Transcript ch01 - Chemistry and Measurements

Chapter 1
Chemistry and Measurements
1
1.1
Chemistry and Chemicals
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
What Is Chemistry?
2
Chemistry
• is the study of composition,
structure, properties, and
reactions of matter
• happens all around you
everyday
Antacid tablets undergo a
chemical reaction when
dropped in water.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chemistry
3
Matter is another word for all substances that make up our world.
•
Antacid tablets are matter.
•
Water is matter.
•
Glass is matter.
•
Air is matter.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Branches of Chemistry
4
The field of chemistry is divided into branches, such as
• organic chemistry, the study of substances that contain
carbon
• inorganic chemistry, the study of all substances except
those that contain carbon
• general chemistry, the study of the composition, properties
and reactions of matter
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Chemistry + Other Sciences
5
Chemistry is often combined
with other sciences:
Geology + Chemistry
=
Geochemistry
Biology + Chemistry
=
Biochemistry
Physical Science + Chemistry =
Physical Chemistry
Biochemists analyze lab
samples.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Chemicals
6
Chemicals are
• substances that have the
same composition and
properties wherever found
• often substances made by
chemists that you use
everyday
Toothpaste is a combination
of chemicals.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Chemicals Commonly Used in Toothpaste
7
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chemicals Used When Cooking
8
Many substances found in the kitchen are obtained using chemical reactions.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
9
Which of the following contains chemicals?
A. sunlight
B. fruit
C. milk
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
10
Which of the following contains chemicals?
A. Sunlight is energy given off by the Sun and therefore
does not contain chemicals.
B. Fruit contains chemicals that have the same
composition and properties wherever found.
C. Milk contains chemicals that have the same
composition and properties wherever found.
Therefore, only B. fruit and C. milk contain chemicals.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
11
Which of the following activities should be included in your
study plan for learning chemistry?
A. Skipping lectures will not help you to understand the
concepts.
B. Forming a study group will be helpful for learning
chemistry.
C. Reviewing Learning Goals before reading will help you
understand and remember the concepts.
D. Becoming an active learner will help you understand the
concepts.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
12
1.2
A Study Plan for Learning Chemistry
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Features in This Text Help You Study
13
1. Periodic Table
- inside front cover
2. Tables
- inside back cover
3. Looking Ahead
- beginning of each chapter
4. Learning Goals
- beginning of each section
5. Glossary and Index
- end of text
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Before Reading
14
Before you begin to read the chapter
• obtain an overview of the chapter by reading Looking
Ahead
• look at the section title and rephrase it to be a question
• review the Learning Goal that tells you what to expect
in that section
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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As You Read
15
As you read the chapter
• try to answer the question you made from the
section title
• review Concept Checks that help you understand
key ideas
• work through the Sample Problems and review visual
Guide to Problem Solving
• try the Questions and Problems that allow you to apply
problem solving to new problems
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Throughout the Chapter
16
Throughout the chapter there are tools that help you connect
the chemical concepts you are learning to real life.
• Chemistry Link to Health
• Chemistry Link to Industry
• Chemistry Link to the Environment
• Chemistry Link to History
• Macro-to-Micro Illustrations
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Figures and Diagrams
17
Many figures and diagrams use micro-to-macro illustrations
• to depict atomic level of organization
• to illustrate the concepts in the text
• to allow you to see the world in a microscopic way
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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End of Chapter
18
At the end of the chapter there are study aids that complete
the chapter, such as
• Concept Maps, which show connections between important
concepts
• Chapter Reviews, which provide a summary
• Key Terms, which are listed with their definitions
• Understanding Concepts, which help you visualize
concepts
• Additional Questions and Problems and Challenge
Problems, which provide a means to assess your
understanding
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Active Learning Steps
19
You can become an active learner and enhance your learning
process by
• reading each Learning Goal for an overview
• forming a question from the section title and looking for
answers as you read
• self-testing by working Concept Checks, Sample
Problems, and Study Checks and checking answers
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Study Check
20
Which of the following activities should be included in your
study plan for learning chemistry?
A. skipping lectures
B. forming a study group
C. reviewing Learning Goals before reading
D. becoming an active learner
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
21
Which of the following activities should be included in your
study plan for learning chemistry?
A. Skipping lectures will not help you to understand the
concepts.
B. Forming a study group will be helpful for learning
chemistry.
C. Reviewing Learning Goals before reading will help you
understand and remember the concepts.
D. Becoming an active learner will help you understand the
concepts.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
22
1.3
Units of Measurement
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Units of Measurement
23
Scientists use the metric system of measurement and have
adopted a modification of the metric system called the
International System of Units as a worldwide standard.
International System of Units (SI) is an official system of
measurement used throughout the world for units in length,
volume, mass, temperature, and time.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Units of Measurement, Metric and SI
24
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Length, Meter (m) and Centimeter (cm)
25
1 m = 100 cm
1 m = 39.4 in.
1 m = 1.09 yd
2.54 cm = 1 in.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Volume, Liter (L) and Milliliter (mL)
26
1L
= 1000 mL
1L
= 1.06 qt
946 mL = 1 qt
We use graduated cylinders to
measure small volumes.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Mass, Gram (g) and Kilogram (kg)
27
1 kg = 1000 g
1 kg = 2.20 lb
454 g = 1 lb
The mass of a nickel is
5.01 g on an
electronic scale.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Temperature, Celsius (oC) and Kelvin (K)
28
Water freezes:
32 oF
0 oC
The Kelvin scale for temperature
begins at the lowest possible
temperature, 0 K.
A thermometer is used to measure
temperature.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Time, Second (s)
29
The second is the correct metric
and SI unit for time.
The standard measure for 1 s is an
atomic clock.
A stopwatch is used to measure the
time of a race.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Learning Check
30
What are the SI units for the following?
A. volume
B. mass
C. length
D. temperature
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
31
What are the SI units for the following?
A. The SI unit for volume is the
cubic meter, m3.
B. The SI unit for mass is the
kilogram, kg.
C. The SI unit for length is the
meter, m.
D. The SI unit for temperature is
Kelvin, K.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Chapter 1
Chemistry and Measurements
32
1.4
Scientific Notation
People have an average of 1 x 105 hairs on their scalp. Each hair
is about 8 x 10−6 m wide.
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Writing a Number in Scientific Notation
33
Numbers written in scientific notation have three parts:
coefficient
power of 10
unit
To write 2400 m in correct scientific notation:
• the coefficient is 2.4
• the power of 10 is 3
• the unit is "m"
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Writing a Number in Scientific Notation
34
2400. m = 2.4 x 1000 = 2.4 x 103 m
 3 places
0.00086 g =
4 places 
coefficient x power of 10 unit
8.6
= 8.6 x 10−4
10,000
g
coefficient x power of 10 unit
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Some Powers of 10
35
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Some Powers of 10
36
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Measurements in Scientific Notation
37
Diameter chickenpox virus
= 0.0000003 m
= 3 x 10−7 m
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Some Measurements Written in Scientific
Notation
38
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Scientific Notation and Calculators
39
Number to enter:
Enter:
Display:
4 x 106
4 EXP (EE) 6
4 06 or 406 4 E06
Number to enter:
Enter:
Display:
2.5 x 10−4
2.5 EXP (EE) +/− 4
2.5 −04 or 2.5−04 2.5 E−04
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Learning Check
40
Write each of the following in correct scientific notation:
A. 64,000 g
B. 0.021 m
C. 138 mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
41
Write each of the following in correct scientific notation:
A. 64,000 g
B. 0.021 m
C. 138 mL
6.4 x 104 g
2.1 x 10−2 m
1.38 x 102 mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
42
1.5
Measured Numbers and
Significant Figures
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Measured Numbers
43
Measured numbers are the numbers obtained when you
measure a quantity such as your height, weight, or
temperature.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Writing Measured Numbers
44
To write a measured number,
• observe the numerical values of marked lines
• estimate value of number between marks
• the estimated number is the final number in your
measured number
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Writing Measured Numbers for Length
45
The lengths of the objects are
measured as
(a) 4.5 cm
(b) 4.55 cm
(c) 3.0 cm
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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A Number Is Significant When
46
A number is a significant figure (SF) if it is
Example
a. not a zero
4.5 g
b. a zero between digits
205 m
c. a zero at the end of a
50. L
decimal number
d. in the coefficient of a
4.8 x 105 m
number written in
scientific notation
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
2 SF
3 SF
2 SF
2 SF
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A Number Is NOT Significant When
47
A number is not significant if it is
Example
a. at the beginning of
a decimal number
b. used as a placeholder
in a large number
without a decimal point
0.0004 s
1 SF
850 000 m
2 SF
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
48
Identify the significant and nonsignificant zeros in each of the
following numbers:
A. 0.002 650 m
B. 43.026 g
C. 1 044 000 L
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
49
Identify the significant and nonsignificant zeros in each of the
following numbers:
A. 0.002 650 m
• The zeros preceding 2 are not significant.
• The digits 2, 6, 5 are significant.
• The zero in last decimal place is significant.
4 SF
B. 43.026 g
• The zeros between nonzero digits or at the
end of decimal numbers are significant.
5 SF
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
50
Identify the significant and nonsignificant zeros in each of the
following numbers:
C. 1 044 000 L
• The zeros between nonzero digits are significant.
• The zeros at end of a number with no decimal
are not significant.
4 SF
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Exact Numbers
51
Exact numbers are
• numbers obtained by counting
• in definitions that compare two units
in the same measuring system
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
8 cookies
6 eggs
1 qt = 4 cups
1 kg = 1000 g
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Learning Check
52
Identify the numbers below as measured or exact and give the
number of significant figures in each measured number:
A. 3 coins
B. the diameter of a circle is 7.902 cm
C. 60 min = 1 h
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
53
Identify the numbers below as measured or exact and give the
number of significant figures in each measured number:
A. 3 coins is a counting number and therefore is an
exact number.
B. The diameter of a circle is 7.902 cm. This is a measured
number and the zero is significant, so it contains 4 SF.
C. 60 min = 1 h is a definition and therefore an exact
number.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
54
1.6
Significant Figures in Calculations
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Rules for Rounding Off
55
1. If the first digit to be dropped is 4 or less, then it and all the
following digits are dropped from the number.
2. If the first digit to be dropped is 5 or greater, then the last
retained digit of the number is increased by 1.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Examples of Rounding
56
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Learning Check
57
Select the correct value when 3.1457 g is rounded to:
A. three significant figures
B. two significant figures
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
58
Select the correct value when 3.1457 g is rounded to:
A. To round 3.1457 to three significant figures,
• drop the final digits, 57
• increase the last remaining digit by 1.
The answer is 3.15 g.
B. To round 3.1457 g to two significant figures,
• drop the final digits 457.
• do not increase the last number by 1 since the first of these
digits is 4.
The answer is 3.1 g.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Rules for Multiplication and Division
59
In multiplication or division, the final answer is written so it has
the same number of significant figures as the measurement
with the fewest significant figures (SFs).
Example 1: Multiply the following measured numbers:
24.66 cm x 0.35 cm
= 8.631 (calculator display)
= 8.6 cm2 (2 significant figures)
Multiplying 4 SFs by 2 SFs gives us an answer with 2 SFs.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Multiplication and Division with SFs
60
Example 2: Multiply and divide the following measured
numbers:
21.5 cm x 0.30 cm =
1.88 cm
Put the following into your calculator:
21.5 x 0.30 ÷ 1.88 = 3.430851063
= 3.4 cm (2 significant figures)
Multiplying 4 SFs by 2 SFs gives us an answer with 2 SFs.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Multiplication and Division with SFs
61
Example 3: Multiply and divide the following measured
numbers:
6.0 g =
2.00 g
Put the following into your calculator:
6.0 ÷ 2.00 = 3 (calculator display)
= 3.0 g (2 significant figures)
Add one zero to give 2 significant figures.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
62
Perform the following calculation of measured numbers. Give
the answer in the correct number of significant figures.
5.00 cm x 3.408 cm =
2.00 cm
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
63
Perform the following calculation of measured numbers. Give
the answer in the correct number of significant figures.
5.00 cm x 3.408 cm =
2.00 cm
(3 SF x 4 SF ÷ 3 SF )
= 8.52 cm calculator display and correct
significant figures.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Addition and Subtraction
64
In addition or subtraction, the final answer is written so it has the
same number of decimal places as the measurement with the
fewest decimal places.
Example 1: Add the following measured numbers:
2.012 g three decimal places
61.09 g two decimal places
+ 3.0
g one decimal place
66.102 g (calculator display)
= 66.1 g answer rounded to one decimal place
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Addition and Subtraction with SFs
65
Example 2: Subtract the following measured numbers:
65.09 g two decimal places
− 3.0 g one decimal place
62.09 g (calculator display)
= 62.1 g answer rounded to one decimal place
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
66
Add the following measured numbers:
82.409 mg
+ 22.0
mg
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
67
Add the following measured numbers:
82.409 mg three decimal places
+ 22.0
mg one decimal place
104.409 mg (calculator display)
= 104.4
mg answer rounded to one decimal place
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
68
1.7
Prefixes and Equalities
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Prefixes
69
A special feature of the SI as well as the metric system is that a
prefix can be placed in front of any unit to increase or
decrease its size by some factor of ten.
For example, the prefixes milli and micro are used to make the
smaller units:
milligram
(mg)
microgram
(μg)
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Prefixes and Equalities
70
•
The relationship of a prefix to a unit can be expressed by
replacing the prefix with its numerical value.
•
For example, when the prefix kilo in kilometer is replaced with
its value of 1000, we find that a kilometer is equal to 1000
meters.
kilometer
=
1000 meters
kilogram
=
1000 grams
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Prefixes That Increase Unit Size
71
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Prefixes That Decrease Unit Size
72
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Learning Check
73
Fill in the blanks with the correct prefix:
A. 1000 m = 1 ___m
B. 1 x 10−3 g = 1 ___g
C. 0.01 m = 1 ___m
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Solution
74
Fill in the blanks with the correct prefix:
A. 1000 m = 1 ___m
The prefix for 1000 is kilo; 1000 m = 1 km
B. 1 x 10−3 g = 1 ___g
The prefix for 1 x 10−3 is milli; 1 x 10−3 g = 1 mg
C. 0.01 m = 1 ___m
The prefix for 0.01 is centi; 0.01 m = 1 cm
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Measuring Length
75
Each of the following equalities describes the same length
in a different unit.
1m
1m
1 cm
= 100 cm
= 1000 mm
= 10 mm
= 1 x 102 cm
= 1 x 103 mm
= 1 x 101 mm
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Measuring Length
76
The metric length of 1 meter is the same as 10 dm, 100 cm,
or 1000 mm.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Measuring Volume
77
Examples of Some Volume Equalities
1 L = 10 dL
1 L = 1000 mL
1 dL = 100 mL
= 1 x 101 dL
= 1 x 103 mL
= 1 x 102 mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
The Cubic Centimeter
78
The cubic centimeter (abbreviated as cm3 or cc) is the
volume of a cube whose dimensions are 1 cm on each
side.
A cubic centimeter has the same volume as a milliliter, and
the units are often used interchangeably.
1 cm3 = 1 cc = 1 mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
The Cubic Centimeter
79
1 cm3 = 1 cc = 1 mL
10 cm x 10 cm x 10 cm = 1000 cm3 = 1000 mL = 1 L
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Measuring Mass
80
Examples of Some Mass Equalities
1 kg = 1000 g
1 g = 1000 mg
1 g = 100 cg
1 mg = 1000 μg
= 1 x 103 g
= 1 x 103 mg
= 1 x 102 cg
= 1 x 103 μg
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
81
Identify the larger unit in each of the following:
A. mm or cm
B. kilogram or centigram
C. mL or μL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
82
Identify the larger unit in each of the following:
A. mm or cm
A mm is 0.001 m, smaller than a cm, 0.01 m.
B. kilogram or centigram
A kilogram is 1000 g, larger than a centigram or 0.01 g.
C. mL or μL
A milliliter is 0.001 L, larger than a μL, 0.000 001 L.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
83
1.8
Writing Conversion Factors
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Equalities
84
Equalities
• use two different units to describe the same measured
amount
• are written for relationships between units of the metric
system, U.S. units, or between metric and U.S. units
For example,
1m
=
1 lb
= 16 oz
2.20 lb =
1000 mm
1 kg
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Exact and Measured Numbers in
Equalities
85
Equalities between units in
• the same system of measurement are definitions that use
exact numbers
• different systems of measurement (metric and U.S.) use
measured numbers that have significant figures
Exception:
The equality 1 in. = 2.54 cm has been defined as an exact
relationship and therefore 2.54 is an exact number.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Some Common Equalities
86
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Equalities on Food Labels
87
The contents of
packaged foods
• in the U.S. are listed
•
in both metric and
U.S. units
indicate the same
amount of a substance
in two different units
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Conversion Factors
88
A conversion factor is
• obtained from an equality and written in the form of a
fraction with a numerator and denominator
Equality: 1 in. = 2.54 cm
• inverted to give two conversion factors for every equality
1 in.
2.54 cm
and
2.54 cm
1 in.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
89
Write conversion factors from the equality for each of the
following:
A. L and mL
B. hours and minutes
C. meters and kilometers
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
90
Write conversion factors from the equality for each of the
following:
A. 1 L = 1000 mL
1L
and 1000 mL
1000 mL
1L
B. 1 h = 60 min
1h
and
60 min
60 min
1h
C. 1 km = 1000 m
1 km
1000 m
1000 m
1 km
and
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Conversion Factors in a Problem
91
A conversion factor
• may be obtained from information in a word problem
• is written for that problem only
Example 1:
The price of one pound (1 lb) of red peppers is $2.39.
1 lb red peppers
and
$2.39
$2.39
1 lb red peppers
Example 2:
The cost of one gallon (1 gal) of gas is $2.89.
1 gal gas
$2.89
and
$2.89
1 gal gas
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Percent as a Conversion Factor
92
A percent factor
gives the ratio of the parts to the whole
% = parts x 100
whole
• uses the same unit in the numerator and denominator
• uses the value of 100
• can be written as two factors
Example: A food contains 30% (by mass) fat:
30 g fat
and
100 g food
100 g food
30 g fat
•
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Percent Factor in a Problem
93
The thickness of the skin fold at
the waist indicates 11% body
fat. What factors can be written
for percent body fat (in kg)?
Percent factors using kg:
11 kg fat and
100 kg mass
100 kg mass
11 kg fat
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Smaller Percents: ppm and ppb
94
Small percents are given as ppm and ppb.
•
•
Parts per million (ppm) =
mg part
kg whole
Example: The EPA allows 15 ppm cadmium in food colors.
15 mg of cadmium = 1 kg of food color
Parts per billion (ppb) = μg part
kg whole
Example: The EPA allows 10 ppb arsenic in public water.
10 μg of arsenic = 1 kg of water
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
95
Write the conversion factors for 10 ppb arsenic in public
water from the equality.
10 μg of arsenic = 1 kg of water
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
96
Write the conversion factors for 10 ppb arsenic in public
water from the equality.
10 μg of arsenic = 1 kg of water
Conversion Factors
10 μg arsenic
1 kg water
and
1 kg water
10 μg arsenic
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Study Tip: Conversion Factors
97
An equality
• is written as a fraction (ratio)
• provides two conversion factors that are the inverse of
each other
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Learning Check
98
Write the equality and conversion factors for each of the
following.
A. meters and centimeters
B. jewelry that contains 18% gold
C. one gallon of gas is $3.40
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
99
A. meters and centimeters:
1 m = 100 cm
1m
and 100 cm
100 cm
1m
B. jewelry that contains
18% gold: 100 g of jewelry = 18 g of gold
18 g gold and 100 g jewelry
100 g jewelry
18 g gold
C. one gallon of gas is $3.40: 1 gal gas = $3.40
1 gal gas and
$3.40
$3.40
1 gal gas
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Risk-Benefit Assessment
100
A measurement of toxicity is
•
LD50 or “lethal dose”
•
the concentration of the substance that causes death
in 50% of the test animals
•
in milligrams per kilogram (mg/kg or ppm) of body mass
•
in micrograms per kilogram (g/kg or ppb) of body mass
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
101
The LD50 for aspirin is 1100 ppm. How many grams of
aspirin would be lethal in 50% of persons with a body
mass of 85 kg?
A. 9.4 g
B. 94 g
C. 94 000 g
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
102
The LD50 for aspirin is 1100 ppm. How many grams of
aspirin would be lethal in 50% of persons with a body
mass of 85 kg?
85 kg x 1100 mg
1 kg
x
1g
1000 mg
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
= 94 g of aspirin
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Chapter 1
Chemistry and Measurements
103
1.9
Problem Solving
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Given and Needed Units
104
To solve a problem,
• identify the given unit
• identify the needed unit
Example:
A person has a height of 2.0 meters. What is that height in
inches?
The given unit is the initial unit of height.
given unit = meters (m)
The needed unit is the unit for the answer.
needed unit = inches (in.)
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Study Tip: Problem Solving Using GPS
105
The steps in the
Guide to
Problem Solving (GPS)
are useful in setting up
a problem with conversion
factors.
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Setting Up a Problem
106
How many minutes are in 2.5 hours?
Solution:
Step 1 State the given and needed quantities.
Given unit:
2.5 hours
Needed unit: min
Step 2 Write a unit plan.
Plan:
hours
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
min
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Solving a Problem
107
How many minutes are in 2.5 hours?
Step 3 State equalities and conversion factors to cancel units.
60 min = 1 h
60 min and
1h
1h
60 min
Step 4 Set up problem to cancel units.
Given
Conversion Needed unit
unit
factor
2.5 h x 60 min = 150 min (2 SF)
1h
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
108
A rattlesnake is 2.44 m long. How many centimeters long is
the snake?
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
109
A rattlesnake is 2.44 m long. How many centimeters long is
the snake?
Step 1 State the given and needed quantities.
Given unit:
2.44 m
Needed unit: cm
Step 2 Write a unit plan.
Plan:
meters
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
centimeters
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Solution
110
A rattlesnake is 2.44 m long. How many centimeters long is
the snake?
Step 3 State equalities and conversion factors to cancel units.
1 m = 102 cm
102 cm and
1m
1m
102 cm
Step 4 Set up problem to cancel units.
Given Conversion Needed unit
unit
factor
2.44 m x 102 cm = 244 cm (3 SF)
1m
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
111
How many minutes are in 1.4 days?

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
112
How many minutes are in 1.4 days?
Step 1 State the given and needed quantities.
Given unit: 1.4 days
Needed unit: minutes
Step 2 Write a unit plan.

Factor 1
Plan:
days
Factor 2
h
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
min
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Solution
113
How many minutes are in 1.4 days?
Step 3 State equalities and conversion factors to cancel units.
1 day = 24 hours
24 hours and
1 day
1 day
24 hours
1 hour = 60 minutes 60 min
and 1 h
1h
60 min
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
114
How many minutes are in 1.4 days?
Step 4 Set up problem to cancel units.
Given Conversion Conversion
unit
factor
factor
Needed unit

1.4 days x 24 h x 60 min = 2.0 x 103 min
1 day
1h
(rounded)
2 SF
Exact
Exact = 2 SF
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Study Tip: Check Unit Cancellation
115
•
•
Be sure to check the unit cancellation in the setup.
The units in the conversion factors must cancel to give the
correct unit for the answer.
What is wrong with the following setup?
1.4 day
x 1 day x 1 h
24 h
60 min
=
day2/min is not the unit needed.
Units don’t cancel properly.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
116
If your pace on a treadmill is 65 meters per minute, how many
minutes will it take for you to walk a distance of 7.5
kilometers?
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
117
If your pace on a treadmill is 65 meters per minute, how many
minutes will it take for you to walk a distance of 7.5
kilometers?
Step 1 State the given and needed quantities.
Given units: 7.5 km, 65 meters per minute
Needed unit: minutes
Step 2 Write a unit plan.
Factor 1
Plan: kilometers
Factor 2
meters
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
min
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Solution
118
If your pace on a treadmill is 65 meters per minute, how many
minutes will it take for you to walk a distance of 7.5
kilometers?
Step 3 State equalities and conversion factors to cancel units.
1 kilometer = 103 meters
103 meters and
1 kilometer
1 kilometer
103 meters
65 meters = 1 minute 65 meters and
1 minute
1 minute
65 meters
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
119
If your pace on a treadmill is 65 meters per minute, how many
minutes will it take for you to walk a distance of 7.5 kilometers?
Step 4 Set up problem to cancel units.
Given
unit
Conversion
factor
Conversion Needed unit
factor
7.5 kilometers x 103 meters x 1 minute = 120 minutes
1 kilometer 65 meters (rounded)
2 SF
Exact
2 SF
= 2 SF
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Percent Factor in a Problem
120
If the thickness of the skin fold at
the waist indicates 11% body fat,
how much fat is in a person with
a mass of 86 kg?
Percent factor:
11 kg fat
100 kg mass
percent factor
86 kg mass x 11 kg fat
100 kg mass
= 9.5 kg of fat
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
121
How many pounds of sugar are in 120 g of candy if the
candy is 25% (by mass) sugar?
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
122
How many pounds of sugar are in 120 g of candy if the candy is
25% (by mass) sugar?
Step 1 State the given and needed quantities.
Given units: 120 g of candy
25% by mass sugar
Needed unit: pounds sugar
Step 2 Write a unit plan.
Conversion
factor
Plan: grams
Percent
factor
pounds candy
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
pounds sugar
Copyright © 2012 by Pearson Education, Inc.
Solution
123
How many pounds of sugar are in 120 g of candy if the
candy is 25% (by mass) sugar?
Step 3 State equalities and conversion factors to cancel
units.
1 pound = 454 grams
454 g and
1 lb
1 lb
454 g
25 pounds of sugar = 100 pounds of candy
25 lb sugar and 100 lb candy
100 lb candy
25 lb sugar
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
124
How many pounds of sugar are in 120 g of candy if the candy is
25% (by mass) sugar?
Step 4 Set up problem to cancel units.
Given
unit
Conversion
factor
Percent
factor
120 g candy x 1 lb candy x 25 lb sugar
454 g candy 100 lb candy
= 0.066 lb of sugar
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Copyright © 2012 by Pearson Education, Inc.
Chapter 1
Chemistry and Measurements
125
1.10
Density
Objects that sink in water are more dense than
water; objects that float in water are less dense.
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Copyright © 2012 by Pearson Education, Inc.
Density
126
Density
• compares the mass of an object to its volume
• is the mass of a substance divided by its volume
Density Expression
Density = mass = g or
volume mL
g
or g/cm3
cm3
Note: 1 mL = 1 cm3
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Densities of Common Substances
127
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Calculating Density
128
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
129
Osmium is a very dense metal. What is its density in g/cm3
if 50.0 g of osmium has a volume of 2.22 cm3?
1) 2.25 g/cm3
2) 22.5 g/cm3
3) 111 g/cm3
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
130
Step 1 State the given and needed quantities.
Given: 50.0 g; 22.2 cm3 Need: density, g/cm3
Step 2 Write the density expression.
D = mass
volume
Step 3 Express mass in grams and volume in mL or cm3.
Mass = 50.0 g Volume = 22.2 cm3
Step 4 Substitute mass and volume into the density expression
and calculate.
D = 50.0 g = 22.522522 g/cm3
2.22 cm3
= 22.5 g/cm3 (rounded to 3 SFs)
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Volume by Displacement
131
A solid
•
completely submerged
in water displaces its
own volume of water
•
has a volume calculated
from the volume
difference
45.0 mL − 35.5 mL
= 9.5 mL
= 9.5 cm3
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Density Using Volume Displacement
132
The density of the zinc object is
calculated from its mass
and volume.
Density =
mass = 68.60 g = 7.2 g/cm3
volume
9.5 cm3
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
133
What is the density (g/cm3) of a 48.0-g sample of a metal if the
level of water in a graduated cylinder rises from 25.0 mL to 33.0
mL after the metal is added?
1) 0.17 g/cm3
25.0 mL
2) 6.0 g/cm3
3) 380 g/cm3
33.0 mL
object
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
134
Step 1 State the given and needed quantities.
Given: 48.0 g
Volume of water = 25.0 mL
Volume of water + metal = 33.0 mL
Need: Density
Step 2 Write the density expression.
Density = mass of metal
volume of metal
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
135
Step 3 Express mass in grams and volume in mL or cm3.
Mass = 48.0 g
Volume of the metal is equal to the volume of water
displaced.
Volume of water + metal
− Volume of water
Volume of metal
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
= 33.0 mL
= 25.0 mL
= 8.0 mL
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Solution
136
Step 4 Substitute mass and volume into the density expression
and calculate the density.
Density = 48.0 g = 6.0 g = 6.0 g/mL
8.0 mL
1 mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Sink or Float
137
•
Ice floats in water
because the
density of ice is
less than the
density of water.
•
Aluminum sinks
because its density
is greater than the
density of water.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Learning Check
138
Which diagram correctly represents the liquid layers in
the cylinder? Karo syrup (K) (1.4 g/mL); vegetable oil
(V) (0.91 g/mL); water (W) (1.0 g/mL)
1
2
3
V
W
K
W
K
V
K
V
W
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
139
1)
V
W
K
vegetable oil 0.91 g/mL
water 1.0 g/mL
Karo syrup 1.4 g/mL
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Problem Solving Using Density
140
Density can be written as an equality.
• For a substance with a density of 3.8 g/mL, the equality is
3.8 g = 1 mL
• From this equality, two conversion factors can be written for
density.
Conversion 3.8 g
factors
1 mL
and
1 mL
3.8 g
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Problem Solving Using Density
141
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
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Learning Check
142
The density of octane, a component of gasoline, is 0.702
g/mL. What is the mass, in kg, of 875 mL of octane?
A. 0.614 kg
B. 614 kg
C. 1.25 kg
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.
Solution
143
The density of octane, a component of gasoline, is 0.702
g/mL. What is the mass, in kg, of 875 mL of octane?
Step 1 State the given and needed quantities.
Given:
Density of octane = 0.702 g/mL
Volume = 875 mL
Needed:
Mass of octane
Step 2 Write a plan to calculate the needed quantity.
Density
Plan: milliliters
Conversion
factor
grams
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
kilograms
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Solution
144
The density of octane, a component of gasoline, is 0.702 g/mL.
What is the mass, in kg, of 875 mL of octane?
Step 3 Write equalities and their conversion factors including
density.
density
0.702 g = 1 mL
and
1 kg = 1000 g
Step 4 Set up problem to calculate the needed quantity.
875 mL x 0.702 g x 1 kg = 0.614 kg
1 mL
1000 g
Answer is A, 0.614 kg.
Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition
Copyright © 2012 by Pearson Education, Inc.