NaNO3 in enough water to make 250 mL of solution. Check: As a ball-park estimate, the mass should be given by [molarity (mol/L)  volume (L)  moles  molar mass (g/mol)  grams]. Let us round the molarity to 1 M and the molar mass to 80 g, because we are simply making an estimate. This gives [1 mol/L  (1/4)L  80 g  20 g]. This is close to our answer of 15.0 g. 4.55

mol M L 60.0 mL  0.0600 L mol MgCl 2 =

4.56

0.100 mol MgCl2 ´ 0.0600 L soln = 6.00 ´ 10-3 mol MgCl 2 1 L soln

Since the problem asks for grams of solute (KOH), you should be thinking that you can calculate moles of solute from the molarity and volume of solution. Then, you can convert moles of solute to grams of solute. ? moles KOH solute =

5.50 moles solute ´ 35.0 mL solution = 0.193 mol KOH 1000 mL solution

The molar mass of KOH is 56.11 g/mol. Use this conversion factor to calculate grams of KOH. ? grams KOH = 0.193 mol KOH ´

4.57

56.11 g KOH = 10.8 g KOH 1 mol KOH

Molar mass of C2H5OH  46.07 g/mol; molar mass of C12H22O11 342.3 g/mol; molar mass of NaCl  58.44 g/mol. (a)

? mol C2 H5OH = 29.0 g C2 H5OH ´ Molarity =

(b)

1 mol C2 H5 OH = 0.629 mol C2 H5OH 46.07 g C2 H5OH

0.629 mol C2 H5OH mol solute = = 1.15 M L of soln 0.545 L soln

? mol C12 H 22 O11 = 15.4 g C12 H 22 O11 ´

1 mol C12 H 22 O11 = 0.0450 mol C12 H 22 O11 342.3 g C12 H 22O11

CHAPTER 4: REACTIONS IN AQUEOUS SOLUTIONS

Molarity =

(c)

4.58

(a)

1 mol CaCl2 = 0.0937 mol CaCl2 110.98 g CaCl2

0.0937 mol CaCl2 = 0.426 M 0.220 L 1 mol C10 H8 = 0.0610 mol C10 H8 128.2 g C10 H8

0.0610 mol C10 H8 = 0.716 M 0.0852 L

First, calculate the moles of each solute. Then, you can calculate the volume (in L) from the molarity and the number of moles of solute. (a)

? mol NaCl = 2.14 g NaCl ´ L soln =

(b)

(c)

1 mol NaCl = 0.0366 mol NaCl 58.44 g NaCl

mol solute 0.0366 mol NaCl = = 0.136 L = 136 mL soln Molarity 0.270 mol/L

? mol C2 H5OH = 4.30 g C2 H5OH ´ L soln =

1 mol C2 H5OH = 0.0933 mol C2 H5 OH 46.07 g C2 H5 OH

0.0933 mol C2 H5OH mol solute = = 0.0622 L = 62.2 mL soln Molarity 1.50 mol/L

? mol CH3COOH = 0.85 g CH3COOH ´ L soln =

4.60

1 mol CH3OH = 0.205 mol CH3OH 32.04 g CH3OH

0.205 mol CH3OH = 1.37 M 0.150 L

? mol C10 H8 = 7.82 g C10 H8 ´ M =

4.59

mol solute 0.154 mol NaCl = = 1.78 M L of soln 86.4 ´ 10-3 L soln

? mol CaCl2 = 10.4 g CaCl2 ´ M =

(c)

1 mol NaCl = 0.154 mol NaCl 58.44 g NaCl

? mol CH3OH = 6.57 g CH 3OH ´ M =

(b)

0.0450 mol C12 H 22 O11 mol solute = = 0.608 M L of soln 74.0 ´ 10-3 L soln

? mol NaCl = 9.00 g NaCl ´ Molarity =

1 mol CH3COOH = 0.014 mol CH3COOH 60.05 g CH3COOH

0.014 mol CH3COOH mol solute = = 0.047 L = 47 mL soln Molarity 0.30 mol/L

A 250 mL sample of 0.100 M solution contains 0.0250 mol of solute (mol M L). The computation in each case is the same: