Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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compare an object on Earth to the same object on the Moon. The

weaker gravitational force on the Moon would cause the object to

weigh less, but it would still have the same mass (Figure 1.18).

For many of our water quality measurements, we might only

need a gram of water. How much is a gram? A raisin’s mass is ap-

proximately 1 g. That is a pretty small amount, but 1000 raisins

have about the same mass as a kilogram.A kilogram weighs about

the same as 2.2 lb on Earth. However, masses with which you will

become familiar in the laboratory are much smaller than the kilo-

gram. In fact, most masses you will record in the laboratory will be

grams or milligrams. In the chemical process industries, which

produce large quantities of the chemicals used to manufacture

consumer goods, masses are often measured in tons (2000 lb) or

metric tons (1000 kg, 1 Mg, or 2200 lb).

Another important piece of information to include with our

water quality measurements would be the distance to the farthest

city that is served by the desalination plant. We would record this

distance using the SI unit of length, the

meter (m). The meter is

about 10% longer than a yard.

1 m = 1.094 yd

18 Chapter 1 The World of Chemistry

To measure long distances, we often report them in kilometers (km).

1 km = 1 × 10

m = 1000 m

One kilometer is equal to 1000 m, or slightly more than 0.6 mi. A runner entered

in a “5K” (5-km) road race will run 5000 m (a little more than 3 miles). In chem-

ical analysis, though, scientists measure lengths much smaller than a meter. Be-

cause the particles of chemistry are extremely tiny compared to objects in the

everyday world, their sizes are often measured in such tiny units as picometers,

nanometers, and micrometers. One nanometer is equal to 0.000000001 m, so

there are 1 billion nanometers (10

nm) in 1 m.

1 nm = 1 × 10

−9

m = 0.000000001 m

How long it takes to desalinate enough water for the people served by the plant

is also a key quantity. To report this value, we could use the SI unit of time, the

second (s). Luckily, the second has been used in virtually all measurement systems,

so we are pretty familiar with it. The units of hour, minute, day, and year are often

used in measurements of time. These are not SI units, but because they are so

widely used, scientists don’t often afﬁx preﬁxes to the unit second to describe large

time spans. For example, instead of reporting 3600 s as 3.6 ks (which is perfectly

appropriate), the scientist would be more likely to indicate 1 hour (1 h),

3600 s = 3.6 × 10

s = 3.6 ks

3600 s = 1 h

Preﬁxes, however, are commonly used to indicate fractions of a second. One mil-

lisecond (1 ms) is one-thousandth of a second.

Temperature is a vital part of our day-to-day choices about what clothing to

wear, what liquid to drink, and a host of other decisions. The SI base unit of tem-

perature is the

kelvin (K). Notice that no “°” symbol is used with the abbreviation

K. Two other temperature units are much more common, however. The English

system uses the Fahrenheit scale, in which the unit of measure is the

degree

Earth

Moon

FIGURE 1.18

A person’s weight on the Moon is differ-

ent from his or her weight on Earth. The

mass of the person hasn’t changed. Only

the force of gravity is different.

The mass of a raisin is about 1 g.

A meter and a yard.

Fahrenheit (°F). The metric system uses the Celsius scale, in which the unit of

measure is the

degree Celsius (°C). All three systems can be used to accurately re-

port temperatures, but they are based on different physical phenomena. The

Fahrenheit scale was created by the German physicist Gabriel Daniel Fahrenheit

in 1714. Fahrenheit used a mercury-based thermometer to deﬁne 0°F as the cold-

est temperature he could make from a mixture of water and ammonium chlo-

ride. He called his body temperature 100°F. (Linus Pauling, a chemist whose work

we will discuss at many points later in this text, mused that Fahrenheit might have

had a slight fever.) Using his two reference points of cold and warm, he found the

boiling point of water to be 212°F.

The Swedish astronomer Anders Celsius developed the Celsius scale in 1742,

based on dividing the difference in temperature from the freezing point, 0°C, to the

boilingpoint,100°C,of waterinto100individualunits(degrees).In1848,theBritish

physicistLord Kelvin (Figure 1.19) inventedthe temperaturescale named for him in

1954 in which, fortunately, one kelvin is equal in magnitude to 1 degree Celsius,

although the scales just startat different points.Although Kelvindid not recognizeit

at the time,the point we now call zero on the Kelvin scale equals –273 on the Celsius

scale,so a temperature in Celsius can be converted into Kelvin by adding 273.

= T

+ 273 and T

= T

− 273

More information and the exact deﬁnition of the Kelvin scale is presented later.

Converting between the Celsius and Fahrenheit scales is a little more compli-

cated. The formula shows that the temperature of an object changes by 9 Fahren-

heit degrees for every 5 Celsius degrees. This means that if you use a Fahrenheit

thermometer to measure an 18-degree drop in temperature (9

× 2), a ther-

mometer based on the Celsius scale will note a 10-degree drop (5

× 2). A

Kelvin-based thermometer would also show a 10-degree drop in temperature,

because 1 K equals 1°C (see Figure 1.20). Figure 1.20 also indicates the different

1.4 Units and Measurement 19

FIGURE 1.19

Lord Kelvin (1824–1907) was born

William Thomson in Belfast. He was

knighted in 1866 and given the title of

Baron Kelvin of Largs in 1892.

180

Fahrenheit

degrees

Boiling point

of water

32°F

212°F

200°F

90°C

220°F

190°F

Freezing point

of water

Fahrenheit

100

Celsius

degrees

0°C

100°C

Celsius

273.15 K

373.15 K

Kelvin

100

kelvins

100°C

360 K

370 K

373.15 K

–

40°F

–

40°C

233.15 K

FIGURE 1.20

The Fahrenheit, Celsius, and Kelvin

temperature scales.

locations of the freezing point of water on the scales. Here are the formulas re-

quired for converting between temperatures on the Celsius and Fahrenheit scales:

= (T

− 32)/1.8 T

= 1.8(T

) + 32

Another property of our water sample, or of any other substance or mixture

of substances, is the number of particles of each kind we have. These particles

represent the amount of each substance in the sample. The amount of a sub-

stance is the number of entities, such as atoms, molecules, or ions (particles con-

taining an electrical charge), present in a given sample of the substance. Deter-

mining the amount of each substance in a glass of tap water is just one part of a

series of analyses that chemists do to make sure that a sample of water is whole-

some to drink. For example, the analysis for the element arsenic in drinking water

really deals with determining how many arsenic atoms are present in a known

quantity of water. This idea is explored fully in Chapter 3.

The base unit for the amount of a substance is the

mole (mol). As a preview of

Chapter 3, we will simply point out that 1 mole corresponds to a speciﬁc very

large number of entities and that this number is approximately 6.02214 × 10

Thus when we talk of 1 mole of atoms, we mean 6.02214 × 10

atoms; 1 mole of

molecules is 6.02214 × 10

molecules, and so on.

Other SI units may also be needed during our water quality measurements.

For instance, the purity of desalinated water is inversely related to its ability

to carry an electric current: The lower the electrical conductivity of the water,

the purer the water (see Figure 1.21). The SI base unit of electric current is the

ampere (A), often called the amp. This unit is used a great deal in chemistry be-

cause some chemical reactions can be used to generate an electric current (within

batteries, for example). Conversely, electricity can be used to make certain chem-

ical reactions happen. We’ll learn more about the ampere in the chapter on elec-

trochemistry (Chapter 19).

The last SI base unit is the

candela (cd). This unit describes the luminous inten-

sity

, or brightness, of something under study. For example, the Sun, which radi-

ates the light energy that powers photosynthesis, has an estimated luminous

intensity of 10

candelas. Light bulbs produce a vastly smaller luminous inten-

sity. We won’t deal with luminous intensity in this text.

We will refer to the world of individual atoms, molecules, and ions as the

nanoworld, in contrast to the everyday, large-scale macroworld (“big world”) with

which we are familiar. There is a huge change in the units that we use to report

quantities as we travel between the nanoworld and the macroworld. To make the

trip smoothly, we must be able to work comfortably with exponential notation

and related mathematical operations. Feel free to reinforce your understanding of

exponential notation by consulting the appendix at the back of the book.

20 Chapter 1 The World of Chemistry

FIGURE 1.21

The conductivity of water solutions

gives an indication of their purity.

One of the most important aspects of learning about chemistry is becoming

familiar with the nanoworld of chemicals and being able to visualize what

happens at that scale. The fundamental particles of chemistry are atoms

(Figure 1.22). These range in size from a diameter of approximately 75 picome-

ters (75

×10

–12

m = 7.5 ×10

–11

m) for a hydrogen atom to about 500 picometers

(5.00

× 10

–10

m) for an atom of francium. How can we get some idea of what

these numbers mean?

Suppose we scale things up in our minds so that atoms be-

come the size of pebbles that we can hold in our hand—some small, some larger,

but averaging around 2 cm in diameter. A real pebble scaled up in size to the same

extent would now be approximately 1000 kilometers (about 600 miles) in diam-

eter. That’s big enough to cover the entire Great Lakes (Figure 1.23). Even with

these comparisons, this extraordinary difference in scale is difﬁcult to visualize.

Yet working with such comparisons should begin to give you some feeling for the

scale of the nanoworld of chemistry. Atoms, molecules, and ions are very small

indeed.

1.4 Units and Measurement 21

Francium

(540 pm)

Hydrogen

(75 pm)

FIGURE 1.22

The relative size of two atoms, hydrogen

(75 pm) and francium (540 pm).

U.S.A.

CANADA

Lake Erie

Lake Ontario

Colorado Springs

Fargo

Green Bay

Marquette

Minot

Norfolk

Rapid City

Rochester

Scottsbluff

St. Joseph

Wichita

Brandon

Thunder Bay

Buffalo

Dodge City

Duluth

Grand Island

Hastings

Hot Springs

Huron

Ironwood

Mankato

Miles City

Sioux City

Sioux Falls

Williston

Winfield

Cleveland

Detroit

Baltimore

Chicago

Louisville

Minneapolis

Philadelphia

St. Louis

Montreal

Buffalo

Kansas City

New York

Norfolk

Omaha

Pittsburgh

Quebec

Winnipeg

Albany

Ames

Augusta

Bismarck

Boston

Charleston

Columbus

Denver

Des Moines

Frankfort

Harrisburg

Hartford

Indianapolis

Lansing

Lincoln

Montpelier

Richmond

Springfield

St. Paul

Tr ent on

Toronto

Concord

Madison

Topeka

Washington, DC

Ottawa

500 Miles0

Scale

Atlantic

Ocean

FIGURE 1.23

If an atom were scaled up to the size of

a pebble on a beach, the beach pebble,

scaled up to the same extent, would be

600 miles wide.

EXERCISE 1.4 Temperature Conversions

The chemical reactions within the human body tend to proceed optimally at a tem-

perature of around 37°C. Express this temperature using the Kelvin and Fahrenheit

scales.

Solution

= T

+ 273 so 37°C = 37 + 273 K = 310 K

= 1.8T

+ 32 so 37°C = 1.8(37) + 32°F = 99°F

Normal body temperature is commonly cited as equal to 98.6

F, which has more

signiﬁcant ﬁgures (see Section 1.6) than are justiﬁed in a comparison to 37

PRACTICE 1.4

Water is fed into the vaporization stage of a desalination plant at 220

F. Express this

temperature using the Celsius and Kelvin scales.

See Problems 37c, 37f, 38c, 38f, 39c, 39f, 40c, 40f, and 53–54.

Derived Units

We often measure quantities other than the seven SI base quantities shown in

Table 1.2. Our analysis of the sample of water we obtain during our daily check of

water quality at the desalination plant or any municipal water treatment system

has several types of measurements associated with it. We’ve

recorded the mass, the time spent to purify the water in the reverse

osmosis column, the distance (using the SI unit of length) of the re-

gion served by the plant, and the temperature of the water, and now

we also note the volume and density of the sample. These last two

measures, along with many other common determinations, use

de-

rived units

, formed by the combination of SI base units. Some of the

more common derived units are shown in Table 1.4.

The volume of a water sample obtained from the desalination

plant is a measure the space occupied by the sample. We could mea-

sure the volume of the sample in SI base units as

cubic meters (m

Speciﬁcally, a cubic meter is the volume described by a cube of mat-

ter measuring 1 m on each side (see Figure 1.24). The SI unit for

volume, the cubic meter, is derived from the SI unit for length, the

meter.

Although the cubic meter is the SI unit for volume, this derived

unit is often much too large to reﬂect the sample sizes that we would

obtain. Chemists tend to measure volumes using smaller derived units, such as

the

cubic decimeter (dm

), which is commonly used in most parts of the world,

with the notable exception of the United States. As Figure 1.25 shows, a cubic

decimeter is the volume of a cube that measures 10 cm on each side. This means

that there are 1000 dm

in every cubic meter. Because we use the cubic decimeter

in most of our measurements, a different name has been given to this unit. The

cubic decimeter is also known as the

liter (L), which is the common volume mea-

surement used by scientists in the United States.

Another common derived unit for volume is the cubic centimeter (cm

). This

unit describes a cube measuring 1 cm on each side (see Figure 1.25). In the health

professions, one cubic centimeter is often abbreviated as 1 cc. There are 1000 cc

in 1 L, so the cubic centimeter is also known as the

milliliter (mL):1 cm

= 1 cc =

1 mL. Note that the unit for liters is a capital L, not a lowercase l.

1 L = 1000 mL = 1000 cc = 1 dm

= 0.001 m

Judith Fairclough-Baity, in her report on the characterization of three types of

silicone tubing, indicated the

density of the tubes she made. Density reports the

mass of a substance that is present in a given volume of the substance. During our

analysis of the quality of water from a desalination plant, we might also report the

density of the sample. The SI unit for measuring density is

kilograms per cubic

meter (kg/m

), although scientists often use smaller units, such as grams per cubic

centimeter (g/cm

). Because 1 cm

equals 1 mL, units of density can be reported

for liquids and many solids in grams per milliliter. The tubing with which

Fairclough-Baity worked had a density of between 1.1 and 1.2 g/mL.

Density

mass

volume

22 Chapter 1 The World of Chemistry

Liter (L)

1 dm

1 m

Cubic meter (m

) = 1000 dm

= 1000 L

1 m

Volume: 1 dm

Volume: 1 cm

1 mL

“1 liter” = 1 L = 1 dm

1000 cm

= 1000 mL

Selected Derived Units

Physical Quantity Derived Unit Name of Derived Unit

Volume m

cubic meter

Density kg

–3

kilograms per cubic meter

Force kg

–2

newton (N)

Pressure kg

–1

–2

) pascal (Pa)

Energy kg

–2

m) joule (J)

Velocity m

–1

meters per second

TABLE 1.4

FIGURE 1.24

The cubic meter, the SI unit for volume.

FIGURE 1.25

One cubic centimeter is equal to

a milliliter.

Video Lesson: CIA

Demonstration: Differences in

Density Due to Temperature

An interesting example of the use of density is in the compari-

son of three cubes of material. As shown in Figure 1.26, each block

of material looks the same. However, one is silver-tinted wax, one

is lead, and the third is aluminum.

How can we tell the wax, lead,

and aluminum apart?

There are many physical properties that differ

among the three blocks, such as color and hardness. Density is an-

other. The wax has a density of 0.95 g/cm

, the lead has a density of

11.4 g/cm

, and the aluminum has a density of 2.70 g/cm

. There-

fore, although the volumes of all the cubes are identical, they differ

in mass.

Extensive Versus Intensive Properties

Why does a log ﬂoat on water? Mistakenly, we might say that the

log is “lighter” than the water. Why does a pebble sink to the bot-

tom of a lake? Again, we might mistakenly say that it is “heavier”

than water. The heaviness or lightness of an object is a measure of

the weight of the object. It is an

extensive property, one that is de-

pendent on how much sample you have. Length is another exten-

sive property. If the sample is a precious metal, such as gold or plat-

inum, then weight and mass are related to total cost, and all of

these are extensive properties.

On the other hand, density, the mass of the metal per cubic

centimeter (or milliliter) is an

intensive property, because it re-

mains the same irrespective of the amount of sample you have.

Gold will still cost about $400 per ounce whether if you have

1 ounce (1 oz) or 1000 oz. The cost per ounce is an intensive prop-

erty. The density of the gold, 19.3 g/mL, is the same no matter how

much gold you have. However, the cost of a chunk of gold is an ex-

tensive property because it depends on the size of the chunk. Large

chunks cost more than small chunks (Figure 1.27). Table 1.5 lists

several intensive and extensive properties.

1.4 Units and Measurement 23

FIGURE 1.26

A block of silver-colored wax, a block of lead, and a block of

aluminum. They look the same but are chemically quite different.

We can tell which is which by measuring their densities.

$212,000 $311,000 $54,000

Mass, length, and cost are extensive properties.

FIGURE 1.27

Extensive and intensive properties of gold. The weight and cost of a chunk of gold

depend on the amount (extensive properties); the color and density of the gold remain

independent of the amount (intensive properties).

Intensive and

Extensive Properties

Intensive Extensive

Properties Properties

Color Mass

Density Length

Temperature Volume

Odor Cost

Physical state Energy

TABLE 1.5

Balsa wood

Hexane

Oak wood

Chloroform

Lead

Water

10.00 mL

Experimental determination of density. A picnometer, which contains an exact volume, can

be used to measure the density of a liquid.

EXERCISE 1.5 Extensive Versus Intensive Properties

During an expedition to the top of a mountain, a camper wishes to cook some noo-

dles for the entire group. In order to heat the noodles, the camper must boil an en-

tire pot of water. Is the temperature at which water boils an intensive or an extensive

property?

First Thoughts

Initially, this problem seems confusing because we know it will take longer to bring

a pot of water to a boil than to heat a cup of water to its boiling point. Let’s consider

the question this way: Does the amount of water make a difference in how hot it

must get before it boils?

Solution

The boiling point of water is an intensive property. The amount of water does not

make a difference in how hot it must get in order to boil. In other words, the pot of

boiling water and the cup of boiling water have the same temperature.

Further Insight

Why, then, does it take longer to boil a pot of water? Answering this question reveals

another extensive property. The amount of water determines how much heat must

be used to boil the water. The amount of heat used to boil the water is an extensive

property of the water.

PRACTICE 1.5

Is the viscosity (the resistance of a liquid to ﬂow, or the syrupiness of a liquid) of

molasses an intensive or an extensive property?

See Problem 115.

We can apply this to our “ﬂoating log and sinking rock” discussion by noting

that the extensive property of weight—that of the log, the rock, or the water—is

not the reason why the log ﬂoats or the rock sinks. The explanation for these

observations is related to the density—an intensive property—of the objects

compared to the density of the water. Those objects with a density greater than

the water will sink; those with a lesser density will ﬂoat. Figure 1.28 illustrates this

effect for six different objects.

24 Chapter 1 The World of Chemistry

FIGURE 1.28

The densities of six different substances

are represented in this density jar.

EXERCISE 1.6 Identifying a Metal via Density

A chunk of metal has a mass of 81.76 g and

is added to a graduated cylinder (a type of

glassware that measures volume) contain-

ing 25.0 mL of water. The total volume of

the metal and water rises to 34.2 mL. What

is the identity of the metal?

First Thoughts

In this exercise, we are given only enough information to tell the metals apart by

their densities. The real question, then, is “What is the density of the unknown

metal?” The metal has a mass of 81.76 g. What volume does it occupy? The volume

of water in the graduated cylinder increases from 25.0 mL to 34.2 mL when the

metal chunk is added. The volume of the metal must then be 34.2 − 25.0 =9.2 mL.

We have mass and volume. We can now solve for density.

Solution

Density =

mass

volume

81.76 g

9.2mL

= 8.9g/mL = 8.9g/cm

The metal has the same density as nickel.

Further Insights

Knowing the density of metals is of great practical importance in manufacturing

and commerce. For example, titanium is used as a component of expensive, light-

weight bicycle frames because of its strength and low density of 4.50 g/mL. The ac-

celeration that a rider can achieve on a bicycle is another derived unit, equal to force

(itself a derived unit; see Table 1.4) per unit mass:

Acceleration =

force

mass

kg·m·s

−2

= m·s

−2

The less mass (or weight) the bicycle has, the greater is the acceleration the rider can

achieve using a constant force.

PRACTICE 1.6

How much water would be displaced if 81.76 g of aluminum, rather than of nickel,

were added to the graduated cylinder discussed in this exercise?

See Problems 45–48 and 59–62.

1.5 Conversions and Dimensional Analysis

The SI units typically used by scientists are often quite different from some of the

units in everyday use. For instance, visitors to Canada from the United States

might wish to ﬁll their cars with gasoline (“petrol” in Canada). Initially, they

might be surprised that the gasoline is so inexpensive. Part of that has to do with

the currency exchange rate: around $1.30 Canadian per $1.00 U.S. Another key is

that the price of Canadian petrol is reported in liters, whereas in the United

States, gasoline is sold by the gallon. Converting liters into gallons solves the dif-

ﬁculty caused by these differences. To do the conversion, we need an expression

that relates one unit to another. Such a

conversion factor is a mathematical ex-

pression of the ratio of one unit to another. For example, on Earth, a 2.2046-lb

mass is equivalent to a 1-kg mass.

2.2046 lb = 1 kg

1.5 Conversions and Dimensional Analysis 25

Metal Density in g/cm

Aluminum 2.7

Nickel 8.9

Tin 7.3

Zinc 7.1

One kilogram weighs 2 lb, 3.30 oz

(2.20 lb).

26 Chapter 1 The World of Chemistry

Common

Conversion Factors

Length 1 mi  1.609 km

1 in  2.54 cm

12 in  1 ft

1 mi  5280 ft

Volume 1 qt  0.9464 L

1 gal  3.785 L

1 gal  4 qt

1ﬂoz 29.57 mL

Mass 2.2046 lb  1.0 kg

1 lb  453.6 g

1 oz  28.35 g

TABLE 1.6

Because these two values are equal, we can use them as a conversion factor

that can be multiplied by a number to convert its units. The operation is identical

to multiplying the original number by 1, because the numerator and denominator

are equal to each other. This equality can be written as a conversion factor in two

ways:

1kg

2.2046 lb

= 1

2.2046 lb

1kg

= 1

For instance, suppose we are interested in learning the mass (in kilograms) of

a book that weighs 3.5 lb. We can multiply the weight of the book by one of our

conversion factors.

3.5lb×

1kg

2.2046 lb

= 1.6kg

Note that we chose to multiply the weight of the book by the conversion factor so

that the unwanted units canceled. If we had multiplied by the other conversion

factor, the answer would have been different, and the units would have been

different, too! In this case, the units would not have canceled, and we would have

been left with a meaningless answer:

3.5lb×

2.2046 lb

1kg

= 7.7

This extremely important method of calculation can be used to solve a great

variety of problems in chemistry. In fact, we will use this method for many of the

calculations that we will do. It is known as

dimensional analysis (or, sometimes, as

the unit-conversion method, or the factor-label method) because canceling out the

units (or “dimensions” or “factors”) associated with each number enables us to

arrive at the proper answer. Table 1.6 lists some of the more common conversion

factors for mass, length, volume, and time. Others may be found on the inside

back cover.

Dimensional analysis can be applied to much more complex problems. For

instance, suppose we would like to know the length of our textbook in millime-

ters, given that someone has measured its length to be 10.25 inches (in). We can

perform a dimensional analysis and convert the units of inches into millimeters.

However, examination of Table 1.6 doesn’t show a direct conversion from inches

to millimeters. But the table does show a conversion factor from inches to cen-

timeters, and there is another conversion factor from centimeters to millimeters

(which we know from Section 1.4). Here’s our plan of attack:

inches to centimeters

−−−−−−−−−−−→cm

centimeters to millimeters

−−−−−−−−−−−−−→ mm

Let’s use this plan to construct a dimensional analysis of the units in one long

problem. Although we could do it stepwise, it is often just as easy to do a single

calculation. We’ll just have to make sure that our unwanted units cancel at each

step.

10.25 in ×

2.54 cm

1in

10 mm

1cm

= 260.4mm

These problems can even get longer, but they don’t necessarily have to get

harder. For example, suppose Judith Fairclough-Baity measured the density of

her tubing as 1.154 g/mL. Assume she was interested in reporting this density in

pounds per gallon. We can use conversion factors to solve this problem by con-

verting one unit at a time into the units we desire. Here’s our plan of attack:

grams to kilograms

−−−−−−−−−→

kilograms to pounds

−−−−−−−−−→

milliliters to liters

−−−−−−−−−→

liters to gallons

−−−−−−−−−→

gal

Video Lesson: Dimensional

Analysis

Tutorial: Unit Conversions

Starting with the initial density in grams per milliliter, we set up the problem,

taking care that each unwanted unit will cancel when we’re done:

1.154 g

1kg

1000 g

2.2046 lb

1kg

1000 mL

3.785 L

1gal

9.629 lb

gal

The entire calculation converts the grams into pounds and the milliliters into

gallons.

EXERCISE 1.7 Practice with Dimensional Analysis

Manufacturing on a global scale requires working with immense quantities of raw

materials. As a close approximation of the actual amount, let’s assume that 96 bil-

lion aluminum cans are produced in a given year. If each can requires 15 g of

aluminum (Al), how many total kilograms of aluminum are used each day (assume

365 days in a year) to make aluminum cans?

First Thoughts

From the problem we can obtain these conversion factors:

365 days

1 year

9.6 ×10

Al cans

year

15 g Al

Al can

1kgAl

1000 g Al

If we map out our steps, there are several routes we can follow. Here is just one:

cans

year

cans to grams

−−−−−−−−−→

year

grams to kilograms

−−−−−−−−−→

year

years to day

−−−−−−−−−→

day

The key with dimensional analysis is to show how each step allows you to cancel

units so that you can end with the units you want. It doesn’t matter what order we

pick for the terms, as long as the units cancel.

Solution

9.6 ×10

cans

year

15 g Al

1 can

1kgAl

1000 g Al

1 year

365 day

3.9 ×10

kg Al

day

Further Insights

One of the most important questions we need to ask is Does our answer make

sense?

That is, did we calculate an answer that we “should be” getting? Our answer

is quite large. Should it be so, or does it make more sense to use, perhaps, 0.001 kg

of aluminum per day? If we prepare billions of cans per year, it seems reasonable

that we should use quite a bit of aluminum each day, and our answer conﬁrms that.

This doesn’t mean that the math was necessarily correct, but the answer is at least

reasonable. Anticipating roughly what an answer should be, and then conﬁrming

that your answer is at least nearby, is certainly part of “thinking like a chemist.”

PRACTICE 1.7

a. What is the mass in grams of 6.50 gallons of water that ﬂow into a desalination

plant every second? Assume the density of the water in this problem is

1.00 g/mL.

b. How many gallons of gasoline are in 58.6 L of gasoline?

c. How much would 11.2 gallons of gasoline cost in Canada if the price of

gasoline in the United States is $1.34 per gallon and the exchange rate is

$1.30 CN = $1.00 U.S.? (Assume the price of gasoline is the same in

the United States and Canada.)

See Problems 65–69 and 71–80.

1.5 Conversions and Dimensional Analysis 27