Chemistry student solutions manual ( 10e 2013 ISBN 9781133933526 ) kenneth w whitten, raymond e davis, larry peck, george g stanley

Student Solutions Manual

Chemistry
TENTH EDITION

Kenneth W. Whitten
University of Georgia, Athens

Raymond E. Davis
University of Texas at Austin

M. Larry Peck
Texas A&M University

George G. Stanley
Louisiana State University

Prepared by
Wendy Keeney-Kennicutt
Texas A&M University

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

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Foreword to the Students
This Solutions Manual supplements the textbook, General Chemistry, tenth edition, by Kenneth W. Whitten,
Raymond E. Davis, M. Larry Peck and George Stanley. The solutions of the 1441 even-numbered problems at
the end of the chapters have been worked out in a detailed, step-by-step fashion.
Your learning of chemistry serves two purposes: (1) to accumulate fundamental knowledge in chemistry which

you will use to understand the world around you, and (2) to enhance your ability to make logical deductions in
science. This ability comes when you know how to reason in a scientific way and how to perform the
mathematical manipulations necessary for solving certain problems. The excellent textbook by Whitten, Davis,
Peck and Stanley provides you with a wealth of chemical knowledge, accompanied by good solid examples of
logical scientific deductive reasoning. The problems at the end of the chapters are a review, a practice and, in
some cases, a challenge to your scientific problem-solving abilities. It is the fundamental spirit of this
Solutions Manual to help you to understand the scientific deductive process involved in each problem.
In this manual, I provide you with a solution and an answer to the numerical problems, but the emphasis lies on
providing the step-by-step reasoning behind the mathematical manipulations. In some cases, I present as many
as three different approaches to solve the same problem, since we understand that each of you has your own
unique learning style. In stoichiometry as well as in many other types of calculations, the "unit factor" method
is universally emphasized in general chemistry textbooks. I think that the over-emphasis of this method may
train you to regard chemistry problems as being simply mathematical manipulations in which the only objective
is to cancel units and get the answer. My goal is for you to understand the principles behind the calculations
and hopefully to visualize with your mind's eye the chemical processes and the experimental techniques
occurring as the problem is being worked out on paper. And so I have dissected the "unit factor" method for
you and introduced chemical meaning into each of the steps.
I gratefully acknowledge the tremendous help over the years provided by Frank Kolar in the preparation of this
manuscript.

Wendy L. Keeney-Kennicutt
Department of Chemistry
Texas A&M University

iii

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Table of Contents
1

The Foundations of Chemistry

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Chemical Formulas and Composition Stoichiometry

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13

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Chemical Equations and Reaction Stoichiometry

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29

4

The Structure of Atoms .

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49

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Chemical Periodicity

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69

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Some Types of Chemical Reactions .

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Chemical Bonding .

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94

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Molecular Structure and Covalent Bonding Theories

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108

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Molecular Orbitals in Chemical Bonding .

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126

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Reactions in Aqueous Solutions I: Acids, Bases, and Salts .

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138

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Reactions in Aqueous Solutions II: Calculations

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150

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Gases and the Kinetic-Molecular Theory .

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167

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Liquids and Solids .

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188

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Solutions .

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209

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Chemical Thermodynamics .

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228

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Chemical Kinetics .

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250

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Chemical Equilibrium .

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270

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Ionic Equilibria I: Acids and Bases .

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289

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Ionic Equilibria II: Buffers and Titration Curves

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306

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Ionic Equilibria III: The Solubility Product Principle

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328

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Electrochemistry

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343

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Nuclear Chemistry .

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366

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Organic Chemistry I: Formulas, Names and Properties .

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378

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Organic Chemistry II: Shapes, Selected Reactions and Biopolymers .

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394

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Coordination Compounds

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404

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Metals I: Metallurgy

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416

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Metals II: Properties and Reactions .

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424

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Some Nonmetals and Metalloids .

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432

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v

Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.


1

The Foundations of Chemistry

1-2.

Refer to the Introduction to Chapter 1 and a dictionary.

(a) Organic chemistry is the study of the chemical compounds of carbon and hydrogen and a few other elements.
(b) Forensic chemistry deals with the chemistry involved in solving crimes, including chemical analyses of crime
scene artifacts, such as paint chips, dirt, fluids, blood, and hair.
(c) Physical chemistry is the study of the part of chemistry that applies the mathematical theories and methods of
physics to the properties of matter and to the study of chemical processes and the accompanying energy
changes.
(d) Medicinal chemistry is the study of the chemistry and biochemistry dealing with all aspects of the medical field.
1-4.

Refer to the Sections 1-1, 1-4, 1-8, 1-13 and the Key Terms for Chapter 1.

(a) Weight is a measure of the gravitational attraction of the earth for a body. Although the mass of an object
remains constant, its weight will vary depending on its distance from the center of the earth. One kilogram of
mass at sea level weighs about 2.2 pounds (9.8 newtons), but that same one kilogram of mass weighs less at the
top of Mt. Everest. In more general terms, it is a measure of the gravitational attraction of one body for another.
The weight of an object on the moon is about 1/7th that of the same object on the earth.
(b) Potential energy is the energy that matter possesses by virtue of its position, condition, or composition. Your
chemistry book lying on a table has potential energy due to its position. Energy is released if it falls from the
table.
(c) Temperature is a measurement of the intensity of heat, i.e. the "hotness" or "coldness" of an object. The
temperature at which water freezes is 0qC or 32qF.
(d) An endothermic process is a process that absorbs heat energy. The boiling of water is a physical process that
requires heat and therefore is endothermic.
(e) An extensive property is a property that depends upon the amount of material in a sample. Extensive properties
include mass and volume.
1-6.

Refer to the Section 1-1 and the Key Terms for Chapter 1.

A reaction or process is exothermic, in general, if heat energy is released, but other energies may be released.
(a) The discharge of a flashlight battery in which chemical energy is converted to electrical energy is referred to as
being exothermic the chemical reaction occurring in the battery releases heat.
(b) An activated light stick produces essentially no heat, but is considered to be exothermic because light is emitted.
1-8.

Refer to Sections 1-1 and 1-5, and the Key Terms for Chapter 1.

(a) Combustion is an exothermic process in which a chemical reaction releases heat.
(b) The freezing of water is an exothermic process. Heat must be removed from the molecules in the liquid state to

cause solidification.

1

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(c) The melting of ice is an endothermic process. The system requires heat to break the attractive forces that hold
solid water together.
(d) The boiling of water is an endothermic process. Molecules of liquid water must absorb energy to break away
from the attractive forces that hold liquid water together in order to form gaseous molecules.
(e) The condensing of steam is an exothermic process. The heat stored in water vapor must be removed for the
vapor to liquefy. The condensation process is the opposite of boiling which requires heat.
(f) The burning of paper is an exothermic process. The heat generated can be used to light the wood in a fireplace.
1-10.

Refer to Section 1-1.

Einstein's equation, written as E = mc2, tells us that the amount of energy released when matter is transformed into
energy is the product of the mass of matter transformed and the speed of light squared. From this equation, we see
that energy and matter are equivalent. Known as the Law of Conservation of Matter and Energy, we can use this
equation to calculate the amount of energy released in a nuclear reaction because it is proportional to the difference
in mass between the products and the reactants. The energy released (in joules) equals the mass difference (in
kilograms) times the square of the speed of light (in m/s).
1-12.

Refer to Section 1-1.

Electrical motors are less than 100% efficient in the conversion of electrical energy into useful work, since a part of

that energy is converted into frictional heat which radiates away.
However, the Law of Conservation of Energy still applies:
electrical energy = useful work + heat
1-14.

Refer to Section 1-3 and Figures 1-7 and 1-8.

Solids:

are rigid and have definite shapes;
they occupy a fixed volume and are thus very difficult to compress;
the hardness of a solid is related to the strength of the forces holding the particles of a solid together; the
stronger the forces, the harder is the solid object.
Liquids: occupy essentially constant volume but have variable shape;
they are difficult to compress;
particles can pass freely over each other;
their boiling points increase with increasing forces of attraction among the particles.
Gases:
expand to fill the entire volume of their containers;
they are very compressible with relatively large separations between particles.
The three states are alike in that they all exhibit definite mass and volume under a given set of conditions. All
consist of some combination of atoms, molecules or ions. The differences are stated above. Additional differences
occur in their relative densities:
gases <<< liquids < solids.
Molecular representations of these three phases can be seen in Figure 1-8. Note that water is an exceptional
compound. The density of the liquid is greater than the solid phase. That is why solid ice floats in liquid water
1-16.

Refer to Section 1-6 and the Key Terms for Chapter 1.

(a) A substance is a kind of matter in which all samples have identical chemical composition and physical
properties, e.g., iron (Fe) and water (H2O).

2

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(b) A mixture is a sample of matter composed of two or more substances in variable composition, each substance
retaining its identity and properties, e.g., soil (minerals, water, organic matter, living organisms, etc.) and
seawater (water, different salts, dissolved gases, organic compounds, living organisms, etc.).
(c) An element is a substance that cannot be decomposed into simpler substances by chemical means, e.g., nickel
(Ni) and nitrogen (N).
(d) A compound is a substance composed of two or more elements in fixed proportions. Compounds can be
decomposed into their constituent elements by chemical means. Examples include water (H2O) and sodium
chloride (NaCl).
1-18.

Refer to Section 1-6.

(a) Gasoline is a homogeneous liquid mixture of organic compounds distilled from oil.
(b) Tap water is a homogeneous liquid mixture, called an aqueous solution, containing water, dissolved salts, and
gases such as chlorine and oxygen.
(c) Calcium carbonate is a compound, CaCO3, consisting of the elements Ca, C and O in the fixed atomic ratio,
1:1:3.
(d) Ink from a ball-point pen is a homogeneous mixture of solvent, water and dyes.
(e) Vegetable soup is a heterogeneous mixture of water, vegetables and the compound, NaCl (table salt), depending
on the recipe.
(f) Aluminum foil is composed of the metallic element, Al.

1-20.

Refer to Section 1-6.

The coin is a heterogeneous mixture of gold and copper because it consists of two distinguishable elements that can
be recognized on sight.
1-22.

Refer to Section 1-4.

(a) Striking a match, causing it to burst into flames, is a chemical property, since a change in composition is
occurring of the substances in the match head and new substances including carbon dioxide gas and water
vapor, are being formed.
(b) The hardness of steel is a physical property. It can be determined without a composition change.
(c) The density of gold is a physical property, since it can be observed without any change in the composition of
the gold.
(d) The ability of baking soda to dissolve in water with the evolution of carbon dioxide gas is a chemical property
of baking soda, since during the reaction, its composition is changing and a new substance is being formed.
(e) The ability of fine steel wool to burn in air is a chemical property of steel wool since a compositional change in
the steel wool occurs and heat is released.
(f) The ripening of fruit is a chemical property. When the temperature of the fruit decreases when put into a
refrigerator, the rate of the chemical reaction slows. So, the lowering of the fruit’s temperature is a physical
change, but temperature has a definite effect on the chemical properties of the fruit.

3

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1-24.

Refer to Section 1-5.

The observations that identify chemical properties are: (c) ultraviolet light converts ozone into oxygen, (e) sodium
metal reacts violently with water, and (f) CO2 does not support combustion.
Some chemists think that dissolution is a chemical process, since it is actually very complex, so some chemists
would include (a).
1-26.

Refer to Section 1-1 and the Key Terms for Chapter 1.

(b), (d) and (e) are examples of potential energy. An inflated balloon (b) possesses energy which will be released if
it is popped. The stored chemical energy in a flashlight battery (d) will convert to electrical energy, then into
kinetic energy once it is put to use. A frozen lake (e) is stored energy. Once spring comes, the water molecules
will be free to move, the lake will be circulating and the energy will convert to kinetic energy. However, a lake
can also be a source of potential energy that can be converted into kinetic energy if the water is released via a
dam.
(a), (c) and (f) are all examples of kinetic energy due to their motion.
1-28.

Refer to Section 1-5.

When the sulfur is heated, some of it obviously became a gas. However, there is not enough information to tell
whether or not this was the result of a physical or a chemical change.
Hypothesis 1:

Solid sulfur could be changing directly into gaseous sulfur. This is a physical change called
sublimation.

Hypothesis 2:

Solid sulfur could be reacting with oxygen in the air to form a gaseous compound consisting of
sulfur and oxygen. This would be a chemical change. The sharp odor may indicate the presence
of SO2, but the smell test is not conclusive.

To verify which hypothesis is correct, we need to identify the gas that is produced.
1-30.

Refer to Appendix A.

(a) 423.006 mL = 4.23006 x 102 mL
(b) 0.001073040 g = 1.073040 x 103 g

(6 significant figures)

(c) 1081.02 pounds = 1.08102 x 10 pounds

(6 significant figures)

3

1-32.

(7 significant figures)

Refer to Appendix A.

(a) 50600

(c) 0.1610

(e) 90000.

(b) 0.0004060

(d) 0.000206

(f) 0.0009000

1-34.

Refer to Appendix A.

? volume (cm3) = 252.56 cm x 18.23 cm x 6.5 cm = 29927 = 3.0 x l04 cm3 (2 significant figures based on 6.5 cm)

4

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1-36.

Refer to Section 1-9, the conversion factors from Tables 1-6 and 1-8, and Examples 1-3 and 1-4.

1 km
(a) ? km = 453.4 m x 1000 m = 0.4534 km
1000 m
(b) ? m = 36.3 km x 1 km = 3.63 x 104 m

1000 g
(c) ? g = 487 kg x 1 kg = 4.87 x l05 g
1000 mL
= 1.32 x l03 mL
(d) ? mL = 1.32 L x
1L
1L
(e) ? L = 55.9 dL x 10 dL = 5.59 L
(f) ? cm3 = 6251 L x
1-38.

1000 cm3
6
3
1 L = 6.251 x l0 cm

(Note: 1 cm3 = 1 mL)

Refer to Section 1-9, the conversion factors listed in Table 1-8, and Example 1-9.

$3.119 1 gal 1.057 qt 100 cents
? cents/L = 1 gal x 4 qt x 1 L x
= 82.42 cents/L
$1
1-40.

Refer to Section 1-10, the conversion factors from Table 1-8, and Examples 1-7 and 1-9.

(12 in)3 (2.54 cm)3
1L

(a) ? L = 0.750 ft3 x (1 ft)3 x (1 in)3 x 1000 cm3 = 21.2 L
(b) ? pints = 1.00 L

1.057 qt 2 pt
1 L x 1 qt = 2.11 pt

km 1 mile 1.609 km 1 gal 1.057 qt
km
(c) ? L = 1 gal x 1 mile x 4 qt x 1 L = 0.4252
L
Therefore, to convert miles per gallon to kilometers per liter, one multiplies the miles per gallon by the factor,
0.4252.
1-42.

Refer to Appendix A.

Average =
1-44.

58.2 + 56.474
= 57.337 = 57.3 % since the answer must be rounded to the tenths place
2

Refer to Section 1-9, Appendix A, the conversion factors from Table 1-8 and Example 1-9.

1 qt
(a) 18 pints x 2 pints = 9.0 qt
(b)

55.0 miles 1.609 km

x 1 mile = 88.5 km/hr
hr

(c) 15.45 s + 2.2 s + 55 s = 72.65 s = 73 s

since the answer must be rounded to the one’s place.

5

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1-46.

Refer to Section 1-11, and Examples 1-11 and 1-12.

m
6.080 mg
Density (mg/mm3) = V = (2.20 mm x 1.36 mm x 1.23 mm) = 1.65 mg/mm3
1.65 mg
1g
(10 mm)3
Density (g/cm3) = 1 mm3 x 1000 mg x (1 cm) 3 = 1.65 g/cm3
1-48.

Refer to Section 1-11 and Example 1-12.

(a) Method 1:

m
m (g)
443 g
D = V ; V (cm3) = D (g/cm3) = 10.5 g/cm3 = 42.2 cm3

since 0.443 kg { 443 g

Method 2:

Dimensional Analysis
1000 g 1 cm3
? cm3 silver = 0.443 kg x 1 kg x 10.5 g = 42.2 cm3
3

3

42.2 cm3 = 3.48 cm
1 in.
(c) length of each edge (in.) = 3.48 cm x 2.54 cm = 1.37 in.
(b) length of each edge (cm) =

1-50.

V=

Refer to Section 1-11.

Plan:

(1) Find the volume of the aluminum wire, assuming that 10-lb spool contains 10.0 lb of aluminum

(2) Calculate the radius of the wire in meters.
(3) Solve for the length of wire in meters, using V = ʌ r2l
453.6 g Al 1 cm3 Al
1 m3 Al
(1) ? V = 10.0 lb Al x 1 lb Al x 2.70 g Al x (100 cm)3 Al = 1.68 x 10–3 m3 Al
0.0808 in. 2.54 cm
1m
(2) ? radius, r = diameter/2 =
x 1 in. x 100 cm = 1.03 x 10–3 m
2
V
1.68 x 10–3 m3
(3) ? length, l = ʌ r2 = 3.1416(1.03 x 10–3 m)2 = 504 m
1-52.

Refer to Sections 1-10 and 1-11.

(1)
(2)
(3)
Plan: L solution Ÿ mL solution Ÿ g solution Ÿ g iron(III) chloride
Using 3 unit factors,
(1) Convert liters to milliliters using 1000 mL = 1 liter,
(2) Convert mL of solution to mass of solution using density, then
(3) Convert mass of solution to mass of iron(III) chloride using the definition of % by mass.
1000 mL soln 1.149 g soln 11 g iron(III) chloride
2
= 3.2 x l0 g
? g iron(III) chloride = 2.50 L soln x 1 L soln x 1 mL soln x
100 g soln

1-54.

Refer to Appendix A, Section 1-12, and Examples 1-16 and 1-17.

In determining the correct number of significant figures, note that the following values are exact: 32qF, 1°C/1.8°F,
and 1°C/1 K and have an infinite number of significant figures.
(a) ? °C =

1qC
x (15qF - 32qF) = 9.4qC
1.8qF

6

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1°C
(b) ? °C = 1.8°F x (32.6°F - 32.0°F) = 0.6°C (1 sig. fig. due to subtraction rules)
1K
? K = 1°C x (0.6°C + 273.2°C) = 273.8 K

since 0°C = 273.15 K

1°C
(c) ? °C = 1 K x (328 K - 273 K) = 55°C
1.8°F
? °F = §55°C x 1°C · + 32°F = 130°F (2 sig. figs.)
©

¹
1.8°F
(d) ? °F = §11.3°C x 1°C · + 32°F = 52.3°F
©
¹
Refer to Section 1-12.

1-56.

Celsius Scale
Fahrenheit Scale
Re'amur Scale

Freezing Point of Water (FP)

Boiling Point of Water (BP)

0qC

100qC

32qF

212qF

0qR

80qR

BPwater - FPwater on Celsius Scale 100qC - 0qC 100qC 1.0qC 5qC

(a) BP
=
=
=
=
80qR 0.8qR 4qR
water - FPwater on Re'amur Scale 80qR - 0qR
Therefore, since both scales set the freezing point of water = 0q, then ? qC = §xqR x

©

(b)

5qC·
4qR¹

BPwater - FPwater on Fahrenheit Scale 212qF - 32qF 180qF 9qF
BPwater - FPwater on Re'amur Scale = 80qR - 0qR = 80qR = 4qR
Therefore, ? qF = §xqR x

9qF ·
+ 32qF
4qR¹
Note that we must add 32qF to account for the fact that 0qR is equivalent to 32qF.

©

(c) From (a), ? qC = §xqR x

©

5qC·
4qR¹

BPmercury (qR) = 356.6qC x
1-58.
For Al:

©

4qR·
5qC¹

4qR
= 285.3qR
5qC

Refer to Section 1-12 and Examples 1-16 and 1-17.
1qC
? qC = 1 K x (933.6 K - 273.2 K) = 660.4qC
? qF = §660.4qC x

©

For Ag:

Rearranging, we have ? qR = §xqC x

1.8qF·
+ 32qF = 1221qF

1qC ¹

1qC
? qC = 1 K x (1235.1 K - 273.2 K) = 961.9qC
? qF = §961.9qC x

©

1.8qF·
+ 32qF = 1763qF
1qC ¹

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Refer to Section 1-12 and Examples 1-16 and 1-17.

1-60.

1qC
x (102.0qF - 32.0qF) = 38.9qC
1.8qF
? K = 38.9 qC + 273.2 q = 312.1 K
? qC =

1-62.

Refer to Section 1-13, and Examples 1-18 and 1-19.

amount of heat gained (J) = (mass of substance)(specific heat)(temp. change)
= 45.3 g x 0.895 J/g˜qC x (62.5qC - 27.0qC)
= 1440 J (3 sig. figs.)
1-64.

Refer to Section 1-13.

