9 Atoms, Molecules, and the Mole

80



c h a p t er 2   Atoms, Molecules, and Ions



2.9 Atoms, Molecules, and the Mole

Module 4: The Mole covers concepts in

this section.



• The “Mole”  The term mole was introduced about 1895 by Wilhelm

Ostwald (1853–1932), who derived the

term from the Latin word moles, meaning a “heap” or a “pile.”



One of the most exciting aspects of chemical research is the discovery of some new

substance, and part of this process of discovery involves quantitative experiments.

When two chemicals react with each other, we want to know how many atoms or

molecules of each are used so that formulas can be established for the reaction’s

products. To do so, we need some method of counting atoms and molecules. That

is, we must discover a way of connecting the macroscopic world, the world we can

see, with the particulate world of atoms, molecules, and ions. The solution to this

problem is to define a unit of matter that contains a known number of particles.

That chemical unit is the mole.

The mole (abbreviated mol) is the SI base unit for measuring an amount of a

substance (◀ Table 1, page 25) and is defined as follows:

A mole is the amount of a substance that contains as many elementary entities

(atoms, molecules, or other particles) as there are atoms in exactly 12 g of the

carbon-12 isotope.



The key to understanding the concept of the mole is recognizing that one mole always

contains the same number of particles, no matter what the substance. One mole of sodium

contains the same number of atoms as one mole of iron and the same as the number of molecules in one mole of water. How many particles? Many, many experiments over the years have established that number as

1 mole = 6.0221415 × 1023 particles



This value is known as Avogadro’s number (symbolized by NA) in honor of Amedeo

Avogadro, an Italian lawyer and physicist (1776–1856) who conceived the basic idea

(but never determined the number).



Atoms and Molar Mass

• A Difference Between the Terms



Amount and Quantity  The terms

“amount” and “quantity” are used in a

specific sense by chemists. The amount

of a substance is the number of moles

of that substance. In contrast, quantity

refers, for example, to the mass or volume of the substance. See W. G. Davies

and J. W. Moore. Journal of Chemical

Education, Vol. 57, p. 303, 1980. See

also http://physics.nist.gov



The mass in grams of one mole of any element (6.0221415 × 1023 atoms of that element) is the molar mass of that element. Molar mass is conventionally abbreviated

with a capital italicized M and has units of grams per mole (g/mol). An element’s

molar mass is the amount in grams numerically equal to its atomic weight. Using sodium

and lead as examples,

Molar mass of sodium (Na) = mass of 1.000 mol of Na atoms





= 22.99 g/mol







= mass of 6.022 × 1023 Na atoms

Molar mass of lead (Pb) = mass of 1.000 mol of Pb atoms







= 207.2 g/mol







= mass of 6.022 × 1023 Pb atoms



Figure 2.25 shows the relative sizes of a mole of some common elements. Although each of these “piles of atoms” has a different volume and different mass,

each contains 6.022 × 1023 atoms.

The mole concept is the cornerstone of quantitative chemistry. It is essential to be able to

convert from moles to mass and from mass to moles. Dimensional analysis, which is

described in Let’s Review, page 39, shows that this can be done in the following way:

MASS

Moles to Mass

grams

Moles ×

= grams

1 mol

molar mass



kotz_48288_02_0050-0109.indd 80



MOLES CONVERSION

Mass to Moles

Grams ×



1 mol

= moles

grams



1/molar mass



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81



2.9 Atoms, Molecules, and the Mole







FIGURE2.25 One mole of

common elements. (Left to right)

Sulfur powder, magnesium chips, tin,

and silicon. (Above) Copper beads.



© Cengage Learning/Charles D. Winters



Copper

63.546 g



Sulfur

32.066 g



Magnesium

24.305 g



Tin

118.71 g



Silicon

28.086 g



For example, what mass, in grams, is represented by 0.35 mol of aluminum? Using the molar mass of aluminum (27.0 g/mol), you can determine that 0.35 mol of

Al has a mass of 9.5 g.

0.35 mol Al ϫ



27.0 g Al

= 9.5 g Al

1 mol Al



Molar masses of the elements are generally known to at least four significant

figures. The convention followed in calculations in this book is to use a value of the

molar mass with at least one more significant figure than in any other number in the

problem. For example, if you weigh out 16.5 g of carbon, you use 12.01 g/mol for

the molar mass of C to find the amount of carbon present.

16.5 g C ×



1 mol C

12.01 g C



= 1.37 mol C



Note that four significant figures are used in the

molar mass, but there are three in the sample mass.



Using one more significant figure for the molar mass means the accuracy of this

value will not affect the accuracy of the result.



Amedeo Avogadro and His Number



Amedeo Avogadro, conte di

Quaregna, (1776–1856) was

an Italian nobleman and a lawyer. In about

1800, he turned to science and was the first

professor of mathematical physics in Italy.

Avogadro did not himself propose the

notion of a fixed number of particles in a

chemical unit. Rather, the number was

named in his honor because he had performed experiments in the 19th century that

laid the groundwork for the concept.



kotz_48288_02_0050-0109.indd 81



Just how large is Avogadro’s number? One

mol of unpopped popcorn kernels would

cover the continental United States to a depth

of about 9 miles. One mole of pennies divided

equally among every man, woman, and child

in the United States would allow each person

to pay off the national debt ($11.6 trillion or

11.6 × 1012 dollars) and there would still be

about $8 trillion left over per person! A mole



E. F. Smith Collection/

Van Pelt Library/

University of Pennsylvania



A CLOSER LOOK



of pennies does not go

as far as it used to!

Is the number a

unique value like π? No.

It is fixed by the definition of the mole as exactly 12 g of carbon-12.

If one mole of carbon were defined to have

some other mass, then Avogadro’s number

would have a different value.



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c h a p t er 2   Atoms, Molecules, and Ions



EXAMPLE 2.6



Mass, Moles, and Atoms



Problem  Consider two elements in the same vertical column of the periodic table, lead and tin.

(a) What mass of lead, in grams, is equivalent to 2.50 mol of lead (Pb)?



© Cengage Learning/Charles D. Winters



(b) What amount of tin, in moles, is represented by 36.6 g of tin (Sn)? How many atoms of tin

are in the sample?

What Do You Know?  You know the amount of lead and the mass of tin. You also know, from the

periodic table in the front of the book, the molar masses of lead (207.2 g/mol) and tin

(118.7 g/mol). For part (b) Avogadro’s number is needed (and is given in the text and inside the

back cover of the book).

Strategy  The molar mass is the quantity in grams numerically equal to the atomic weight of the

element.

Lead.  A 150-mL beaker containing

2.50 mol or 518 g of lead.



Part (a)  Multiply the amount of Pb by the molar mass.

Part (b)  Multiply the mass of tin by (1/molar mass). To determine the number of atoms, multiply

the amount of tin by Avogadro’s number.

Solution

(a) Convert the amount of lead in moles to mass in grams.

2.50 mol Pb ϫ



207.2 g

ϭ 518 g Pb

1 mol Pb



(b) Convert the mass of tin to the amount in moles,

36.6 g Sn ϫ





1 mol Sn

ϭ 0.308 mol Sn

118.7 g Sn



and then use Avogadro’s number to find the number of atoms in the sample.

0.308 mol Sn ϫ



6.022 ϫ 1023 atoms Sn

ϭ 1.86 × 1023 atoms Sn

1 mol Sn



© Cengage Learning/Charles D. Winters



Think about Your Answer  To find a mass from an amount of substance, use the conversion factor (mass/mol) (= molar mass). To find the amount from mass of a substance, use the conversion

factor (mol/mass) (= 1/molar mass). You can sometimes catch a mistake in setting up a conversion factor upside down if you think about your answers. For example, if we had made this mistake in part b, we would have calculated that there was less than one atom in 0.308 mol of Sn,

clearly an unreasonable answer.

Check Your Understanding



Tin.  A sample of tin having a mass of

36.6 g (or 1.86 × 1023 atoms).



The density of gold, Au, is 19.32 g/cm3. What is the volume (in cubic centimeters) of a piece

of gold that contains 2.6 × 1024 atoms? If the piece of metal is a square with a thickness of

0.10 cm, what is the length (in centimeters) of one side of the piece?



Molecules, Compounds, and Molar Mass

The formula of a compound tells you the type of atoms or ions in the compound

and the relative number of each. For example, one molecule of methane, CH4, is

made up of one atom of C and four atoms of H. But suppose you have Avogadro’s

number of C atoms (6.022 × 1023) combined with the proper number of H atoms.

