Solution manual for chemistry the molecular science 5th edition by moore

Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

1

Chapter 1: The Nature of Chemistry
Teaching for Conceptual Understanding
Chemistry – The Molecular Science encourages students to think about chemistry at three different levels: the
macroscopic, the particulate or nanoscale, and the symbolic. Science education research has shown that an
understanding of just one level does not imply an understanding of the others. Whenever possible use all three
levels in your teaching and assessment of student learning. Figure 1.17 illustrates the macroscopic and particulate
levels for solid ice, liquid water, and water vapor. Have students visualize or draw other simple molecules, such as
bromine or mercury, at the particulate level in three different states.
Section 1-3 discusses how science is done, i.e., how observations lead to a hypothesis, how the hypothesis leads to
more observations or experimentation, which in turn can lead to a law or theory. Students sometimes get the idea
that science is a step-by-step procedure that takes place in a laboratory instead of a process that people can use
everyday to solve problems and understand the world around them. It would be helpful to explain the process of
science by using examples with which students can identify. Relevant examples will change over time, so be
diligent in coming up with new examples from year to year.
Learning chemistry is like learning a foreign language because of the extensive use of new terminology. Students
will make quicker and stronger associations between terms and concepts when the root or origin of the terms is
explained. For example, in this chapter the terms nanoscale and microscale are introduced. This is a good time to
reinforce the size of the metric prefixes “micro-” and “nano-” as explanations for the terminology.

Suggestions for Effective Learning
Keep in mind that students are excited and ready to work the first day of class. It is best not to waste this valuable
moment by simply reviewing the syllabus and dismissing the students. Have demonstrations, multimedia material,
and activities ready to engage them in learning.
If you plan on using demonstrations during the course, show some of your favorites during the first class. A few
easy demonstrations that always grab students’ attention are: Igniting hydrogen filled balloons, freezing flowers in
liquid nitrogen, and dropping pieces of sodium or potassium into water.

In addition to the cooperative learning activities suggested below, consider having the students write briefly on their
impressions of chemistry, or what properties solids, liquids, and gases have, or how they use the scientific process in
solving their everyday problems.
Finally, take a few minutes to explain something about yourself. The students will respect and engage with you
better when they feel they understand who you are, why you’re there, and what you care about. Their college
experience as well as their relationship with you is enriched by a small amount of personal information.

Cooperative Learning Activities
Whether you intend to use cooperative learning activities during class time or not, it is very important for students to
get acquainted with each other. Set aside at least five minutes during your first class for students to meet others
seated around them. In addition to exchanging personal information, e.g., name, hometown, major, have them share
their feelings about taking a college chemistry course. Some students feel anxious about taking chemistry; however,
knowing how others feel can help them see that they are not alone.
Questions, problems, and topics that can be used for Cooperative Learning Exercises and other group work are:
•

Questions for Review and Thought: 4, 5, 7-8, 11-12, 37-38, 47-48, 63-64, 98-99, 117, 118

•

Conceptual Challenge Problems: CP1.A, CP1.B, CP1.C, CP1.D, CP1.E, CP1.F, CP1.G, and CP1.H

•

Choose one or two statements from Section 1-1 that are most relevant to your students. Have the students
spend five minutes writing down their ideas, then have them share those ideas with others around them in a
short discussion. Another group activity is to have students list three to five ways that they have used
chemicals today.

Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
2

Chapter 1: The Nature of Chemistry

End-of-Chapter Solutions for Chapter 1
Summary Problem
×1015

Result: 3
J; water is a compound, none are mixtures; See nanoscale diagram below,
CO2 (g) + 2 H2 (g)
CH4 (g) + O2 (g)
Analyze and Plan: Given the number of electric vehicles, the average miles each drives per year, the relationship
between miles and kilometers, the energy efficiency of each vehicle, and some relationships between energy units,
calculate the total energy needed for these vehicles.
Execute:

⎛ 10, 000 mi / vehicle ⎞ ⎛ 1.6 km ⎞ ⎛ 1 kW h ⎞ ⎛ 1000 W h ⎞ ⎛ 3.6 PJ ⎞
e ×
e ×⎜
⎟ = 3 PJ
300, 000 vehicles × 1 yr × ⎜
⎟× ⎜
⎟× ⎜
⎟ ⎜
⎟
yr
⎠ ⎝ 1 mi ⎠ ⎝ 6 km ⎠ ⎝ 1 kWe h ⎠ ⎜⎝ 1012 We h ⎟⎠
⎝

Because 1PJ = 1 ×1015 J, this is 3 × 1015 J of energy.

Reasonable Result Check: The answer in petajoules is a convenient number, 3, which is consistent with PJ being
commonly used to measure such energies.
Explanation: Given information about reactions involving hydrogen, water, and oxygen, determine if these are
elements, compounds, or mixtures.
Because the substance called water is formed using hydrogen and hydrogen is recreated when water is decomposed,
it is certain that water is a compound. The information provided does not make it clear whether the substances
labeled hydrogen or oxygen are compounds or elements, but they are not mixtures because each time one destroyed
or created they can be isolated and characterized as a pure substance.
Explanation: Given chemical formulas and information about a reaction write nanoscale and symbolic
representations of the reaction that are consistent with modern atomic theory.
The parts of modern atomic theory that applies to this example are: Compounds are formed by the chemical
combination of two or more different kinds of atoms and a chemical reaction involves joining, separating, or
rearranging atoms.
A molecule of methane is composed of one atom of carbon combined with four atoms of hydrogen. A molecule of
oxygen is composed of two atoms of oxygen combined together. These two substances together represent the
reactants in the first nanoscale box and on the left side of the symbolic representations shown below. A molecule of
carbon dioxide is composed of one atom of carbon combined with two atoms of oxygen. A molecule of hydrogen is
composed of two atoms of hydrogen combined together. One carbon dioxide molecule and two hydrogen molecules
represent the products in the second nanoscale box and on the right side of the symbolic representations shown
below. Because the reactants include four atoms of hydrogen, there are two molecules of H2 represented in the
products to account for all the hydrogen atoms present initially.

methane and oxygen gas
CH4(g) + O2(g)

carbon dioxide gas and hydrogen gas
CO2(g) + 2 H2(g)

Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

3

End-of-Chapter Solutions for Chapter 1
Questions for Review and Thought
Review Questions
1.

Result/Explanation: The structure of a molecule refers to the way atoms are connected together in the molecule
and to the three dimensional arrangement of the atoms relative to one another. Structure is important because it
is the key to the properties and reactivity of a molecule.

2.

Result/Explanation: Quantitative observations of a piece of electronics must include numerical information;
whereas qualitative observations do not involve numerical details. For example, a television: Quantitative
observations might include the dimensions of the screen and case, the various control settings (volume level,
channel selection, etc.), the screen resolution, the power supply requirements, its position in the room, on-off,
etc. The qualitative observations might include the color of the case, the types and locations of control buttons
or knobs, the quality of the audio output, the quality of the video image, and the type of connection to a video
signal (antenna, cable, satellite, internet, etc.).

3.

Result/Explanation: A scientific law (a) summarizes and explains a wide range of experimental results, (b) has
not been contradicted by experiments, and (c) is capable of predicting unknown results. Some laws are
described in Atomic Theory. Two of these laws are: the law of conservation of mass (i.e., there is no detectable
change in mass during an ordinary chemical reaction.) and the law of constant composition (i.e., A chemical

compound always contains the same elements in the same proportion by mass.).

4.

Result/Explanation: A theory is a unifying principle that explains a body of facts and the laws based on them –
hence a theory is our reason for believing in the law; whereas a law gives just a summary conclusion of a wide
range of experimental results. Models are used to make theories more concrete, often in physical or
mathematical form – hence a model is our way of looking at the theory in detail.

5.

Result/Explanation: Chemists build the bridge between the nanoscale world into the microscale world. The
details of the nanoscale world often profoundly affect the activity of a chemical in the micro- and macroscopic
worlds. A specific example is the determination of the mechanism of paclitaxel activity in the search for
improved anticancer drugs.

6.

Result/Explanation: (Described in Section 1-7) Two examples of when purity of a substance is important: The
purity of the elemental silicon is important to the production of electronic chips. To properly characterize the
properties of a substance, it is necessary to test the substance in its pure state.

Topical Questions
Why Care About Chemistry? (Section 1-1)
7.

Result/Explanation: Important issues change from year to year, and sometimes even from month to month or
even day to day. A good website for seeking up-to-date information on a wide range of scientific topics is
In summer of 2013, the following issues were big:
• forensics

• energy technology, renewable energy, alternative fuels
• nanotechnology
• environmental issues: climate, air quality, pollution, global warming, acid rain
• cancer, stem cells, brain tumors, heart disease, medical technology
• transportation sciences
• nature of water
• materials in electronic devices (batteries, display, functionality, etc.)

8.

Result/Explanation: Several questions related to chemistry and science phenomena are given in Section 1-1.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
4

Chapter 1: The Nature of Chemistry

How Science is Done (Section 1-3)
9.

Result: (a) Qualitative (b) Quantitative and qualitative (c) Quantitative and qualitative (d) Qualitative
Explanation:
(a) The details of the appearance of a substance (silvery-white) and information about the specific element it
contains (sodium) are both qualitative.
(b) The temperature at which a solid melts (660 °C) is quantitative information. Information about what
element it is (aluminum) is qualitative.
(c) The mass percentage of an element in the human body (about 23%) is quantitative information.
Information about what element it is (carbon) is qualitative.
(d) The allotropic forms of an element (graphite, diamond, and fullerenes) and information about what element

it is (carbon) are both qualitative.

10. Result: (a) Quantitative and qualitative (b) Qualitative (c) Quantitative and qualitative (d) Qualitative
Explanation:
(a) The atomic mass of an element (12.011 atomic mass units) is quantitative information. Information about
what element it is (carbon) is qualitative.
(b) The purity, the details of the appearance of a substance (silvery-white), the lack of magnetic capabilities
(nonmagnetic), the relative density (low), the inability to produce sparks when struck, and information
about the specific element it contains (aluminum) are all qualitative.
(c) The density of a substance (0.968 g/mL) is quantitative information. Information about what element it is
(sodium) is qualitative.
(d) The identity of an extracellular element (sodium), the element’s ionic form (cation), and the biological
importance of the element (nerve function) are all qualitative.

Identifying Matter: Physical Properties (Section 1-4)
11. Result/Explanation: Bromine is a reddish-brown liquid. Sulfur is a chalky yellowish solid. They appear to have
no property in common. The physical phase, shape, color, and appearance are different.
12. Result/Explanation: Calcite is a clear, colorless, cubic solid. Calcium appears as a metallic solid with a white
coating. Carbon is a black, granular solid. The property they have in common is that they are solids. The
shape, color, and appearance of all three are different.
13. Result: The solid will melt because your body temperature of 37°C is above the melting point of 29.76°C.
Analyze and Plan: Many Americans only remember the human body temperature in the Fahrenheit scale. That
is 98.6 °F. If that is the case, we can quickly apply the °F to °C conversion equation, so we can compare it to
the melting point.
Execute:

C = 5 × ⎛⎜
F − 32 ⎞⎟ = 5
× 98.6 − 32 = 37.0
C

9 ⎝

⎠

9

(

)

If the sample melts at a temperature of 29.76 °C and your hand is 37 °C, the liquid will boil when exposed to
the heat energy emitted by your hand when you hold the sample.
14. Result: Charlotte is the warmest city. Montreal is the coldest city.
Analyze: We use these questions about temperature using different scales to help us become familiar with the
Celsius scale. So, it is useful to try estimating the answer before using equations using some key connections
between the Celsius scale and the Fahrenheit scale
Plan: Compare two thermometers side-by-side, one calibrated with a Celsius scale and one calibrated with a
Fahrenheit scale using body temperature (37 °C, 98.9 °F) and the freezing point of water (0 °C, 32 °F) as
reference points.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

5

Execute:

98.7 °F

37 °C

Body temperature

20 °C
40 °F

0 °C

32 °F
28 °F

Water freezes

–10 °C
This comparison shows that Charlotte (at 20 °C) is the warmest city and Montreal (at – 10 °C) is the coldest.
To be quantitative, convert 40 °F to °C to make sure it’s lower than 20 °C, and convert 28 °F to °C to make sure
it’s higher than –10 °C.


5
9
5
9

5 ⎛
⎞
C = × ⎜
F − 32 ⎟
⎠
9 ⎝

⎛
⎞

× ⎜ 40
F − 32⎟ = 4
C lower than 20 °C (confirms estimate)
⎝
⎠

⎛
⎞
× ⎜28
F − 32⎟ = − 2
C higher than – 10 °C (confirms estimate)
⎝
⎠

Measurements, Units, and Calculations (Section 1-5)
15. Result: 0.00283 kg, Ca and F
Analyze and Plan: Given the mass of the crystal as 2.83 grams, find the mass in kilograms. Using the
conversion factor between grams and kilograms, determine the mass in kilograms.
Execute:

2.83 g ×

1 kg
= 0.00283 kg
1000 g

The symbols for the elements in this crystal are Ca (calcium) and F (fluorine).

Reasonable Result Check: Kilograms are larger units than grams, so the number of kilograms should be
smaller than the number of grams.
16. Result: 7.8 × 102 m3, 7.8 × 108 cm3, 7.8 × 105 L
Analyze: Given the length, width, and height of a room, determine the volume in cubic meters, cubic
centimeters, and liters.
Plan: Using V = (length) × (width) × (height) calculate the volume in cubic meters. Use conversion factors
between centimeters and meters to determine the volume in cubic centimeters. Use the conversion factor

between cubic centimeters and milliliters and between milliliters and liters to determine the volume in liters.
Execute:

V = (length) × (width) × (height) = (18 m) × (15 m) × (2.9 m) = 7.8 × 102 m3

⎛ 100 cm ⎞3
8
3
7.8 × 10 2 m3 × ⎜
⎟ = 7.8 × 10 cm
⎝ 1m ⎠
7.8 × 10 8 cm3 ×

1 mL
1 cm3

×

1L
= 7.8 × 105 L
1000 mL


Reasonable Result Check: A liter is smaller than a cubic meter but larger than a cubic centimeter, so the
number of m3 should be the smallest number, then the number of liters, then the number of cm3.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
6

Chapter 1: The Nature of Chemistry

17. Result: No, 23.4 mi/hr < 25 mi/hr
Analyze: Given the length of the track and the time to run its length, determine the miles per hour and compare
to 25 mi/hr.
Plan: Divide the meters by the seconds, then use conversion factors between meters and centimeters,
centimeters and inches, inches and feet, and feet and miles to determine the distance in miles. And use the
conversion factor between second and hours to determine the time in hours.
Execute:

100 m ⎛ 100 cm ⎞ ⎛ 1 in ⎞ ⎛ 1 ft ⎞ ⎛ 1 mi ⎞ ⎛ 3600 s ⎞
mi
×⎜
⎟× ⎜
⎟× ⎜
⎟× ⎜
⎟×⎜
⎟ = 23.4
9.58 s ⎝ 1 m ⎠ ⎝ 2.54 cm ⎠ ⎝ 12 in ⎠ ⎝ 5280 ft ⎠ ⎝ 1 hr ⎠
hr
No, this runner could not be arrested for exceeding the 25 mi/hr speed limit, since 23.4 mi/hr < 25 mi/hr.

Reasonable Result Check: No one on foot could chase and catch up with a car going 25 mph.
18. Result: (a) BMI = 21.4 kg/m2 (b) No, because 21.4 kg/m2 < 25 kg/m2
Analyze: (a) Given mass in pounds and height in inches, use conversion factors to calculate mass in kg and
height in meters, then calculate the body mass index (BMI) as described in the question.
Plan: Using conversion factors relating pounds to grams and grams to kilograms, determine the mass in kg.
Use inches to centimeters and centimeters to meters to determine the height in meters. BMI = (kg)/[(m)]2
Execute:

⎛ 453.59 g ⎞ ⎛ 1 kg ⎞
176 lb × ⎜

⎟× ⎜
⎟ = 79.8 kg
⎝ 1 lb ⎠ ⎝ 1000 g ⎠
⎛ 2.54 cm ⎞ ⎛ 1 m ⎞
76.0 in × ⎜
⎟× ⎜
⎟ = 1.93 m
⎝ 1 in ⎠ ⎝ 100 cm ⎠
BMI =

79.8 kg

(1.93 m)

2

= 21.4

kg
m2

(b) No, this person should not be considered overweight, since 21.4 kg/m2 < 25 kg/m2.

Reasonable Result Check: This person is fairly tall (over six feet), so his 176 lbs is distributed properly.
19. Result: (a) Four (b) Three (c) Four (d) Four (e) Three (f) Four
Analyze: Given several measured quantities, determine the number of significant figures.
Plan: Use rules given in Section 1-5, summarized here: All non-zeros are significant. Zeros that precede (sit
to the left of) non-zeros are never significant (e.g., 0.003). Zeros trapped between non-zeros are always
significant (e.g., 3.003). Zeros that follow (sit to the right of non-zeros are (a) significant, if a decimal point is
explicitly given (e.g., 3300.) OR (b) not significant, if a decimal point is not specified (e.g., 3300).
Execute:

(a) 1374 kg has four significant figures. The 1, 3, 7, and 4 digits are each significant.
(b) 0.00348 s has three significant figures. The 3, 4, and 8 digits are each significant. The zeros are all before
the first non-zero-digit 3 and therefore they are not significant.
(c) 5.619 mm has four significant figures. The 5, 6, 1, and 9 digits are each significant.
(d) 2.475 × 10–3 cm has four significant figures. The 2, 4, 7, and 5 digits are each significant.
(e) 33.1 mL has three significant figures. The 3, 3, and 1 digits are each significant.
(f) 2300. m has four significant figures. The 2, 3, 0, and 0 digits are each significant.

Reasonable Result Check: Only 2 answers have zeros in them. Those in (b) are to the left of the first nonzero digit, so none of the zeros there were significant. Those in (f) are in a number with a decimal point, so all
of them were significant.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

7

20. Result: (a) Four (b) Two (c) Two (d) Three (e) Three (f) Two
Analyze and Plan: Given several measured quantities, determine the number of significant figures. Use the
rules given in Section 1-5, summarized in the solution to Question 19.
Execute:
(a) 1.022 × 102 km has four significant figures (The 1, 0, 2, and 2 digits are each significant. The zero is
trapped between the non-zeros 1 and 2 and therefore is significant.)
(b) 34 m2 has two significant figures (The 3 and 4 are each significant.)
(c) 0.042 L has two significant figures (The 4 and 2 digits are each significant. The zeros are all before the
first non-zero-digit 4 and therefore they are not significant.)
(d) 28.2 °C has three significant figures (The 2, 8, and 2 digits are each significant.)
(e) 323. mg has three significant figures (The 3, 2, and 3 digits are each significant.)
(f) 420 mg has two significant figures (The 4 and 2 digits are each significant. The zero follows the non-zerodigits in a number with no decimal point. The ones place is unknown.)

Reasonable Result Check: Three answers had zeros in them. In (a), it was trapped and significant; in (c),
they were to the left of the first non-zero digit, so not significant; in (f), the number has a zero that is not

significant because there is no decimal point.
21. Result: (a) 1.9 g/mL (b) 291.2 cm3 (c) 0.0217 (d) 5.21 × 10–5
Analyze: Given numbers combined in calculations, determine the result with proper significant figures.
Plan: Perform the mathematical steps according to order of operations, applying the proper significant figures
(addition and subtraction retains the least number of decimal places in the result; multiplication and division
retain the least number of significant figures in the result). Notice: if operations are combined that use different
rules, we must stop and determine the intermediate result any time the rule switches.
Execute:
(a)

4.850 g − 2.34 g
1.3 mL
The numerator uses the subtraction rule. The first number has three decimal places (the 8, 5, and 0 are all
decimal places -- digit that follow the decimal point to the right) and the second number has two decimal
places (the 3 and the 4 are both decimal places), so the result of the subtraction has two decimal places.

2.51 g
1.3 mL
The ratio uses the division rule. The numerator has three significant figures and the denominator has two
significant figures, so the answer will have two significant figures. Therefore, the answer is 1.9 g/mL.
(b)

V =4/3 πr3 = 4/3 × (3.1415926) × (4.112 cm)3
This whole calculation uses the multiplication rule, with four significant figures, limited by the
measurement of r. The numerals 4 and 3 are exact, in this context. The value of π must be carried to more
than four significant figures, such as 3.1415926… The answer comes out 291.2 cm3.

(c)

(4.66 × 10–3) × 4.666

This calculation uses the multiplication rule. The first number, 4.66 × 10–3, has three significant figures and
the second number, 4.666, has four significant figures, so the answer has three significant figures 0.0217.

(d)

0.003400
65.2
This calculation uses the division rule. The numerator has four significant figures and the denominator has
three significant figures, so the answer has three significant figures 0.0000521 or 5.21 × 10–5.


Reasonable Result Check: The significant figures, size, and units of the answers are appropriate.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
8

Chapter 1: The Nature of Chemistry

22. Result: (a) 1.20×
× 103 (b) 0.690 (c) 81.7 cm2 (d) 1.7×102
Analyze: Given some numbers in calculations, determine the result with proper significant figures.
Plan: Perform the mathematical steps according to the order of operations, applying the proper significant
figures (Remember: addition and subtraction retains the least number of decimal places in the result;
multiplication and division retain the least number of significant figures in the result). Notice: if operations are
combined that use different rules, it is important to stop and determine the intermediate result any time the rule
switches.
Execute:
(a)

3256.5

3.20
The ratio uses the division rule. The numerator has five significant figures and the denominator has three
significant figures, so the result of division will have three significant figures. Therefore, we get:
2221.05 −

2221.05 − 1.02 × 103 = 2.22105 × 103 − 1.02 × 103
The next calculation uses the subtraction rule. To count decimal places, both numbers must have the same
power of ten:

2.22105 × 103 − 1.02 × 103
In the ×103 power of ten, the first number has five decimal places (the 2, 2, 1, 0, and 5 are all decimal
places) and the second number has two decimal places (the 0 and the 2 are both decimal places). The result
× 103.
of the subtraction has two decimal places, in the × 103 power of ten: 1.20×
Another way to look at this is write both numbers 2221.05 – 1020 without scientific notation. The
uncertainty in the second number is in the tens place, so the answer must be rounded to the tens place.
(b)

(343.2) × (2.01×10–3)
This calculation uses the multiplication rule. The first number, 343.2, has four significant figures and the
second number, 2.01×10–3, has three significant figures, so the answer has three significant figures 0.690.

(c)

S = 4πr2 = 4 × (3.1415926) × (2.55 cm)2
This whole calculation uses the multiplication rule, with three significant figures, limited by the
measurement of r. The numeral 4 is exact, in this context. The value of π should be carried to more than
three significant figures, such as 3.1415926… The answer comes out 81.7 cm2.

(d)

2802
− 0.0024 × 10, 000.
15
The ratio uses the division rule. The numerator has four significant figures and the denominator has two
significant figures, so the result has two significant figures. The multiplication also uses the same rule.
The first number has two significant figures and the second number has 5 significant figures. So, the result
of the multiplication will have two significant figures.

(

)

1.9 × 102 – 25
The difference uses the subtraction rule. We need these numbers in the same power of ten to be able to
count decimal places.
1.9 × 102 – 0.25 × 102
In the ×102 power of ten, the first number has one decimal places (the 9) and the second number has two
decimal places (the 2 and the 5 are both decimal places). The result of the subtraction (1.65 × 102) has to be
rounded to one decimal place in the ×102 power of ten: 1.7 × 102. Another way to look at this is that the
uncertainty in the first number (190) is in the tens place, so the answer (165) must be rounded to the tens
place, 170, with the zero being non-significant.

Reasonable Result Check: The significant figures, size, and units of the answers are appropriate.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

9

23. Result: Copper
Analyze and Plan: We have the mass of the metal and some volume information. We need to determine the
density. Use the initial and final volumes to find the volume of the metal piece, then use the mass and the
volume to get the density. The metal piece displaces the water when it sinks, making the volume level in the
graduated cylinder rise.
Execute: The metal piece volume is the difference between the starting volume and the final volume:
Vmetal = Vfinal – Vinitial = (37.2 mL) – (25.4 mL) = 11.8 mL

d=

m 105.5 g
g
=
= 8.94
V 11.8 mL
mL

According to Table 1.1, this is very close to the density of copper (d = 8.93 g/mL).

Reasonable Result Check: The metal piece sinks, so the density of the metal piece must be higher than water.
(Table 1.1 gives water density as 0.998 g/mL.)
24. Result: 30.9 mL
Analyze and Plan: We have the mass of the lead and the initial volume information. We need to determine the
final volume after the lead is submerged. The metal piece displaces the water when it sinks, making the volume
level in the graduated cylinder rise. The difference between the starting volume and the final volume, must be
the volume of the lead. Since we know the metal is made of lead, look up the density. Use the density and the
mass information to find the volume of the lead, then use the volume of the lead and the initial volume of the
ethanol to get the final volume.
Execute: The density of lead is 11.34 g/mL. Use conversion factors to convert the lead mass into mL of lead:
1 mL lead
10.0 g lead ×

= 0.882 mL lead
11.34 g lead
The difference between the starting volume and the final volume must be the volume of the metal piece, so
adding the volume of the metal piece to the starting volume gives the final volume:
Vfinal = Vmetal + Vinitial = (30.0 mL) + (0.882 mL) = 30.882 mL
Rounding to one decimal place, the limit of the uncertainty, gives a final volume of 30.9 mL.

Reasonable Result Check: The sample mass is close to the density (the number of grams in one mL), so the
ethanol level should rise about one mL.
25. Result: Aluminum
Analyze: We have the three linear dimensions of a regularly shaped piece of metal.
10.0 cm long
1.0 cm thick
2.0 cm wide
We also have its mass. We have a table of densities (Table 1.1). We need to determine the identity of the
metal.
Plan: Use the three linear dimensions to find the volume of the metal piece. Use the volume and the mass to
find the density. Use the table of densities to find the identity of the metal.
Execute: V = (thickness) × (width) × (length) = (1.0 cm) × (2.0 cm) × (10.0 cm) = 20. cm
Using conversion factors, find the volume in mL: 20. cm3 ×
Find the density:

3

1 mL

= 20. mL
1 cm3
g
m 54.0 g
= 2.7

d= =
mL
V 20. mL

According to Table 1.1, the density that most closely matches this one is aluminum (d = 2.70 g/mL).

Reasonable Result Check: The mass is larger than the volume. so d is larger than 1.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
10

Chapter 1: The Nature of Chemistry

26. Result: 15.7 mL
Analyze and Plan: We have the mass of the sample, and the identity. We are asked to find the volume.
Use density of bromobenzene and mass to find the volume.
Execute: Density of bromobenzene is 1.49 g/mL.

23.4 g ×

1 mL
= 15.7 mL
1.49 g


Reasonable Result Check: A density larger than 1g/mL tells us that the volume of a sample of
bromobenzene will be smaller than the mass.
27. Result: 3.9 × 103 g
Analyze: We have the three linear dimensions of a regularly shaped sodium chloride crystal:
12 cm long
10. cm thick

15 cm wide
We have a table of densities (Table 1.1). We need to determine the mass of the crystal.
Plan and Execute: Use the three linear dimensions to find the volume of the crystal. Use the volume and
density (Table 1.1) to find the mass.
V = (thickness) × (width) × (length) = (10. cm) × (15 cm) × (12 cm) = 1800 cm3 = 1.8 × 103 cm3
Using conversion factors, find the volume in mL:

1 mL

1.8 × 103 cm3 ×

3

= 1.8 × 103 mL

1 cm
Find the mass using the density:

1.8 × 10 3 mL ×

2.16 g
= 3.9 × 10 3 g
1 mL

Notice: We’re carrying two significant figures since the length data was only that precise.

Reasonable Result Check: This crystal is pretty large. So, while the mass calculated is a large number, the
volume is still about half the mass, consistent with a density around two.
28. Result: 0.551 mL
Analyze: We have the mass of a liquid. We need to determine the volume.

Plan: Use the mass and density (Table 1.1) to find the volume.
Execute: Table 1.1 gives the density of iron (d = 7.86 g/mL). Find the volume using the density as a ratio
(mL/g) so that the mass units (g) will cancel when multiplied by the mass in grams:

4.33 g ×

1 mL
= 0.551 mL
7.86 g


Reasonable Result Check: The density is less than 1, so the volume should be larger than the mass. Here,
the calculated volume of 4.92 mL is a larger number than the original mass (4.33 g).

Chemical Change and Chemical Properties (Section 1-6)
29. Result: (a)

Physical (b) Chemical (c) Chemical (d) Physical

Explanation:
(a) The normal color of bromine is a physical property. Determining the color of a substance does not change


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

11

its chemical form.
(b) The fact that iron can be transformed into rust is a chemical property. Iron undergoes a transformation,
from its elemental metallic state to become a part of the compound identified as rust.

(c) The fact that dynamite can explode is a chemical property. The dynamite is chemically changed when it is
observed to explode.
(d) Observing the shininess of aluminum does not change it, so this is a physical property. Melting aluminum
does not change it to a different substance, though it does change its physical state. It is still aluminum, so
melting at 660 °C is a physical property of aluminum.
30. Result: (a)

Physical (b) Chemical (c) Physical (d) Chemical

Explanation:
(a) Subliming Dry Ice (converting it from a solid directly into a gas without going through the liquid state,
first) does not change it to a different substance, though the physical state does change. It is still the same
chemical material. Therefore, subliming at – 78 °C is a physical property of Dry Ice.
(b) The fact that methanol burns in air is a chemical property. The methanol compound is converted into
combustion products when it is observed burning.
(c) Sugar’s solubility in water is a physical property. Dissolving sugar does not change it to a different
substance, though it does change its physical state. It is still sugar. Evaporating the water would recreate
the sugar crystals.
(d) Hydrogen peroxide’s ability to decompose into oxygen and water is a chemical property. The hydrogen
peroxide compound is converted into oxygen and water when it is observed decomposing.
31. Result: (a)

Chemical (b) Chemical (c) Physical

Explanation:
(a) Bleaching clothes from purple to pink is a chemical change. The purple substance in the clothing reacts
with the bleach to make a pink substance. The purple color cannot be brought back nor can the bleach.
(b) The burning of fuel in the space shuttle (hydrogen and oxygen) to form water and energy is a chemical
change. The two elements react to form a compound.
(c) The ice cube melting in the lemonade is a physical change. The H2O molecules do not change to a

different form in the physical state change.
32. Result: (a)

Physical (b) Chemical (c) Physical (d) Chemical

Explanation:
(a) Salt dissolving when you add water is a physical change. The salt crystals would be reformed if the water
were evaporated off.
(b) Food is digested and metabolized in your body by way of a chemical change. The molecules composing
the food are decomposed to provide energy for activities and functions in your body.
(c) Sugar stays sugar when it is ground from a crystalline state into a powdered state, so this is a physical
change.
(d) When potassium is put in water, there is a purplish-pink flame and the water becomes basic (alkaline).
This is a chemical change of the elemental potassium and water into a compound in the water that makes it
basic.
33. Result: (a) Forcing a chemical reaction to occur (b) Causing work to be done (c) Causing work to be
done (d) Forcing a chemical reaction to occur
Explanation:
(a) The conversion of excess food into fat molecules is the body’s way of storing energy for doing work later.
So, this represents an outside source of energy (from the food we eat) forcing a chemical reaction to
occur (the production of fat).
(b) Sodium reacts with water rather violently. It produces a lot of heat energy and causes work to be done.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
12

Chapter 1: The Nature of Chemistry

(c) Sodium azide in an automobile’s airbag decomposes causing the bag to inflate. This uses a chemical

reaction to release energy and cause work to be done (inflation of the air bag).
(d) The process of hard-boiling an egg on your stove uses energy from the stove to cause a chemical reaction
to occur (the coagulation of the white and yolk of the egg).
34. Result/Explanation: Many people may not think of all of these things.
•

Gathering the wood caused you to sweat and feel warm. That was due to metabolic processes involving
complex biochemical changes resulting from your body expending energy. Water may have been a product
of those metabolic processes– hence, a chemical change.

•

Your body also loses heat energy when the sweat evaporates off your skin – a physical change involving
the transfer of heat energy.

•

Once you stopped collecting wood, you started feeling cold again, since your warm body heat energy was
escaping into the cold night air– again a physical change involving the transfer of heat energy – so you lit a
fire.

•

The wood, when lighted, underwent a chemical change known as combustion. This energy-producing
chemical change produced heat energy that warmed up the air nearby, a physical change. Some of the heat
energy was absorbed by you, also a physical change, to replenish the heat energy you are losing to the cold
night air. That made you feel warmer.

•

Later, you stopped the combustion reaction in the fire pit by separating the reactants of the combustion
reaction. Adding water to the coals – a physical change –prevented it from continuing to interact with the
oxygen.

•

In addition, the cold water absorbed the heat energy from the fire, warming the water up – a physical
change. Some of the water absorbed so much heat energy that it was converted from liquid form to vapor
form – a physical change. That was the steam you saw.

•

When you got into your sleeping bag, the heat energy lost from your body was retained near you by the
reflective fabric of the bag – a physical change. That reduced the loss of heat energy from your warm body
into the cold night air, so you were able to keep yourself warm all night.

Classifying Matter: Substances and Mixtures (Section 1-7)
35. Result/Explanation: It is clear by visual inspection that the mixture is non-uniform (heterogeneous) at the
macroscopic level. Iron could be separated from sand using a magnet, since iron is attracted to magnets and
the sand is not.
36. Result: (a) Heterogeneous (b) Homogenous (c) Heterogeneous (d) Homogeneous
Explanation: The terminology “heterogeneous” and “homogeneous” is somewhat subjective. In Section 1-7,
the terms are described. A heterogeneous mixture is described as one whose uneven texture can be seen without
magnification or with a microscope. A homogeneous mixture is defined as one that is completely uniform,
wherein no amount of optical magnification will reveal variations in the mixture’s properties. Notice there is a
“gap” between these two terms. The question is: “how close do I look?” The bottom line is this: These terms
were designed to HELP us classify things, not to create trick questions. If you explain your answer with a valid
defense, the answer ought to be right. Think about what you would SEE. Identify whether what you see has
variations, then make a case for the proper term.
(a) An asphalt blacktop is classified as a heterogeneous mixture. Driving down the road, the blacktop looks

pretty uniform, but if you look at it closely, the mixture is made up of a black goopy material along with
other substances that are more like black or dark dull rocks. These visual variations lead us to conclude
that the mixture is heterogeneous.
(b) Ocean water, if it were clear, would be considered to be a homogeneous mixture. Dissolved in the water,
we know that there is salt and other dissolved minerals. However, if the solution has seaweed or silt
suspended in it – we might be tempted to call it heterogeneous.
(c) Ice tea with ice cubes is definitely heterogeneous. The portions representing ice are far different from the
aqueous solution of tea particles, both in composition and in physical state. It is clear that this is an
example of a heterogeneous mixture.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

13

(d) This filtered apple juice is a homogeneous solution of sugar, other flavors, and water. If the solution is
filtered sufficiently to remove all the pulp, which might qualify it as heterogeneous, it is probably safe to
call this mixture homogeneous.
37. Result/Explanation: Sometimes, it is necessary to try some tests to see if different parts of the mixture respond
differently to physical separation techniques. It may require some experimentation, such as testing whether
samples of the pure substances dissolve in water or are attracted to a magnet. Such information may also be
available on the internet.
(a) Table salt dissolved in water can be separated by evaporating the water, which would leave the dry salt.
(b) Testing shows that iron filings are attracted to a magnet, but magnesium pieces are not. Using a magnet,
the iron filings can be lifted out of the mixture, leaving behind the magnesium pieces.
(c) This mixture will have shiny silver metal pieces in a white crystalline powder, so the first thing we could
try is use tweezers or forceps to pick out the shiny metal zinc pieces from the white sugar crystals.
Solubility testing shows that sucrose dissolves in water, but zinc does not. Therefore, put the mixture in
water, dissolve the sucrose, separate the solution from the solid zinc using a filter or a sieve, then

evaporate the water.
38. Result/Explanation: Sometimes, it is necessary to try some tests to see if different parts of the mixture respond
differently to physical separation techniques. It may require some experimentation, such as testing whether
samples of the pure substances dissolve in water or are attracted to a magnet. Such information may also be
available on the internet.
(a) Table sugar dissolved in water can be separated by evaporating the water, which would leave the dry
sucrose.
(b) This mixture will have bright yellow pieces in a white crystalline powder, so the first thing we could try is
use tweezers or forceps to pick out the yellow sulfur pieces from the white sugar crystals. Solubility testing
shows that table salt dissolves in water, but sulfur does not. Therefore, put the mixture in water, dissolve
the salt, separate the solution from the solid sulfur using a filter or a sieve, then evaporate the water.
(c) Testing shows that iron filings are attracted to a magnet, but granular zinc is not. Using a magnet, the iron
filings can be lifted out of the mixture, leaving behind the granular zinc.

Classifying Matter: Elements and Compounds (Section 1-8)
39. Result/Explanation:
(a) A blue powder turns white and loses mass. The loss of mass is most likely due to the creation of a gaseous
product. That suggests that the original material was a compound that decomposed into the white
substance (a compound or an element) and a gas (a compound or an element).
(b) If three different gases were formed, that suggests that the original material was a compound that
decomposed into three compounds or elements.
40. Result/Explanation:
(a) A reddish metal is placed in a flame. The material turns black and the black material has a higher mass
than the original reddish material. That suggests that the reddish substance combined with something in the
air to form a new compound.
(b) A white solid is heated in oxygen and forms two gases. The mass of the gases produced is the same as the
solid and the oxygen. Oxygen is a reactive element and readily combines with other elements to form
compounds. That suggests that the product gases are compounds and the elemental oxygen combined with
another element or compound to form one or two compounds. However, there is no way to absolutely
rule out the possibility that one of the product gases might be an element instead of a compound.

41. Result: (a) Heterogeneous mixture (b) Pure compound (c) Element (d) Homogeneous mixture
Explanation:
(a) Chunky peanut butter is definitely a heterogeneous mixture. The uncrushed peanut chunks do not have
the same properties as the smooth, sweetened part of the mixture.
(b) Distilled water is a pure compound. The distillation process removes other minerals and substances from


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
14

Chapter 1: The Nature of Chemistry

water, leaving it just water.
(c) Platinum is an element with the symbol “Pt”.
(d) Air is usually considered to be a homogeneous mixture. Now, sometimes air has enough variable
properties to qualify as heterogeneous, such as near the tailpipe of a diesel truck; however, most of the
time, the gases in a sample of air are sufficiently well mixed such that there is no visible difference in the
properties of various regions of that air sample.
42. Result: (a) Compound (b) Compound (c) Heterogeneous mixtures (d) Element
Explanation:
(a) Table salt (sodium chloride) is a compound. Some brands of table salt are “iodized” which suggests that
some of the crystals of salt in the mixture are NaI along with NaCl. In those instances, it is appropriate to
call table salt a mixture; however, table salt is generally considered to be just the one compound, NaCl.
(b) Methane is a compound. The combustion products (carbon dioxide and water) suggest that methane must
contain at least carbon and hydrogen, making it a compound.
(c) Chocolate chip cookies are heterogeneous mixtures. It is clear that the properties of the regions including
a chocolate chip are quite different from the regions including the cooked cookie dough.
(d) Silicon is an element, with the symbol “Si”.
43. Result: (a) No (b) Maybe
Explanation:

(a) The black substance was both the source of the element that contributes to the red-orange substance and the
source of the oxygen in the water.
(b) The red-orange substance may be a combination of two or more elements including possibly hydrogen or
oxygen, or it maybe an elemental substance, since the water produced could account for the hydrogen and
the oxygen in the products.
44. Result: (a) No (b) Yes
Explanation:
(a) It may be an element, but it may be a compound, there is insufficient information to tell.
(b) The increase in mass and the color change suggests the formation of a compound.

Nanoscale Theories and Models (Section 1-9)
45. Result/Explanation: Using Figure 1.16 to help define scales. The scale bar defines the image scale to be in
nanometers. The images of electrons from the scanning tunneling microscope are at the nanoscale.
46. Result/Explanation: Using Figure 1.16 to help define scales. The tobacco mosaic virus is microscale.
47. Result/Explanation: When we open a can of a carbonated beverage, the carbon dioxide gas expands rapidly as it
rushes out of the can. At the nanoscale, this can be explained as large number of carbon dioxide molecules
crowded into the unopened can. When the can is opened, the molecules that were about to hit the can surface
where the hole was made continue forward through the hole. A large number of the carbon dioxide particles
that were contained within the can escape very quickly through the same hole.
48. Result/Explanation: Wet clothes dry in the sun. At the nanoscale, this is described by the relatively slow liquid
water molecules being made to move more quickly by the heat energy from the sun, as well as wind taking
molecules off the surface of the liquid. Once the molecules get free of the liquid state, they can’t often return to
the condensed phase. Once sufficient warming or evaporation occurs, the clothes feel dry. The water
undergoes only a change in physical state, thus the change is a physical change, not a chemical one.
49. Result/Explanation: The atoms in the solid sucrose molecules start off at a relatively low energy and compose a
rather complex molecule. A significant amount of heat energy must be added to increase the motion of these
atoms so that they are able to break free of the bonds that hold them together in the sugar molecule and to
interact with each other to make the “caramelized” products.
50. Result/Explanation: The melting point is the temperature at which a solid changes into a liquid. In the

Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

15

nanoscale picture, the liquid molecules are moving faster and they are farther apart. That motion spreads out
the molecules and makes the volume larger, even though the collection of molecules has the same mass. Hence,
the density of the liquid is smaller than the density of the solid.
The boiling point is the temperature at which the liquid molecules change into gas-phase molecules. In the
nanoscale picture, the gas molecules are moving faster and they are farther apart. That motion spreads out the
molecules and makes the volume larger, even though the collection of molecules have the same mass. Hence,
the density of the gas is smaller than the density of the liquid.

The Atomic Theory (Section 1-10)
51. Result/Explanation: Conservation of mass is easy to see from the point of view of the atomic theory. A
chemical change is described as the rearrangement of atoms. Because the atoms in the starting materials must
all be accounted for in the substances produced, and because the mass of each atom does not change, there
would be no change in the overall mass.
52. Result/Explanation: Atomic theory can explain the constant composition of chemical compounds. Compounds
are composed of a specific number and type of atoms. Atoms of each element all weigh the same. The sum of
all the atoms’ masses is the mass of the compound. The percentage of mass represented by one element in the
compound must be constant, since all the atoms of that element are the same mass, and the compound contains
a fixed number of those element’s atoms.
53. Result/Explanation: The law of multiple proportions says that if two compounds contain the same elements and
samples of those two compounds both contain the same number of atoms of one element, then the ratio of the
atoms of the other elements will be a small whole number.
54. Result: Yes, they do, due to simple whole-number ratios of Cr mass in each sample
Analyze: These three compounds all contain chromium and oxygen. The mass of chromium in each compound
is given for 100-gram samples of the respective compounds. Determine if these data conform to the law of

multiple proportions.
Plan: The law of multiple proportions says: if two compounds contain the same elements and samples of those
two compounds both contain the same mass of one element, then the ratio of the masses of the other elements
will be a small whole number. Find the masses of Cr and O in 100.0-gram samples. Scale the sample sizes of
each compound so that all three of them contain the same amount of oxygen. Determine the mass of chromium
in each of those samples. Take ratios of the chromium masses to see if they can be represented by ratios of
small whole numbers.
Execute: The masses of Cr in 100.0-gram samples are given. The mass of O must be the difference between
the total compound mass and the mass of Cr:
Compound 1:

100.0 g of compound 1 – 52.0 g Cr = 48.0 g O

Compound 2:

100.0 g of compound 2 – 68.4 g Cr = 31.6 g O

Compound 3:

100.0 g of compound 3 – 76.5 g Cr = 23.5 g O

Determine the scale factor by determining a multiplier that makes every sample have the same oxygen mass as
the first 100.0-gram sample:
⎛ 48.0 g ⎞
100.0 g of compound 1 has 48.0 g O
⎜
⎟ = 1.00
⎝ 48.0 g ⎠
⎛ 48.0 g ⎞
100.0 g of compound 2 has 31.6 g O

⎜
⎟ = 1.52
⎝ 31.6 g ⎠
⎛ 48.0 g ⎞
100.0 g of compound 3 has 23.5 g O
⎜
⎟ = 2.04
⎝ 23.5 g ⎠
Scale the samples by these factors to determine the mass of Cr in each sample.
100.0 g of compound 1 has 48.0 g O and 52.0 g Cr
(1.52) × 100.0 g of compound 2 has (1.52) × 31.6 g O and (1.52) × 68.4 g Cr
So, 151.9 g of compound 2 has 48.0 g O and 104 g Cr


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
16

Chapter 1: The Nature of Chemistry

(2.04) × 100.0 g of compound 3 has (2.04) × 23.5 g O and (2.04) × 76.5 g Cr
So, 204 g of compound 3 has 48.0 g O and 156 g Cr
Now, take ratios of the chromium masses:

g Cr in compound 2 104 g
2
=
= 2.00 =
g Cr in compound 1 52.0 g
1
g Cr in compound 3

=

156.3 g
3
= 3.01 =
52.0 g
1

=

156.3 g
3
= 1.50 =
104 g
2

g Cr in compound 1
g Cr in compound 3
g Cr in compound 2

These simple whole-number ratios show that these data do conform to the law of multiple proportions.

Reasonable Result Check: A law is a summary conclusion from a large number of different experiments. It
is satisfying that these results also uphold the law.

Communicating Chemistry: Symbolism (Section 1-11)
55. Result/Explanation: Formula for each substance and nanoscale picture:
(a) Water

H 2O

(b) Nitrogen

N2

(c) Neon

Ne

(d) Chlorine

Cl2

56. Result/Explanation: Formula for the substance and nanoscale picture:
(a) Iodine

I2

(b) ozone

O3


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

(c) Helium

He

57. Result/Explanation:

(d) Carbon dioxide

S (s) + O2 (g)

Sulfur solid and
oxygen gas
58. Result/Explanation:

CO 2

SO2 (g)

Sulfur dioxide gas

2 CO (g) + O2 (g)

carbon monoxide and oxygen gas
59. Result/Explanation: I2 (s)

liquid bromine

carbon dioxide gas

I2 (g)

solid iodine
60. Result/Explanation: Br2 ()

2 CO2 (g)

iodine gas
Br2 (g)

bromine gas

17


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
18

Chapter 1: The Nature of Chemistry

The Chemical Elements (Section 1-12)
61. Result/Explanation: Many pairs of responses are equally valid. Below are a few common examples. These lists
are not comprehensive; many other answers are also right. The periodic table on the inside cover of the
textbook is color coded to indicate metals, non-metals and metalloids.
(a) Common metallic elements: iron, Fe; gold, Au; lead, Pb; copper, Cu; aluminum, Al
(b) Common non-metallic elements: carbon, C; hydrogen, H; oxygen, O; nitrogen, N
(c) Metalloids: boron, B; silicon, Si; germanium, Ge; arsenic, As; antimony, Sb; tellurium, Te
(d) Elements that are diatomic molecules: nitrogen, N2; oxygen, O2; hydrogen, H2; fluorine, F2; chlorine, Cl2;
bromine, Br2; iodine, I2
62. Result/Explanation: Many pairs of responses are equally valid here. Below are a few common examples.
These lists are not comprehensive; many other answers are also right.
(a) Some elements that are gases at room temperature: N2, nitrogen; O2, oxygen; H2, hydrogen; He, helium;
Ne, neon; Ar, argon; Kr, krypton.
(b) Some elements that are solids at room temperature: Fe, iron; Au, gold; Pb, lead; Cu, copper; Al, aluminum;
C, carbon; B, boron; Si, silicon.

(c) Some elements that are not molecules: Fe, iron; Au, gold; Pb, lead; Cu, copper; Al, aluminum; C, carbon;
B, boron; Si, silicon.
(d) Some elements that have different allotropic forms: O oxygen (O2 and O3); C carbon (graphite, diamond,
and fullerenes).

The Periodic Table (Section 1-13)
63. Result/Explanation: Possible elements are Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answers will vary depending
upon the source chosen. For example, the New College Edition of the American Heritage Dictionary of the
English Language (©1978) gives the following information:
Titanium: A strong, low-density, highly corrosion resistant, lustrous white metallic element that occurs widely
in igneous rocks and is used to alloy aircraft metals for low weight, strength, and high-temperature stability.
Atomic number 22, atomic weight 47.90, melting point 1,675°C, boiling point 3,260°C, specific gravity 4.54,
valences 2, 3, 4. [New Latin, from Greek Titan, TITAN, So named by Klaproth, who had also named uranium
after the Planet Uranus. Uranus, in Greek Mythology, is the father of the Titans.](page 1348)
Chromium: A lustrous, hard, steel-gray metallic element, resistant to tarnish and corrosion, and found
primarily in chromite. It is used as a catalyst, to harden steel alloys, to produce stainless steel, in corrosionresistant decorative platings, and as a pigment in glass. Atomic number 24, atomic weight 51.996, melting
point 1,890°C, boiling point 2,482°C, specific gravity 7.18, valences 2, 3, 6. [New Latin, from French Chrome,
CHROM(E).] (page 240)
Iron: A silvery-white, lustrous, malleable, ductile, magnetic or magnetizable, metallic element occurring
abundantly in combined forms, notably in hematite, limonite, magnetite, and taconite, and used alloyed in a
wide range of important structural materials. Atomic number 26, atomic weight 55.847, melting point 1,535°C,
boiling point 3,000°C, specific gravity 7.847 (20°C), valences 2, 3, 4, 6. [Middle English yren, yron, iren, Old
English iren, earlier isern, isen] (page 691)
Copper: A ductile, malleable, reddish-brown metallic element that is an excellent conductor of heat and
electricity and is widely used for electrical wiring, water piping, and corrosion-resistant parts either pure or in
alloys such as brass and bronze. Atomic number 29, atomic weight 63.54, melting point 1,083°C, boiling point
2,595°C, specific gravity 8.96, valences 1, 2. [Middle English coper, Old English coper, copor, from common
Germanic kupar (unattested), from Late Latin cuprum, from Latin Cyprium, “(copper) of Cyprus (Cyprus was
known in ancient times as the source of the best copper)”, from Cyprus, of Cyprus, from Greek Kuprios, from
Kupros, CYPRUS] (page 294)

64. Result/Explanation: New College Edition of the American Heritage Dictionary of the English Language
(©1978) gives the following information:


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

19

Chlorine: A highly irritating, greenish-yellow gaseous halogen, capable of combining with nearly all other
elements, produced principally by electrolysis of sodium chloride and used widely to purify water, as a
disinfectant, a bleaching agent, and in the manufacture of many important compounds including chloroform and
carbon tetrachloride. Atomic number 17, atomic weight 35.45, freezing point –100.98°C, boiling point
–34.6°C, specific gravity 1.56 (– 33.6°C), valences 1, 3, 5, 7. [CHLORO- + -INE] (page 236)
Fluorine: A pale-yellow, highly corrosive, highly poisonous, gaseous halogen element, the most
electronegative and most reactive of all elements. It is used in a wide variety of industrially important
compounds. Atomic number 9, atomic weight 18.9984, freezing point –219.62°C, boiling point –188.94°C,
specific gravity of liquid 1.108, valences 1. [French, from New Latin fluor, generic name for a group of
minerals used as fluxes, FLUOR] (page 506)
65. Result/Explanation: There are currently six elements in Group 4A of the periodic table. They are non-metal:
carbon (C), metalloids: silicon (Si) and germanium (Ge), and metals: tin (Sn), lead (Pb) and flerovium (Fl).
66. Result/Explanation: The fourth period of the periodic table has eighteen elements. They are the following
metals: potassium (K), calcium (Ca), scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese
(Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), zinc (Zn), gallium (Ga); the following metalloids:
germanium (Ge) and arsenic (As); and the following non-metals selenium (Se), bromine (Br), and krypton (Kr).
67. Result: (a) I (b) In (c) Ir (d) Fe
Explanation: Look up the four elements on Figure 1.26 in Section 1-13.
(a) I, iodine, is the halogen (because it is in Group 7A)
(b) In, indium, is a main group metal (because it is a metal found in an A group—Group 3A)
(c) Ir, iridium, is a transition metal (colored blue on Figure 1.26) in period 6. The period number 6 is given to

the far left of the row in the periodic table next to the element Cs.
(d) Fe, iron, is a transition metal (colored blue on Figure 1.26) in period 4. The period number 4 is given to the
far left of the row in the periodic table next to the element K.
68. Result: (a) S (b) Na (c) Ag (d) Si
Explanation: Look up the four elements on Figure 1.26 in Section 1-13.
(a) S, sulfur, is the solid non-metal. It is the only non-metal (colored violet on Figure 1.26) given.
(b) Na, sodium, is the alkali metal (because it is in Group 1A).
(c) Ag, silver, is the transition metal (colored blue on Figure 1.26).
(d) Si, silicon, is the metalloid (colored orange on Figure 1.26).
69. Result: (a) Mg (b) Na (c) C (d) S (e) I (f) Mg (g) Kr (h) O (i) Ge Notice: There are multiple
Results to (a), (b) and (i) in this Question. The ones given here are only examples.
Analyze: Use the periodic table and information given in Section 1-13.
(a) An element in Group 2A is magnesium (Mg).
(b) An element in the third period is sodium (Na).
(c) The element in the second period of Group 4A is carbon (C).
(d) The element in the third period in Group 6A is sulfur (S).
(e) The halogen in the fifth period is iodine (I).
(f) The alkaline earth element in the third period is magnesium (Mg).
(g) The noble gas element in the fourth period is krypton (Kr).
(h) The non-metal in Group 6A and the second period is oxygen (O).
(i) A metalloid in the fourth period is germanium (Ge).
70. Result: (a) Zn (b) Xe (c) Pb (d) P (e) Na (f) Xe (g) Se (h) Te Notice: There are multiple Results to
(a), (b) and (h) in this Question. The ones given here are only examples.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
20

Chapter 1: The Nature of Chemistry

Analyze: Use the periodic table and information given in Section 1-13.
(a) An element in Group 2B is zinc (Zn).
(b) An element in the fifth period is xenon (Xe) (or any other element with atomic number between 37-54).
(c) The element in the sixth period in Group 4A is lead (Pb).
(d) The element in the third period in Group 5A is phosphorus (P).
(e) The alkali metal in the third period is sodium (Na).
(f) The noble gas in the fifth period is xenon (Xe).
(g) The element in Group 6A and the fourth period is selenium (Se). It is a non-metal.
(h) A metalloid in Group 5A is antimony (Sb) or arsenic (As).

Biological Periodic Table (Section 1-14)
71. Result/Explanation: One compound makes up about 60% of the human body and is nearly 90% oxygen. This
compound is H2O. Other compounds that have significant oxygen contribution are phosphate and carbonate
compounds. Carbohydrates have more oxygen than fats and oils.
72. Result/Explanation:
(a) Metals are found in the body as ions.
(b) Two uses for metals in the body are calcium (Ca2+) in bones and teeth and Fe2+ in hemoglobin used in the
transport of oxygen in blood. There are many other answers.
73. Result/Explanation: Macrominerals are also called “major minerals” which are more abundant in the human
body (see Figure 1.27). Major minerals are present in greater than 0.01% of body weight (100 mg per kg).
Microminerals are also called “trace minerals” which are less plentiful. Trace minerals are present in less than
0.01% of body weight (100 mg per kg), sometimes far less.
74. Result/Explanation: An essential trace mineral that is poisonous at high levels is arsenic, as discussed in
Section 1-2. Most heavy metals such as iron and copper and nonmetals such as iodine are safe and necessary,
but they are toxic at high concentrations.

General Questions
75. Result/Explanation:
(a) The mass of the compound (1.456 grams) is quantitative and relates to a physical property. The color
(white), the fact that it reacts with a dye, and the color change in the dye (red to colorless) are all

qualitative. The colors are related to physical properties. The reaction with the dye is related to a
chemical property.
(b) The mass of the metal (0.6 grams) is quantitative and relates to a physical property. The identity of the
metal (lithium) and the identities of the chemicals it reacts with and produces (water, lithium hydroxide,
and hydrogen) are all qualitative information. The fact that a chemical reaction occurs when the metal is
added to water is qualitative information and related to a chemical property.
76. Result/Explanation:
(a) The volume of the water (one liter) is quantitative and relates to a physical property. The identities of the
materials (water, charcoal) are qualitative information. The fact that the charcoal adsorbs the dye is
qualitative and related to a physical property, since the adsorption process is reversed later and the dye’s
color (the one property we have to identify it) is retained.
(b) All of the information in these statements is qualitative. The colors, the fact that the white powder
dissolved, the container used to hold the solution, the relative temperature change, and the physical state of
the substance formed are related to physical properties. The fact that there was a transformation of the
dissolved substance into an insoluble substance is related to a chemical property.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

21

77. Result: Garden requires 3.0 ft3, which is more than 1.45 ft3 bag.
Analyze and Plan: Given the linear dimensions of the plot and the depth expected, calculate the volume
needed, then compare to the volume in the bag.
6 ft × 6 ft × 1 in ×

Execute:

1 ft

= 3.0 ft 3
12 in

Garden requires 3.0 ft3, which is more than 1.45 ft3 bag.

Reasonable Result Check: Perhaps the company meant to indicate ½” depth.
78. Result: 255 mL
Analyze: A known mass of sulfuric acid is in a solution with a given density and mass percentage. Determine
the volume in mL.
Plan: Start with the sample—here, the mass of sulfuric acid. Use the mass percentage as a conversion factor
between grams of sulfuric acid and grams of solution. Then use the density of the solution as a conversion
factor between grams and cubic centimeters. Then convert the cubic centimeters into milliliters.
Execute: 100 grams of the solution contains 30.08 grams H2SO4. 1 cubic centimeter of the solution weighs
1.285 grams.

125g H2SO4 ×

100 gsolution
1cm3 solution 1mLsolution
×
×
= 255 mLsolution
38.08 g H2SO4 1.285gsolution 1cm3 solution


Reasonable Result Check: The units “g H2SO4” cancel properly, as do the units “g solution”. Multiplying a
number by 100, then dividing by approximately 50 (= 38 × 1.3), should give an answer about twice as big.
79. Result: 0.197 nm, 197 pm
Analyze: A distance is given in angstroms (Å), which are defined. Determine the distance in nanometers and
picometers.
Plan: Use the given relationship between angstroms and meters as a conversion factor to get from angstroms to
meters. Then use the metric relationships between meters and the other two units to find the distance in

nanometers and picometers.
Execute:

1.97 Å ×

1 nm
1× 10−10 m
×
= 0.197 nm
1Å
1× 10−9 m

1.97 Å ×

1 pm
1× 10−10 m
×
= 197 pm
1 Å
1× 10−12 m


Reasonable Result Check: The unit nanometer is larger than an angstrom, so the distance in nm should be
smaller. The unit picometer is smaller than an angstrom, so the distance in pm should be larger.
80. Result: 0.995 g Pt
Strategy and Explanation: Given the mass of a sample of cisplatin along with its percentage platinum,
determine the grams of platinum.
Always start with the sample. Convert the mass of compound to mass of platinum using the percentage
platinum as a conversion factor. 100 grams of cisplatin contains 65.0 grams of Pt.

1.53 g cisplatin ×

65.0 g Pt
= 0.995 g Pt
100 g cisplatin


Reasonable Result Check: The mass of Pt should be about

2
3

of the mass of the compound cisplatin.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
22

Chapter 1: The Nature of Chemistry

81. Result/Explanation: If the density of solid calcium is almost twice that of solid potassium, but their masses are
approximately the same size, then the volume must account for the difference. This suggests that the atoms of
calcium are smaller than the atoms of potassium:

solid calcium
smaller atoms
closer packed
smaller volume

solid potassium
larger atoms
less closely packed

larger volume

82. Result/Explanation: The volume of a gas is independent of the identity of the gas molecule, because there is so
much empty space between one molecule and another in the gas state. Therefore, the density of a gas is directly
proportional to the mass of the substance of which the gas is composed. In this case, because the mass of an
argon atom is almost exactly ten times the mass of a helium atom, we see that the density of the argon gas is
almost exactly ten times the density of helium gas.
83. Result: 508 m
Analyze: We have the mass of a spool of aluminum wire with known diameter. Assuming the wire is a
cylinder, find the length () of wire in meters. Use the density to find the volume from the mass, then use the
given volume equation and the known diameter to find the length.

10.0 lb ×

453.59 g 1 mL 1 cm3
×
×
= 1680 cm3
2.70 g 1 mL
1 lb

The radius is half the diameter. Determine the radius in centimeters:
R=

1
2.54 cm
× 0.0808 in ×
= 0.103 cm
2
1 in

(

)

Rearrange V = πr2 to solve for , plug in the known values and convert to meters:


=

V
πr 2

=

1680 cm3

(

)

(3.14159) × 0.103 cm

2

×

1m
= 508 m
100 cm

Reasonable Result Check: It makes sense that a quantity of wire that weighs 10 pound would be several
hundred feet long.
84. Result: 1.7 × 10 –3 cm
Analyze: We have the length, width, and mass of a sample of aluminum foil. Find the thickness of the foil in
centimeters.
Plan and Execute: Use the density to find the volume from the mass:

0.83 kg ×

1000 g 1 mL 1 cm3
×
×
= 307.4 cm3 (2 sig figs)
2.70 g 1 mL
1 kg

Convert the known linear dimensions to centimeters:

length = 66

2
3 ft
12 in 2.54 cm
yards ×
×
×
= 6096 cm (3 sig figs)
3
1 yard 1 ft
1 in

width = 12 in ×

2.54 cm
= 30.48 cm (2 sig figs)
1 in


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

23

Then use V = (thickness) × (width) × (length), to find the thickness in centimeters:
(thickness) =

V
307.4 cm3
= 1.7 × 10 –3 cm
=
(width) × (length) 30.48 cm × 6096 cm

(

) (

)


Reasonable Result Check: Aluminum foil is quite thin, so this small number for thickness makes sense.
85. Result/Explanation: The highest density materials will sink to the bottom, with increasingly less dense

materials floating on top.
The solid material with the highest density is the Teflon plastic pieces (d = 2.3 g/cm3), so those pieces will be
found at the bottom of the graduated cylinder sitting in the liquid perfluorohexane, which has the highest
density of the liquids (d = 1.669 g/cm3).
Floating on the surface of the perfluorohexane will be the liquid water (d = 1.00 g/cm3).
Floating on the water will be the pieces of HDPE plastic (d = 0.97 g/cm3) and the liquid hexane
(d = 0.766 g/cm3).

Hexane
HDPE pieces
Water
Perfluorohexane
Teflon pieces
86. Result/Explanation: The density of the HDPE is 0.97 g/mL. It will float in any liquid with a density greater
than 0.97 g/mL. Examining the given chart, HDPE will float in ethylene glycol (d = 1.113 g/mL), water
(d = 0.9982 g/mL), acetic acid (1.0498 g/mL) and glycerol (d = 1.2611 g/mL).
87. Result: He, Ne, Ar, Kr, Xe, Rn, H2 , N2, O2, Cl2, F2; found at top and/or right side of the periodic table
Explanation: There is not one place in the textbook where all the gaseous elements at room temperature are
identified, but throughout the text there are a number of places where we can find the identity of these eleven
gaseous elements. Some are identified in Section 1-12 when discussing examples of gaseous non-metals: H2,
N2, and Cl2 and the noble gases: He, Ne, Ar, Kr, Xe, and Rn. Hydrogen, H 2, is identified as a gas in
Section 1-6, Figure 1.6, and Section 1-8. Oxygen, O2, is shown as a gas in Figure 1.22. The halogen chlorine,
Cl2, is depicted as a gas in Figure 1.24, and because halogen fluorine, F2, is an even smaller molecule than
chlorine, it is logical to believe it will be a gas, also. Lastly, it is possible to use the boiling points for various
elements given in Table 9.2 in Chapter 9 to confirm that several of these elements (Ne, Ar, Xe, F2, and Cl2) are
above their boiling point and therefore gaseous at room temperature. Except hydrogen, all these elements are
located at the far top right side of the periodic table.
88. Result: (a) K (b) Ar (c) Cu (d) Ge (e) H (f) Ca (g) Br (h) P
Explanation: Use the periodic table and information given in Section 1-13.
(a) K is an alkali metal. (Group 1A)

(b) Ar is a noble gas. (Group 8A)
(c) Cu is a transition metal. (Group 1B)
(d) Ge is a metalloid. (Group 4A)


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
24

Chapter 1: The Nature of Chemistry

(e) H is a group 1 nonmetal.
(f) Ca is an alkaline earth metal. (Group 2A)
(g) Br is a halogen. (Group 7A)
(h) P is a nonmetal that is a solid. (Group 5A)
89. Result/Explanation: Look at the periodic table given.
(a) A colorless gas is a non-metal. Those gases are found in the lavender area.
(b) A solid that is ductile and malleable are metals. Metals are found in the gray and blue areas.
(c) Non-metals and metalloids are poor electrical conductors. Solids with this characteristic are found in the
orange and lavender areas.
90. Result/Explanation: Relate the characteristics of elements to locations on the periodic table.
(a) A shiny solid that conducts electricity would be a metal, so it will be in the grey and blue areas.
(b) A gas whose molecules consist of single atoms will be a nonmetal, so it will be in the lavender area.
(c) An element that is a semiconductor will be a metalloid, so it will be in the orange area.
(d) A yellow solid with a low electrical conductivity will be a nonmetal, so it will be in the lavender area.
91. Result/Explanation: Se and S have the greatest similarities in physical and chemical properties because they are
both in the same periodic group (Group 6A).
92. Result/Explanation: Mg and Sr have the greatest similarities in physical and chemical properties because they
are both in the same periodic group (Group 2A).
93. Result/Explanation: A substance that can be broken down is not an element. A series of tests will result in a
confirmation with one positive test. To prove that something is an element requires a battery of tests that all

have negative results. A hypothesis that the substance is an element and cannot be broken down is more
difficult to prove. (Section 1-3)
94. Result: (a) Mixture; homogeneous (b) 58.1 kg (c) See explanation below
(a) Explanation: Fat is a mixture, solution of lye is a mixture of water (a substance that is a compound) and lye
(sodium hydroxide, a compound). The mixtures are probably both homogeneous.
(b) Analyze: Conservation of mass says that the soap will have a mass that is the sum of all three component
substances.
Plan: Convert the volume to kilograms and add all the masses.
Execute:

50.0 L water ×

1000 mL 0.998 g water
1 kg
×
×
= 49.9 kg
1L
1 mL water
1000 g

3.24 kg + 49.9 kg + 5.0 kg = 58.1 kg
(c) Explanation: Physical processes that occur include the mixing of substances to make solutions. Chemical
processes involved are the formation of carboxylate salts (soap molecules) and glycerol from fat and lye.

Reasonable Result Check: The density of water is close to one, so the mass and the volume are close to the
same value. If everything used is converted to soap, it make sense that the mass is additive.
95. Result: (a) Nickel, lead and magnesium (b) Titanium
Explanation:
(a) According to the table of densities, a metal will float if the density is lower. That means that nickel, lead
and magnesium will float on liquid mercury.

(b) The more different the densities, the smaller the fraction of the floating element will be below the surface.
That means that titanium will float highest on mercury.


Solution Manual for Chemistry The Molecular Science 5th Edition by Moore
Chapter 1: The Nature of Chemistry

25

96. Result: 6.02 × 10 –29 m3
Analyze and Plan: We have the length of the edge of a cube. Find the volume of the cube in m3.
392 pm long
392 pm thick

392 pm wide
Use the linear dimensions to find the volume, then convert the volume to m3 using metric conversions.
Execute:
V = (thickness) × (width) × (length) = (392 pm) × (392 pm) × (392 pm) = 6.02 × 107 pm3
Using conversion factors, find the volume in m3:

⎛ −12 ⎞3
10
m⎟
6.02 × 10 pm × ⎜
= 6.02 × 10−29 m3
⎜ 1 pm ⎟
⎝
⎠
7

3


Reasonable Result Check: The cube is from the nanoscale, so it makes sense that it would be a very small
volume using a macroscale unit of measure.
97. Result: 566 pm
Analyze and Plan: Given the volume of the cube in cm3, find the length of a side in picometers:
 pm long
 pm thick
 pm wide
Use the volume to find the three linear side dimensions of the cube, then convert the volume to pm using
metric conversions.
Execute: V = 3
3
 = 3 V = 1.81 × 10−22 cm3 = 5.66 × 10−8 cm

Using conversion factors, find the length in picometers:

5.66 × 10−8 cm ×

10−2 m
1 pm
×
= 566 pm
1 cm
10−12 m


Reasonable Result Check: The cube is from the nanoscale, so it makes sense that it would be a number
with convenient size using a nanoscale unit of measure.

Applying Concepts

98. Result/Explanation: Separating the components of a mixture containing sand, sugar, and sulfur requires
employing variable physical properties. Adding the mixture to water, would dissolve all the sugar. The sugar
can be reclaimed by separating the solution from the un-dissolved solid using filtration, then evaporating the
water. The remaining solid could be added to liquid carbon disulfide to dissolve the sulfur. The sulfur could be
reclaimed by separating the solution from the un-dissolved sand via filtration, then evaporating off the liquid
carbon disulfide. Drying the sand leaves it separate from the sugar and sulfur.
99. Result/Explanation: Separating the components of a mixture containing sand, salt, and naphthalene requires
employing variable physical properties. Adding the mixture to water, would dissolve all the salt. The salt could
be reclaimed by separating the solution from the un-dissolved solid using filtration, then evaporating the water.
The remaining solid could be added to liquid benzene to dissolve the naphthalene. Naphthalene could be