(a) amount of heat gained (J) = (mass of substance)(specific heat)(temp. change)
= (69,700 g)(0.818 J/g˜qC)(41.0qC - 25.0qC)
= 9.12 x 105 J
(b) Note that we will follow the convention of representing temperature (°C) as t and temperature (K) as T.
In any insulated system, the Law of Conservation of Energy states:
the amount of heat lost by Substance 1 = amount of heat gained by Substance 2
As will be discussed in later chapters, "heat lost" is a negative quantity and "heat gained" is a positive quantity.
However, the "amount of heat lost" and the "amount of heat gained" quoted here call for absolute quantities
without a sign associated with them. In other words, because we are using the words “lost” and “gained” the
heat involved is positive and the differences in temperature are positive values as well in this exercise.
~the amount of heat lost by Substance 1~ = ~amount of heat gained by Substance 2~
~(mass)(Sp. Ht.)(temp. change)~1 = ~(mass)(Sp. Ht.)(temp. change)~2
In this exercise,
~(mass)(Sp. Ht.)(temp. change)~limestone = ~(mass)(Sp. Ht.)(temp. change)~air
Since any "change" is always defined as the final value minus the initial value, we have
(temp. change)limestone = (30.0qC - 41.0qC) and (temp. change)air = (tfinal - 10.0qC)
for the limestone,

~30.0qC - 41.0qC~ = ~negative value~ = (41.0qC - 30.0qC) = 11.0qC

for the interior air,

~tfinal - 10.0qC~ = ~positive value~ = (tfinal - 10.0qC)

Before we start, we must first calculate the mass of air inside the house:
1000 mL 1.20 x 105 g
? g air = 2.83 x 105 liters x 1 L x
= 3.40 x 105 g
1 mL
69,700 g limestone x 0.818 J/g˜qC x (41.0qC - 30.0qC) = 3.40 x 105 g air x 1.004 J/g˜qC x (tfinal - 10.0qC)
6.27 x 105 J = (3.41 x 105 x tfinal) J - 3.41 x 106 J
4.04 x 106 J = (3.41 x 105 J/oC) x tfinal
tfinal = 11.8qC

8

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1-66.

Refer to Section 1-13 and Example 1-19.
~the amount of heat lost by Substance 1~ = ~amount of heat gained by Substance 2~
~(mass)(Sp. Ht.)(temp. change)~metal = ~(mass)(Sp. Ht.)(temp. change)~water

50.0 g x (Sp. Ht.) x (75.0qC - 18.3qC) = 100. g x 4.18 J/gqC x (18.3qC - 15.0qC)
(Sp. Ht.) x 2835 (remember: it has only 3 sig. figs.*) = 1379 (only 2 sig. figs.)
Solving, Sp. Ht. of the metal = 0.49 J/gqC (2 significant figures set by the temperature change of the water)
* Note: it is better to carry all the numbers in your calculator and do your rounding to the correct number of
significant figures at the end.

1-68.

Refer to Sections 1-9 and 1-10.

100 tons ore
(a) ? tons ore = 5.79 tons hematite x 9.24 tons hematite = 62.7 tons ore
100 kg ore
(b) ? kg ore = 6.40 kg hematite x 9.24 kg hematite = 69.3 kg ore
1-70.

Refer to Appendix A, Section 1-9 and the conversion factors from Table 1-8.

12 in. 2.54 cm
1m
? m = 23.5 ft x 1 ft x 1 in x 100 cm = 7.16 m
1-72.

Refer to Section 1-9 and Table 1-8.

453.6 g body wt 1 kg body wt
1.5 mg drug
? lethal dose = 165 lb body wt x 1 lb body wt x 1000 g body wt x 1 kg body wt = 110 mg drug (2 sig. figs.)
1-74.

Refer to Sections 1-10 and 1-11.

(1)
(2)
Plan: g ammonia Ÿ g solution Ÿ mL solution
Using 2 unit factors,

(1) Convert mass of ammonia to mass of solution using the definition of % by mass, then
(2) Convert mass of solution to volume (in mL) of solution using density

1 mL soln
100 g soln
? L solution = 25.8 g ammonia x 5 g ammonia x 1.006 g soln = 500 mL (1 significant figure due to 5% ammonia)
1-76.

Refer to Sections 1-3 and 1-11, Example 1-2, and Figure 1-7.

(a) Box (i) represents the very ordered, dense solid state.
(b) Box (iii) represents the less ordered, slightly less dense liquid state.
(c) Box (ii) represents the disordered, much less dense gaseous state.
(d) The physical states rank from least dense to most dense: gaseous state << liquid state < solid state

9

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1-78.

Refer to Sections 1-4 and 1-5.

Physical properties:

zinc metal is a gray and shiny solid
zinc metal piece can be cut with scissors

copper chloride solution is blue in color
the new product is brown and granular

Physical changes:

the zinc pieces reduced in size when cut with scissors
the zinc pieces reduced in size during the reaction
the solution became colorless and became warmer

Chemical changes:

some of the zinc disappeared. It must have reacted, because zinc metal is not soluble in water
a new brown granular product formed
the reaction is exothermic and heat was released, making the flask warm to the touch

1-80.

Refer to Sections 1-4 and 1-5, and Exercise 1-79.

Water is more dense than ice at 0oC because a cube of ice (less dense) will float in a glass of water (more dense).
The first drawing shows liquid water molecules that are disorganized and slightly closer together, whereas the
second drawing depicts the water molecules in a very rigid, ordered structure. When a sample has more mass per
unit volume, it is more dense, so liquid water is more dense than solid water because its molecules are closer
together.
1-82.

Refer to your life story.

Chemical vocabulary and understanding can come from many experiences, besides the classroom. Perhaps you
visited a science museum, or had a chemistry “magic show” come to your school. You may have been given a

chemistry set as a present. There are many science-related shows on television and the internet has many, many
links to science pages. Use your own life experiences to answer this question.
1-84.

Refer to Appendix A, Table 1-8 for conversion factors, and Example 1-4.
o

o

Each cesium atom has a diameter = 2 x 2.65 A = 5.30 A
o

? Cs atoms = 1.00 inch x

1-86.

1A
2.54 cm
1m
1 atom
7
o = 4.79 x 10 atoms
1 in x 100 cm x 1010 m x
5.30 A

Refer to Section 1-5 and your common sense.

As a student writes out an End-of-Chapter Exercise, the direct chemical changes that occur include
(1) reactions (including irreversible adsorption) of the ink in the pen with the paper,
(2) the body's biochemical reactions,

(3) the creation of new neural pathways in the student's brain due to the new information she/he is learning.
More indirect chemical changes include the burning of coal or natural gas to provide the power for electricity, heat
and light. If the student is doing a problem outside on a beautiful day, chemical changes might involve
photosynthesis occurring in the plants around her/him providing oxygen for the student to breathe and the fusion
reactions in the sun which provide heat and light, etc.

10

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The complete answer is limited only by the student's imagination and understanding of the meaning of chemical
changes. So, definitely yes, the answer involves knowledge not covered in Chapter 1.
1-88.

Refer to Section 1-12 and Example 1-16.

? qC of iron =

1qC
x (65qF - 32qF) = 18qC
1.8qF

Therefore, the water sample at 65qC has a higher temperature than the iron sample at only 18qC.
1-90.

Refer to Section 1-2.

From left to right: NO, NO2, N2O, N2O3, N2O4 and N2O5.

1-92.

Refer to Section 1-2, Figures 1-3 and 1-4, and Example 1-1.

At room temperature, sulfur (rhombic) is a solid with formula, S8, oxygen is a diatomic gas, O2 and sulfur dioxide is
a gas, SO2.

Sulfur, S8(s)

Oxygen, O2(g)

Sulfur dioxide, SO2(g)

Mixture of S8 and O2

One similarity between S8 and O2 is that they are both elements composed of molecules. However, S8 is a solid,
with the molecular units arranged close together in a systematic way and O2 is a gas, with its diatomic molecules
relatively far apart.
The compound, SO2, and the sample of S8 mixed with O2 both contain the elements, sulfur and oxygen, but SO2
sample contains S and O in the definite ratio of 1:2 in each molecule and the individual gaseous SO2 molecules are
far apart. The mixture of S8 and O2 contains solid sulfur and molecular oxygen and the ratio of S to O can be
variable. The mixture is heterogeneous, because S8(s) and O2(g) are present in different phases.
1-94.

Refer to Section 1-11 and Appendix A.

The calculation only involves multiplying and dividing. The number of significant figures in the answer is then set
by the value with the least number of significant figures. Since density (=8.92 g/mL) has only 3 significant figures,
the answer can only have 3 significant figures, which includes the first doubtful digit. The answer is V = 475 cm3
and “5” is the first doubtful digit.

1-96.

Refer to Section 1-9 and Appendix A.

Many calculations in chemistry can be done in different ways. Consider the conversion of 3475 cm to miles.
1 ft
1 mile
1 in.
(1) ? miles = 3475 cm x 2.54 cm x 12 in. x 5280 ft = 0.021592649 miles or 0.02159 miles

11

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Note: The following conversions are exact: 1 in. = 2.54 cm, 1 ft = 12 in., 1 mile = 5280 ft, so 2.54, 12, and
5280 have infinite numbers of significant figures. The number of significant figures in the answer is then
set by the data: 4.
1 km
1 mile
1m
(2) ? miles = 3475 cm x 100 cm x 1000 m x 1.609 km = 0.021597265 miles or 0.02160 miles
Note: Exact conversions: 1 m = 100 cm, 1 km = 1000 m. Inexact conversion: 1 mile = 1.609 km to 4
significant figures. The number of significant figures in the answer is set by the data (4 sig. figs.) but the
answer has extra source of error since the conversion from kilometers to miles is only good to 4 sig. figs.
Method (1) uses all exact conversions and will give a more accurate answer than Method (2). If you really wanted
to use Method (2), be sure that the inexact conversion contains more significant figures than your data. For
example, if you used 1 mile = 1.6093 km, your answer would have been 0.021593239, and to 4 significant figures,
both methods would have given essentially the same answer, differing only in the doubtful digit.

1-98.

Refer to Sections 1-12 and 1-13, and the Key Terms for Chapter 1.

Students often get the terms, heat, specific heat and temperature confused. Here are the formal definitions:
Heat: A form of energy that flows between two samples of matter because of their difference in temperature,
measured in joules (J).
Specific heat: The amount of heat required to raise the temperature of one gram of a substance one degree Celsius.
Its units are J/g·oC.
Temperature: A measure of the intensity of heat, that is, the hotness or coldness of a sample or object. Temperature
also refers to molecular motion. The warmer a substance is, the more its molecules are moving. Scientists
usually work in oC or K.
If two samples of the same element are at different temperatures, their atoms have different kinetic energies and are
moving at different average speeds. If the two samples touch, energy (heat) will transfer from the hotter to the
colder element until their temperatures are the same and the average speed of their respective molecules are the
same.
Different substances require different amounts of heat to change their temperatures. Specific heat is the constant
that gives that information. It has units of J/g·oC and is the amount of heat required (in joules) to heat up 1 gram of a
substance by 1oC.
As a final note, consider a 5.0 gram block of iron and a 15 gram block of iron, both at 25oC. They are both at the
same temperature, so if they came into contact, neither would change temperature. However, the 15 g iron block
contains three times more heat than the 5.0 gram block. In other words, three times more heat is required to change
the temperature of the 15 gram block of iron to 26oC, as the 5.0 gram block of iron.

12

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2 Chemical Formulas and Composition
Stoichiometry
2-2.

Refer to Section 2-1 and the Key Terms for Chapter 2.

Allotropes are defined as different forms of the same element in the same physical state. Two examples of
allotropes are:
(1) oxygen, O2 (a diatomic molecule) and ozone, O3 (a triatomic molecule), and
(2) carbon as graphite, Cgraphite, and carbon as diamond, Cdiamond.
2-4.

Refer to Section 2-1 and Figure 2-1.

The structural formulas and ball-and-stick models of water and ethanol are given in Figure 2-1. You can see
that the general shape and bond angles are similar around the oxygen atom.
2-6.

Refer to Section 2-1 and Figure 2-1.

Organic compounds can be distinguished from inorganic compounds because organic compounds contain C−C
or C−H bonds or both. Refer to Figure 2-1. According to this definition, water, H2O, hydrogen peroxide,
H2O2, and carbon tetrachloride, CCl4, are considered inorganic molecules, whereas ethanol, C2H5OH, is an
organic molecule.
2-8.

Refer to Section 2-1, Table 2-1, and Figure 1-5.

Ball-and-stick model of ethane,
CH3CH3:

2-10.

Refer to Section 2-1 and Table 2-1.

(a) O3, HNO3, SO3

(b) H2, H2O, H2O2, H2SO4

(d) CH3COOH, C2H6

(e) CH3CH2CH3, CH3CH2CH2OH

2-12.

Refer to Sections 2-1 and 2-2, and Tables 2-1 and 2-2.

(a) HNO3 nitric acid
2-14.

(c) H2O2, NH3, SO3

(b) C5H12 pentane

(c) NH3 ammonia

(d) CH3OH methanol

Refer to Section 2-2 and Table 2-2.

(a) Mg2

monatomic cation

(b) SO32

polyatomic anion

(d) NH4

polyatomic cation

(e) O2

monatomic anion

(c) Cu

monatomic cation

13

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2-16.

Refer to Section 2-3, Table 2-2 and Examples 2-2 and 2-3.

2-18.

SO42– sulfate ion
NO3– nitrate ion
CH3COO– acetate ion

Ba2+ barium ion
Mg2+ magnesium ion
Na+ sodium ion

barium sulfate
BaSO4
magnesium nitrate Mg(NO3)2
sodium acetate
NaCH3COO

Refer to Section 2-3, Table 2-2, and Examples 2-2 and 2-3.

(a) CuCO3

copper(II) carbonate

(b) SrBr2

strontium bromide

(c) (NH4)2CO3

ammonium carbonate

(d) ZnO

zinc oxide

(e) Fe2(SO4)3

iron(III) sulfate

2-20.

Refer to Sections 2-2 and 2-3, Table 2-2, and Examples 2-2 and 2-3.

(a) Na2CO3
2-22.

(c) Zn(OH)2

(d) (NH4)2S

(e) NaI

Refer to Sections 2-2 and 2-3, Tables 2-1 and 2-2, and Examples 2-2 and 2-3.

(a) NaBr
(c) SO2
(e) K2S
2-24.

(b) MgCl2

sodium bromide
sulfur dioxide or SO3 sulfur trioxide
potassium sulfide

(b) MgBr2
(d) CaO
(f) AlBr3

magnesium bromide
calcium oxide
aluminum bromide

Refer to Section 2-4.

The mass ratio of a rubidium atom (85.4678 amu) to a bromine atom (79.904 amu) is 85.4678/79.904 = 1.0696
(to 5 significant figures) or 1.070 (to 4 significant figures).
2-26.

Refer to Section 2-4 and the Key Terms for Chapter 2.

(a) The atomic weight of an element is the weighted average of the masses of all the element’s constituent
isotopes.
(b) Atomic weights can be referred to as relative numbers, because all atomic weights are determined relative
to the mass of a particular carbon isotope, called carbon-12. The atomic mass unit (amu) is defined as
exactly 1/12 of the mass of the carbon-12 isotope.
2-28.

Refer to Section 2-6, Example 2-8 and the Periodic Table.

(a) bromine, Br2

2 x Br = 2 x 79.904 amu = 159.808 amu

(b) hydrogen peroxide, H2O2

2 x H = 2 x 1.008 amu = 2.016 amu
2 x O = 2 x 15.999 amu = 31.998 amu
formula weight = 34.014 amu

(c) saccharin, C7H5NSO3

7 x C = 7 x 12.011 amu
5 x H = 5 x 1.008 amu
1 x N = 1 x 14.007 amu
1 x S = 1 x 32.06
amu
3 x O = 3 x 15.999 amu
formula weight

= 84.077
= 5.040
= 14.007
= 32.06
= 47.997
= 183.18

amu
amu
amu
amu

amu
amu

14

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(d) potassium chromate, K2CrO4

2-30.

2 x K = 2 x 39.0983 amu
1 x Cr = 1 x 51.9961 amu
4 x O = 4 x 15.999 amu
formula weight

= 78.1966
= 51.9961
= 63.996
= 194.189

amu
amu
amu
amu

Refer to Section 2-6 and Example 2-8.

All atomic weights are rounded to 2 decimal places.
(a) hydrogen sulfide, H2S

2 x H = 2 x 1.01
amu =
1 x S = 1 x 32.06 amu =
formula weight =

(b) phosphorus trichloride, PCl3

1 x P = 1 x 30.97 amu = 30.97 amu
3 x Cl = 3 x 35.45 amu = 106.35 amu*
formula weight = 137.3 amu

(c) hypochlorous acid, HClO

1 x H = 1 x 1.01
amu
1 x Cl = 1 x 35.45 amu
1 x O = 1 x 16.00 amu
formula weight

(d) hydrogen iodide, HI

1 x H = 1 x 1.01
amu =
1.01 amu
1 x I = 1 x 126.90 amu = 126.90 amu
formula weight = 127.91 amu

=
=
=
=

2.02 amu
32.06 amu
34.08 amu

1.01
35.45
16.00
52.46

amu
amu
amu
amu

* The number was not rounded to the correct number of significant figures until after addition.
2-32.

Refer to Section 2-6.

Use the units of formula weight to derive a formula relating grams, moles and formula weight:
grams of compound
g
formula weight, FW §mol· = moles of compound
© ¹

Method 1:

Therefore, grams of compound = moles of compound x FW
(1) ? g CCl4 = 2.371 mol CCl4 x 153.8 g/mol = 364.7 g CCl4
1 kg
(2) ? kg CCl4 = 374.7 g CCl4 x 1000 g = 0.3647 kg CCl4
Method 2: Dimensional Analysis
153.8 g CCl4
1 kg
(2) ? kg CCl4 = 2.371 mol CCl4 x 1 mol CCl x 1000 g = 0.3647 kg CCl4
4
2-34.

Refer to Section 2-6, and Examples 2-10 and 2-11.

The molecular mass of C3H8 is 44.1 g/mol. Each C3H8 molecule contains 8 hydrogen atoms.
Plan: g C3H8 Ÿ mol C3H8 Ÿ molecules C3H8 Ÿ atoms H
1 mol C H 6.02 x 1023 C3H8 molecules
8 H atoms
? H atoms = 167 g C3H8 x 44.1 g C3 H8 x
x 1 C H molecule = 1.82 x 1025 H
1 mol C3H8
3 8
3 8
atoms

15

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2-36.

Refer to Section 2-6 and Example 2-10.

Method 1:

Use the units of formula weight to derive a formula relating grams, moles and formula weight:

grams of substance
g
formula weight, FW §mol· = moles of substance

©

¹

grams of substance
Therefore, moles of substance = formula weight (g/mol)
12.50 g
? mol NH3 = 17.03 g/mol = 0.7340 mol NH3
(Note: be sure you use at least as many significant figures in the formula weight as you have
significant figures in your data.)
Method 2: Dimensional Analysis
1 mol NH
? mol NH3 = 12.50 g NH3 x 17.03 g NH3 = 0.7340 mol NH3
3

2-38.

Refer to Section 2-6 and Example 2-10.

(1)
(2)
Plan: g substance Ÿ moles substance Ÿ molecules substance
Method 1:

Recall:

g substance
mol substance = formula weight

and

Avogadro's Number, N = 6.02 x 1023

molecules/mol
As an example:
g CO2
31.6 g
(a) (1) ? mol CO2 = FW CO = 44.0 g/mol = 0.718 mol CO2
2
(2) ? molecules CO2 = 0.718 mol CO2 x (6.02 x 1023 molecules/mol) = 4.32 x 1023 molecules CO2
Method 2:

Dimensional Analysis. Each unit factor corresponds to a step in the Plan.

Step 1
Step 2

1 mol CO2 6.02 x 1023 molecules CO2
= 4.32 x 1023 molecules CO2
(a) ? molecules CO2 = 31.6 g CO2 x 44.0 g CO x
1 mol CO2
2
1 mol N 6.02 x 1023 molecules N2
= 6.79 x 1023 molecules N2
(b) ? molecules N2 = 31.6 g N2 x 28.0 g N2 x
1 mol N
2

2

1 mol P 6.02 x 1023 molecules P4
= 1.53 x 1023 molecules P4
(c) ? molecules P4 = 31.6 g P4 x 124 g P4 x
1 mol P
4

4

1 mol P 6.02 x 1023 molecules P2
(d) ? molecules P2 = 31.6 g P2 x 62.0 g P2 x
= 3.07 x 1023 molecules P2
1 mol P
2

2

4 atoms P

(e) ? atoms P in (c) = 1.53 x 1023 molecules P4 x 1 P molecule = 6.12 x 1023 atoms P in (c)
4

2 atoms P
? atoms P in (d) = 3.07 x 1023 molecules P2 x 1 P molecule = 6.14 x 1023 atoms P in (d)
2
Yes, there is the same number of P atoms in 31.6 g of pure phosphorus, regardless of whether the
phosphorus is in the form of P4 or P2. The difference is due to rounding error only.

16

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