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

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This is also the average formula mass (or formula weight) of adrenaline. Formula

mass is a more general term than molecular mass, because it also covers com-

pounds that do not exist in the form of individual molecules, such as ionic

compounds. For example, the ionic compound sodium chloride (NaCl) has an

average formula mass of 58.44 amu, found by adding the atomic mass of sodium,

22.99 amu, to the atomic mass of chlorine, 35.45 amu.

The formula mass, or molecular mass, of adrenaline is the average mass of

one molecule of adrenaline. Each molecule of the substance contains exactly the

same number of each type of atom, but different individual molecules can have

different masses, depending on which isotopes of carbon, oxygen, or hydrogen

they contain.

EXERCISE 3.1 Calculating Formula Masses

1. The formula for aspirin is C

. What is the formula mass of aspirin?

2. Many of us use the stimulant properties of the chemical caffeine in coffee

to remain alert, but even just the smell of coffee can seem to perk us up

in anticipation. One of the molecules that produce the aroma of coffee

is C

OS (2-furylmethanethiol). Calculate the formula mass of

2-furylmethanethiol.

Solution

1. Aspirin = C

9 atoms of carbon @ 12.01 amu = 108.09

8 atoms of hydrogen @ 1.008 amu = 8.064

4 atoms of oxygen @ 16.00 amu = 64.00

––––––––––

Formula mass = 180.154 = 180.2 amu

2. 2-Furylmethanethiol =C

5 atoms of carbon @ 12.01 amu = 60.05

6 atoms of hydrogen @ 1.008 amu = 6.048

1 atom of oxygen @ 16.00 amu = 16.00

1 atom of sulfur @ 32.07 amu = 32.07

––––––––––

Formula mass = 114.168 = 114.17 amu

PRACTICE 3.1

Calculate the formula masses of these compounds:

a. Codeine, a painkiller with the formula C

b. Magnesium sulfate, found in many medical ointments, with the formula

MgSO

c. Trinitrotoluene (TNT), an explosive with the formula C

See Problems 1–4.

NaCl

Na 22.99 amu

Cl 35.45 amu

Total 58.44 amu

88 Chapter 3 Introducing Quantitative Chemistry

Tutorial: Formula Masses

3.2 Counting by Weighing

When we do chemistry, we generally begin by measuring our chemicals using

masses. However, we need to convert these masses into some measure of the

numbers of atoms, molecules, or ions in order to relate the masses to what is ac-

tually happening in the chemical nanoworld.

Why can’t we take a sample of adren-

aline, which comes as a white powder, and use a pair of tweezers to separate out the

number of individual molecules we need?

Adrenaline molecules, like all molecules,

are stunningly small—on the order of 1 nm across. This means that you would

need to line up 100 million adrenaline molecules with your tweezers to get a sam-

ple as long as your index ﬁnger (about 10 cm)! This is why we need to establish

the link between weighing things and counting them. Weighing a sample of adren-

aline gives us insight into how many molecules of adrenaline we have.

We can see how this is done on a more familiar size scale by using one tablet

of vitamin C that weighs 0.75 g. If all the tablets in a jar are placed onto a balance,

and if they weigh a total of 45.0 g, how many tablets are in the jar? As shown in

Figure 3.2, there are a total of 60 tablets in the jar.

45 g

1 vitamin C tablet

0.75 g

= 60 vitamin C tablets

We can do exactly the same thing with the atoms, molecules, and ionic com-

pounds, the only difference being that the numbers are most often extraordinarily

large or small.

Suppose you had just taken a pain reliever that contained aspirin, also known

as acetylsalicylic acid (see Figure 3.3). The chemical formula for aspirin is

. If the tablet you used contained 500 mg (0.500 g) of aspirin, could you

calculate how many molecules of pain reliever you had just ingested? You would

3.2 Counting by Weighing 89

NNO

NCH

Codeine TNT

Vitamin C (ascorbic acid)

One vitamin C tablet

weighs 0.75 g.

All the tablets in the

jar weigh 45 g.

45 g × = 60 tablets

1 tablet

0.75 g

FIGURE 3.2

Counting by weighing—the basic logic.

Aspirin (acetylsalicylic acid)

FIGURE 3.3

need to recall from Chapter 2 that 1 amu =1.6605 × 10

−24

g (1.6605 × 10

−27

kg).

Here is the calculation:

The average molecular mass of aspirin, calculated in Exercise 3.1, is 180.2 amu.

180.2amu

1 molecule

1.66 ×10

−24

1amu

= 2.99 × 10

−22

g per molecule

0.500 g ×

1 molecule

2.99 ×10

−22

= 1.67 × 10

molecules

This quantity is approximately 300 billion times more than the current human

population of the earth! The comparison serves as a reminder of just how tiny

molecules must be if that many are required to make up half a gram.

Knowing that the average molecular mass of aspirin is 180.2 amu, a chemist

at a pharmaceutical plant could weigh out a 180.2-g sample. How many mole-

cules would that sample contain? Let’s do the calculations in order to prove that

the number of molecules of aspirin (molecular mass = 180.2 amu) in a 180.2-g

sample must be the same as the number of amu’s in 1 g (see Figure 3.4). How many

amu’s are in 1 g? The mass of 1 amu = 1.6605 × 10

−24

g, so we can set up this

equation and solve it:

1amu

1.6605 ×10

−24

x amu

x = 6.022 × 10

amu in 1 g of any substance.

What a huge number that is!

90 Chapter 3 Introducing Quantitative Chemistry

There are

6.022 × 10

amu in 1 g.

amu per molecule = grams per mole

6.022 × 10

aspirin

molecules (known as

1 mol of aspirin)

One aspirin molecule,

too small to see!

FIGURE 3.4

The link between masses in amu and

in grams.

How many molecules of aspirin are in 180.2 g of aspirin? We’ll use our di-

mensional analysis method to solve the problem. First we create a ﬂowchart

that will help us arrive at our answer.

g aspirin

amu

grams

−−−−−→amu aspirin

molecules

amu

−−−−−→molecules aspirin

Then we perform the calculation.

180.2 g aspirin ×

1 amu aspirin

1.6605 ×10

−24

g aspirin

1 molecule aspirin

180.2 amu aspirin

= 6.022 × 10

molecules aspirin

This is the same huge number as the number of amu’s in 1 g of a substance!

By choosing to weigh out a mass in grams that has the same numerical value

as the formula mass in amu’s (180.2 grams and 180.2 amu, in the case of aspirin)

we are able to ensure that the sample contains just as many molecules as there

are amu’s in 1 g. This gives us a very useful automatic link between masses and

numbers of molecules (or atoms or ions) that works for any chemical.

This is a difﬁcult construct, and it is entirely possible to read the preceding

discussion several times and still feel uncertain about the mass-to-amu relation-

ships. To help us reinforce the idea, let’s do the same calculation with an element,

rather than a compound. The element calcium is crucial to the manufacture

of healthy bones and teeth. The average atomic mass of the element calcium

is 40.08 amu, so we need to check how many calcium atoms are in 40.08 g of

calcium:

40.08 g ×

1amu

1.6605 ×10

−24

1atom

40.08 amu

= 6.022 × 10

atoms

That number 6.022 × 10

appears again! This is a very important and useful

number for us, because if we weigh out a mass in grams that is the same as an

element’s atomic mass or a molecule’s formula mass in amu, we will always have

6.022 × 10

atoms or formula units. The number 6.022 × 10

is known as

Avogadro’s number (N

) (or the Avogadro constant), after the Italian physicist

Amadeo Avogadro (1776−1856; Figure 3.5) and is abbreviated as N

EXERCISE 3.2 How Many Molecules?

Codeine, with the formula C

, is often used as an analgesic (painkiller) for

intense pain. How many molecules would be in a 0.10-g sample of codeine?

First Thoughts

Our ﬁrst thoughts lead us to determining the average molecular mass of codeine.

Once this mass is known, we should be able to determine the number of molecules

in that mass, using the same strategy we employed to get the number of atoms of

aspirin and the number of atoms of copper.

Solution

The molecular mass of codeine, C

, is 299.4 amu.

0.10 g codeine ×

1 amu codeine

1.6605 ×10

−24

g codeine

1 molecule codeine

299.4 amu codeine

= 2.0 × 10

molecules codeine

Further Insight

How many molecules are there in 299.37 g of codeine? Again, this would give

Avogadro’s number as the answer. This number must be rather important or we

3.2 Counting by Weighing 91

FIGURE 3.5

Lorenzo Romano Amadeo Carlo Avo-

gadro, conte di Quaregna e di Cerreto

(1776–1856), was born into a family of

well-established lawyers in Italy. He, too,

prepared for a legal career, obtaining his

law degree in 1792 when he was only

16 years old. However, in 1800 he began

studying mathematics and physics pri-

vately, completing his ﬁrst research pro-

ject on electricity with his brother Felice

in 1803. He was hired in 1809 at the Col-

lege of Vercelli, where he began his most

inﬂuential work in chemistry. He eventu-

ally obtained an appointment as a pro-

fessor of mathematical physics at the

University of Turin, where he worked

until he retired in 1850.

Codeine

Video Lesson: The Mole and

Avogadro’s Number

92 Chapter 3 Introducing Quantitative Chemistry

wouldn’t keep running into it! In fact, in chemistry this number is vital to many of

the calculations you will accomplish; accordingly, we give it a special name when we

do calculations (see the next section).

PRACTICE 3.2

How many atoms are there in 100.0 g of gold-197?

See Problems 9–16.

The Quantity We Call the Mole

An amount of any substance that contains 6.022 × 10

“elementary entities”

such as molecules, atoms, or ions is given its own name in chemistry. We call it

one

mole (often abbreviated mol) of the substance. Perhaps it seems odd to give a

special name to what is, in effect, just a number, but this is actually a very familiar

practice:

1 mole corresponds to 6.022 × 10

just as

1 dozen corresponds to 12

The mole is the basic counting unit used by chemists to indicate how many atoms,

molecules, or ions they are dealing with. A chemist counts moles of particles just as

a baker counts dozens of cupcakes.

To work out the mass of one mole of a chemical, you just look up its formula

mass and convert the units of that number from amu into grams. The resulting

quantity, expressed in grams per mole (

mol

), is known as the

molar mass—the

mass of one mole of a substance. Remember these relationships (illustrated by the

example in Figure 3.4):

■

The mass of 1 mol of any chemical is its formula mass expressed in units of

grams. This is the molar mass of the substance. The molar mass of aspirin is

180.2g

mol

■

1 mol corresponds to 6.022 × 10

entities, such as atoms, molecules, or ions.

■

6.022 ×10

is known as Avogadro’s number. It is often abbreviated as N

Although you should learn to think of the mole as simply equal to 6.022 × 10

it is formally deﬁned in a more complicated way by reference to the carbon-12

isotope. This formal deﬁnition states that a mole of a substance is that amount

of the substance that has the same number of atoms as exactly 12 g of the isotope

carbon-12.

Does this make sense in terms of what we have already said about the mole? We

can check that the deﬁnition ﬁts with the “1 mol = 6.022 ×10

” rule by calcu-

lating the number of particles in exactly 12 g of the isotope carbon-12. The

atomic mass of carbon-12 is exactly 12 amu.

12.00 g ×

1amu

1.6605 ×10

−24

1atom

12 amu

= 6.022 ×10

atoms

3.3 Working with Moles

Molecules to Moles and Back Again

Figure 3.6 illustrates the quantities of 1 mol of water, 1 mol of sodium chloride

(“table salt”), and 1 mol of aspirin—three different chemicals of great signiﬁ-

cance to life and medicine. Each has a different mass and occupies a different

The “dozen” is a counting unit. Its use in

counting groups is similar to a pair (2), a

ream (500), and a mole (6.022 × 10

Video Lesson: Introducing

Conversions of Masses, Moles,

and Number of Particles

volume, but each contains the same number of formula units of the compounds

concerned. The containers in the ﬁgure contain 6.022 × 10

molecules of water,

6.022 × 10

molecules of aspirin and 6.022 × 10

formula units of NaCl. (Note:

Because each formula unit of NaCl includes one Na

ion and one Cl

−

ion, our

sample contains 6.022 × 10

ions and 6.022 × 10

−

ions!)

EXERCISE 3.3 Molecules and Moles

Really large numbers such as 4.33 × 10

can be quite confusing. Let’s show how

using moles can simplify our work.

a. How many moles of water are there in 4.33 × 10

molecules of water?

b. How many moles of codeine (C

) are there in 4.33 × 10

molecules

of codeine?

Solution

4.33 ×10

molecules ×

1 mol

6.022 ×10

molecules

= 7.19 mol

This number is much easier to work with.

b. Note that the formula of the compound doesn’t change the number of

molecules that exist in 1 mol of a substance. Therefore, the answer is the same.

4.33 × 10

molecules of any substance is equal to 7.19 mol, whether we are

considering aspirin, adrenaline, copper, or codeine.

PRACTICE 3.3

How many moles are there in 2.88 × 10

formula units of sodium benzoate

COONa), a food preservative? How many formula units are there in 1.5 mol

of sodium benzoate?

See Problems 19b, 19d, 20b, 20d, 23b, and 24b.

Grams to Moles and Back Again

Many of us sweeten our drinks by adding some “table sugar,” the com-

pound sucrose, with the formula C

. If you added 10.0 g of sugar

to a glass of tea, how many moles did you add? To answer this question,

we must ﬁrst determine the formula mass of sucrose, which tells us the

mass of 1 mol and so gives us the mass-to-moles conversion factor that

we need. Here is the calculation:

12 mol C ×

12.01 g

1 mol C

= 144.12 g

22 mol H ×

1.008 g

1 mol

= 22.176 g

11 mol O ×

16.00 g

1 mol

= 176.0 g

Molar mass of sucrose = 342.296 = 342.3 g

This means that the mass of 1 mol of sucrose is 342.3 g of sucrose. Math-

ematically, our moles-to-mass conversion factor is

1 mol sucrose

342.3 g sucrose

.Now,

our sample of 10.0 g can be converted to moles:

10.0 g sucrose

1 mol sucrose

342.3 g sucrose

= 0.02921 = 0.0292 mol sucrose

3.3 Working with Moles 93

FIGURE 3.6

One mole of water (top), one mole of

sodium chloride (“salt”) (bottom),

and one mole of aspirin (center). 1 mol =

Avogadro’s number = 6.022 x 10

Sodium benzoate

OH CH

HOH

OH H

Sucrose

We can also determine how many molecules of sucrose have been added to our

glass of tea. We do this using Avogadro’s number, the number we call the mole,

6.022 × 10

0.02921 mol sucrose

6.022 ×10

molecules sucrose

1 mol sucrose

= 1.759 × 10

molecules sucrose

= 1.76 × 10

molecules sucrose

We would have stirred this number of sugar molecules into our tea:

17,600,000,000,000,000,000,000

EXERCISE 3.4 Calculating Masses Corresponding to Moles

Hydrochlorothiazide (HCT) is a medicine used to lower blood pressure and

deal with other illnesses related to ﬂuid retention. Its molecular formula is

ClN

a. In a small-scale production process, a company needs 63.4 mol of the

medicine. How many grams of hydrochlorothiazide is this?

b. A common individual dosage is 3.37 × 10

−5

mol. How many milligrams are

in this individual dose?

First Thoughts

We have two very different amounts to calculate. The ﬁrst represents a small-scale

industrial process, the second an individual dose. But the strategy is still the same.

We can use the molar mass of the compound to go from moles to mass. The molar

mass of the compound in grams per mole will be the same as the average molecular

mass in amu.

Solution

The molar mass of hydrochlorothiazide is 297.72 g per mole. Again, we ﬁrst write a

ﬂowchart to assist us in our calculations:

mol HCT

g HCT

mol HCT

−−−−−→g HCT

63.4 mol HCT

297.72 g HCT

1 mol HCT

= 1.89 ×10

g HCT

mol HCT

g HCT

mol HCT

−−−−−→g HCT

−−−−−→mg HCT

3.37 × 10

−5

mol HCT ×

297.72 g HCT

1 mol HCT

1 mg HCT

1 ×10

−3

g HCT

= 10.0 mg HCT

Further Insights

Do the answers make sense? We expect the industrial synthesis to result in much

more of the product than is in the individual dose, so this answer makes sense. You

may have noted from other medicines you have taken that a dose of perhaps 10 to

500 mg is standard. Therefore, our answers are reasonable.

PRACTICE 3.4

Calculate the mass in grams of 26 mol and of 0.0025 mol of each of these com-

pounds: table salt (NaCl),water (H

O), and the sweetener aspartame (C

See Problems 25 and 26.

94 Chapter 3 Introducing Quantitative Chemistry

HCT

Aspartame

EXERCISE 3.5 Calculating Moles Corresponding to Masses

Methylphenidate hydrochloride, C

·HCl, is a medical drug used to stimu-

late the central nervous system. It is known as Ritalin, and its use in the treatment of

attention deﬁcit hyperactivity disorder (ADHD) has been a subject of controversy

for years. A company sent a 2.00 × 10

-g shipment to be processed into individual

20.0-mg tablets. How many moles of Ritalin are in the shipment and in each tablet?

How many molecules of Ritalin are in each tablet? Note that the dot in the formula

is used to indicate small molecules that are chemically part of the overall formula.

Therefore, the formula methylphenidate hydrochloride could be written as

Cl.

Solution

For the entire shipment, we calculate the chemical amount in moles as follows, after

determining that the molar mass of Ritalin is 269.76 g/mol.

2.00 × 10

g Ritalin ×

1 mol Ritalin

269.8 g Ritalin

= 74.1 mol Ritalin

We calculate the number of moles in each 20.0-mg tablet using the same general

strategy, being careful to add the extra conversion factor that changes milligrams to

grams before ﬁnding moles.

mg Ritalin

tablet

−−−−−→

g Ritalin

tablet

mole

g Ritalin

−−−−−→

mol Ritalin

tablet

20.0 mg Ritalin

tablet

1 g Ritalin

1000 mg Ritalin

1 mol Ritalin

269.8 g Ritalin

= 7.413 × 10

−5

= 7.41 × 10

−5

mol Ritalin per tablet

We can convert from moles to molecules of Ritalin in a 20.0-mg tablet:

mol Ritalin

tablet

molecules

mole

−−−−−→

molecules Ritalin

tablet

7.413 ×10

−5

mol Ritalin

tablet

6.022 ×10

molecules Ritalin

1 mol Ritalin

4.46 ×10

molecules Ritalin

tablet

The individual dose has considerably fewer moles than the industrial shipment, so

the answer makes sense. The number of molecules in the individual dose is exceed-

ingly high, which is also reasonable because molecules are so small.

PRACTICE 3.5

Calculate the number of moles in 5.3 g and 0.0022 g these compounds: lactic acid,

, and sulfuric acid, used in car batteries (H

See Problems 19a, 19c, 20a, and 20c.

As researchers have investigated the causes of heart-related health problems,

one particular molecule, cholesterol, has been found to be associated with certain

types of circulatory problems. Giving a blood sample for the analysis of your cho-

lesterol level is a routine part of a medical checkup. The situation is not as simple

as cholesterol being “bad.” Cholesterol is essential for life—without it, you would

die. Cholesterol plays a crucial role in maintaining the thin membranes that

enclose every cell of your body. It also aids in the production of hormones,

3.3 Working with Moles 95

C C

Lactic acid Sulfuric acid

Ritalin

Application

HEMICAL ENCOUNTERS:

Cholesterol in the Blood

96 Chapter 3 Introducing Quantitative Chemistry

including estrogen and testosterone. So again, we ﬁnd that the quantity of a

chemical is crucial. The chemical structure of cholesterol is shown in Figure 3.7.

Its formula is C

O. If a blood analysis showed a cholesterol level of 195 mg

per deciliter of blood, which is equal to 1950 mg per liter, how many molecules

of cholesterol would be present per liter?

In this case you can see that the starting measurement is in milligrams, but

the question asks about the number of molecules. Remember that the “nano”

(molecules, in this case) and “macro” (moles, here) worlds are connected via

Avogadro’s number. The strategy is to express the mass as a quantity of moles of

cholesterol and then convert moles to molecules. Our dimensional analysis ﬂow-

chart for the task looks like this:

mg cholesterol

1000 mg

−−−−−→

g cholesterol

mol chol

g chol

−−−−−→

mol cholesterol

molecules

mol

−−−−−→

molecules cholesterol

When we have ﬁnished our calculation, our answer should make sense. What sort

of answer should we expect?

We know that there are a great many molecules in

even the smallest sample of cholesterol-containing blood.We would therefore ex-

pect a very large number of molecules. Answers such as 10

–24

molecule/L or even

molecules/L would not make sense.

Because cholesterol has a known molecular formula, C

O, we can deter-

mine its molar mass as 386.6 g per mole (check that you agree). We will follow

our ﬂowcharted strategy to get the answer.

1950 mg C

1gC

1000 mg C

1 mol C

386.6gC

6.022 ×10

molecules C

1 mol C

3.04 ×10

molecules C

Does the answer make sense? The quantity 195 mg is much less than a mole of

cholesterol (386.6 g per mole), so we would have expected an answer far less than

Avogadro’s number, but because we are counting our sample in molecules, the

Cholesterol

FIGURE 3.7

PRACTICE 3.6

How many molecules are there in 1.50 g of water? Would you need a bathtub, a

swimming pool, or a cup to hold 5.33 ×10

molecules of water?

See Problems 23a and 24a.

3.4 Percentages by Mass

Just like humans, plants undergo chemical changes that are a part of life. They

have genetic material that directs the synthesis of proteins, some of which are

part of cell structure and some of which make possible the chemical reactions

that sustain life. Nitrogen is a vital element in the genetic material and the pro-

teins in plants, so in order to grow, plants need plenty of nitrogen. For farms

throughout the world, fertilizer supplies that nitrogen in the form of manure or

industrially produced products.

Some plants need more nitrogen than others. Environmental

factors, such as temperature, rainfall, and the type of soil, also af-

fect the ability of plants to use the nitrogen in fertilizer. This

means that a farmer must apply the correct amount of nitrogen

for each particular situation. Among the most important factors

in deciding what fertilizer to use is the percent by mass,or

mass

percent

, of nitrogen in the product (see Figure 3.8). Mass percent

values are determined by dividing the mass of the component of

interest by the total mass of the entire sample. The result, multi-

plied by 100%, is the mass percent of the component in the sam-

ple. The component can be a compound (such as ammonium

nitrate, NH

, which is found in many liquid and solid fertil-

izers) or it can be an ion, atom, or element.

Mass percent =

total mass component

total mass whole substance

× 100%

Which of the following two compounds would supply plants

with the more concentrated source of nitrogen: ammonium

nitrate, NH

, or ammonium sulfate, (NH

)

or (NH

)

We can answer the question by calculating the mass percent

of nitrogen present in each compound. We need to express the

mass of nitrogen in each one as a percentage of the total mass.

number should still be quite large. Our value meets these conditions, so our

answer makes sense.

EXERCISE 3.6 Sugar Is Sweet

Sucrose (C

) is a wonderful molecule used to sweeten our cup of coffee. If a

packet of sucrose contains 1.50 g of sucrose, how many molecules is this?

Solution

3.4 Percentages by Mass 97

Application

HEMICAL ENCOUNTERS:

Nitrogen in Fertilizers

FIGURE 3.8

The mass percent of nitrogen

is often reported on the

ingredients label of a bag

of fertilizer.

1.50 g sucrose ×

1 mol sucrose

342.3 g sucrose

6.022 ×10

molecules of sucrose

1 mol sucrose

= 2.639 × 10

= 2.64 × 10

molecules of sucrose