For CH4 this means there are 4 × 6.022 × 1023 H atoms per mole of C atoms. What

masses of atoms are combined, and what is the mass of this many CH4 molecules?

C



kotz_48288_02_0050-0109.indd 82



+



4H



⎯→



CH4



6.022 × 1023 C atoms



4 × 6.022 × 1023 H atoms



6.022 × 1023 CH4 molecules



  = 1.000 mol of C



  = 4.000 mol of H atoms



  = 1.000 mol of CH4 molecules



  = 12.01 g of C atoms



  = 4.032 g of H atoms



  = 16.04 g of CH4 molecules



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2.9  Atoms, Molecules, and the Mole







83



Because we know the number of moles of C and H atoms, we can calculate

the masses of carbon and hydrogen that combine to form CH4. It follows that the

mass of CH4 is the sum of these masses. That is, 1 mol of CH4 has a mass equal to

the mass of 1 mol of C atoms (12.01 g) plus 4 mol of H atoms (4.032 g). Thus, the

molar mass, M, of CH4 is 16.04 g/mol. The molar masses of some substances are:

Molar and Molecular Masses

Substance

O2

NH3

H 2O

NH2CH2CO2H (glycine)



Molar Mass, M

(g/mol)



Average Mass of One Molecule

(g/molecule)



32.00

17.03

18.02

75.07



5.314 × 10–23

2.828 × 10–23

2.992 × 10–23

1.247 × 10–22



Ionic compounds such as NaCl do not exist as individual molecules. Thus, for

ionic compounds we write the simplest formula that shows the relative number of

each kind of atom in a “formula unit” of the compound, and the molar mass is calculated from this formula (M for NaCl = 58.44 g/mol). To differentiate substances

like NaCl that do not contain molecules, chemists sometimes refer to their formula

weight instead of their molar mass.

Figure 2.26 illustrates 1-mol quantities of several common compounds. To find

the molar mass of any compound, you need only to add up the atomic masses for

each element in the compound, taking into account any subscripts on elements. As

an example, let us find the molar mass of aspirin, C9H8O4. In one mole of aspirin

there are 9 mol of carbon atoms, 8 mol of hydrogen atoms, and 4 mol of oxygen

atoms, which add up to 180.2 g/mol of aspirin.

12.01 g C

ϭ 108.1 g C

1 mol C

1.008 g H

Mass of H in 1 mol C 9H8O4 ϭ 8 mol H ϫ

ϭ 8.064 g H

1 mol H

16.00 g O

Mass of O in 1 mol C 9H8O4 ϭ 4 mol O ϫ

ϭ 64.00 g O

1 mol O

Mass of C in 1 mol C 9H8O4 ϭ 9 mol C ϫ



Total mass of 1 mol of C 9H8O4 ϭ molar mass of C 9H8O4 ϭ 180.2 g



FIGURE 2.26   One-mole quantities of some compounds.  The



© Cengage Learning/Charles D. Winters



second compound, CuCl2 ∙ 2H2O, is

called a hydrated compound because

water is associated with the CuCl2 (see

page 94). Thus, one “formula unit”

consists of one Cu2+ ion, two Cl− ions,

and two water molecules. The molar

mass is the sum of the mass of 1 mol of

Cu, 2 mol of Cl, and 2 mol of H2O.

H2O

18.02 g/mol



Aspirin, C9H8O4

Copper(II) chloride Iron(III) oxide, Fe2O3

180.2 g/mol dihydrate, CuCl2 ∙ 2 H2O

159.7 g/mol

170.5 g/mol



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c h a p t er 2   Atoms, Molecules, and Ions



• Aspirin Formula  Aspirin has the

molecular formula C9H8O4 and a molar

mass of 180.2 g/mol. Aspirin is the

common name of the compound acetylsalicylic acid.



As was the case with elements, it is important to be able to convert between

amounts (moles) and mass (grams). For example, if you take 325 mg (0.325 g) of

aspirin in one tablet, what amount of the compound have you ingested? Based on a

molar mass of 180.2 g/mol, there is 0.00180 mol of aspirin per tablet.

0.325 g aspirin ϫ



1 mol aspirin

ϭ 0.00180 mol aspirin

180.2 g aspirin



Using the molar mass of a compound it is possible to determine the number of

molecules in any sample from the sample mass and to determine the mass of one

molecule. For example, the number of aspirin molecules in one tablet is

0.00180 mol aspirin ϫ



and the mass of one molecule is



O



CH3

O



C

O



C



C



H



180.2 g aspirin

1 mol aspirin

ϫ

ϭ 2.99 ϫ 10Ϫ22 g/molecule

1 mol aspirin

6.022 ϫ 1023 molecules



C OH

C



C

C



H

  Interactive EXAMPLE 2.7 Molar Mass and Moles



C



H



6.022 ϫ 1023 molecules

ϭ 1.08 ϫ 1021 molecules

1 mol aspirin



H



Problem  You have 16.5 g of oxalic acid, H2C2O4.

(a) What amount is represented by 16.5 g of oxalic acid?

(b) How many molecules of oxalic acid are in 16.5 g?

(c) How many atoms of carbon are in 16.5 g of oxalic acid?



Strategy Map 2.7

PROBLEM



Find amount of oxalic acid in a

given mass. Then find number of

molecules and number of C atoms

in the sample.



DATA/INFORMATION KNOWN



• Mass of sample

• Formula of compound

• Avogadro’s number



What Do You Know?  You know the mass and formula of oxalic acid. The molar mass of the compound can be calculated based on the formula.

Strategy  The strategy is outlined in the strategy map.

•



The molar mass is the sum of the masses of the component atoms.



•



Part (a) Use the molar mass to convert mass to amount.



•



Part (b) Use Avogadro’s number to calculate the number of molecules from the amount.



•



Part (c) From the formula you know there are two atoms of carbon in each molecule.



Solution

Calculate molar mass

of oxalic acid.

STEP 1.



Molar mass of oxalic acid (g/mol)



(a) Moles represented by 16.5 g





12.01 g C

= 24.02 g C per mol H2C2O4

1 mol C

1.008 g H

= 2.016 g H per mol H2C2O4

2 mol H per mol H2C2O4 ϫ

1 mol H

16.00 g O

4 mol O per mol H2C2O4 ϫ

= 64.00 g O per mol H2C2O4

1 mol O



2 mol C per mol H2C2O4 ϫ



Use molar mass to

calculate amount (multiply

mass by 1/molar mass).

STEP 2.



Amount (mol) of oxalic acid



Molar mass of H2C2O4 = 90.04 g per mol H2C2O4



Multiply by Avogadro’s

number.



STEP 3.



Number of molecules

STEP 4. Multiply by number

of C atoms per molecule.



Let us first calculate the molar mass of oxalic acid:







Now calculate the amount in moles. The molar mass (expressed here in units of 1 mol/90.04 g)

is used in all mass-mole conversions.

16.5 g H2C2O4 ϫ



1 mol

ϭ 0.183 mol H2C2O4

90.04 g H2C2O4



Number of C atoms in sample



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2.10 Describing Compound Formulas







85



(b) Number of molecules

Use Avogadro’s number to find the number of oxalic acid molecules in 0.183 mol of H2C2O4.

0.183 mol ϫ

(c)



6.022 ϫ 1023 molecules

ϭ 1.10 × 1023 molecules

1 mol



Number of C atoms

Because each molecule contains two carbon atoms, the number of carbon atoms in 16.5 g of

the acid is

1.10 ϫ 1023 molecules ϫ



2 C atoms

ϭ 2.21 × 1023 C atoms

1 molecule



Think about Your Answer The mass of oxalic acid was 16.5 g, much less than the mass of a

mole, so check to make sure your answer reflects this. The number of molecules of the acid

should be many fewer than in one mole of molecules.

Check Your Understanding

If you have 454 g of citric acid (H3C6H5O7), what amount (moles) does this represent? How many

molecules? How many atoms of carbon?



rEvIEW & cHEcK FOr SEctIOn 2.9

1.



What is the molar mass of calcium nitrate?

(a)



150.08 g/mol



(c)



(b) 102.08 g/mol

2.



24 g of O2



(b) 4.0 g of H2



19 g of F2



(d) 28 g of N2



0.50 mol Na



(b) 0.20 mol K



(c)



0.40 mol Ca



(d) 0.60 mol Mg



Which quantity represents the largest amount (moles) of the indicated substance?

(a)



6.02 × 1023 molecules of H2O



(b) 22 g of CO2

5.



(c)



Which of the following has the largest mass?

(a)



4.



(d) 68.09 g/mol



Which of the following contains the largest number of molecules?

(a)



3.



164.09 g/mol



(c)



3.01 × 1023 molecules of Br2



(d) 30 g of HF



How many atoms of oxygen are contained in 16 g of O2?

(a)



6.0 × 1023 atoms of O



(b) 3.0 × 1023 atoms of O



(c)



1.5 × 1023 atoms of O



(d) 1.2 × 1024 atoms of O



2.10 DescribingCompoundFormulas

Given a sample of an unknown compound, how can its formula be determined?

The answer lies in chemical analysis, a major branch of chemistry that deals with the

determination of formulas and structures.



Percent Composition

Any sample of a pure compound always consists of the same elements combined in

the same proportion by mass. This means molecular composition can be expressed

in at least three ways:

1.

2.



in terms of the number of atoms of each type per molecule or per formula

unit—that is, by giving the formula of the compound

in terms of the mass of each element per mole of compound



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86



c h a p t er 2   Atoms, Molecules, and Ions



3. in terms of the mass of each element in the compound relative to the total mass

of the compound—that is, as a mass percent

Suppose you have 1.0000 mol of NH3 or 17.031 g. This mass of NH3 is composed

of 14.007 g of N (1.0000 mol) and 3.0237 g of H (3.0000 mol). If you compare the mass

of N to the total mass of compound, 82.246% of the total mass is N (and 17.755% is H).

Mass of N per mole of NH3 ϭ

Mass percent N in NH3 ϭ

ϭ



1 mol N

14 .0 0 7 g N

ϫ

ϭ 14 .007 g N/1 mol NH3

1 mol NH3

1 mol N

mass of N in 1 mol NH3

ϫ 100%

mass of 1 mol NH3

14 .007 g N

ϫ 100 %

17 .031 g NH3



ϭ 82 .244 % (or 82.244 g N in 100.000 g NH3)



• Molecular Composition  Molecular

composition can be expressed as a percent (mass of an element in a 100-g

sample × 100%). For example, NH3 is

82.244% N. Therefore, it has 82.244 g

of N in 100.000 g of compound.



Mass of H per mole of NH3 ϭ



H



N

H



H



17.755% of NH3 mass

is hydrogen.



ϫ



1.0079 g H

1 mol H



ϭ 3.0237 g H/1 mol NH3

Mass percent H in NH3 ϭ

ϭ



82.244% of NH3 mass

is nitrogen.



3 mol H

1 mol NH3



mass of H in 1 mol NH3

ϫ 100%

mass of 1 mol NH3

3.0237 g H

ϫ 100%

17.031 g NH3



ϭ 17.755% (or 17.755 g H in 100.000 g NH3)



These values represent the mass percent of each element, or percent composition by

mass. They tell you that in a 100.00-g sample there are 82.244 g of N and 17.755 g of H.



EXAMPLE 2.8



Using Percent Composition



Problem  What is the mass percent of each element in propane, C3H8? What mass of carbon is

contained in 454 g of propane?

What Do You Know?  You know the formula of propane. You will need the atomic weights of

C and H to calculate the mass percent of each element.

Strategy

•



Calculate the molar mass of propane.



•



From the formula, you know there are 3 moles of C and 8 moles of H per mole of C3H8.

Determine the mass of C and H represented by these amounts. The percent of each element

is the mass of the element divided by the molar mass.



•



The mass of C in 454 g of C3H8 is obtained by multiplying this mass by the % C/100, (that is,

by the decimal fraction of C in the compound).



Solution

(a) The molar mass of C3H8 is 44.097 g/mol.

(b) Mass percent of C and H in C3H8:

3 mol C

12.01 g C

ϫ

ϭ 36.03 g C/1 mol C 3H8

1 mol C 3H8

1 mol C

Mass percent of C in C 3H8 ϭ



36..03 g C

ϫ 100% ϭ

44.097 g C 3H8



81.71% C



1.008 g H

8 mol H

ϫ

ϭ 8.064 g H/1 mol C 3H8

1 mol H

1 mol C 3H8

Mass percent of H in C 3H8 ϭ



kotz_48288_02_0050-0109.indd 86



8.064 g H

ϫ 100% ϭ

44.097 g C 3H8



18.29% H



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2.10  Describing Compound Formulas







87



(c) Mass of C in 454 g of C3H8:

454 g C 3H8 ϫ



81.71 g C

ϭ 371 g C

100.0 g C 3H8



Think about Your Answer  Once you know the percent C in the sample, you could calculate the

percent H from it using the formula %H = 100% − %C.

Check Your Understanding

1.



Express the composition of ammonium carbonate, (NH4)2CO3, in terms of the mass of each

element in 1.00 mol of compound and the mass percent of each element.



2.



What is the mass of carbon in 454 g of octane, C8H18?



Empirical and Molecular Formulas from Percent Composition

Now let us consider the reverse of the procedure just described: using relative mass

or percent composition data to find a molecular formula. Suppose you know the

identity of the elements in a sample and have determined the mass of each element

in a given mass of compound by chemical analysis [▶ Section 4.4]. You can then

calculate the relative amount (moles) of each element, which is also the relative

number of atoms of each element in the formula of the compound. For example,

for a compound composed of atoms of A and B, the steps from percent composition

to a formula are as follows:

ST EP 1 .



ST EP 2 .



STEP 3.



Convert

mass percent

to mass



Convert

mass

to moles



Find

mole ratio



%A



gA



x mol A



%B



gB



y mol B



Convert to

whole-number

ratio of A to B

x mol A

y mol B



AaBb



As an example, let us derive the formula for hydrazine, a compound used to

remove oxygen from water in heating and cooling systems. It is composed of

87.42% N and 12.58% H and is a close relative of ammonia.

Step 1:  Convert mass percent to mass. The mass percentages of N and H in hydrazine

tell us there are 87.42 g of N and 12.58 g of H in a 100.00-g sample.

Step 2:  Convert the mass of each element to moles. The amount of each element in the

100.00-g sample is

87.42 g N ϫ



1 mol N

ϭ 6.241 mol N

14.007 g N



12.58 g H ϫ



1 mol H

ϭ 12.48 mol H

1.0079 g H



• Deriving a Formula  Percent composition gives the mass of an element in

100 g of a sample. However, in deriving

a formula, any amount of sample is appropriate if you know the mass of each

element in that sample mass.



Step 3:  Find the mole ratio of elements. Use the amount (moles) of each element in the

100.00-g of sample to find the amount of one element relative to the other. To do

this divide the larger amount by the smaller amount. For hydrazine, this ratio is

2 mol of H to 1 mol of N,

12.48 mol H

2.00 mol H

ϭ

→ NH2

6.241 mol N

1.00 mol N



showing that there are 2 mol of H atoms for every 1 mol of N atoms in hydrazine.

Thus, in one molecule, two atoms of H occur for every atom of N; that is, the formula is NH2. This simplest, whole-number atom ratio of atoms in a formula is called

the empirical formula.



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88



c h a p t er 2   Atoms, Molecules, and Ions



Percent composition data allow us to calculate the atom ratios in a compound.

A molecular formula, however, must convey two pieces of information: (1) the relative

numbers of atoms of each element in a molecule (the atom ratios) and (2) the total

number of atoms in the molecule. For hydrazine there are twice as many H atoms

as N atoms, so the molecular formula could be NH2. Recognize, however, that percent composition data give only the simplest possible ratio of atoms in a molecule.

The empirical formula of hydrazine is NH2, but the true molecular formula could

be NH2, N2H4, N3H6, N4H8, or any other formula having a 1∶2 ratio of N to H.

To determine the molecular formula from the empirical formula, the molar mass must be obtained from experiment. For example, experiments show that the molar mass of hydrazine

is 32.0 g/mol, twice the formula mass of NH2, which is 16.0 g/mol. Thus, the molecular formula of hydrazine is two times the empirical formula of NH2, that is, N2H4.



  Interactive EXAMPLE 2.9 Calculating a Formula from Percent

Composition

Problem  Many soft drinks contain sodium benzoate as a preservative. When you consume the

sodium benzoate, it reacts with the amino acid glycine in your body to form hippuric acid, which

is then excreted in the urine. Hippuric acid has a molar mass of 179.17 g/mol and is 60.33% C,

5.06% H, and 7.82% N; the remainder is oxygen. What are the empirical and molecular formulas

of hippuric acid?

Strategy Map 2.9

PROBLEM



Determine empirical and

molecular formulas based on

known composition and known

molar mass.

DATA/INFORMATION KNOWN



• Molar mass

• Percent composition



What Do You Know?  You know the mass percent of C, H, and N. The mass percent of oxygen is

not known but is obtained by difference. You also know the molar mass and will need atomic

weights of these four elements for the calculation.

Strategy  Assume the mass percent of each element is equivalent to its mass in grams, and convert each mass to moles. The ratio of moles gives the empirical formula. The mass of a mole of

compound having the calculated empirical formula is compared with the actual, experimental

molar mass to find the true molecular formula.

Solution  The mass of oxygen in a 100.0-g sample of hippuric acid is

100.00 g = 60.33 g C + 5.06 g H + 7.82 g N + mass of O



Assume each atom % is

equivalent to mass in grams in

100 g sample.

STEP 1.



Mass of each element in a 100 g

sample of the compound

Use atomic weight

of each element to calculate

amount of each element in

100 g sample

(multiply mass by mol/g).

STEP 2.



Amount (mol) of each element in

100 g sample

STEP 3. Divide the amount of

each element by the amount of

the element present in the least

amount.



Whole-number ratio of the amount

of each element to the amount of

element present in the least amount

= empirical formula

Divide known molar mass

by empirical formula mass.

STEP 4.



Molecular formula



kotz_48288_02_0050-0109.indd 88



Mass of O = 26.79 g O

The amount of each element is

60.33 g C ϫ



1 mol C

ϭ 5.023 mol C

12.011 g C



5.06 g H ϫ



1 mol H

ϭ 5.02 mol H

1.008 g H



7.82 g N ϫ



1 mol N

ϭ 0.558 mol N

14.01 g N



26.79 g O ϫ



1 mol O

ϭ 1.674 mol O

15.999 g O



To find the mole ratio, the best approach is to base the ratios on the smallest number of moles

present—in this case, nitrogen.

mol C

5.023 mol C

9.00 mol C

ϭ

ϭ

ϭ 9 mol C/1 mol N

mol N

0.558 mol N

1.00 mol N

mol H

5.02 mol C

9.00 mol C

ϭ

ϭ

ϭ 9 mol H/1 mol N

mol N

0.5558 mol N

1.00 mol N

mol O

1.674 mol O

3.00 mol O

ϭ

ϭ

ϭ 3 mol O/1 mol N

mol N

0.558 mol N

1.000 mol N

Now we know there are 9 mol of C, 9 mol of H, and 3 mol of O per 1 mol of N. Thus,

 the empirical formula is C9H9NO3 .

The experimentally determined molar mass of hippuric acid is 179.17 g/mol. This is the same as

the empirical formula weight, so the  molecular formula is C9H9NO3 .



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2.10  Describing Compound Formulas







89



Think about Your Answer  There is another approach to finding the molecular formula here.

Knowing the percent composition of hippuric acid and its molar mass, you could calculate that in

179.17 g of hippuric acid there are 108.06 g of C (8.997 mol of C), 9.06 g of H (8.99 mol of H),

14.01 g N (1.000 mol N), and 48.00 g of O (3 mol of O). This gives us a molecular formula of

C9H9NO3. However, you must recognize that this approach can only be used when you know both

the percent composition and the molar mass.

Check Your Understanding

1.



What is the empirical formula of naphthalene, C10H8?



2.



The empirical formula of acetic acid is CH2O. If its molar mass is 60.05 g/mol, what is the

molecular formula of acetic acid?



3.



Isoprene is a liquid compound that can be polymerized to form natural rubber. It is composed of 88.17% carbon and 11.83% hydrogen. Its molar mass is 68.11 g/mol. What are its

empirical and molecular formulas?



4.



Camphor is found in camphor wood, much prized for its wonderful odor. It is composed

of 78.90% carbon and 10.59% hydrogen. The remainder is oxygen. What is its empirical

formula?



Hippuric Acid, C9H9NO3  This

substance, which can be isolated as

white crystals, is found in the urine of

humans and of herbivorous animals.



Determining a Formula from Mass Data

The composition of a compound in terms of mass percent gives us the mass of each

element in a 100.0-g sample. In the laboratory we often collect information on the

composition of compounds slightly differently. We can

1. Combine known masses of elements to give a sample of the compound of

known mass. Element masses can be converted to amounts (moles), and the

ratio of amounts gives the combining ratio of atoms—that is, the empirical

formula. This approach is described in Example 2.10.

2. Decompose a known mass of an unknown compound into “pieces” of known

composition. If the masses of the “pieces” can be determined, the ratio of moles

of the “pieces” gives the formula. An example is a decomposition such as

Ni(CO)4(ℓ) n Ni(s) + 4 CO(g)



 he masses of Ni and CO can be converted to moles, whose 1∶4 ratio would reT

veal the formula of the compound. We will describe this approach in Chapter 4.



  Interactive EXAMPLE 2.10 Formula of a Compound from

Combining Masses

Problem  Oxides of virtually every element are known. Bromine, for example, forms several

oxides when treated with ozone. Suppose you allow 1.250 g of bromine, Br2, to react with ozone

and obtain 1.876 g of BrxOy. What is the formula of the product?

What Do You Know?  You know that you begin with a given mass of bromine and that all of the

bromine becomes part of bromine oxide of unknown formula. You also know the mass of the

product, and because you know the mass of Br in this product, you can determine the mass of O

in the product.

Strategy

•



The mass of oxygen is determined as the difference between the product mass and the mass

of bromine used.



•



Calculate the amounts of Br and O from the masses of each element.



•



Find the lowest whole number ratio between the moles of Br and moles of O. This defines

the empirical formula.



kotz_48288_02_0050-0109.indd 89



Strategy Map 2 . 1 0

PROBLEM



Determine empirical formula based

on masses of combining elements.



DATA/INFORMATION KNOWN



• Mass of one element in

binary compound

• Mass of product

Find mass of second

element by difference.



STEP 1.



Mass of each element in a sample

of the compound

Use atomic weight of

each element to calculate

amount of each element in

sample (multiply mass by mol/g).



STEP 2.



Amount (mol) of each element

in sample

S T E P 3 . Divide the amount of

each element by the amount of

the element present in the least

amount.



Whole-number ratio of the amount

of each element to the amount of

element present in the least amount

= empirical formula



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90



c h a p t er 2 Atoms, Molecules, and Ions



PrOBLEM SOLvInG tIP 2.3

•



•



The experimental data available to find a

formula may be in the form of percent

composition or the masses of elements

combined in some mass of compound. No

matter what the starting point, the first

step is always to convert masses of elements to moles.

Be sure to use at least three significant

figures when calculating empirical formulas. Using fewer significant figures can

give a misleading result.



Finding Empirical and Molecular Formulas

•



•



When finding atom ratios, always divide

the larger number of moles by the

smaller one.

Empirical and molecular formulas can differ for molecular compounds. In contrast,

there is no “molecular” formula for an

ionic compound; all that can be recorded

is the empirical formula.



•



•



Determining the molecular formula of a

compound after calculating the empirical

formula requires knowing the molar mass.

When both the percent composition and

the molar mass are known for a compound, the alternative method mentioned

in Think about Your Answer in Example 2.9

could be used.



Solution You already know the mass of bromine in the compound, so you can calculate the mass

of oxygen in the compound by subtracting the mass of bromine from the mass of the product.

1.876 g product − 1.250 g Br2 = 0.626 g O

Next, calculate the amount of each reactant. Notice that, although Br2 was the reactant, we need

to know the amount of Br in the product.

1.250 g Br2 ϫ

0.007822 mol Br2 ϫ

0.626 g O ϫ



1 mol Br2

ϭ 0.007822 mol Br2

159.81 g

2 mol Br

ϭ 0.01564 mol Br

1 mol Br2

1 mol O

ϭ 0.0391 mol O

16.00 g O



Find the ratio of moles of O to moles of Br:

Mole ratio ϭ



0.0391 mol O

2.50 mol O

ϭ

1.00 mol O

0.01564 mol Br



The atom ratio is 2.5 mol O/1.0 mol Br. However, atoms combine in the ratio of small whole

numbers, so we double this to give a ratio of 5 mol O to 2 mol Br. