Personal Tutor
to accompany
Chemistry
a project of the
American Chemical Society
W.H. Freeman and Company
Wayne Morgan
1
Table of Contents
0.0
Introduction to the Personal Tutor (pg.2)
1.0
Understanding the Uses of Numbers (pg.3)
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
2.0
Measurement and the Metric System (pg. 3)
Derived Units (pg. 6)
Dimensional Analysis (pg. 8)
Precision and Accuracy in Measurement (pg. 10)
Percent Error (pg. 12)
Significant Figures (pg. 13)
Using Significant Figures in Calculations (pg. 16)
Scientific Notation (pg. 18)
Understanding and Interpreting Graphs and Tables (pg. 20)
2.1
Types of Graphs (pg. 20)
2.1.1
2.1.2
2.1.3
2.1.4
2.1.5
2.2
2.3
3.0
What is a mole? (pg. 35)
Molar Mass (pg. 35)
MassMole Conversions (pg. 37)
VolumeMole Conversions (pg. 40)
Understanding Mass Relationships in Chemical Reactions (pg. 42)
4.1
4.2
4.3
4.4
5.0
Basic Graphing Rules (line and x-y graphs) (pg. 31)
Interpreting Tables (pg. 32)
Understanding the Mole (pg. 35)
3.1
3.2
3.3
3.4
4.0
Pie Charts (pg. 20)
Bar Graphs (pg. 21)
Multiple Bar Graphs (pg. 22)
Line Graphs (pg. 24)
x-y Plots/Graphs (pg. 25)
Understanding Chemical Equations (pg. 42)
Balancing Chemical Equations (pg. 43)
Stoichiometry (pg. 45)
Limiting Reactants (pg. 46)
Understanding Lewis Dot and Structural Formulas (pg. 50)
5.1
5.2
5.3
5.4
Steps in Writing the Lewis Structure (pg. 50)
Structural Formulas (pg. 54)
Bonding Patterns in Many Hydrocarbon Molecules (pg. 55)
Formal Charge (pg. 57)
6.0
Understanding Gas Law Problems (pg. 59)
7.0
Answers, Solutions, and Explanations for the Practice Problems (pg. 67)
2
Introduction
A chemistry textbook is a valuable resource. It can help you in developing a deeper understanding of the
material you discuss in class. It can illustrate the most common types of problems you will face and give
examples of how to approach those problems. The textbook cannot anticipate which students will have
difficulty with computations. Even the most comprehensive textbook cannot possibly show every type of
problem that might be encountered. The Personal Tutor is a supplement to the textbook to give you more
guidance and practice with problems and computations. Using the Personal Tutor is much like visiting the
professor during office hours or going to a human tutor. You may need to take advantage of it frequently
or you may only need its assistance a few times during the course. The questions can be a good review
when preparing for the test even if you are confident that you know how to solve the problems or do the
computations.
Even the Personal Tutor is not a comprehensive set of problems. However, the types of problems shown
here should help you get a good foundation in solving problems. It is important to remember that an
examination would not truly test your understanding of concepts and your ability to solve problems if the
professor merely put on the problems you had already seen. If you use your textbook, the Personal Tutor,
and any other sources available you can be successful in this chemistry course.
3
Understanding the Uses of Numbers
Chemists rely on data in order to test theories, solve problems and make decisions. Most of the data is
numerical data accumulated from careful measurements in the laboratory. An important part of learning
chemistry is learning how to interpret and use numerical data.
MEASUREMENT AND THE METRIC SYSTEM
Metric units were first introduced in France more than 100 years ago. A modernized form of the metric
system was internationally adopted in 1960. The system is called “SI,” which is an abbreviation of its
French name, Le Système International d’Unités. SI units are used by scientists in all nations, including
the United States. This system has a small number of base units from which all other necessary units are
derived.
Quantity
SI Unit
kilogram
meter
second
Kelvin
mole
ampere
Mass
Length
Time
Temperature
Amount of substance
Electric current
Abbreviation
kg
m
s
K
mol
A
Figure 1 Fundamental SI Units
The limited number of units in Figure 1 is not sufficient for every type of measurement a chemist might
need to make. For example, the SI unit of length is the meter (symbolized by m). Most doorways are about
two meters high. However, many lengths we may wish to measure are either much larger (distance from the
earth to the sun) or much smaller (width of a dime or the size of an atomic nucleus) than a meter. To handle
such measurements easily, common metric prefixes are used to change the size of the unit. The distance
from the earth to the sun can be expressed in kilometers (km), the width of a dime can be expressed in
terms of millimeters (mm) and the size of an atomic nucleus in terms of picometers (pm) . Figure 2 lists the
most useful metric prefixes and their meanings.
Prefix Abbreviation
M
mega
Meaning
106
kilo
k
103
deci
d
101
centi
c
102
milli
m
103
micro-

106
nano
n
109
pico
p
1012
Example(s)
700 Megabyte = 700,000,000 bytes
(storage on a CD-R)
1.6 kilometers = 1600 meters (about 1
mile)
5.91 deciliters = 0.591 liters (soda in a 20
oz. bottle)
2.4 centimeters = .024 meters (diameter
of a quarter)
355 milliliters = .355 liters (volume of
soda in a 12 oz. can)
25 m = 2.5 X 105 m (size of a cell in
your body)
656 nm = 6.56 X 107 m (wavelength of
the red line in the hydrogen spectrum)
.01 pm = 1 x 1014 m (approximate size
of an atomic nucleus)
Figure 2 Common Metric Prefixes
4
You may me more familiar with the typical English units of measurement that we use in everyday life.
Here are some comparisons that may help you visualize the metric units.
1 kilogram is a little over 2 pounds (1 kilogram = 2.2 lbs)
1 meter is a little over 1 yard (1 meter = 39 inches)
1.6 kilometer is about 1 mile (1600 meters  1 mile)
273 K is the freezing point of water (0 oC or 32 oF)
373 K is the boiling point of water (100 oC or 212 oF)
293 K is about room temperature (20 oC or 68 oF)
One advantage of using the metric system is the ease with which you can convert from one prefix to
another. Changing units within the metric system is a matter of changing decimal places. You can use the
meanings from Figure 2 to help you.
Example: Convert 1456 g to kilograms.
In Figure 2 the meaning of kilo- tells that 1 kilogram would be equal to 103 grams.
Equivalance: 1 kilogram (kg) = 1 X 10 3 grams (g)
The equivalence between kilograms and grams can be used to create a conversion factor. A
conversion factor, often called a unit conversion factor, is one where the numerator is a quantity equal
to the quantity in the denominator. The only difference between the numerator and denominator is the
units. This means that a unit conversion factor is really equal to one. Multiplication by a unit
conversion factor is like multiplying by one. It will result in the units changing but not the value of the
measurement.
Unit Conversion Factor:
1 kg
1 X 10 3 g
1 X 10 3 g
1 kg
or
Since we are given a mass in grams it will be necessary to choose the conversion factor that will allow
grams to be cancelled and to have kilograms left.
Unit Conversion:
1456 grams 
1 kg
 1.456 kg  1.46 kg
1 X 10 3 g
More than one unit conversion factor may be used in a single problem. Sometimes it is easier to make a
conversion in more than one step. There is no real limit to the number of unit conversions that may be put
together. However, never use any more unit conversions that are really necessary to move from the unit
you are given to the unit you desire. The next example shows how to use more than one step to make a unit
conversion.
5
Example: Convert 325 mg to kilograms.
1 g = 1000 mg = 1 X 103 mg
1 kg = 1000 g = 1 X 103 g
Equivalences:
1g
1 X 10 3 mg
Conversion factors:
Multiplication:
325 mg X
1 kg
1 X 10 3 g
1g
1 kg
X
3.25 X 10  4 kg
3
3
1 X 10 mg 1 X 10 g
Converting units or solving problems using the sequential multiplication of conversion factors is called
dimensional analysis or the factor label method. Dimensional analysis can be used to solve many
different types of problems in chemistry. It is a useful tool if you wish to avoid algebraic equations.
In chemistry numbers are all the result of some measurement in the laboratory. Numbers should always be
accompanied by the units appropriate for that measurement. In chemistry the unit is as important as the
number itself.
Practice Problems
1.
Which metric unit and prefix would be most convenient to measure each of the following?
a.
b.
c.
d.
e.
f.
g.
2.
the diameter of a giant sequoia tree
the diameter of a human hair
time necessary to blink your eye
mass of gasoline in a gallon
mass of a cold virus
amount of aspirin in one tablet
mass of concrete needed to pave a parking lot
What word prefixes are used in the metric system to indicate the following multipliers?
a.
1 x 103
b.
1 x 103
c.
0.01
d.
1 x 106
3.
An antacid tablet contains 168 mg of the active ingredient ranitidine hydrochloride. How
many grams of ranitidine hydrochloride are in the tablet?
4.
There are 1.609 km in exactly 1 mile. How many centimeters are there in 1 mile?
5.
A paper clip is 3.2 cm long. What is the length of the paper clip in millimeters?
6.
State at least one advantage of SI units over the customary US units.
6
Derived Units
Frequently we wish to measure quantities that cannot be expressed using one of the basic SI units. In these
situations two or more units are combined to create a new unit. These units are called derived units. For
example, speed is defined as the ratio of distance to time. To measure speed, two units—distance and
time—are combined. For example, the speed of your car would be expressed in the units of kilometers per
hour (km hr1).
NOTE: You may be more familiar with seeing an expression like kilometers per hour written as
km/hr. In chemistry the community of scientists has established rules telling how numbers, names, and
other information should be written for publication. The accepted way to write the “per” expression is to
write the term in the denominator with a 1 exponent. This is the way these mixed units will be written
throughout the personal tutor.
A derived unit frequently used in chemistry is volume. Let us see how these derived unit for volume is
related to fundamental SI units. The volume of a cube is determined by multiplying (length X width X
height). A cube with sides of 10 cm, has a volume of 1000 cm3 (10cm X 10cm X 10cm), which is defined
as a liter.
1000 cm3  1 liter (1L)
In the laboratory the most common unit of volume will likely be the milliliter. The metric prefixes used for
the fundamental units may also be applied to the derived unit, the Liter. 1000 mL = 1 L.
Since 1000 cm3  1 L and 1000 mL = 1 L, a useful equivalence can be stated. 1 cm3 = 1 mL. This
relationship can be very useful in solving chemistry problems and will often be a useful conversion in the
chemistry laboratory.
Density is an important property for determining the identity of a sample of matter. It is defined as the
ratio of mass to volume ( Density 
Mass
). It is a combination of several of the fundamental units so
Volume
it too is a derived unit. Two units are combined—a mass unit and a volume unit. The density of solids and
liquids is usually expressed as g cm3 and the density of gases as g L1.
7
Practice Problems
1.
The average person in the United States uses 340 L of water daily. Convert this to
milliliters.
2.
A quart is approximately equal to 946 mL. How many liters are in 1 quart?
3.
One hundred fifty milliliters of rubbing alcohol has a mass of 120 g. What is the density of
rubbing alcohol?
4.
A ruby has a mass 7.5 g and a volume of 1.9 cm3. What is the density of this ruby?
5.
What is the density of isopropyl alcohol if 5.00 mL weigh 3.93 g?
There are other units that you will encounter as you go through your chemistry course. The calorie and
the joule are units used in expressing energy. The calorie was defined as the amount of energy necessary
to raise the temperature of exactly one gram of water one degree Celsius. The joule comes from physics
and relates energy to forces at work. In chemistry we use joules as our unit of energy measure. In biology
the calorie or the kilocalorie ( 1 kcal = 1000 calories) is still used. 1 calorie = 4.18 J. The nutritional
Calorie (Cal) given on food labels is actual 1 kcal or 1000 calories.
In chemistry there are several units that are used to express the pressure of gases. The normal atmospheric
pressure at sea level is defined as being 1 atmosphere (atm) of pressure. One historical unit for pressure is
the Torr, named for Torrecelli who invented the barometer. Traditionally barometers had a column of
mercury that would rise and fall with changing air pressure. The height of the mercury was measured to
give the pressure in millimeters of mercury (mm Hg). Torr and mmHg of mercury are numerically
identical measurements. 1 atmosphere = 760 Torr = 760 mm Hg. You will find all three of these units for
measuring pressure in many chemistry textbooks and laboratory manuals. The SI unit of pressure,
however, is the pascal (Pa). 1 atmosphere = 101, 325 Pa Since this is such a large number it is often the
case that textbooks will give pressure in kilopascals (kPa). 1 atm = 101.325 kPa.
8
DIMENSIONAL ANALYSIS
Dimensional analysis, also called the factor-label method, is widely used by
scientists to solve a wide variety of problems. You have already used this method to convert one type of
metric unit to another. The method is helpful in setting up problems and also in checking work because if
the unit label is incorrect, the numeric answer to the problem is likely to be incorrect. The use of
dimensional analysis consists of three basic steps:
1.
Identify equivalence relationships in order to create unit conversion
factors.
2.
Identify the given unit and the new unit desired.
3.
Arrange the conversion factor so given units cancel, leaving the new
desired unit. Perform the calculation.
The following example illustrates the use of dimensional analysis.
Example:
In an exercise your laboratory partner measured the length of an object to be
12.2 inches. All other measurements were in centimeters and the answer was to
be reported in cm3. Your lab partner could measure the object again this time
in centimeters, or 12.2 inches could be converted to centimeters. Since you
already put away the ruler you decide to convert inches to centimeters.
Step 1:
Find the equivalence relating centimeters and inches.
NOTE: Conversions factors are frequently found inside the back covers of science
textbooks. Sometimes they can be found in the appendices of the text. If you
cannot find the unit conversions you need in this, or any other text, you can do an
internet search and find them at the many encyclopedia or reference web sites.
2.54 cm = 1 in
Step 2:
Identify the given unit and the “desired” unit
Given unit: inches
“Desired” unit: cm
Step 3:
Choose the fraction with the “given” quantity in the denominator and the “desired”
quantity in the numerator.
12.2 in X
2.54 cm
 30.0 cm
1 in
The units of inches (in) cancel leaving the desired unit of centimeters (cm).
Frequently in chemistry more than one conversion factor will be necessary in order to solve a particular
problem. The next example shows how more than one factor can be used to solve the problem.
9
Example:
How many seconds are there in 24 hours?
Step 1:
Identify the appropriate equivalences:
1 hr = 60 min
1 min = 60 sec
Step 2:
Given unit: hours (hr)
“Desired” unit: seconds (sec)
Step 3:
Arrange the conversion factors so that the given unit will cancel and the “desired” unit
will be the final unit left uncancelled
 60 min
24 hr 
 1 hr
  60 sec 
 
  86,400 sec
  1 min 
Practice Problems
1.
The distance between New York and San Francisco is 4 ,741,000 m. Now, that may sound
impressive, but to put all those digits on a car odometer is slightly inconvenient. (Of course, in
the United States the odometer measures miles, but that is another story.) In this case,
kilometers are a better choice for measuring distance. Change the distance to kilometers.
2.
Convert 7,265 mL to L.
3.
The 1500 meter race is sometimes called the “metric mile.” Convert 1500 m to miles. (1 m =
39.37 in). NOTE: now the mile race is the 1600 meter.
4.
The density of aluminum is 2.70 g cm3. What is the mass of 235 cm3
of aluminum?
5.
How many 250 mL servings can be poured from a 2.0 L bottle of soft drink?
6.
The speed limit in Canada is 100 km/hr. Convert this to meters/second.
7.
The density of helium is 0.17 g/L at room temperature. What is the mass of helium in a 5.4 L
helium balloon?
8.
Liquid bromine has a density of 3.12 g/mL. What volume would 7.5 g of bromine occupy?
9.
An irregularly shaped piece of metal has a mass of 147.8 g. It is placed in a graduated cylinder
containing 30.0 mL of water. The water level rises to 48.5 mL. What is the density of the
metal?
10
PRECISION AND ACCURACY IN MEASUREMENT
Chemistry experiments often require a number of different measurements, and there is always some error in
measurement. How much error depends on several factors, such as the skill of the experimenter, the quality
of the instrument, and the design of the experiment. The reliability of the measurement has two
components: precision and accuracy. Precision refers to how closely measurements of the same quantity
agree. A high-precision measurement is one that produces very nearly the same result each time it is
measured. Accuracy is how well measurements agree with the accepted or true value.
It is possible for a set of measurements to be precise without being accurate. Figure 3
demonstrates different possible combinations of precision and accuracy in an experiment designed to hit
the center of the target.
Precise, not accurate
Precise and accurate
Not precise, not accurate
Accurate, but not precise
Figure 3 Precision and Accuracy
A second example of accuracy and precision is given in Figure 4. The table lists the results of temperature
measurements from a beaker of boiling water. The standard temperature of boiling water is 100 °C. The
data in the table illustrates the different possible combinations of precision and accuracy in an experiment.
Trial
1
2
3
4
Average
Range
Thermometer
1
99.9
100.1
100.0
99.9
100.0
0.2
Temperature Readings
(oC)
Thermometer Thermometer
2
3
97.5
98.3
102.3
98.5
99.7
98.4
100.9
98.7
100.1
98.5
4.8
0.4
Figure 4 Measured Temperature of 100 mL of Boiling Water
11
Thermometer
4
97.5
99.7
96.2
94.4
97.0
5.3
The average value for each set is usually the value compared to the accepted boiling point of water.
Accuracy will be judged by how close the average is to the accepted value. The range—the difference
between the largest and smallest values—is the measure of the agreement among the individual
measurements and will be used to judge precision.
The data taken with Thermometer 1 is accurate and precise, since the average agrees with the accepted
value and the range is small. Thermometer 2 provided data that is accurate but not precise since the range is
relatively large. The data from Thermometer 3 is precise but not accurate. The range is small enough that it
is possible that Thermometer 3 may not have been calibrated properly. Thermometer 4 provides data that is
neither precise nor accurate.
Practice Problems
Groups of students determined the density of an unknown liquid in the laboratory. Calculate the
average and range for each group’s measurements.
1.
Group 1 obtained the following values: 1.34 g mL1, 1.32 g mL1 , 1.36 g mL1. The actual
value is 1.34 g mL1.
2.
Group 2 obtained the same results, but the actual value is 1.40 g mL1 .
3.
Group 3 obtained the following values: 1.66 g mL1, 1.28 g mL1, 1.18 g mL1. The actual
value is 1.34 g mL1.
4.
Group 4 obtained the following values: 1.60 g mL1, 1.70 g mL1, 1.40 g mL1. The actual
value is 1.40 g mL1.
12
Percent Error
Percent error is a measurement of the accuracy of the measurement. It is calculated using the following
formula:
Percent Error 
Experimental Value  Accepted Value
X 100%
Accepted Value
Note: Percent error is a positive number when the experimental value is too high and is a negative
number when the experimental value is too low. Often, though, percent error is reported as a positive
number regardless of whether the experimental value is too high or low. Frequently the absolute value of
the difference between the experimental and accepted values is used. Experimental Value  Accepted
Value
Practice Problems
Calculate the percent error for all four groups in the previous set of practice problems. Use the
average for each group as the experimental value. The actual values are your accepted values.
Here are the data sets you need.
Groups of students determined the density of an unknown liquid in the laboratory. Calculate the
average and range for each group’s measurements.
1.
Group 1 obtained the following values: 1.34 g mL1, 1.32 g mL1 , 1.36 g mL1. The actual
value is 1.34 g mL1.
2.
Group 2 obtained the same results, but the actual value is 1.40 g mL1 .
3.
Group 3 obtained the following values: 1.66 g mL1, 1.28 g mL1, 1.18 g mL1. The actual
value is 1.34 g mL1.
4.
Group 4 obtained the following values: 1.60 g mL1, 1.70 g mL1, 1.40 g mL1. The actual
value is 1.40 g mL1.
13
SIGNIFICANT FIGURES
As discussed earlier, measurements are an integral part of most chemical experimentation. However, the
numerical measurements that result have some inherent uncertainty. This uncertainty is a result of the
measurement device as well as the fact that a human being makes the measurement. No measurement is
absolutely exact. When you use a piece of laboratory equipment, read and record the measurement to one
decimal place beyond the smallest marking on the piece of equipment.
The length of the arrow placed along the centimeter stick is 4.75 cm long. There are no graduation
markings to help you read the last measurement as 5. This is an estimate. As a result this digit is uncertain.
Another person may read this as 4.76 cm. This is acceptable since it is an estimation. There is error
(uncertainty) built into each measurement and cannot be avoided.
If the measurement is reported as 4.75 cm, scientists accept the principle that the last digit has an
uncertainty of 0.01 cm. In other words the length might be as small as 4.74 cm or as large as 4.76 cm. It is
understood by scientists that the last digit recorded is an estimation and is uncertain. It is important to
follow this convention.
There are three processes involving significant figures that we will mention here. The first process is
estimation in measurement. You will encounter this most frequently in the laboratory when measuring
volumes with a graduated cylinder, a buret or perhaps a pipette. Estimation will also be important when
measuring length with a ruler. The process of estimating the last digit will be very similar to the process
described at the beginning of this section.
Another process is examining measurements or data already taken by you or someone else and determining
the number of significant figures in the measurement. The next section provides you with six guidelines
for determining the number of significant figures in a recorded measurement.
The third process is reporting a calculated answer to the proper number of significant figures. The rules for
assigning the appropriate number of significant figures to an answer are covered in the section that begins
on page 16.
Significant figures will be of concern any time you perform a mathematical computation based on data. It
is important to present an answer that reflects the precision of the instruments used to collect the data.
14
Guidelines for Determining Significant Digits
1.
All digits recorded from a laboratory measurement are called significant figures (or
digits).
The measurement of 4.75 cm has three (3) significant figures.
Note:
If you use an electronic piece of equipment, such as a balance, you should record the
measurement exactly as it appears on the display.
2.
All non-zero digits are considered significant.
There are special rules for zeros. Zeros in a measurement fall into three types: leading
zeros, trailing zeros, and middle zeros (zeroes between non-zero digits).
3.
A middle zero is always significant.
303 mm has 3 significant figures: a middle zero—always significant
4.
A leading zero is never significant. It is only a placeholder; not a part of the actual
measurement.
0.0123 kg has 3 significant figures: a leading zero—never significant
5.
A trailing zero is significant when it is to the right of a decimal point. This is not a
placeholder. It is a part of the actual measurement.
23.20 mL has 4 significant figures: a trailing zero—significant to the right of a decimal
point
220 mL has only 2 significant figures: a trailing zero—there is no decimal point so this
zero is NOT significant
6.
All significant figures include units since they are a result of a measurement. A
number without units has little significance.
Measurement
123 g
46.54 mL
0.33 cm
726.8 mmHg
0.0336
Number of Significant Figures
3
4
2
4
3
Figure 5 Significant Figures
The most common errors concerning significant figures are (1) recording all
digits on the calculator readout, (2) failing to include significant trailing zeroes
(e.g., 14.150 g), and (3) considering leading zeroes to be significant (e.g.,
0.002 g: only 1 significant figure, not three).
15
Practice Problems
1.
How many significant figures are in each of the following?
a.
b.
c.
d.
e.
2.
451 000 m
6.626 X 1034 J • s
0.0065 g
4056 V
0.0540 mL
For the centimeter rulers below record the length of the arrow shown.
a.
b.
16
USING SIGNIFICANT FIGURES IN CALCULATIONS
When performing calculations in chemistry you will take measurements and apply some mathematical
operation. You might be adding masses or finding a difference in masses. You might be solving a problem
through multiplication or division. It is easy to plug the numbers into the calculator and get a display that
may have up to ten digits. Not all of those digits can be kept in the final answer! Chemists, and chemistry
students, have to be honest at the end of every calculation and reflect the level of precision in the data.
Here are some rules to help you determine how many digits to keep at the end of a mathematical operation.
Addition and Subtraction
The number of decimal places in the answer should be the same as in the measured quantity with the least
precision. The easiest way to determine this is by examining the number of places, e.g., hundreds, tens,
ones, tenths, hundredths, thousandths, in the measurement. The answer can have as many places in the
answer as the term with the fewest places.
1259.1
2.365
15.34
1277.075
grams
grams
grams
grams = 1277.1 grams
 0.1 gram is least precise measurement
i.e., you can only go to the tenths place
1875 J
950 J
86 J
2911 J = 2910 J
 10 J is the least precise measurement
i.e., you can only go to the tens place
Multiplication and Division
The number of significant figures in the answer should be the same as in the measured quantity with the
smallest number of significant figures.
13.356 g
 1.2817658 g mL1  1.282 g mL1
10.42 mL
13.345 g has 5 significant figures, 10.42 mL has 4 significant figures. The answer can only have 4
significant figures.
17
Practice Problems
1.
Answer the following problems using the correct number of significant figures.
a.
16.27 + 0.463 + 32.1
b.
42.05  3.6
c.
15.1 x 0.032
d.
13.36  0.0468
e.
(13.36  0.046) x 12.6
1.424
2.
In the laboratory a group of students was assigned to determine the density of an unknown
liquid. They used a buret to measure the liquid and found a volume of 2.04 mL. The mass
was determined on an analytical balance to be 2.260 g. How should they report the density
of the liquid?
3.
In the first laboratory activity of the year, students were assigned to find the total area of
three tabletops in the room. To save time, each of the three students grabbed a ruler and
measured the dimensions. They then calculated the area for each tabletop and added them
together. Figure 6 presents the students’ measurements. What is the total area of the three
tabletops?
Student
A
B
C
Length
Width
127 cm
1.30 m
50.0 in
74 cm
0.80 m
29.5 in
Figure 6 Tabletop Dimensions
Note: Only numbers resulting from measurements made using instruments have significant figures.
Exact numbers have an infinite number of significant figures. Exact numbers include numbers derived from
counting or definition.
Examples: 25 desks in a room or 100 cm = 1 meter
18
SCIENTIFIC NOTATION
In chemistry we deal with very small and very large numbers. It is awkward to use many zeros to express
very large or very small numbers, so scientific notation is used. The number is rewritten as the product of a
number between 1 and 10 and an exponential term—10n, where n is a whole number.
Examples
1.
The distance between New York City and San Francisco = 4,741,000
meters:
4,741,000 m = (4.741 X 1,000,000) m, or 4.741 X 10 6 m
2.
The mass of 25 water molecules = 0.000 000 000 000 000 000 000 748
grams:
0.000 000 000 000 000 000 000 748 grams = 7.48 X 0.000 000 000 000 000 000 000 1 g =
7.48 X 1022 grams
It is easier to assess the magnitude and to perform operations with numbers written in scientific notation. It
is also easier to identify the proper number of significant figures.
Addition/Subtraction Using Scientific Notation
1.
2.
3.
Convert the numbers to the same power of ten.
Add (subtract) the nonexponential portion of the numbers.
The power of ten remains the same.
Example: 1.00 X 104 + 2.30 X 105
A good rule to follow is to express all numbers in the problem to the highest power of ten.
Convert 1.00 X 104 to 0.100 X 105.
0.100 X 105 + 2.30 X 105 = 2.40 X 105
Multiplication Using Scientific Notation
1.
2.
3.
Example:
The numbers (including decimals) are multiplied.
The exponents are added.
The answer is converted to proper scientific notation—the product of a number between
1 and 10 and an exponential term.
(4.24 X 102) X (5.78 X 104)
(4.24 x 5.78) X (102+4) = 24.5 X 106
Convert to proper scientific notation
= 2.45 X 107
19
Division Using Scientific Notation
1.
2.
3.
Divide the decimal parts of the number.
Subtract the exponents.
Express the answer in scientific notation.
Example:
(3.78 X 105)  (6.2 X 108)
(3.78  6.2) X (105–8) = 0.61 X 103
Convert to proper scientific notation = 6.1
X 104
Practice Problems
1.
2.
Convert the following numbers to exponential notation.
a.
0.0000369
c.
0.0452
d.
4 520 000
b.
36
1000
e.
365 000
Carry out the following operations:
a.
b.
c.
(1.62 X 103) + (3.4 X 102)
(1.75 X 101)  (4.6 X 102)
(15.1 X 102) x (3.2 X 102)
d.
e.
20
(6.02 X 1023) x (2.0 X 102)
(6.02 X 1023)  (12.0)
Understanding and Interpreting Graphs and Tables
The ability to interpret graphs and tables is a necessary skill in science but also finds use in everyday life.
In articles or textbooks you are likely to find graphs and tables. Understanding the article’s message
depends heavily on being able to interpret many different types of graphs and tables.
In science tables are used to provide information. Frequently one quantity in a table depends upon or is
related to another. Data from tables can be graphed to aid interpretation. Graphs give a visual
representation of the data that helps to reveal regularities and patterns.
TYPES OF GRAPHS
Graphs are of four basic types: pie charts, bar graphs, line graphs, and XY-plots. The type chosen depends
on the characteristics of the data displayed.
Pie Charts
Pie charts show the relationship of parts to a whole. The pie chart in Figure 8
displays the largest contributions to the composition of the human body. This presentation helps the reader
to visualize the magnitude of the differences between various elements which make up the body. Pie charts
are used infrequently in the analysis of data from the chemistry laboratory. There are times when you will
see scientific information presented in a pie chart and will need to interpret it.
Composition of the Human Body
All Other Elements
Hydrogen
Carbon
10%
6%
23%
61%
Oxygen
Figure 7: Elemental Composition of the Human Body
21
Bar Graphs
Bar graphs can be useful to study trends and compare relative values. In this table atomic number and size
are compared.
Atomic Number
Atomic Radius
(nm)
1
7.9
2
4.9
3
2005
4
14
5
11.7
6
9.1
7
7.5
8
6.5
9
5.7
10
5.1
11
22.3
12
17.2
13
18.2
14
14.6
15
12.3
16
10.9
17
9.7
18
8.8
19
27.7
Figure 8: Atomic Number and Atomic Radius for the First Nineteen Elements
Atomic Radius (nm)
Atomic Radius vs. Atomic Number
30
25
20
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Atomic Number
Figure 9: The Effect of Atomic Number on Atomic Radius for the First Nineteen Elements
This graph shows the relationship between atomic number and atomic radius. You can compare the radii of
any of these nineteen elements to each other. Trends and patterns are often brought out when data is
presented in a graph. The key is to find the most effective way of presenting the data so that the most valid
conclusions may be drawn from it.
22
Practice Problems
Use Figure 9 to answer these questions.
1.
What happens to atomic radius as you go across a series on the periodic table from left to
right?
2.
What element(s) have the smallest atomic radius among these 19? Give the name of the
element.
3.
What happens to atomic radius as you go down a group on the periodic table?
4.
What is the second period element with the largest atomic radius?
Multiple Bar Graphs
Multiple bar graphs compare relationships of closely related data sets. Atomic radii plotted against atomic
number (Figure 9) shows trends and relationships. However, it mighty give some additional insight into
atomic structure if another variable were also included in the graph. This data table includes the first
ionization energy for each of the nineteen elements.
Atomic Number
Atomic Radius
First Ionization Energy
(nm)
(eV)
1
7.9
13.598
2
4.9
24.587
3
20.5
5.392
4
14
9.322
5
11.7
8.298
6
9.1
11.26
7
7.5
14.534
8
6.5
13.618
9
5.7
17.422
10
5.1
21.564
11
22.3
5.139
12
17.2
7.645
13
18.2
5.986
14
14.6
8.151
15
12.3
10.486
16
10.9
1036
17
9.7
12.967
18
8.8
15.759
19
27.7
4.341
Figure 10: Atomic Number and Atomic Radius for the First Nineteen Elements
23
Comparison of Atomic Radius and First Ionization Energy
Atomic Radius
(nm)
Ionization Energy
(eV)
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19
Atomic Number
Atomic Radius
Ionization Energy
Figure 11: Comparison of Atomic Radius and First Ionization for the First Nineteen Elements
Practice Problems
1.
What happens to the first ionization energy as you go from left to right across the second
period of the periodic table? How does this compare to the trend you found for atomic
radius?
2.
Does the trend for first ionization energy hold for the third period elements as well?
3.
Which of the first nineteen elements has the largest first ionization energy?
4.
What is the relationship between atomic radius and first ionization energy?
24
Line Graphs
Constructing a line graph is another way to show the relationship between two variables. Consider the data
collected from the titration of 15.00 mL of 0.100 M HCl with 0.100 M NaOH. In this laboratory exercise
the 15.00 mL of hydrochloric acid is placed into a flask and the pH of the solution is recorded after the
addition of each 1.00 mL of sodium hydroxide. A line graph of this data will allow you to visualize what is
happening at various points in the titration.
Volume of NaOH
(mL)
pH
Volume of NaOH
(mL)
0.00
1.00
16.00
1.00
1.06
17.00
2.00
1.12
18.00
3.00
1.18
19.00
4.00
1.24
20.00
5.00
1.30
21.00
6.00
1.37
22.00
7.00
1.44
23.00
8.00
1.52
24.00
9.00
1.60
25.00
10.00
1.70
26.00
11.00
1.81
27.00
12.00
1.95
28.00
13.00
2.15
29.00
14.00
2.46
30.00
15.00
7.00
Figure 12: Data from the Titration of 0.100 M HCl with 0.100 M NaOH
pH
11.51
11.80
11.96
12.07
12.15
12.22
12.28
12.32
12.36
12.40
12.43
12.46
12.48
12.50
12.52
pH
Titration of 0.100 M HCl with 0.100 M
NaOH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0.00
5.00
10.00
15.00
20.00
Volume of NaOH (mL)
Figure 13: Titration of 0.100 M HCl with 0.100 M NaOH
25
25.00
30.00
It is clear from the graph that the pH changes only slowly at first, experiences a rapid increase near 15.00
mL and then only gradually increases after that. The line graph is an excellent choice for looking at how
data varies over time or concentration.
Practice Problems
Alkanes are compounds of carbon and hydrogen with the general formula, C nH2n+2. Suppose that you
did an experiment to determine the heat of combustion of several alkanes and noticed that the heat of
combustion/mole increased as the number of carbons in the alkane increased. The data taken are shown
in Figure 14.
Alkane
Number of Carbon Atoms
Heat of Combustion
(kJ mol1)
Methane, CH4
1
891
Ethane, C2H6
2
1561
Propane, C3H8
3
2219
n-Butane, C4H10
4
2879
n-Pentane, C5H12
5
3509
Figure 14: Heat of Combustion of Some Alkanes
1.
Plot a line graph of the data.
2.
Examine the line graph of the data. What is the relationship between the number of carbon
atoms in an alkane and the heat of combustion?
3.
Predict the value for the heat of combustion for n-hexane, C6H14.
4.
Predict the value for the heat of combustion for a substance with no carbon atoms. Why is the
value not 0? (Hint: Consider what remains in the formula when there are no carbon atoms.)
5
Could you use this same graph to predict the heat of combustion for other kinds of
hydrocarbons? Why or why not?
x-y Plot/Graph
An x-y plot (also called a scatterplot) demonstrates a mathematical relationship between two variables. This
type of plot is especially useful in scientific work. Sometimes it is difficult to decide if a graph is a line
graph or an x-y plot. One difference is that in an x-y plot it is possible to determine a mathematical
relationship between the variables. Sometimes the relationship is the equation for a straight line (y = mx +
b), but other times it is more complex and requires manipulation of the data. To clarify, we will first look at
a straight line, or direct relationship, then proceed to more complex situations.
Example
An entrepreneur was considering investing in a mine that was said to produce gold. Several very small
irregular nuggets were given to a chemist for analysis. The chemist, who was instructed to use
nondestructive methods, decided to determine the density of the small samples. A micro-buret was used to
determine the volume of each nugget, and the mass was determined on an analytical balance. The data
collected are shown in Figure 15.
26
Nugget
Mass
(grams)
0.116
0.251
0.290
0.347
0.386
1
2
3
4
5
Figure 15 Gold Nugget Data
Volume
(mL)
0.006
0.012
0.015
0.018
0.021
Because a mathematical relationship is expected between the mass and volume of an element, the chemist
constructed an x-y plot.
Mass (g)
Mass/Volume Ratio
0.5
0.4
0.3
0.2
0.1
0
0
0.005
0.01
0.015
0.02
0.025
Volume (mL)
Figure 16 Gold Nugget Graph
To connect the plotted points, the best smooth curve is drawn. In this case it appears that the best curve is a
straight line. Programs like Excel will give you the option to put a trend line on your graph. The software
will let you see what type of curve or line best fits the data you have plotted.
Mass (g)
Mass/Volume Ratio
0.5
0.4
0.3
0.2
0.1
0
R2 = 0.9882
y = 17.973x + 0.0192
0
0.005
0.01
0.015
Volume (mL)
Figure 17 Gold Nugget Graph with best fit trend line
27
0.02
0.025
The x-y graph is often used when you are looking to establish a mathematical relationship between two
variables. It is also frequently used to assist in predicting the values of one variable based on the value of
the other.
Example:
Find the slope of the trend line drawn on the Mass/Volume graph.
To find the slope, choose two points on the line. These points do not need to be
ones you plotted. Determine the x and y coordinates of each point.
Calculate the slope using the formula:
Step 1:
slope 
y 2  y1
x 2  x1
Pick out two points from the line
It is usually best to pick points that have one axis value that falls on the grid line.
Two places are marked on the graph where the trend line crosses a grid mark on
the y axis. If we use these two we will know the y value with good precision and
only have to estimate the x value.
The y value of the first point is .20 grams (y1) and the x value of the first point is
about .010 mL (x1). The y value of the second point is .30 grams (y2) and the x
value of the second point is about .0155 mL (x2).
Step 2:
Substitute into the slope formula and solve
y 2  y1
x 2  x1
.3 g  .2 g
.1 g
Slope 

 18 g mL1
.0155 mL  .010 mL .0055 mL
slope 
28
Mass (g)
Mass/Volume Ratio
0.5
0.4
0.3
0.2
0.1
0
R2 = 0.9882
y = 17.973x + 0.0192
0
0.005
0.01
0.015
0.02
0.025
Volume (mL)
Figure 18: Mass/Volume Ration Graph with Trend Line, Equation and R2 Value
If you look at the last graph of the gold nugget data you will see two special features. Many types of
graphing software will not only give you the best fit trend line for your data it will also do some analysis of
how well the line fits the data. In this case the relationship between mass and volume was linear, i.e., the
data points gave a straight line best fit curve. The software was able to give the equation for that straight
line ( y = 17.973x + 0.0192). The equation for a straight line is y = mx + b. The x and y are the variables
you measured and graphed. The m is the slope of the line and b is the y-intercept, the place where the line
would cross the y axis. For the trend line the equation gives a slope of 17.973 which is in close agreement
with the 18 g mL1 we calculated by picking two points on the line and using the slope formula.
You will also notice some other information: R2 = 0.982. This value tells how close the data points lie to
the trend line. The closer the R2 value is to 1 the more perfect the fit between the data points and the trend
line. The graphing software you may use to plot data from lecture or lab will probably have this feature as
well.
The relationship between the two variables we measure and graph will not always be linear. Figure 17
provides the data for the volume of one mole of NH3 gas at various pressures. The two variables measured
in the laboratory are volume, in mL, and pressure, in atmospheres.
Volume
Pressure
(mL)
(atm)
244.5
0.1000
122.2
0.2000
61.02
0.4000
30.44
0.8000
12.17
2.0000
5.975
4.0000
2.925
8.0000
Figrue 19; Data for the Effect of Pressure on Volume of Ammonia Gas
29
Volume (mL)
The Dependence of Volume on Pressure
10
8
6
4
2
0
0
50
100
150
200
250
300
Pressure (atm)
Figure 20: Graph of Pressure-Volume Relationship
It is obvious from this graph that the volume is not related to pressure by a linear, straight line, relationship
like we say with the graph of mass and volume for the gold nugget samples. You can still do a best fit
curve and trend line. If you use graphing software you can experiment with possibilities until you find the
trend line that gives the best fit. Using Excel the best trend line comes from the “power” choice.
Volume (mL)
The Dependence of Volume on
Pressure
10
2
R =1
5
0
0
100
200
300
Pressure (atm)
Figure 21: The Effect of Pressure on the Volume of Ammonia Gas with Trend Line
Since R2 is equal to 1 this is a very good fit between the data points and the trend line. You cannot get a
better fit! This will almost never be the case with data you gather in the laboratory. The data points for this
example were actually calculated from the mathematical relationship between volume and pressure. There
are always things that make the data you collect in the laboratory less precise than the calculated values. In
other words, do not expect to get such a marvelous fit when you analyze your laboratory data.
When the best smooth curve is not a straight line, the data can be manipulated to see if another
mathematical relationship is possible. In this case it appears that as the pressure increases the volume
decreases. So we can calculate the value of 1/V, add another column to the table (Figure 19), and graph that
data (Figure 20).
30
Volume
1/Volume
(mL)
(mL)
244.5
0.00409
122.2
0.00816
61.02
0.0164
30.44
0.0329
12.17
0.0822
5.975
0.167
2.925
0.3419
Figrue 22: Data for the Effect of Pressure on Volume of Ammonia Gas
Pressure
(atm)
0.1000
0.2000
0.4000
0.8000
2.0000
4.0000
8.0000
1/Volume (1/mL)
The Dependence of 1/Volume on Pressure
10
5
0
0
0.1
0.2
0.3
0.4
Pressure (atm)
Figure 23 Graph Showing Inverse Relationship Between Pressure and Volume
1/Volume (1/mL)
The Dependence of 1/Volume on Pressure
10
8
6
4
2
0
R2 = 0.9998
y = 23.418x + 0.0309
0
0.1
0.2
0.3
0.4
Pressure (atm)
Figure 24: Graph of 1/Volume vs. Pressure with Trend Line
This time the graph exhibits a straight line so we know that pressure and volume are inversely related. If
this mathematical manipulation did not result in a straight line, other mathematical changes or analysis
might be considered. Some of the other possible mathematical changes might involve taking the natural
logarithm of one variable or taking the values to a power.
31
BASIC GRAPHING RULES (line and x-y graphs)
1.
First decide where the information will be graphed. The horizontal axis (x-axis) is used for the
quantity that can be controlled or adjusted. This is called the independent variable. The vertical
axis (y-axis) is used for the quantity that responds to the changes in the quantity on the x-axis.
This is called the dependent variable.
2.
Choose the scale so the graph becomes large enough to fill most of the available space on the
paper.
3.
Each regularly spaced division on the graph paper should equal some convenient, constant value.
In general, each interval should have a value that can be easily divided visually such as 1, 2, 5, or
10, rather than a value such as 3, 6, 7, or 9.
4.
An axis does not need to start at zero, particularly if the plotted values cluster in a narrow range
not near zero.
5.
Label each axis with the quantity and unit being graphed. For example an axis might be labeled
“Temperature, °C.”
6.
Plot each point. If you plot more than one curve on the same graph, use a different color or
geometric shape to distinguish each set of data.
7.
For an XY graph, draw a smooth line that lies as close as possible to most of the points. Think of
this drawing as a line that is averaging your data. Do not draw a line that connects one point to the
next one as in a dot-to-dot drawing. If the curve appears to be straight, draw one continuous line
with a ruler.
8.
Title your graph with an informative title
32
INTERPRETING TABLES
Tables can be as simple as listing the value for a single property of a substance or as complex as the one in
Figure 25. The unshaded portion lists the melting points for several substances. The shaded portion of the
chart suggests some additional information to aid in interpretation. You may be asked to look for
relationships in the data.
Substance
Molar Mass
(g mol1)
18
Water
H2O
78
Benzene
C6H6
128
Naphthalene
C10H8
58.5
Sodium chloride
NaCl
16
183
Methane
CH4
1248
62
Magnesium fluoride MgF2
32
97.8
Methanol
CH3OH
Figure 25: Table Showing Selected Properties of Substances
Example:
Formula
Melting Point
(oC)
0
5
80
800
Stucture
Molecular
Molecular
Molecular
Ionic
Molecular
Ionic
Molecular
Polarity of
Molecule
Polar
Nonpolar
Nonpolar
Not applicable
Nonpolar
Not applicable
polar
Find two compounds in the table with similar molar masses. Compare their melting
points. Which of the characteristics listed appears to correlate with the differences
in melting point?
Sodium chloride (MM = 58.5 g mol1) and magnesium fluoride (MM = 62 g mol1) have
very similar molar masses. There melting points are both much higher than other
substances listed on the table. They are also significantly different from each other, 800
and 1248 oC respectively.
Both have ionic structures so this cannot account for the difference in melting point. The only other factor
from the table is the formula. There must be something about their composition that explains the
difference in melting point. NaCl is made up of two ions: Na + and Cl. MgF2 is made up of three ions:
Mg2+ and 2F. We learned early in chapter one that charge was responsible for interaction between
substances. Charge is also important in the interaction between ions in an ionic substance. Coulomb’s law
tells us that the larger the charge, the greater the force of attraction or repulsion. The magnesium fluoride
has a larger force of attraction with the Mg2+ ion than there is with the Na+ ion in the sodium chloride.
This explains the difference in melting points for the two substances. In chapter 2 there is a greater
explanation of the interaction of ions in an ionic solid.
Example:
Compare the molecular compounds with the ionic compounds and make a
generalization about structure and melting point.
The two ionic compounds have melting points of 800 oC and 1248 oC. The molecular
substances have melting points that range from 183 oC to +80 oC. It is always
dangerous to make generalizations from small amounts of data. However, it appears
that ionic compounds, in general, have much higher melting points than molecular
substances. The forces that hold ionic solids together must be much stronger than the
forces that hold molecular substances together. These forces are discussed in chapters
one and two of the textbook.
33
Practice Problems
1.
Compare the characteristics of methane, benzene, and naphthalene. What factor seems to
be responsible for differences in the melting points of these three substances?
2
The previous three questions use only some of the information available in the table. Write
two more questions that might be asked about the table.
3.
It is important to use all of the information available in a table. However, you should not
make sweeping generalizations that are supported by only a small number of facts. Look at
your answer to Question 1 and state what other information you might wish to look up to
support your statement.
Additional Practice Problems
1.
The graph in Figure 26 shows the approximate level of CO2 in the atmosphere from 1900 to 1990
for available decades. Study the graph and answer these questions:
a.
Predict the CO2 levels in 1910, 1950, and 2000.
b.
What other type(s) of graph might also be useful to study this data?
Figure 26: CO2 Levels from 1900 to 1990
34
2.
Graphically determine the density of ethylene glycol for the following data collected in the
laboratory. The density will be the slope of the straight line best fit curve for the data points.
Mass
(g)
11.20
16.72
22.14
17.78
33.42
Volume
(mL)
10.0
15.0
20.0
25.0
30.0
Figure 27: Ethylene Glycol Density Data
3.
The data below was collected when water was heated to its boiling point. Decide which type of
graph to use and graph this data. Answer the questions that follow based upon your graph.
Time
(minutes)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
Temperature
(oC)
23.0
27.0
34.0
43.0
58.0
69.0
75.0
83.0
90.0
94.0
96.0
97.0
98.0
99.0
100.0
100.5
Figure 28: Data from Heating Water to Boiling
a.
b.
c.
d.
e.
What type of graph did you choose to plot? Explain why you chose this type.
Describe the change in temperature with time.
Predict the temperature at 4.3 minutes.
Predict the temperature at 8.5 minutes.
During what time period was there the greatest change in temperature?
35
Understanding the Mole
WHAT IS A MOLE?
The mole is the SI unit of amount. It is used as a counting number in chemistry. Pair, dozen and ream are
other examples of counting numbers. It is unlikely that you would ever purchase computer paper by the
sheet. Instead you would probably purchase a ream, or package of 500 sheets, of paper. At the grocery store
you buy eggs by the dozen. It is convenient to buy groups of things like paper and eggs. When you buy a
package of a dozen, you know you will get twelve objects. When you by a ream it is consistently 500
sheets. A gross is a dozen dozen, 144 objects. In the same way, a mole equals 6.02 X 10 23 objects! Most
often things that are counted in units of moles are very small in size, e.g., atoms, molecules, or electrons.
Counting individual chemical units would be impossible since there are billions of billions of units in even
a relatively small amount of matter. It is much easier to count them in groups of moles. The mole will be
the common unit of amount in chemistry. It is used in many calculations since 1 mole of anything will
always contain 6.02 X 1023 chemical units.
Because it would take an impossibly long time to count 6.02 X 10 23 objects, an indirect method is used. An
analogy may be helpful. If you go into a home improvement store you may find nuts and bolts in bins that
can be bought in bulk. It would be very tedious to count out every single nut if you wanted a large number.
Instead you could weigh out a bag containing a number of nuts. Suppose it weighs 1,000. grams. If you
know how much a single nut weighs we can calculate the number of nuts in the bag. Suppose each nut
weighs 10. grams. If we divide 1,000. grams by 10. grams per nut we find that there are 100 nuts in the
bag. We can do the same thing with chemicals. If we know the mass of a sample we can divide by the
mass of one mole to calculate the number moles in the sample. The mass of one mole is called the molar
mass of a substance and is discussed in the next section.
Molar Mass
The modern definition of a mole is the number of atoms in exactly 12 grams of the carbon-12 (C-12)
isotope. This number itself is named after Amedeo Avogadro, who investigated related concepts but never
determined the number. At least four different types of experiments have determined that the number is
6.02 X 1023. Avogadro’s number is known to ten significant figures, but three will be enough for most of
your calculations.
1 mole = 6.022126736 X 1023 particles
The modern atomic-weight scale is also based on C-12. The relative mass of a hydrogen atom compared to
a carbon atom is 1.008. Therefore, one mole of hydrogen atoms has a mass of 1.008 g, and one mole of
oxygen atoms is 15.9994 g.
36
The number of grams in one mole, or molar mass, of a compound is found by adding the relative atomic
masses of the atoms in the formula.
Example:
What is the molar mass of methanol, CH 3OH?
Step 1: Identify the elements in the compound.
Carbon, Hydrogen, and Oxygen
Step2:
Read the atomic weights of the elements from the periodic table
Carbon:
Hydrogen:
Oxygen:
12.01 amu per atom
1.01 amu per atom
16.00 amu per atom
12.01 g per mole of atoms
1.01 g per mole of atoms
16.00 g per mole of atoms
NOTE: Since we are usually interested in moles of atoms, molecules, ions or formula units
it is common practice to use the grams per mole values. These are numerically equivalent to
the amu from the atomic weights given on the periodic table.
NOTE: Ask your instructor if they have a preference on whether to use the entire atomic
weight given or how many decimal places they prefer you have for your atomic weights.
Step3:
Multiply the number of moles of atoms of each element by the molar mass of each element
Carbon:
Hydrogen
Oxygen
1mole x 12.01g mole1 = 12.01 g
4 moles x 1.01 g mole1 = 4.04 g
1 moles x 16.00 g mole1 = 16.00 g
Step 4: Add together the molar mass contributions of all the elements
12.01 g + 4.04 g+ 16.00 g = 32.05 g for one mole of methanol
Methanol has a molar mass of 32.05 g mol1. This means that in every 32.05 grams of methanol there
are 1 mole or 6.02 X 1023 molecules of methanol.
Practice Problems
Find the molar mass (grams in one mole) of each of the following:
1.
2.
3.
4.
5.
6.
Ethanoic acid (acetic acid), CH3COOH
Methanal, (formaldehyde), HCHO
2-Dodecanol, CH3(CH2)9CH(OH)CH3
Glucose, C6H12O6
Ethanol, C2H5OH
Phosphoric acid, H3PO4
37
Mass-Mole Conversions
Unit conversions, dimensional analysis, can be used to convert between mass and moles.
mole
grams
grams A  moles A
grams
moles
moles A  grams A
Example:
How many moles are there in 100.0 grams of water?
Step 1: Calculate the molar mass of water
H: 1.01 g mole1 x 2 moles = 2.02 g
O: 16.00 g mole1 x 1 mole = 16.00 g
2.02 g + 16.00 g = 18.02 g = 18.02 g mole 1
Step 2: Choose the appropriate conversion factor
1 mole = 18.02 g so
1 mole
18.02 g
or
18.02 g
1 mole
mole
grams
grams A  moles A
Since you are given grams and want moles you will choose
1 mole
18.02 g
Step 3: Multiple the grams given by the conversion factor
100.0 grams H 2 O X
1 moleH 2 O
 5.549 moles H 2 O
18.02 grams H 2 O
38
mole
grams
grams A  moles A
grams
moles
moles A  grams A
Example:
How many grams in 5.50 moles of water?
Step 1: Calculate the molar mass
The molar mass of water, H2O was calculated before and found to be 18.02 g mole 1
Step 2: Choose the appropriate conversion factor
18.02 g
1 mole
grams
moles
moles A  grams A
Step 3: Multiple the moles given by the conversion factor
5.50 moles H 2 O X
18.02 grams H 2 O
 99.1 grams H 2 O
1 mole H 2 O
39
The conversion factors are chosen so that the units given will cancel and leave you with the units you want
for your answer. Sometimes it is necessary to perform more than one unit conversion in solving one of
these problems.
Example:
How many moles are there in 1.25 kg of ethanol, CH 3CH2OH?
Step 1: Calculate the molar mass
C: 12.01 g mole1 x 2 moles = 24.02 g
H: 1.01 g mole1 x 6 moles = 6.06 g
O: 16.00 g mole1 x 1 moles = 16.00 g
24.02 g + 6.06 g + 16.00 g = 46.08 g = 46.08 g mole1
Step 2: Choose the appropriate conversion factors
The first conversion is from kilograms to grams:
1000 grams CH 3CH 2 OH
1 kg CH 3CH 2 OH
The second conversion is from grams to moles:
1 mole CH 3CH 2 OH
46.08 grams CH 3CH 2 OH
Step 3: Multiple by the appropriate conversion factors
1.25 kg CH 3CH 2 OH X
1000 g CH 3CH 2 OH
1 mole CH 3CH 2 OH
X
 27.1 mole CH 3CH 2 OH
1 kg CH 3CH 2 OH
46.08 g CH 3CH 2 OH
Practice Problems
1.
Acetic acid, CH3COOH, and salicylic acid, C7H6O3, combine to form aspirin. If a chemist uses
5.00 g salicylic acid and 10.53 g acetic acid, calculate the number of moles of each compound
used.
2.
2-Dodecanol, CH3(CH2)9CH(OH)CH3 is used in synthesis of wetting agents. A manufacturer
orders 500.0 kg of the compound. How many moles are ordered?
3.
Calcium chloride hexahydrate, CaCl2 • 6 H2O, is sprinkled on sidewalks to melt ice and snow.
How many moles of the compound are in a 5.0 kg sack of the material?
4.
1.5 mol sodium hydroxide, NaOH, are required to prepare a solution. What is the equivalent
number of grams?
5.
The laboratory technician must prepare a solution that requires 0.123 mol silver nitrate, AgNO 3.
How many grams are necessary?
40
Volume – Mole Conversions
Quite often in the chemistry laboratory you will be using solution with concentrations labeled with the
symbol M. The M represents concentration in molarity, moles L1. This is a very convenient unit since it
allows us to calculate moles if we know the volume of the solution in liters.
liters
mole
moles A 
 volume of A solution (L)
mole
liter
volume of A solution (in L) 
 moles A
Example:
How many moles are there in 25.00 mL of a 0.0500 M solution of NaCl?
Step 1: Convert volume to liters
1 L = 1000 mL so
25.00 mL X
1L
 0.02500 L
1000 mL
Step 2: Use the molarity as a conversion factor to solve for moles
mole
liter
volume of A solution (in L) 
 moles A
0.02500 L X
0.0500 moles
1L
 0.00125 moles NaCl
If you know the desired amount of moles you can calculate the volume of solution that will contain that
many moles
Example:
How many milliliters of 0.250 M potassium dichromate are needed to supply
0.040 moles of potassium dichromate?
Use conversion factors to convert from moles to liters and then to milliliters. Notice that the chemical
used makes no difference in these calculations. Remember one mole of any substance contains the
same number of molecules as one mole of any other substance. However, it is always best to include
the chemical identity with the units!
0.040 moles X
1L
1000 mL
X
 1600 mL potassium dichromate
0.250 moles
1L
41
You can even use molarity as a conversion factor along with the molar mass to convert from volume to
mass.
Example:
How many grams of sucrose, C12H22O11, are there in 500.0 mL of a 0.150 M
solution?
There are two conversion factors needed for this calculation: molarity and molar mass.
The molarity is:
0.500 L X
0.150 moles
342.34 grams
. The molar mass is:
.
1L
1 mole
0.150 moles 342.34 grams
X
 26.7 grams sucrose
1L
1 mole
Practice Problems
1.
How many moles are there in:
A.
B.
C.
125 mL of 0.0996 M hydrochloric acid, HCl
250.0 mL of 0.145 M methylamine, CH3NH2
5.00 mL of 0.0100 M acetic acid, CH3COOH
2.
How many milliliters of 0.0500 M sodium hydroxide, NaOH, are needed to provide 0.025
moles of NaOH?
3.
How many milliliters of 0.225 M hydrofluoric acid, HF, are needed to provide 0.0125 moles
of HF?
4.
How many grams of potassium chlorate, KClO3, are there in 25.00 mL of a 0.l250 M
solution?
5.
How many grams of sodium bicarbonate, NaHCO3, are there in 100.00 mL of a 0.100 M ?
42
Understanding Mass Relationships in Chemical Reactions
Equations are the language of chemistry, and they are important in helping us understand how atoms and
molecules form new substances. It has been said many times that chemistry is a quantitative science.
Chemistry is also a very precise science. Atoms combine to form molecules in definite ratios. Substances
react in definite ratios to form new materials. Atoms and molecules are very small and difficult to count
individually, so the idea of a mole was developed.
UNDERSTANDING CHEMICAL EQUATIONS
One way to describe a chemical reaction can be described by writing an English sentence.
When ethanol burns in the presence of sufficient oxygen, the products are carbon dioxide and water.
Another way is to write a word equation using the chemical names of the reactants and products. The arrow
here is often read “yields,” but it can also be thought of as an equal sign.
Ethanol + oxygen  carbon dioxide + water
While a word equation does provide us with additional information, replacing the chemical names with
formulas and indicating the physical states of each substance is even more descriptive and specific.
CH3CH2OH(l) + O2(g)  CO2(g) + H2O(l)
However, a chemical sentence is not an equation until it is balanced. The law of conservation of matter
states that in a chemical reaction, matter is neither created nor destroyed. That means the number of atoms
of each type must be the same on both sides of the equation. The following chemical equation shows a
balanced chemical equation. Count the number of carbon atoms, hydrogen atoms, and oxygen atoms to be
sure.
CH3CH2OH(l) + O2(g)  CO2(g) + H2O(l)
Reactant Side

2 carbon atoms
6 hydrogen atoms
7 oxygen atoms
Product Side
2 carbon atoms
6 hydrogen atoms
7 oxygen atoms
43
BALANCING CHEMICAL EQUATIONS
There are two key ingredients to writing chemical equations. First and foremost, you must write the correct
formula for each substance in the equation. The second ingredient is satisfying the law of conservation of
mass. In chemical equations this means that there must be the same number of atoms of each element in the
reactants as there are in the products. No atoms may be lost and none may be created in a chemical
reaction!
Consider the reaction: hydrogen gas reacts with oxygen gas to form gaseous water
Step 1: Write the formulas for each reactant and product correctly.
Hydrogen gas: H2 (g)
Oxygen gas: O2 (g)
Water: H2O (g)
Step 2: Write out the equation and see if it is balanced
H2 (g) + O2 (g)  H2O (g)
Reactants Side
2 hydrogen atoms
2 oxygen atoms

Products Side
2 hydrogen atoms
1 oxygen atom
The number of oxygen atoms is not the same on both sides of the equation.
Step 3: Place coefficients in front of formulas until the equation is balanced
We could place a 2 in front of the formula for water to give more water molecules since we found
a deficit of oxygen on the product side before
H2 (g) + O2 (g)  2H2O (g)
Reactants Side
2 hydrogen atoms
2 oxygen atoms

Products Side
4 hydrogen atoms
2 oxygen atom
The oxygen atoms are now equal on both sides. However, there are more hydrogen atoms on the
product side than the reactant side. Now we need to place a coefficient in front of the hydrogen
gas formula to help balance the equation. Place a 2 in front of the H 2 and see if the equation is
balanced.
2H2 (g) + O2 (g)  2H2O (g)
Reactants Side
4 hydrogen atoms
2 oxygen atoms

Products Side
4 hydrogen atoms
2 oxygen atom
Now the equation is balanced and the law of conservation of mass is satisfied.
44
This is the fundamental process used for balancing most equations. It is called the inspection method
because you look at the equation and try to place coefficients in front of the correct formulas until you get
the equation balanced. Some books give tricks or procedures for trying to balance equations. Generally,
you will see right away what elements need to be increased and on which side of the equation. Start with
one coefficient at a time and check to see the effect it had on balancing the equation. You can then keep
trying various coefficients in front of formulas until you are successful. This is really a trial and error
process so do not give up too easily.
Practice Problems
Balance these molecular equations
1.
Zn (s)+ HCl (aq)  ZnCl2 (aq) + H2 (g)
2.
Al (s) + O2 (g)  Al2O3 (s)
3.
C4H10 (g) + O2 (g)  CO2 (g) + H2O (g)
4.
KClO3 (s)  KCl (s) + O2 (g)
5.
Fe (s) + H2O (l)  Fe3O4 (s) + H2 (g)
6.
CaC2 (s) + H2O (l)  C2H2 (g) + Ca(OH)2 (aq)
7.
MnO2 (s) + HCl (aq)  MnCl2 (aq) + H2O (l) + Cl2 (g)
8.
Fe2O3 (s) + CO (g)  Fe (s) + CO2 (g)
9.
H2O2 (aq)  H2O (l) + O2 (g)
10.
CH3CH2OH (l) + O2 (g)  CO2 (g) + H2O (l)
Write the balanced molecular equation for the reactions that occur in each of the following situations.
11.
When liquid benzene, C6H6, reacts with oxygen gas, O2, the products are gaseous carbon
dioxide, CO2, and water, H2O.
12.
When photosynthesis takes place in a green plant, carbon dioxide gas and liquid water
combine to produce solid glucose, C6H12O6, and oxygen gas.
13.
Nitroglycerin, C3H5N3O9, a drug used for heart pain problems, is synthesized from glycerin,
C3H8O3, and nitric acid, HNO3, in the presence of a catalyst. What is also a product of the reaction.
14.
Some antacids contain solid aluminum hydroxide, Al(OH) 3, which reacts with the aqueous
hydrochloric acid, HCl, in the stomach to produce aqueous aluminum chloride, AlCl3, and liquid
water.
15.
An antacid remedy contains solid sodium bicarbonate, NaHCO 3, and solid citric acid,
H3C6H5O7, which react to produce gaseous carbon dioxide (the source of the familiar fizz), aqueous
sodium citrate, Na3C6H5O7, and liquid water.
16.
When solid table sugar (sucrose), C12H22O11, is heated, gaseous water and solid elemental
carbon are produced.
45
Stoichiometry
If the number of atoms is conserved in a chemical reaction, the mass must also be conserved as expected
from the Law of Conservation of Mass. In the equation for the formation of water—2 H2(g) + O2(g)  2
H2O(l)—2 molecules of hydrogen and 1 molecule of oxygen combine to form 2 molecules of water. We
could also say that 2 moles of hydrogen molecules react with 1 mole of oxygen molecules to form 2 moles
of water molecules. Using the number of grams in a mole of each substance, the mass relationships in the
table can be determined. The ratio of moles of hydrogen to moles of oxygen to form water will be 2:1. If 10
moles of hydrogen are available, 5 moles of oxygen are required.
2 H2 (g)
2 molecules
2 moles
(2 mol)(2.02 g mol1)
4.04 g
+
O2 (g)

1 molecule
1 mole
(1 mol)(32.00 g mol1)
32.00 g
2 H2O (l)
2 molecules
2 moles
(2 mol)(18.02 g mol1)
36.04 g
Solving problems involving the masses of products and or reactants is conveniently accomplished by
dimensional analysis. All numerical problems involving chemical reactions begin with a balanced equation.
Example
Find the mass of water produced when 10.0 grams hydrogen
reacts with excess oxygen.
2 H2(g) + O2(g)  2 H2O(l)
Step 1: Convert grams of a given substance to moles of a given substance.
mole
grams
grams A  moles A
 1 mole H 2 
  4.95 moles H 2
(10.0 grams H 2 )
 2.02 grams H 2 
Step 2: Convert from moles of the substance given to moles of substance wanted
You will use the coefficients from the balanced equation to convert from moles of hydrogen to
moles of water.
 2 moles H 2 O 
  4.95 moles H 2 O
(4.95 moles H 2 )
2
moles
H
2 

Step 3: Convert from moles of substance wanted to grams of the substance wanted.
grams
moles
moles B  grams B
 18.02 g H 2O 
  89.2 g H 2O
(4.95 moles H 2 O)
 1 mole H 2O 
46
Generally, you will want to put these three steps together in one problem. If you have the conversion
factors set up properly the units will cancel out and leave you with the units you need for the final answer.
 1 mol H 2
(10.0 g H 2 )
 2.02 g H 2
 2 mol H 2 O  18.02 g H 2 O 


  89.2 g H 2 O
2
mol
H
1
mol
H
O
2 
2


NOTE: This is the theoretical yield of this reaction. It represent the amount of product that can be made
if the reaction works perfectly. In practice the yield of product would be something less than this amount.
Practice Problems
1.
Acetylene burns in air to form carbon dioxide and water:
5 C2H2(g) + 2 O2(g)  4 CO2(g) + 2 H2O(l)
How many moles of CO2 are formed from 25.0 moles C2H2?
2.
If insufficient oxygen is available, carbon monoxide can be a product of the combustion of
butane:
9 C4H10(l) + 2 O2(g)  8 CO(g) +10 H2O(l).
What mass of CO could be produced from 5.0 g butane?
3.
15.0 g NaNH2 is required for an experiment. Using the following reaction, what mass of
sodium metal is required to produce the NaNH2?
2 Na(s) + 2 NH3(g)  2 NaNH2(s) + H2(g)?
4.
Ethanol and acetic acid react to produce ethyl acetate according to the reaction C 2H5OH +
CH3C(O)OH  CH3C(O)OC2H5 + H2O. If the reaction is only 35% efficient at the conditions
used, what mass of CH3C(O)OH will be necessary to produce 100. g CH3C(O)OC2H5?
Assume that sufficient ethanol is available.
5.
Heating CaCO3 yields CaO and CO2. Write the balanced equation. Calculate the mass of
CaCO3 consumed when 4.65 g of CaO forms
.
47
Understanding Limiting Reactants
If a combustion problem states that excess oxygen is available, we need not concern ourselves with the
oxygen-to-water ratio. The mass of water produced is predicted from (and limited by) the mass of hydrogen
available. Of course in the laboratory we often deal with specified masses of each reactant, but that requires
an enhanced problem-solving method.
Suppose a family wants to make several chile relleno casseroles to serve at a neighborhood party. The
recipe lists required ingredients, which are itemized in the left column of the table that follows. The right
column represents a survey of pantry and refrigerator contents.
Required Ingredients
1 27-oz can whole green chiles
1 lb Monterrey Jack cheese
1 lb Cheddar cheese
3 eggs
3 Tbsp flour
5-oz canned evaporated milk
Available Ingredients
2 27-oz cans whole green chiles
3 lb Monterrey Jack cheese
3 lb Cheddar cheese
1 doz eggs
5 lb flour
4 5-oz cans evaporated milk
Possible Casseroles
2
3
3
4
Many
4
How many casseroles can be made? Although four casseroles can be made from the available eggs or milk,
there are only enough cans of green chiles for two casseroles. In other words, the number of cans of green
chiles can be called the limiting factor. After the two casseroles are prepared, cheese, eggs, flour, and milk
will remain, but all the green chiles will be used. Therefore, no more than two casseroles can be made.
In this example, the green chiles are the limiting reactant. The limiting reactant is the reactant that is
consumed first and limits the amount of product that can be made. The same principle applies in
determining the quantity of product that can be produced in a chemical reaction. Let’s take another look at
the reaction of hydrogen and oxygen to produce water, then consider what would happen if 2.00 mol
hydrogen and 2.00 mol oxygen were available. How many moles of water can be produced? What is the
limiting reactant? Which reactant will be in excess and by how much?
2 H2(g) + O2(g)  2 H2O(l)
The balanced chemical equation states that 2.00 mol hydrogen react with 1.00 mol oxygen. When the
reaction is complete, 2.00 mol water are produced and 1.00 mol oxygen remains unreacted. This problem is
easy to solve by inspection. Many times it will not be so easy to judge which is the limiting reactant
through inspection. This is especially true when you are given grams of the reactants instead of moles.
One method that is easy and will always help you determine the limiting reactant is to set up the three steps
(grams of given  moles of given, moles of given  moles of wanted, moles of wanted  grams of
wanted) for both reactants as if the other one was present in excess. The reactant that gives the smaller
amount of product will be the limiting reactant. It is the reactant that limits the amount of product that can
be made.
48
Example
How many grams of water can be made if 5.00 g of H 2 (g) and 5.00 g of O2 (g)
are allowed to react? What is the limiting reactant?
Set up the conversion factors for each one and solve.

 2 mol H 2 O  18.02 g H 2 O 


  44.6 g H 2 O
 2.02 g H 2  2 mol H 2  1 mol H 2 O 
5.00 g H 2  1 mol H 2
 1 mol O2  2 mol H 2O  18.02 g H 2O 


  5.63 g H 2O
 32.00 g O2  1 mol O2  1 mol H 2O 
5.00 g O2 
Since the oxygen gas produces the smaller amount of water it must be the limiting reactant. That
means that when 5.00 g of hydrogen and oxygen gases react a total of 5.63 g of water can be produced.
Once we have identified the limiting reactant we can calculate the mass of the excess reactant that
remains unused at the end of the reaction.
First, calculate how much of the excess reactant is used up in the reaction. Start with the limiting
reactant and solve for the mass of the excess reactant.
 1 mol O2  2 mol H 2  2.02 g H 2 


  .631 g H 2 used
32
.
00
g
O
1
mol
O
1
mol
H
2 
2 
2 

5.00 g O2 
It only takes 0.631 g of hydrogen gas to react with the 5.00 grams of oxygen gas. We started out with
5.00 grams of oxygen gas and used up 0.631 grams. That means that 4.37 grams of hydrogen remain
unused at the end of the reaction (5.00 g – 0.631 g = 4.37 g H2).
49
Example:
Aluminum chloride, AlCl3, has many uses including in deodorants and
antiperspirants. It is synthesized from aluminum and chlorine. What mass of
AlCl3 can be produced if 100. g of each reactant are available? What is the
limiting reactant? How many grams of the excess reactant remain?

Al
100. g Al  261.mol
98 g Al

 2 mol AlCl3  133.33 g AlCl3 


  494 g AlCl3
 2 mol Al  1 mol AlCl 
3




 1 mol Cl2  2 mol AlCl3  133.33 g AlCl3 
  125 g AlCl3


 70.90 g Cl2  3 mol Cl2  1 mol AlCl32 
100. g Cl2 
The Cl2 must be the limiting reactant since it produces the smaller amount of AlCl 3. In this reaction 125
g of aluminum chloride can be produced. This is the calculated or theoretical yield of the reaction. It
represents the amount of product that can be made if the limiting reactant is completely used up and the
reaction produces all the product theoretically possible. In actual practice the yield will be less than this
amount.
 1 mol Cl2  2 mol Al

70
.
90
g
O
2  3 mol Cl2

100. g Cl2 
 26.98 g Al

 1 mol Al

  25.4 g Al used

We started with 100. g of aluminum. 25.4 g of the aluminum is used in the reaction leaving 74.6 g of
aluminum unused at the end of this reaction.
Practice Problems
1.
In the synthesis of sodium amide (NaNH2), what is the maximum mass of NaNH2 possible
if 50.0 g of Na and 50.0 g NH3 were used?
2 Na(l) + 2 NH3(g)  2 NaNH2(s) +H2(g)
2.
The fuel methanol, CH3OH, can be made directly from carbon monoxide ( CO) and
hydrogen (H2).
a.
b.
c.
d.
3.
Write a balanced equation for the reaction.
Calculate the maximum mass of methanol if one starts with 5.75 g CO
and 10.0 g H2.
Which reactant is the limiting reactant?
How much of the excess reactant remains?
Aspirin (C9H8O4) is synthesized in the laboratory from salicylic acid ( C7H6O3) and acetic
anhydride (C4H6O3):
C7H6O3(s) + C4H6O3(l)  C9H8O4(s) +CH3COOH(l)
a.
b.
c.
What is the theoretical yield of aspirin if you started with 15.0 g salicylic
acid and 15.0 g acetic anhydride?
Which reactant is the limiting reactant?
What mass of the excess reactant remains?
50
Understanding Lewis Dot and Structural Formulas
There are a number of ways that molecules can be represented. One way is the molecular formula. The
molecular formula identifies the elements in the compound and gives the number of atoms of each element,
e.g., the molecular formula of ethanol is C2H6O. The molecular formula does not give any information
about connectivity, which atoms are bonded to each other in the molecule. In order to determine how the
atoms are bonded to each other a different type of representation will have to be used. The Lewis dot and
structural formula show how the valence electrons are distributed among the atoms of a molecule. Writing
out the Lewis structure will show us connectivity and help us predict the shapes of molecules as well.
Steps in Writing the Lewis Structure
Step 1: Determine the number of valence electrons available
Identify the number of valence electrons for each element
Multiple by the number of valence electrons by the number of atoms of that element
Sum the valence electrons for all the elements
Add in electrons equal to any negative charge on an ion
Subtract electrons equal to the positive charge on an ion
Example:
How many valence electrons in water, H 2O?
H:
O:
(1 valence electron) X (2 hydrogen atoms) = 2 valence electrons
(6 valence electrons) X (1 oxygen atom) = 6 valence electrons
The sum of the valence electrons is 8.
Example:
O:
How many valence electrons in the sulfate ion, SO42
S:
(6 valence electrons) X (1 sulfur atom) = 6 valence electrons
(6 valence electrons) X (4 oxygen atoms) = 24 valence electrons
The sum of the valence electrons is 30.
Since this is a negatively charged ion we need to add in electrons equal to its
charge.
The total valence electrons will be 30 + 2 = 32.
Example:
How many valence electrons in the hydronium ion, H 3O+?
H: (1 valence electron) X (3 hydrogen atoms) = 3 valence electrons
O: (6 valence electrons) X (1 oxygen atom) = 6 valence electrons
The sum of the valence electrons is 9.
Since this is a positive ion we must subtract electrons equal to the charge. The
total number of valence electrons is 9  1 = 8.
51
Step 2: Write the symbols of the elements and connect them to each other with a pair of electrons
NOTE: It is usually good to start by putting the element with the fewest atoms in the formula in
the center and arranging the other elements around it. You would not choose hydrogen as
a central atom for any of the three examples here.
H
O H
O
O
S
H
O
O
O
H
H
Step 3: Distribute electrons to the atoms that are not in the center until they have 8 valence electrons
(4 pairs of electrons). This is called an octet. NOTE: The pair in the bond is counted as one
of the 4 pairs of the octet. Hydrogen will not get an octet. It will only have one pair of
valence electrons.
H
O H
O
O
S
H
O
O
O
H
H
Step 4: If there are any valence electrons remaining they may be placed around the atom in the
center until it has an octet or all the available electrons are used up.
2
1+
H
O H
O
O
S
H
O
O
O
H
H
NOTE: Ions are written with brackets around them and the charge noted at the upper right hand
corner. This assists in the electron counting process. The brackets and charge identify why there
are extra or missing electrons. You can wait until you have completed the Lewis structure to put
on the brackets and charge if you wish.
52
Step 5: Check to see if each atom has an octet of electrons. If the central atom does not have an
octet it may be necessary to share more than one pair of electrons with one of the other
atoms, i.e., form a double or triple bond.
NOTE: C, O, S, N, and P are all capable of sharing two pairs of electrons in a bond, i.e., forming
double bonds. C and N are capable of sharing three pairs of electrons in a bond, i.e.,
forming a triple bond.
Hydrogen and the halogens will not routinely form double or triple bonds.
All three of the compounds shown in the examples for Step 4 have octets around all the atoms
except hydrogen. Therefore, no double bonds are needed and all three Lewis structures are fine as
written.
Example:
Write the Lewis structure for the nitrite ion, NO 2.
Step 1: Total valence electrons
N: (5 valence electrons) X (1 nitrogen atom) = 5 valence electrons
O: (6 valence electrons) X (2 oxygen atoms) = 12 valence electrons
The sum of the valence electrons is 17. However, since it has a 1 charge we have to add 1 more
to the 17 to get a grand total of 18 valence electrons available in the nitrite ion.
Step 2: Bond atoms with a pair of electrons
O
N
O
Step 3: Complete octets of electron on outer atoms
O
N
O
Step 4: Place any remaining electrons around the central atom
O
N
O
Nitrogen does not have an octet after adding the remaining electrons.
Step 5: Complete the octet around the central atom by forming double or triple bonds if possible
2
O
N
O
Sharing two pairs of electrons with the oxygen gives nitrogen an octet.
O
N
O
2
All three atoms now have an octet of electrons.
53
Example:
Write the Lewis structure for methane, CH4
Step 1: C: 4 valence electrons; H: 1 X 4 = 4; total is 8 valence electrons
H
Step 2:
H
C
H
Steps 3 – 5 will not result in any changes to the structure.
H
Example:
Write the Lewis structure for methanal, HC(O)H
Step 1: H: 1 X 2 = 2; C: 4 X 1 = 4; O: 6 X 1 = 6; total valence electrons is 12.
Step 2:
O
H
C
H
NOTE: You may wonder why the carbon atom is placed in the center since there are one atom
each of carbon and oxygen. It would be possible to write a Lewis structure with oxygen as the
central atom. In a later section the concept of formal charge will be introduced as a way to help
determine which of several possible Lewis structures would be more likely to occur. Right now
you could apply the bonding patterns for hydrocarbons and would see that the Lewis structure
with oxygen in the center will not be consistent with the bonding patterns.
Step 3:
O
H
C
H
All 12 valence electrons have been used.
Step 4: There are no valence electrons remaining to place on the carbon atom.
Step 5:
O
H
C
H
Two pairs of electrons must be shared in order for carbon to have an octet.
54
Example:
Write the Lewis structure for carbon dioxide, CO2
Step 1: C: 4 X 1 = 4; O: 6 X 2 = 12; total valence electrons is 16
Step 2:
O
C
O
O
C
O
Step 3:
Step 4: There are no left over valence electrons to be placed around the carbon atom. It does not have an
octet so it will be necessary to share more than one electron with each oxygen.
Step 5:
O
C
O
There will not be a triple bond with one oxygen and a single bond with the other since oxygen is
not one of the two elements likely to form triple bonds. Oxygen does make double bonds.
Structural Formulas
Sometimes you will see the Lewis structure abbreviated by replacing the dots representing shared pairs of
electrons, i.e., bonds, with dashes.
1
NO2 : O
N
CH4:
H
H
C
O
H
H
HC(O)H:
O
H
C
H
CO2: O
C
O
55
Bonding Patterns in Many Hydrocarbon Molecules
In the textbook a number of biologically important molecules are described. There are some simple
bonding patterns that can be applied to many of the hydrocarbon molecules that you will encounter in the
text so that it will be easier to write structural or Lewis formulas. You cannot apply these patterns beyond
hydrocarbon molecules composed of carbon, hydrogen and oxygen without the risk of the patterns breaking
down.
In these hydrocarbon molecules carbon will make four bonds. This could be four single (sigma bonds) or
any combination of single, double or triple bonds that add up to four total bonds. Oxygen will make two
bonds. This could be from two single bonds (sigma bonds) or from a double bond (a sigma bond + a pi
bond). Hydrogen will only form one bond. It will always be a single (sigma bond).
Examples of Bonding in Hydrocarbon Molecules

C


C

C
C
O
O
H
Sometimes the line formula of a compound will give some help in writing the structural formula. It will
give us a general idea about the order of the atoms. You will still have to apply the patterns observed with
hydrocarbons and everything else you know about writing Lewis structures.
Example
Write the structural formula of ethanoic acid, CH3COOH.
There are two ways that we might try to write this formula.
H
H
C
C
O
O
H
H
H
H
O
C
C
H
56
O
H
The first structure is not consistent with the bonding patterns seen in hydrocarbon molecules. The second
carbon only makes two bonds instead of the 4 that carbon normally makes.
The second structure adheres to all the general rules. Hydrogen makes 1 bond, both oxygen atoms make
two bonds, and both the carbon atoms make four bonds each.
NOTE: A concept called formal charge can be used to help you decide among possible structural
formulas. It is discussed in the next section.
Practice Problems
Write the Lewis structures for these molecules and ions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
carbon tetrachloride, CCl4
carbonate ion, CO32
sulfur dioxide, SO2
sulfur trioxide, SO3
methanol, CH3OH
acetone, CH3C(O)CH3
sulfate ion, SO42
phosphate ion, PO43
ammonium ion, NH4+
hydrogen cyanide, HCN
57
Formal Charge
Suppose you have more than one possible way of writing a Lewis structure for a molecule.
OCS:
O=C=S
OSC:
O=S=C
SOC:
S=O=C
The technique of assigning formal charge to each atom within a compound can help determine which of
several structures is most likely. Formal charge is a book keeping method for electrons in a compound.
Formal Charge = Valence Electrons – (1/2 bonding electrons + nonbonding electrons)
OR
Formal Charge = Valence Electrons  (# of bonds + # of nonbonding electrons)
Formal charge is calculated for each atom in a molecule. The sum of the formal charges on the atoms in
any ion must be equal to the charge of the ion. When comparing two structures the one with the most
formal charges of zero is the one you would predict to be more probable.
NOTE: The method of calculating formal charge shown here is an alternative to the method shown in the
book. Chemists often have several methods available to them for solving problems. You need to find the
method that makes sense to you and use it. Even though the methods are not identical each one will yield
the same result.
Example: Which structure is most probable: OCS, OSC, or SOC?
OCS:
FCO = 6 – [1/2(4) + 4] = 0 (The oxygen has 6 valence electrons. Since there is a double bond
there are 4 electrons shared between the oxygen and carbon. The
oxygen also has 4 nonbonding electrons)
FCC = 4 – [1/2(8) + 0] = 0 (Carbon has 4 valence electrons. It forms two double bonds in this
compound for a total of 8 shared/bonded electrons. Carbon does not
have any nonbonding electrons.)
FCS = 6 – [1/2(4) + 4] = 0 (Sulfur has 6 valence electrons. It shares 4 electrons with the carbon
and has 4 nonbonding electrons as well.)
The formal charge on each atom is 0 for this molecule.
OSC: FCO = 6 – [1/2(4) + 4] = 0
FCS = 6 – [1/2(8) + 0] = +2
FCC = 4 – [1/2(4) + 4] = 2
The formal charges are not 0 for the sulfur and carbon. The first formula is more
probable than this one.
58
SOC: FCS = 6 – [1/2(4) + 4] = 0
FCO = 6 – [1/2(8) + 0] = +2
FCC = 4 – [1/2(4) + 4] = 2
The formal charges on two of the atoms are not 0 with this formula.
The first formula OCS is the most probable of these three since the formal charge of each
atom in that formula is 0.
Practice Problems
1.
Assign formal charges to each atom in both of these structural formulas.
H
H
C
C
O
O
H
H
H
H
O
C
C
O
H
H
Does the formal charge help you pick the more probable formula?
2.
There are three structures suggested for the ion NCO.
[ N
C
O ]
[ N
C
O ]
[ N
C
O ]
Assign formal charges to each atom. Use the formal charges to predict which structure is most
probable.
59
Understanding Gas Law Problems
The behavior of gases can be predicted by several laws.
Boyle’s Law:
The volume of a gas is inversely proportional to pressure. As the pressure goes up the
volume goes down if all other variables are constant.
volume 
1
pressure
Charles’ Law: The volume of a gas is directly proportional to Kelvin temperature. As the temperature
goes up so does the volume if all other variables are constant.
volume  temperature
Avogadro’s Law: The volume of a gas is directly proportional to the number of moles of gas particles. As
the number of moles of gas particles goes up so does the volume if all other variables are constant.
volume  moles of gas particles
Gay-Lussac’s Law: The pressure of a gas is directly proportional to the Kelvin temperature. As the
temperature goes up so does the pressure if all other variables are
constant.
pressure  temperature
Ideal Gas Law:
PV = nRT (combines all four gas laws)
P = pressure (in atmospheres)
V = volume (in liters)
N = number of moles (in moles)
R = gas constant (0.08296 L atm mol1 K1)
T = temperature (in Kelvin degrees)
Problems requiring the use of Boyle’s, Charles’, Avogadro’s and Gay-Lussac’s laws always involve
changes in one of the variables: pressure, temperature, number of moles, or volume. The mathematical
formulas for the gas laws are given here.
Boyle’s Law:
Charles’ Law:
Avogadro’s Law:
V1 P2

or V1P1 = V2P2
V2 P1
V1 T1

or V1T2 = V2T1
V2 T2
V1 n1

or V1n2 = V2n1
V2 n 2
60
Gay-Lussac’s Law:
T1 P1

or T1P2 = T2P1
T2 P2
The four gas laws can be put together into a combined gas law to aid calculation when more than one
variable changes.
Combined Gas Law:
P1V1 P2V2

or P1V1n2T2 = P2V2n1T1
n1T1 n2T2
The pressure of a gas is not always given in atmospheres. The most common units are given here and
related to 1 atmosphere of pressure.
1 atmosphere = 760 Torr = 760 mmHg = 101.325 kPa = 101325 Pa = 14.7 psi
These relationships can be used to arrive at the proper conversion factor when changing from one of these
units to atmospheres.
Just like the SI units that have been established to help standardize reporting among scientist there are
standard units for temperature and pressure used in gas problems. Standard temperature and pressure
(STP) are 273 K and 1atmosphere.
61
Example
What volume would a 50.0 mL sample of gas occupy if the pressure were
increased from 2.0 atm to 8.0 atm?
Step 1: Identify the gas law or laws associated with the problem
In this problem you are given pressure and volume. That means that this is a Boyle’s law
problem.
V1 P2

or V1P1 = V2P2
V2 P1
Step 2: Identify the variables and their values
V1 (starting volume) = 50.0 mL
V2 (ending volume) = ?
P1 (starting pressure) = 2.0 atm
P2 (ending pressure) = 8.0 atm
Step 3: Substitute and solve
(50.0 mL)(2.0 atm) = (V2)(8.0 atm); V2 
(50.0 mL)( 2.0 atm)
 12.5 mL  12 mL
(8.0 atm)
NOTE: There can only be two significant figures in our answer since the pressure values
only have two significant figures.
Step 4: Check to see if the answer makes sense
Boyle’s law says that volume is inversely related to pressure. An increase in pressure
should result in a smaller volume if all other variables remain constant. The answer is
consistent with this relationship.
62
Example
A 50.0 mL sample of gas has a pressure of 725 mmHg at 25.0 oC. What would
the pressure be if the temperature were raised by 75.0 oC?
Step 1: Identify the gas law
In this problem you are given pressure and temperature. Gay-Lussac’s law gives the
relationship between temperature and pressure. The volume given in this problem is
irrelevant.
T1 P1

or T1P2 = T2P1
T2 P2
Step 2: Identify the variables and their values
P1 = 735 mmHg
P2 = ?
T1 = 25.0 oC = 298 K
T2 = 100. oC = 373 K
NOTE: The temperature is raised by 75.0 oC and must be in
Kelvin.
Step 3: Substitute and solve
(725 mmHg)(373 K) = (P 2)(298 K); P2 
(725 mmHg )(373 K )
 907 mmHg
(298 K )
Step 4: Check to see if the answer makes sense
Gay-Lussac’s law says that temperature and pressure are directly related. An increase in
temperature should result in an increase in pressure. Our answer is consistent with this
relationship.
63
Example
A 75.0 mL sample of hydrogen gas, H 2 (g), has a pressure of 715 mmHg at
22.0 oC. What volume would the gas occupy at STP? No gas is lost in the
change of temperature and pressure.
Step 1: Identify the gas law
Volume, pressure and temperature are all given in this problem. This requires a
combination of Boyle’s and Charles’ laws. We can use the combined gas law.
P1V1 P2V2

or P1V1n2T2 = P2V2n1T1
n1T1 n2T2
Step 2: Identify the variables and their values
P1 = 715 mmHg
P2 = 760 mmHg (standard pressure in
mmHg)
V1 = 75.0 mL
V2 = ?
n1 = ?
n2 = ?
T1 = 22.0 oC = 295 K
T2 = 273 K (standard temperature)
What about n1 and n2? Since no gas is lost or gained the number of moles remains the
same. You can cancel out the n1 and n2 from the equation or put the same value in for
both n1 and n2, e.g., n1 = n2 = 1.0 mole.
P1V1T2 = P2V2T1
Step 3: Substitute and solve
We can ignore n1 and n2 since the number of moles remains constant.
(715 mmHg)(75.0 mL)(273 K) = (760 mmHg)(V 2)(295 K)
V2 
(715 mmHg )(75.0 mL)( 273 K )
 65.3 mL
(760 mmHg )( 295 K )
an alternative method would be to substitute in the same value for n1 and
n2.
(715 mmHg)(75.0 mL)(1.00 mol)(273 K) = (760 mmHg)(V 2)(1.00 mol)(295 K)
V2 would still be equal to 65.3 mL.
Step 4: Check to see if the answer makes sense
There is a decrease in temperature and an increase in pressure. Both of these changes
would contribute to a decrease in volume. Our answer is consistent with this.
64
A sample of oxygen gas occupies a volume of 125 mL at 25.0 oC and 721.0
mmHg. What volume would the gas occupy at 35.0 oC and 850.0 mmHg?
Example
Step 1: Identify the gas law
Since more than one variable changes we will use the combined gas law.
P1V1 P2V2

or P1V1n2T2 = P2V2n1T1
n1T1 n2T2
Step 2: Identify the variables and their values
P1 = 721.0 mmHg
V1 = 125.0 mL
n1= n2
T1 = 25.0 oC = 298 K
P2 = 850.0 mmHg
V2 = ?
n2 = n1
T2 = 35.0 oC = 308 K
Step 3: Substitute and solve
(721.0 mmHg)(125.0 mL)(308 K) = (850.0 mmHg)(V 2)(298 K)
V2 
(721.0 mmHg )(125.0 mL)(308 K )
 110. mL
(850.0 mmHg )( 298 K )
Step 4: Check to see if the answer makes sense
A temperature increase will contribute to an increase in volume. A pressure increase will
contribute to a decrease in volume. The pressure increase is proportionally larger than the
increase in temperature so a net decrease in volume would be expected. Our answer is
consistent with this reasoning.
65
Example
What volume would be occupied by 2.5 moles of nitrogen gas at
725.0 mmHg and 22.0 oC?
Step 1: Identify the gas law
Only one value is given for each variable. The Ideal Gas Law deals with one
value for each of the variables.
PV = nRT
Step 2: Identify the variables and their values
P = (725.0 mmHg)(1 atm/760 mmHg) = .9539 atm
V=?L
n = 2.5 moles
R = 0.08206 L atm mol1 K1
T = 22.0 oC = 295 K
Step 3: Substitute and solve
(.9539 atm)(V) = (2.5 moles)( 0.08206 L atm mol 1 K1)(295 K)
V = 63 L
Example
How many moles of gas would be found in 5.00 L of gas at STP?
Step 1: Identify the gas law
The Ideal Gas Law will be used since there is only one set of values for the variables.
Step 2: Identify the variables and their values
P = 1 atm
V = 5.00 L
n=?
R = 0.08206 L atm mol1 K1
T = 273 K
Step 3: Substitute and solve
(1.0 atm)(5.00 L) = (n)( 0.08206 L atm mol 1 K1)(273 K)
n = .223 moles
66
Example
What is the volume of 1.00 mole of hydrogen gas at 25.0 oC and 725
mm Hg?
Step 1: Identify the gas law
The Ideal Gas Law will be used since there is only one set of values for the variables.
Step 2: Identify the variables and their values
P=
 1 atm 
 = .954 atm
(725 mmHg )
 760 mmHg 
V = ?L
n = 1.00 moles
R = 0.08206 L atm mol1 K1
T = 298 K
Step 3: Substitute and solve
(.954 atm)(V) = (1.00 moles)( 0.08206 L atm mol 1 K1)(298 K)
V = 25.6 L
NOTE: You may have learned that the volume of 1 mole of gas is 22.4 L. This is only true for
1 mole of an ideal gas at STP.
Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
The volume of a sample of gas is expanded from 2.5 L to 4.5 L. If the original pressure
was 855 mm Hg, what is the pressure after the gas is expanded?
The volume of a sample of gas is compressed from 7.5 L to 1.0 L. If the initial
temperature is 24.0 oC, what is the temperature of the gas after it has been compressed?
A sample of gas occupying a volume of 75.0 mL contains 0.0250 moles. Gas is released
until the volume reaches 50.0 mL. How many moles of gas remain in the container?
The pressure of a gas is decreased from 950.0 mmHg to 720.0 mmHg. If the temperature
is 26.5 oC initially, what is the temperature after the pressure has been reduced?
A 120.0 mL sample of nitrogen gas, N2, has a pressure of 1.25 atm at 45.0 oC. What
volume will it occupy at STP?
A 75.0 mL sample of hydrogen gas, H2, has a pressure of 0.850 atm at 20.0 oC. What
volume will it occupy at 2.25 atm and 60.0 oC?
What would be the pressure of a 45.0 mL sample of hydrogen gas containing 0.750 moles
at 31.0 oC?
What would the temperature of 500.0 mL of oxygen gas be if there were 1.10 moles of gas
at a pressure of 850.0 mmHg?
67
Answers, Solutions and Explanations for the Practice Problems
Even though there are example problems in the Personal Tutor and Worked Examples in the textbook you
may not yet feel fully confident with your work on every problem. The solutions and explanations for most
of the problems have been included here so you can compare your work and your answer to this key.
Hopefully, you will be able to examine your work and see where is differs from the solutions given here.
Who knows, you may be able to uncover some errors that we have made! The answers, like the examples
elsewhere, are only of benefit if you actually use them. Be sure to work the problems and then check your
answers. The reward for doing the work and then revising it if need be will be far greater than if you
simply look at the problems and their answers. The original questions are also included here to help you.
Good luck!
Measurement and the Metric System (page 5)
1.
2.
Which metric unit and prefix would be most convenient to measure each of the following?
a.
meter
b.
micrometer
c.
millisecond
d.
kilogram
e.
nanogram
f.
milligram
g.
megagram
What word prefixes are used in the metric system to indicate the following multipliers?
a.
kilo
b.
milli
c.
centi
d.
micro
3.
An antacid tablet contains 168 mg of the active ingredient ranitidine hydrochloride. How many
grams of ranitidine hydrochloride are in the tablet?
 1g 
  0.168 g
168 mg
 1000 mg 
4.
There are 1.609 km in exactly 1 mile. How many centimeters are there in 1 mile?
 1000 m  100 cm 

  1.609 x 10 5 cm
1.609 km
 1 km  1 m 
5.
A paper clip is 3.2 cm long. What is the length of the paper clip in millimeters?
 10 mm 
  32 mm
3.2 cm 
 1 cm 
6.
State at least one advantage of SI units over the customary US units.
The advantage to the SI units is that they are based on multiples of ten. This makes it easy to vary
the size of the units by changing the power of ten. Other advantages might include wide
international use, common trade pricing, or the flexibility of the system.
68
Derived Units (page 7)
1.
The average person in the United States uses 340 L of water daily. Convert this to
milliliters.
 1000 mL 
  340,000 mL  3.0 x 10 5 mL
340 L 
1
L


2.
A quart is approximately equal to 946 mL. How many liters are in 1 quart?
 1L 
  .946 L
946 mL 
 1000 mL 
3.
One hundred fifty milliliters of rubbing alcohol has a mass of 120 g. What is the density of
rubbing alcohol?
120 g
 0.80 g mL1
150 mL
4.
A ruby has a mass 7.5 g and a volume of 1.9 cm3. What is the density of this ruby?
7.5 g
 3.9 g cm 3
3
1.9 cm
5.
What is the density of isopropyl alcohol if 5.00 mL weigh 3.93 g?
3.93 g
 0..786 g mL1
5.00 mL
Dimensional Analysis (page 9)
1.
The distance between New York and San Francisco is 4 ,741,000 m. Now, that may sound
impressive, but to put all those digits on a car odometer is slightly inconvenient. (Of course,
in the United States the odometer measures miles, but that is another story.) In this case,
kilometers are a better choice for measuring distance. Change the distance to kilometers.
 1 km 
  4741 km
4,741,000 m 
1000
m


2.
Convert 7,265 mL to L.
 1L 
  7.265 L
7,265 mL 
1000
mL


3.
The 1500 meter race is sometimes called the “metric mile.” Convert 1500 m to miles. (1 m =
39.37 in). NOTE: now the mile race is the 1600 meter.
69
 39.37 in   1 ft   1 mile 
 
 
  0.93 mile
1500 m 
 1 m   12 in   5280 ft 
4.
The density of aluminum is 2.70 g cm3. What is the mass of 235 cm3
of aluminum?
 2.70 g 
  634 g
235 cm 3 
3 
 1 cm 
5.
How many 250 mL servings can be poured from a 2.0 L bottle of soft drink?
 1000 mL   1 serving 
 
  8.0 servings
2.0 L 
 1 L   250 mL 
6.
The speed limit in Canada is 100 km/hr. Convert this to meters/second.
100 km  1000 m   1 hr


1 hr  1 km   60 min
7.
  1 min
 
  60 s

  27.8 m s 1

The density of helium is 0.17 g/L at room temperature. What is the mass of helium in a 5.4 L
helium balloon?
 0.17 g 
  0.92 g
5.4 L 
 1L 
8.
Liquid bromine has a density of 3.12 g/mL. What volume would 7.5 g of bromine occupy?
 1 mL 
  2.4 mL
7.5 g 
3
.
12
g


9.
An irregularly shaped piece of metal has a mass of 147.8 g. It is placed in a graduated
cylinder containing 30.0 mL of water. The water level rises to 48.5 mL. What is the density
of the metal?
48.5 mL 30.0 mL = 18.5 mL (This is the volume of the piece of metal.)
147.8 g
7.99 g mL1
18.5 mL
70
Precision and Accuracy (page 11)
Groups of students determined the density of an unknown liquid in the laboratory. Calculate the average
and range for each group’s measurements.
1.
Group 1 obtained the following values: 1.34 g mL1, 1.32 g mL1 , 1.36 g mL1. The actual
value is 1.34 g mL1.
The sample data is precise and accurate. The average equals 1.34 g mL1, which is equal to the
actual value. The range is only 0.02 g mL1.
2.
Group 2 obtained the same results, but the actual value is 1.40 g mL 1 .
The data is precise (range is only 0.02 g mL1), but the accuracy is lower. The average is
1.34 g mL1, but significantly varied from the actual value of 1.40 g mL1.
3.
Group 3 obtained the following values: 1.66 g mL1, 1.28 g mL1, 1.18 g mL1. The actual
value is 1.34 g mL1.
The average is 1.37 g mL1 and the actual value is 1.34 g mL1, so the accuracy is good. The range
is 0.48 g mL1, so the precision is not very good
4.
Group 4 obtained the following values: 1.60 g mL1, 1.70 g mL1, 1.40 g mL1. The actual
value is 1.40 g mL1.
The average is 1.57 g mL1 and the range is 0.30 g mL1, so the measurements are neither very
accurate or precise.
Percent Error (page 12)
Calculate the percent error for all four groups in the previous set of practice problems. Use the average for
each group as the experimental value. The actual values are your accepted values.
1.
experimental value = 1.34 g mL1
accepted value = 1.34 g mL1
% Error 
2.
1.34 g mL1  1.34 g mL1
x 100 %  0.00 %
1.34 g mL1
experimental value = 1.34 g mL1
accepted value = 1.40 g mL1
% Error 
1.34 g mL1  1.40 g mL1
x 100 %   4.29%
1.40 g mL1
NOTE: The percent error is negative because the experimental value is less than the accepted
value.
71
3.
experimental value = 1.37 g mL1
accepted value = 1.34 g mL1
1.37 g mL1  1.34 g mL1
% Error 
x 100 %  2.24 %
1.34 g mL1
4.
experimental value = 1.57 g mL1
accepted value = 1.40 g mL1
1.57 g mL1  1.40 g mL1
% Error 
x 100 %  12.1 %
1.40 g mL1
Significant Figures (page 15)
1.
How many significant figures are in each of the following?
a.
b.
c.
d.
e.
2.
451 000 m
6.626 X 1034 J • s
0.0065 g
4056 V
0.0540 mL
3 significant figures
4 significant figures
2 significant figures
4 significant figures
3 significant figures
For the centimeter rulers below record the length of the arrow shown.
a.
The ruler is marked to the 0.1 cm. It is clear that the tip of the arrow is past
seven. The tip appears to be right on the eight. We can say with certainty that
the measurement is 8.0 cm. We are allowed to estimate one digit farther than
the markings on the ruler. The ruler is marked to the tenths of a centimeter so
we can estimate the hundredths place. Since the tip of the arrow seems to be
exactly on the 8 that makes the measurement 8.00 cm.
8.00 cm
b.
10.50 cm
Using Significant Figures in Calculations (page 17)
1.
Answer the following problems using the correct number of significant figures.
a.
16.27 + 0.463 + 32.1
48.833 = 48.8
Since this is an addition problem the answer can only be as precise as the term with the least
precision. The least precise measurement is 32.1 with a precision of  .1 so the answer must be
rounded to the nearest tenth.
72
b.
42.05  3.6
38.45 = 38.5
Since this is a subtraction problem the answer can only be as precise as the term with the least
precision. The term with the least precision is 3.6 which has a precision of  .1 (the precision of
42.05 is  .01). The answer must be rounded to the nearest tenth.
c.
15.1 x 0.032
0.4832 = 0.48
Since this a multiplication problem the answer can only have as many significant figures as the
term in the problem with the fewest significant figures. In this case 15.1 has 3 significant figures
and 0.032 has 2 significant figures. The answer can only have 2 significant figures since this is
the smaller number of significant figures.
d.
13.36  0.0468
285.470086 = 285
Since this is a division problem the answer can only have as many significant figures as the term
in the problem with the fewest significant figures. In this case 13.36 has 4 significant figures and
0.0468 has 3 significant figures. The answer can only have 3 significant figures since this is the
smaller number of significant figures among the terms in this problem.
e.
(13.36  0.046) x 12.6
1.424
118.6558 = 119
This problem is a combination of addition, multiplication and division. The rules for order of
operation tell us that the calculation within the parenthesis is done first. The calculation within the
parenthesis is addition and the answer to that part of the overall computation can have an answer
that is only as precise as the least precise term. That means the answer for the calculation within
the parenthesis would be 13.41 since the precision of the least precise term is  0.01. The rules for
multiplication and division are the same. The final answer can only have as many significant
figures as the term with the fewest number of significant figures. In this case 12.6 is the term with
the fewest significant figures. That means the final answer can only be expressed to 3 significant
figures.
2.
In the laboratory a group of students was assigned to determine the density of an unknown
liquid. They used a buret to measure the liquid and found a volume of 2.04 mL. The mass
was determined on an analytical balance to be 2.260 g. How should they report the density of
the liquid?
Density is mass over volume, so
2.260 g
 1.107843 g mL1  1.11 g mL1
2.04 mL
Since this is a division problem the answer may have as many significant figures as the term with
the fewest significant figures. In this case the volume (2.04 mL) has 3 significant figures and the
mass (2.260 g) has 4 significant figures. The answer should have 3 significant figures.
73
3.
In the first laboratory activity of the year, students were assigned to find the total area of
three tabletops in the room. To save time, each of the three students grabbed a ruler and
measured the dimensions. They then calculated the area for each tabletop and added them
together. Figure 6 presents the students’ measurements. What is the total area of the three
tabletops?
Student
A
B
C
Length
Width
127 cm
1.30 m
50. in
74 cm
0.80 m
29.5 in
Figure 6 Tabletop Dimensions
Area = length x width
One thing the students forgot to do was decide upon one unit of measurement. It will be necessary
to convert the units before the values can be added. You cannot add values with different units
together. In this case it might be easiest to convert all the measurements to centimeters (cm) and
then find the areas.
Student A’s measurements are already in centimeters so they do not need any conversion.
Student B’s measurements are in meters. 1 m = 100 cm.
 100 cm 
  130 m
1.3 m 
 1m 
 100 cm 
  80. cm
0.80 m 
 1m 
Student C’s measurements are in inches. 1 in = 2.54 cm
 2.54 cm 
  127 cm
50.0 in 
1
in


 2.54 cm 
  74.9 cm
29.5 in 
1
in


Table 1 (Student A): 127 cm x 74 cm = 9398 cm2 = 9400 cm2
(only 2 significant figures)
Table 2 (Student B): 130 cm x 80. cm = 10400 cm2 = 1.0 x 104 cm2 (2 significant figures)
Table 3 (Student C): 127 cm x 74.9 cm = 9512.3 cm2 = 9510 cm2 (3 significant figures)
The total area is the sum of these three values.
9,400 cm2 + 10,000 cm2 + 9,510 cm2 = 28,910 cm2 = 29,000 cm2
Since this is an addition problem the answer can only be as precise as the least precise
measurement. In other words, the answer’s last digit must be the same place value as the
term with the last digit in the highest place value. The second term in this group has the
last digit in the thousands place (that was why it was written 1.0 X 104 cm2). It is the
least precise measurement. Student A’s value has a precision of  100 cm2, Student B’s
value has a precision of  1000 cm2, and Student C’s value has a precision of  10 cm2.
That means our answer must have the last digit in the thousands place. It can only have a
precision of  1000 cm2.
74
Scientific Notation (page 19)
1.
2.
Convert the following numbers to exponential notation.
3.69 x 105
a.
0.0000369
b.
36
1000
c.
0.0452
4.52 x 102
d.
4 520 000
4.52 x 106
e.
365 000
3.65 x 105
3.6 x 102
Carry out the following operations:
a.
(1.62 x 103) + (3.4 x 102) = (1.62 x 103) + (0.34 x 103) = 1.96 x 103
b.
(1.75 x101)  (4.6 x 102) = (1.75 x 101)  (0.46 x 101) = 1.29 x 101
c.
(15.1 x 102) x (3.2 x 102) = 48.32 = 48 or 4.8 x 101
d.
(6.02 x 1023) x (2.0 x 102) = 1.204 x 1026 = 1.2 x 1026
e.
(6.02 x 1023)  (12.0) = 7.224 x 1024 = 7.22 x 1024
75
Bar Graphs (page 22)
1.
What happens to atomic radius as you go across a series on the periodic table from left to
right?
The atomic radius decreases as you from left to right across a series on the periodic table. The
arrows help show the pattern of decreasing atomic radius in the first three periods of the periodic
table. (NOTE: Horizontal rows on the periodic table may be called either series or periods.
Vertical columns are called either groups or families.)
Atomic Radius (nm)
Atomic Radius vs. Atomic Number
30
25
20
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Atomic Number
2.
What element(s) have the smallest atomic radius among these 19? Give the name of the
element.
Helium, He, and Neon, Ne, have the smallest atomic radii.
Elements 2 and 10 have the smallest atomic radii among the first nineteen elements. Element 2 is
Helium and element 10 is neon.
Atomic Radius (nm)
Atomic Radius vs. Atomic Number
30
25
20
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Atomic Number
76
3.
What happens to atomic radius as you go down a group on the periodic table?
Atomic radius seems to increase as you go from top to bottom in a group/family on the periodic
table.
The ovals show the radii of elements within group 1 (the alkali metals) on the periodic table. The
atomic radii increase as you go farther down the group. The rectangles illustrate the trend with
group 2 (alkaline earth) elements.
Atomic Radius (nm)
Atomic Radius vs. Atomic Number
30
25
20
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Atomic Number
4.
What is the second period element with the largest atomic radius?
Lithium, Li
The second period elements are elements 3 to 10, i.e., the elements from lithium to neon in the
second row of elements on the periodic table. Since atomic radius decreases as you go from left to
right the largest radius should be found in the first element of the series. This is consistent with
the graph. Element 3, lithium, has the largest atomic radius of the elements in the second period.
Atomic Radius (nm)
Atomic Radius vs. Atomic Number
30
25
20
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Atomic Number
77
Multiple Bar Graphs (page 23)
1.
What happens to the first ionization energy as you go from left to right across the second period of
the periodic table? How does this compare to the trend you found for atomic radius?
First ionization energy generally increases as you go from left to right across the second period of
the periodic table. There are two exceptions to this, but the general trend is for the first ionization
energy to increase. This is just the opposite of the trend with the atomic radius. While first
ionization energy increases from left to right the atomic radius decreases.
Comparison of Atomic Radius and First Ionization Energy
Atomic Radius
(nm)
Ionization Energy
(eV)
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19
Atomic Number
Atomic Radius
2.
Ionization Energy
Does the trend for first ionization energy hold for the third period elements as well?
The trend across the period holds for the third period elements as well.
Comparison of Atomic Radius and First Ionization Energy
Atomic Radius
(nm)
Ionization Energy
(eV)
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19
Atomic Number
Atomic Radius
3.
Ionization Energy
Which of the first nineteen elements has the largest first ionization energy?
78
Helium, He, has the largest first ionization energy of the first nineteen elements.
Comparison of Atomic Radius and First Ionization Energy
Atomic Radius
(nm)
Ionization Energy
(eV)
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19
Atomic Number
Atomic Radius
4.
Ionization Energy
What is the relationship between atomic radius and first ionization energy?
The larger an element’s atomic radius, the smaller its first ionization energy. There is an indirect
relationship between atomic radius and first ionization energy. This means as one variable gets
larger, the other variable gets smaller.
Comparison of Atomic Radius and First Ionization Energy
Atomic Radius
(nm)
Ionization Energy
(eV)
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19
Atomic Number
Atomic Radius
Ionization Energy
x-y Graphs (page 25)
Alkanes are compounds of carbon and hydrogen with the general formula, C nH2n+2. Suppose that you
did an experiment to determine the heat of combustion of several alkanes and noticed that the heat of
combustion/mole increased as the number of carbons in the alkane increased. The data taken are
shown in Figure 14.
79
Alkane
Number of Carbon Atoms
Methane, CH4
1
Ethane, C2H6
2
Propane, C3H8
3
n-Butane, C4H10
4
n-Pentane, C5H12
5
Figure 14: Heat of Combustion of Some Alkanes
1.
Heat of Combustion
(kJ mol1)
891
1561
2219
2879
3509
Plot a line graph of the data.
Heat of Combustion as a Function of
the Number of Carbon Atoms
Heat of
Combustion
(kJ/mol)
4000
3000
2000
1000
0
0
1
2
3
4
5
6
Number of Carbon Atoms
2.
Examine the line graph of the data. What is the relationship between the number of carbon
atoms in an alkane and the heat of combustion?
The larger the number of carbon atoms the higher the heat of combustion according to the graph of
the data.
3.
Predict the value for the heat of combustion for n-hexane, C6H14.
80
Heat of Combustion
(kJ/mol)
Heat of Combustion as a Function of
the Number of Carbon Atoms
4500
4000
3500
3000
2500
2000
1500
1000
500
0
y = 655.4x + 245.6
0
2
4
6
Number of Carbon Atoms
4200 kJ mol1
When a trend line is added to the graph the value for hexane, C6H14, can be estimated. The trend
line intersects the line for 6 carbons just barely less than half way between the two gridlines. The
lower grid line is 4000 kJ mol 1 and the higher grid line is 4500 kJ mol1. Exactly halfway
between them would be 4250 kJ mol1. Since it is just a bit less than this we might estimate it to
be 4200 kJ mol1.
4.
Predict the value for the heat of combustion for a substance with no carbon atoms. Why is
the value not 0? (Hint: Consider what remains in the formula when there are no carbon
atoms.)
250 kJ mol1
The same trend line we used in number 3 can be used to estimate the heat of combustion with zero
carbons. Using the formula CnHn+2 a value of zero for n would still give 2 hydrogen atoms in a
molecule. What we are estimating from the trend line is the heat of combustion of H2.
You may also notice that the formula for the trend line is given on the graph. The formula of a
straight line is y = mx + b. The x and y are the two variables being graphed, m is the slope of the
line, and b is the y-intercept. The y-intercept, b, is where the trend line crosses a vertical line
where the number of carbon atoms would be zero. The value of b in the equation is 245.6 kJ
mol1. This is in close agreement with our estimate from the trend line. You can utilize the power
of your graphing software to help you analyze your data.
81
5
Could you use this same graph to predict the heat of combustion for other kinds of
hydrocarbons? Why or why not?
It is possible that this graph could be used to predict the heat of combustion for other straight
chain hydrocarbons of greater length. However, it is likely that the farther out we estimate from
this data the more inaccurate it will become. It is not possible to use this graph to predict the heat
of combustion of other types of hydrocarbons. If the hydrocarbon is branched, cyclic, or there are
elements besides carbon and hydrogen present this graph would be inappropriate for making any
predictions. A graph can only be used to make predictions and estimates that are appropriate. Do
not generalize to far from a limited set of data in a graph!
Interpreting Tables (page 33)
Substance
Formula
Water
H2O
Benzene
C6H6
Naphthalene
C10H8
Sodium chloride
NaCl
Methane
CH4
Magnesium fluoride MgF2
Methanol
CH3OH
1.
Melting Point
(oC)
0
5
80
800
183
1248
97.8
Molar Mass
(g mol1)
18
78
128
58.5
16
62
32
Stucture
Molecular
Molecular
Molecular
Ionic
Molecular
Ionic
Molecular
Polarity of
Molecule
Polar
Nonpolar
Nonpolar
Not applicable
Nonpolar
Not applicable
polar
Compare the characteristics of methane, benzene, and naphthalene. What factor seems to be
responsible for differences in the melting points of these three substances?
With these three substances molar mass seems to be the factor that explains the differences in
melting point. All three are nonpolar molecules. The melting points increase as the molar mass
increases.
2
The previous three questions use only some of the information available in the table. Write
two more questions that might be asked about the table.
The questions you write might look at comparisons of polarity or composition and their
relationship to melting point.
3.
It is important to use all of the information available in a table. However, you should not make
sweeping generalizations that are supported by only a small number of facts. Look at your answer
to Question 1 and state what other information you might wish to look up to support your
statement.
Information that might be good to add to this table are: boiling point, density, intermolecular
forces in solids, and shape of molecule.
82
Graphing Problems (page 33)
1.
The graph in Figure 26 shows the approximate level of CO 2 in the atmosphere from 1900 to
1990 for available decades. Study the graph and answer these questions:
a.
Predict the CO2 levels in 1910, 1950, and 2000.
There seems to be a consistent increase in CO2 levels from 1900 to 1940. The estimate
for 1910 would be intermediate between the values for 1900 and 1920 or approximately
290 ppm.
The values for 1940 and 1960 are nearly identical. Although it would not be
unreasonable to assume that the value for 1950 could be slightly lower or higher, we have
no reason to assume that is was anything other than the approximately 320 ppm found in
the decade before and after.
There is a consistent increase in CO2 levels from 1960 to 1990. Extending a trend line
gives us an estimate of approximately 370 ppm for 2000.
b.
What other type(s) of graph might also be useful to study this data?
An x-y graph with a mathematically derived trend line would be very useful.
Figure 26: CO2 Levels from 1900 to 1990
2.
Graphically determine the density of ethylene glycol for the following data collected in the
laboratory. The density will be the slope of the straight line best fit curve for the data points.
Mass
(g)
11.20
16.72
22.14
17.78
33.42
Volume
(mL)
10.0
15.0
20.0
25.0
30.0
Figure 27: Ethylene Glycol Density Data
83
Mass (g)
Mass/Volume Ratio (Density) of
ethylene glycol
40
35
30
25
20
15
10
5
0
B
A
0
5
10
15
20
25
30
35
Volume (mL)
slope 
y 2  y1
x 2  x1
Point A: 22 mL, 25 g
Point B: 31 mL, 35 g
slope 
35 g  25 g
10 g

 1.111 g mL1  1.1 g mL1
31 mL  22 mL 9 mL
In this case using a trend line mathematically determined by your graphing software might lead to
a problem. There is one data point which does not seem to match the others. It is likely that this
is a result of experimental error. Ideally your analysis should include all the data. However, it is
often the case in the laboratory for a chemistry course that time will be limited so you will not be
able to do extra trials in the case of inconsistent results and some erroneous data points may be
eliminated from your analysis. In a research setting all the data would be included and further
trials would be done to minimize the impact of one stray data point.
3.
The data below was collected when water was heated to its boiling point. Decide which type
of graph to use and graph this data. Answer the questions that follow based upon your
graph.
Time
(seconds)
0
0.5
1.0
1.5
2.0
2.5
Temperature
(oC)
23.0
27.0
34.0
43.0
58.0
69.0
84
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
75.0
83.0
90.0
94.0
96.0
97.0
98.0
99.0
100.0
100.5
Figure 28: Data from Heating Water to Boiling
Temperature
(oC)
Time vs. Temperature for Heating
Water
120
100
80
60
40
20
0
0
1
2
3
4
5
6
7
8
Time (min)
Temperature (oC)
Time vs. Temperature for Heating
Water
120
110
100
90
80
70
60
50
40
30
20
10
0
0
1
2
3
4
Time (min)
85
5
6
7
8
a.
What type of graph did you choose to plot? Explain why you chose this type.
A line graph or an x-y graph are the most likely choices here since you are asked to make
some predictions about temperatures. A bar graph could also be used. Regardless of the
type of graph time would be best placed on the x-axis and temperature on the y-axis.
b.
Describe the change in temperature with time.
Temperature increases very rapidly during the first three minutes. From three to five
minutes the temperature still increases but at a slower rate of increase. After 5 minutes
the temperature changes only slightly.
c.
Predict the temperature at 4.3 minutes.
The temperature at 4.3 minutes is about 90. oC from the trend line. Averaging the values
between 4.0 and 4.5 minutes would give an estimate of 92 oC.
d.
Predict the temperature at 8.5 minutes.
Extending the trend line out to 8.5 minutes gives a value of approximately 102 oC.
e.
During what time period was there the greatest change in temperature?
The largest increase in temperature comes between 1.5 and 2.0 minutes
Molar Mass (page 36)
Find the molar mass (grams in one mole) of each of the following:
1.
Ethanoic acid (acetic acid), CH3COOH
C: 12.01 g mol1 x 2 moles of carbon atoms = 24.02 g
H: 1.01 g mol1 x 4 moles of hydrogen atoms = 4.04 g
O: 16.00 g mol1 x 2 moles of oxygen atoms = 32.00 g
24.02 g + 4.04 g + 32.00 g = 60.06 g is the mass of 1 mole of ethanoic acid (60.06 g mol 1)
2.
Methanal, (formaldehyde), HCHO
H: 1.01 g mol1 x 2 moles of hydrogen atoms = 2.02 g
C: 12.01 g mol1 x 1 mole of carbon atoms = 12.01 g
O: 16.00 g mol1 x 1 mole of oxygen atoms = 16.00 g
2.02 g + 12.01 g + 16.00 g = 30.03 g is the mass of 1 mole of methanal (30.03 g mol1)
3.
2-Dodecanol, CH3(CH2)9CH(OH)CH3
C: 12.01 g mol1 x 12 moles of carbon atoms = 144.12 g
H: 1.01 g mol1 x 26 moles of hydrogen atoms = 26.26 g
O: 16.00 g mol1 x 1 mole of oxygen atoms = 16.00 g
144.12 g + 26.26 g + 16.00 g = 186.38 g is the mass of 1 mole of 2-dodecanol (186.38 g mol1)
86
4.
Glucose, C6H12O6
C: 12.01 g mol1x 6 moles of carbon atoms = 72.06 g
H: 1.01 g mol1 x 12 moles of hydrogen atoms = 12.12 g
O: 16.00 g mol1 x 6 moles of oxygen atoms = 96.00 g
72.06 g + 12.12 g + 96.00 g = 180.18 g is the mass of 1 mole of glucose (180.18 g mol 1)
5.
Ethanol, C2H5OH
C: 12.01 g mol1 x 2 moles of carbon atoms = 24.02 g
H: 1.01 g mol1 x 6 moles of hydrogen atoms = 6.06 g
O: 16.00 g mol1 x 1 mole of oxygen atoms = 16.00 g
24.02 g + 6.06 g + 16.00 g = 46.08 g is the mass of 1 mole of ethanol (46.08 g mol 1)
6.
Phosphoric acid, H3PO4
H: 1.01 g mol1 x 3 moles of hydrogen atoms = 3.03 g
P: 30.97 g mol1 x 1 mole of phosphorus = 30.97 g
O: 16.00 g mol1 x 4 moles of oxygen atoms = 64.00 g
3.03 g + 30.97 g + 64.00 g = 98.00 g is the mass of 1 mole of phosphoric acid (98.00 g mol 1)
MassMole Conversions (page 39)
1.
Acetic acid, CH3COOH, and salicylic acid, C7H6O3, combine to form aspirin. If a chemist
uses 5.00 g salicylic acid and 10.53 g acetic acid, calculate the number of moles of each
compound used.
 1 mol 
  0.0362 mol C 7 H 6 O3
5.00 g C 7 H 6 O3 
 138.13 g 
 1 mol 
  0.1753 mol CH 3 COOH
10.53 g CH 3 OOH 
60
.
06
g


2.
2-Dodecanol, CH3(CH2)9CH(OH)CH3 is used in synthesis of wetting agents. A manufacturer
orders 500.0 kg of the compound. How many moles are ordered?
 1000 g   1 mol 
 
  2682 mol C12 H 26 O
500.0 kg C12 H 26 O 
1
kg
186
.
38
g



87
3.
Calcium chloride hexahydrate, CaCl2 • 6 H2O, is sprinkled on sidewalks to melt ice and
snow. How many moles of the compound are in a 5.0 kg sack of the material?
NOTE: When finding the molar mass of hydrates you must also include the mass of the water
molecules that are included as water of crystallization with the formula.
The molar mass of CaCl2 is 110.98 g mol1 and the mass of six water molecules is 108.12 g. That
makes the molar mass of the calcium chloride hexahydrate, CaCl26H2O 219.10 g mol1.
 1000 g   1 mol 
 
  23 mol CaCl2  6H 2 O
5.0 kg CaCl2  6H 2 O 
 1 kg   219.10 g 
4.
1.5 mol sodium hydroxide, NaOH, are required to prepare a solution. What is the equivalent
number of grams?
 40.00 g 
  60. g NaOH
1.5 mol NaOH 
 1 mol 
NOTE: It is necessary to put the decimal point in to show that there are 2 significant figures in the
answer.
5.
The laboratory technician must prepare a solution that requires 0.123 mol silver nitrate,
AgNO3. How many grams are necessary?
 169.91 g 
  20.9 g AgNO3
0.123 mol AgNO3 
 1 mol 
VolumeMole Conversions (page 41)
1.
How many moles are there in:
A.
125 mL of 0.0996 M hydrochloric acid, HCl
 1 L   0.0996 mol 
 
  0.0124 mol HCl
125 mL HCl 
1L
 1000 mL  

B.
250.0 mL of 0.145 M methylamine, CH 3NH2
 1 L   0.145 mol 
 
  0.03625 mol CH 3 NH 2
250.0 mL CH 3 NH 2 
 1000 mL   1 L

C.
5.00 mL of 0.0100 M acetic acid, CH3COOH
 1 L   0.0100 mol 
 
  5.00 x 10 5 mol CH 3 COOH
5.00 mL CH 3 COOH 
1L
 1000 mL  

88
2.
How many milliliters of 0.0500 M sodium hydroxide, NaOH, are needed to provide 0.025
moles of NaOH?

  1000 mL 
1L
 
  5.0 x 10 2 mL NaOH
0.025 mol NaOH 
 0.0500 mol   1 L 
3.
How many milliliters of 0.225 M hydrofluoric acid, HF, are needed to provide 0.0125 moles
of HF?
 1L
  1000 mL 
 
  55.6 mL HF
0.0125 mol HF 
0
.
225
mol
1
L



4.
How many grams of potassium chlorate, KClO 3, are there in 25.00 mL of a 0.l250 M
solution?
 1 L   0.1250 mol   122.55 g 
 
 
  .3830 g KClO3
25.00 mL KClO3 
1000
mL
1
L
1
mol




5.
How many grams of sodium bicarbonate, NaHCO 3, are there in 100.00 mL of a 0.100 M ?
 1 L   0.100 mol   84.01 g 
 
 
  0.840 g NaHCO3
100.00 mL NaHCO3 
1L
 1000 mL  
  1 mol 
Balancing Equations (page 44)
Balance these molecular equations
1.
Zn (s)+ HCl (aq)  ZnCl2 (aq) + H2 (g)
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
2.
Al (s) + O2 (g)  Al2O3 (s)
4 Al (s) + 3 O2 (g)  2 Al2O3 (s)
3.
C4H10 (g) + O2 (g)  CO2 (g) + H2O (g)
2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (g)
4.
KClO3 (s)  KCl (s) + O2 (g)
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
89
5.
Fe (s) + H2O (l)  Fe3O4 (s) + H2 (g)
3 Fe (s) + 4 H20 (l)  Fe3O4 (s) + 4 H2 (g)
6.
CaC2 (s) + H2O (l)  C2H2 (g) + Ca(OH)2 (aq)
CaC2 (s) + 2 H20 (l)  C2H2 (g) + Ca(OH)2 (aq)
7.
MnO2 (s) + HCl (aq)  MnCl2 (aq) + H2O (l) + Cl2 (g)
MnO2 (s) + 4 HCl (aq)  MnCl2 (aq) + 2 H20 (l) + Cl2 (g)
8.
Fe2O3 (s) + CO (g)  Fe (s) + CO2 (g)
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g)
9.
H2O2 (aq)  H2O (l) + O2 (g)
2 H2O2 (aq)  2 H2O (l) + O2 (g)
10.
CH3CH2OH (l) + O2 (g)  CO2 (g) + H2O (l)
CH3CH2OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (l)
Write the balanced molecular equation for the reactions that occur in each of the following
situations.
11.
When liquid benzene, C6H6, reacts with oxygen gas, O2, the products are gaseous carbon
dioxide, CO2, and water, H2O.
2 C6H6 (l) + 15 O2 (g)  12 CO2 (g) + 6 H2O (l)
12.
When photosynthesis takes place in a green plant, carbon dioxide gas and liquid water
combine to produce solid glucose, C6H12O6, and oxygen gas.
6 CO2 (g) + 6 H2O (l)  C6H12O6 (s) + 6 O2 (g)
13.
Nitroglycerin, C3H5N3O9, a drug used for heart pain problems, is synthesized from
glycerin, C3H8O3, and nitric acid, HNO3, in the presence of a catalyst. What is also a
product of the reaction.
C3H8O3 + 3 HNO3  C3H5N3O9 + 3 H20
14.
Some antacids contain solid aluminum hydroxide, Al(OH)3, which reacts with the aqueous
hydrochloric acid, HCl, in the stomach to produce aqueous aluminum chloride, AlCl3, and
liquid water.
Al(OH)3 (s) + 3 HCl (aq)  AlCl3 (aq) + 3 H2O (l)
90
15.
An antacid remedy contains solid sodium bicarbonate, NaHCO3, and solid citric acid,
H3C6H5O7, which react to produce gaseous carbon dioxide (the source of the familiar fizz),
aqueous sodium citrate, Na3C6H5O7, and liquid water.
3 NaHCO3 (s) + H3C6H5O7 (s)  3 CO2 (g) + Na3C6H5O7 (aq) + 3 H2O (l)
16.
When solid table sugar (sucrose), C12H22O11, is heated, gaseous water and solid elemental
carbon are produced.
C12H22O11 (s)  11 H2O (g) + 12 C (s)
Stoichiometry (page 46)
1.
Acetylene burns in air to form carbon dioxide and water:
5 C2H2(g) + 2 O2(g)  4 CO2(g) + 2 H2O(l)
How many moles of CO2 are formed from 25.0 moles C2H2?
 4 mol CO2
25.0 mol C 2 H 2 
 5 mol C 2 H 2

  20.0 mol CO2

NOTE: The coefficients in the balanced equation are treated as exact numbers. That means they
have an infinite number of significant figures. In other words, the coefficients will not be used to
determine the number of significant figures in the answer.
2.
If insufficient oxygen is available, carbon monoxide can be a product of the combustion of
butane:
9 C4H10(l) + 2 O2(g)  8 CO(g) +10 H2O(l).
What mass of CO could be produced from 5.0 g butane?
 1 mol C 4 H 10   8 mol CO   28.01 g CO 
 
 
  2.1 g CO
5.0 g C 4 H 10 
 58.14 g C 4 H 10   9 mol C 4 H 10   1 mol CO 
3.
15.0 g NaNH2 is required for an experiment. Using the following reaction, what mass of
sodium metal is required to produce the NaNH2?
2 Na(s) + 2 NH3(g)  2 NaNH2(s) + H2(g)?
 1 mol NaNH 2
15.0 g NaNH 2 
 39.02 g NaNH 2
  2 mol Na
 
  2 mol NaNH 2
91
  22.99 g Na 
 
  8.84 g Na
1
mol
Na


4.
Ethanol and acetic acid react to produce ethyl acetate according to the reaction C2H5OH +
CH3C(O)OH  CH3C(O)OC2H5 + H2O. If the reaction is only 35% efficient at the
conditions used, what mass of CH3C(O)OH will be necessary to produce 100 g
CH3C(O)OC2H5? Assume that sufficient ethanol is available.
First calculate the theoretical yield as if the reaction is 100 % efficient.
 1 mol   1 mol CH 3 COOH
 
100. g CH 3 C (O)OC 2 H 5 
 88.12 g   1 mol CH 3 C (O)OC 2 H 5
  60.06 g CH 3 COOH
 
  1 mol CH 3 COOH

  68.2 g CH 3 COOH

Since the reaction is only 35% efficient the starting amount of acetic acid must be larger than 68.2
grams. In fact 68.2 g is 35% of the amount of acetic acid actually needed for this reaction. The
actual starting material multiplied by 0.35 (the decimal equivalent of 35%) will equal 68.2 g.
Solving for the starting material will involving dividing both sides by 0.35.
? x 0.35 = 68.2 g CH3COOH
? = 68.2 g  0.35 = 195 g CH3COOH
5.
Heating CaCO3 yields CaO and CO2. Write the balanced equation. Calculate the mass of
CaCO3 consumed when 4.65 g of CaO forms.
CaCO3 (s)  CaO (s) + CO2 (g)
 1 mol CaO   1 mol CaCO3   100.09 g CaCO3 
  8.30 g CaCO3
 
 
4.65 g CaO 
 56.08 g CaO   1 mol CaO   1 mol CaCO3 
Understanding Limiting Reactants (page 49)
1.
In the synthesis of sodium amide (NaNH2), what is the maximum mass of NaNH 2 possible if
50.0 g of Na and 50.0 g NH3 were used?
2 Na(l) + 2 NH3(g)  2 NaNH2(s) +H2(g)
 1 mol Na   2 mol NaNH 2
 
50.0 g Na 
 22.99 g Na   2 mol Na
  39.02 g NaNH 2
 
  1 mol NaNH 2
 1 mol NH 3   2 mol NaNH 2
 
50.0 g NH 3 
 17.04 g NH 3   2 mol NH 3

  84.9 g NaNH 2

  39.02 g NaNH 2
 
  1 mol NaNH 2

  114 g NaNH 2

Starting with 50.0 grams of each reactant the maximum amount of product that can be
theoretically made is 84.9 g. There is only enough sodium, Na, present to produce this much
product.
NOTE: In the last unit conversion the molar mass of NaNH2 is used. It is still the number of
grams per one mole regardless of the coefficients in the balanced equation.
92
2.
The fuel methanol, CH3OH, can be made directly from carbon monoxide (CO) and
hydrogen (H2).
a.
Write a balanced equation for the reaction.
CO (g) + 2 H2 (g)  CH3OH (l)
b.
Calculate the maximum mass of methanol if one starts with 5.75 g CO
and 10.0 g H2.
 1 mol CO   1 mol CH 3 OH
 
5.75 g CO 
 28.01 g CO   1 mol CO
 1 mol H 2
10.0 g H 2 
 2.02 g H 2
  1 mol CH 3 OH
 
  2 mol H 2
  32.05 g CH 3 OH
 
  1 mol CH 3 OH
  32.05 g CH 3 OH
 
  1 mol CH 3 OH

  6.58 g CH 3 OH


  79.3 g CH 3 OH

Starting with 5.75 g carbon monoxide, CO, and 10.0 g hydrogen, H 2, the maximum
amount of product that can be made is 6.58 g.
c.
Which reactant is the limiting reactant?
The carbon monoxide, CO, is the limiting reactant. It limits the amount of product that
can be made is would be used up completely in this reaction.
d.
How much of the excess reactant remains?
 1 mol CO   2 mol H 2
 
5.75 g CO 
 28.01 g CO   1 mol CO
  2.02 g H 2
 
  1 mol H 2

  0.829 g H 2 used

10.0 g of H2 (initially in the reaction)  0.829 g H2 (used in reaction) = 9.2 g H2 (remaining)
3.
Aspirin (C7H8O4) is synthesized in the laboratory from salicylic acid (C7H6O3) and acetic
anhydride (C4H6O3):
C7H6O3(s) + C4H6O3(l)  C9H8O4(s) +CH3COOH(l)
a.
What is the theoretical yield of aspirin if you started with 15.0 g salicylic
acid and 15.0 g acetic anhydride?
 1 mol C 7 H 6 O3   1 mol C 9 H 8 O4
 
15.0 g C 7 H 6 O3 
 138.13 g C 7 H 6 O3   1 mol C 7 H 6 O3
 1 mol C4 H 6O3   1 mol C9 H8O4
 
15.0 g C4 H 6O3 
 102.10 g C4 H 6O3   1 mol C4 H 6O3
  180.17 g C9 H 8 O4
 
  1 mol C9 H 8 O4

  19.6 g C 9 H 8 O4

  180.17 g C9 H8O4 
 
  26.5 g C9 H8O4
  1 mol C9 H8O4 
Start with 15.0 g each of the two reactions the maximum amount of aspirin that can be
synthesized is 19.6 g.
93
b.
Which reactant is the limiting reactant?
The salicylic acid, C7H6O3, is the limiting reactant and will be used up completely in this
reaction.
c.
What mass of the excess reactant remains?
 1 mol C7 H 6O3   1 mol C4 H 6O3
 
15.0 g C7 H 6O3 
 138.13 g C7 H 6O3   1 mol C7 H 6O3
  101.10 g C4 H 6O3 
 
  11.1 g C4 H 6O3 used
  1 mol C4 H 6O3 
15.0 g C4H6O3 (initial)  11.1 g C4H6O3 (used) = 3.9 g C4H6O3 (remaining)
Writing Lewis Structures (page 56)
Write the Lewis structures for these molecules and ions.
1.
carbon tetrachloride, CCl4
C: 4 valence electrons; Cl: 7 valence electrons
32 total valence electrons to distribute
Once the 32 valence electrons are distributed among these five atoms all them has an octet of
electrons.
2.
carbonate ion, CO32
C: 4 valence electrons; O: 6 valence electrons
22 valence electrons are available from the carbon and oxygen plus 2 more since it is a divalent
anion, i.e., has a 2 charge; 24 total electrons to distribute
After all the electrons have been distributed the three oxygen atoms each have an octet of electrons.
However, the carbon atom only has 6 electrons around it. It is necessary to make a double
bond, i.e., share two pairs of electrons, between one of the oxygen atoms and the carbon
atom. Once the double bond is made all four atoms have an octet of electrons.
94
3.
sulfur dioxide, SO2
S: 6 valence electrons; O: 6 valence electrons
18 valence electrons to distribute
After the 18 valence electrons are distributed among the three atoms the sulfur atom remains
without an octet of electrons. The formation of a double bond between one of the oxygen atoms
and the sulfur atoms solves this problem.
4.
sulfur trioxide, SO3
S: 6 valence electrons; O: 6 valence electrons
24 valence electrons to distribute
It is necessary to form a double bond between one of the oxygen atoms and the sulfur atom in
order to have an octet of electrons for all four atoms.
5.
methanol, CH3OH
C: 4 valence electrons; H: 1 valence electron; O: 6 valence electrons
14 valence electrons to distribute
Here is a situation where the bonding patterns for hydrocarbons can be applied. Knowing that
hydrogen will form only one bond it is clearly not put between two other atoms. Since carbon can
make four bonds it is logical to put it as the central atom. If carbon is placed as the central atom
the oxygen logically is bonded to the carbon and one of the hydrogen atoms. The condensed
formula that was given, CH3OH, helps guide the bonding as well. It would have been more
challenging if the molecular formula, CH4O, would have been given. However, applying the
95
bonding patterns for hydrocarbons and the other rules for writing Lewis structures would have still
lead to a successful representation of methanol.
6.
acetone, CH3C(O)CH3
C: 4 valence electrons; H: 1 valence electron; O: 6 valence electrons
24 valence electrons to distribute
In this case there must be a double bond between one of the carbon atoms and the oxygen atom so
that the octet rule can be satisfied for every atom in the molecule. This is also consistent
with the bonding patterns for hydrocarbons. In the final Lewis structure all the hydrogen
atoms make one bond, the three carbon atoms make four bonds each and the oxygen atom
makes two bonds. Remember that a double bond counts as two bonds in the bonding
pattern. This is appropriate since the double bond is a sigma and a pi bond between those
two atoms. Chapter 5 gives more details about the bonding between atoms.
7.
sulfate ion, SO42
S: 6 valence electrons; O: 6 valence electrons
30 valence electrons are available from sulfur and oxygen plus 2 more since the ion has a 2 charge;
32 total electrons to distribute
Remember that the Lewis structures for polyatomic ions should be written within brackets and the
charge of the ion should be placed at the upper right-hand corner of the brackets.
96
8.
phosphate ion, PO43
P: 5 valence electrons; O: 6 valence electrons
29 valence electrons from the phosphorus and oxygen atoms plus 3 more since the ion has a 3
charge; 32 total electrons to distribute
The phosphate ion is a polyatomic anion with a 3 charge. The brackets are used to indicate that
this is an ion and the charge in the upper right-hand corner tells how many addition electrons are
present in this anion.
9.
ammonium ion, NH4+
N: 5 valence electrons; H: 1 valence electron
9 valence electrons from the nitrogen and hydrogen atoms minus 1 since the ammonium ion is a
cation with a 1+ charge; 8 total electrons to distribute
The brackets and charge are once again used to indicate that the ammonium ion is a polyatomic
cation with a 1+ charge.
10.
hydrogen cyanide, HCN
H: 1 valence electron; C: 4 valence electrons; N: 5 valence electrons
10 valence electrons to distribute
With only 10 electrons to distribute it is necessary to make a triple bond between the carbon and
nitrogen atoms. Hydrogen will only make one bond in the hydrocarbon molecules. Carbon makes
four bonds and the nitrogen makes three bonds with the triple bond between them. A triple bond
is a sigma bond and two pi bonds between two atoms. Chapter 5 in the textbook gives more
information about the bonding in this molecule.
97
Formal Charge (page 58)
1.
Assign formal charges to each atom in both of these structural formulas.
A
H
H
C
C
O
O
H
H
H
H
O
B
C
C
O
H
H
Does the formal charge help you pick the more probable formula?
Structural formula A:
Structural formula B:
FChydrogen = 1  ((2) + 0) = 0
FCcarbon = 4  ((8) + 0) = 0
FCcarbon = 4  ((4) + 4) = 2
FCoxygen = 6  ((4) + 4) = 0
FCoxygen = 6  ((4) + 4 ) = 0
FChydrogen = 1  ((2) + 0) = 0
FChydrogen = 1  ((2) + 0) = 0
FCcarbon = 4  ((8) + 0) = 0
FCcarbon = 4  ((8) + 0) = 0
FCoxygen = 6  ((4) + 4) = 0
FCoxygen = 6  ((4) + 4) = 0
FChydrogen = 1  ((2) + 0) = 0
The formal charges of all the atoms in the structural formula B are zero. This would indicate that
this structural formula is the more likely of the two given here. The second structural formula is
also consistent with the bonding patterns seen for hydrocarbons.
2.
There are three structures suggested for the ion NCO.
[ N
C
A
O ]
[ N
C
O ]
[ N
C
O ]
C
B
Assign formal charges to each atom. Use the formal charges to predict which structure is
most probable.
Formula A
Formula B
FCnitrogen = 5  ((2) + 6) = 2
FCcarbon = 4  ((8) + 0) = 0
FCoxygen = 6  ((6) + 2) = +1
0
FCnitrogen = 5  ((4) + 4) = 1
FCcarbon = 4  ((8) + 0) = 0
FCoxygen = 6  ((4) + 4 ) =
Formula C
FCnitorgen = 5  ((6) + 2) = 0
FCcarbon = 4  ((8) + 0) = 0
FCoxygen = 6  ((2) + 6) = 1
The formal charges add up to 1 for all three structural formulas. Since this is an anion with a 1
charge this is appropriate and does not help determine which is the most likely structural formula
of the three. Formula A is probably least likely since it has two atoms with formal charges that are
98
not zero and the formal charge of oxygen is +1. In this case assigning formal charge will not help
us determine the more likely structural formula between Formula B and C. Further knowledge
about bonding is necessary to make the determination. It is clear that both of those structural
formula satisfy all the rules for writing Lewis structures and give reasonable values for formal
charge. Look in Chapter 6 to get more insight into the structure of molecules.
Understanding Gas Law Problems (page 66)
1.
The volume of a sample of gas is expanded from 2.5 L to 4.5 L. If the original pressure was
855 mm Hg, what is the pressure after the gas is expanded?
This is a Boyle’s law problem.
V1 P2

or V1P1 = V2P2
V2 P1
V1 = 2.5 L
V2 = 4.5 L
P1 = 855 mm Hg
P2 = ?
P2 
2.
P1 V1 (855 mm Hg ) (2.5 L)

 475 mm Hg
V2
4.5 L
The volume of a sample of gas is compressed from 7.5 L to 1.0 L. If the initial temperature is
24.0 oC, what is the temperature of the gas after it has been compressed?
This is a Charles’ law problem.
V1 T1

or V1T2 = V2T1
V2 T2
V1 = 7.5 L
V2 = 1.0 L
T1 = 24.0 oC = 297 K
T2 = ?
NOTE: Temperatures must always be expressed in Kelvin! K = oC + 273
T2 
3.
V2 T1 (1.0 L) (297 K )

 39 K   230 o C
V1
7.5 L
A sample of gas occupying a volume of 75.0 mL contains 0.0250 moles. Gas is released until
the volume reaches 50.0 mL. How many moles of gas remain in the container?
This is an Avogadro’s law problem.
V1 n1

or V1n2 = V2n1
V2 n 2
V1 = 75.0 mL
V2 = 50.0 mL
n1 = 0.0250 mol
n2 = ?
99
n2 
4.
V2 n1 (50.0 mL) (0.0250 mol)

 0.0167 mol
V1
(75.0 mL)
The pressure of a gas is decreased from 950.0 mmHg to 720.0 mmHg. If the temperature is
26.5 oC initially, what is the temperature after the pressure has been reduced?
This is a problem requiring Gay-Lussac’s law.
T1 P1

or T1P2 = T2P1
T2 P2
P1 = 950.0 mm Hg
P2 = 720.0 mm Hg
T1 = 26.5 oC = 299.8 K
T2 = ?
T2 
5.
T1 P2 (299.5 K ) (720.0 mm Hg )

 227 K   46.0 o C
P1
(950.0 mm Hg )
A 120.0 mL sample of nitrogen gas, N2, has a pressure of 1.25 atm at 45.0 oC. What volume
will it occupy at STP?
This problem requires the use of the combined gas law.
P1 = 1.25 atm
V1 = 120.0 mL
T1 = 45.0 oC = 318.0 K
V2 
6.
P1V1 P2V2

or P1V1n2T2 = P2V2n1T1
n1T1 n2T2
P2 = 1 atm (standard pressure is defined as exactly 1 atm)
V2 = ?
T2 = 273 K (standard pressure is defined as exactly 273 K)
P1 V1 T2 (1.25 atm) (120.0 mL) (273 K )

 129 mL
T1 P2
(318.0 K ) (1 atm)
A 75.0 mL sample of hydrogen gas, H2, has a pressure of 0.850 atm at 20.0 oC. What volume
will it occupy at 2.25 atm and 60.0 oC?
This is another problem that will use the combined gas law.
P1 = .850 atm
V1 = 75.0 mL
T1 = 20.0 oC = 293.0 K
V2 
P2 = 2.25 atm
V2 = ?
T2 = 60.0 oC = 333.0 K
P1 V1 T2 (0.850 atm) (75.0 mL) (333.0 K )

 32.2 mL
T1 P2
(293.0 K ) (2.25 atm)
100
7.
What would be the pressure of a 45.0 mL sample of hydrogen gas containing 0.750 moles at
31.0 oC?
This problem will use the ideal gas law. PV = nRT
P=?
V = 45.0 mL = 0.0450 L
N = 0.750 mole
R = 0.08206 L atm mol1 K1
T = 31.0 oC = 304.0 K
NOTE: If this value of R is used the volume must be expressed in liters and the pressure in
atmospheres. There are other values of R that can be used, e.g., 62.364 L Torr mol 1 K1 or
8.31451 L kPa mol1 K1. In every case volume is still expressed in liters. Temperature is always
expressed in Kelvin. The only thing that varies among the possible values of the gas constant R is
the units of pressure.
P
8.
n R T (0.750 mol) (0.08206 L atm mol 1 K 1 ) (304.0 K )

 416 atm
V
(0.0450 L)
What would the temperature of 500.0 mL of oxygen gas be if there were 1.10 moles of gas at
a pressure of 850.0 mmHg?
The ideal gas law is most appropriate for solving this problem.
PV = nRT
P = 850.0 mm Hg = 1.118 atm
V = 500.0 mL = 0.5000 L
n = 1.10 mol
R = 0.08206 L atm mol1 K1
T=?
NOTE: The units of pressure must be changed from mm Hg to atmospheres when using this value
of R. 1 atm = 760. mm Hg = 760. Torr = 101.325 kPa = 101,325 Pa = 14.7 psi These
equivalences will be helpful in making conversions between units of pressure. These are defined
equalities and are not usually considered when determining the proper number of significant digits
in a numerical answer. The unit conversion for changing mm Hg to atm is:
T
PV
(1.118 atm) (0.5000 L)

 6.19 K
n T (1.10 mol) (0.08206 L atm mol 1 K 1
101
1 atm
.
760 mm Hg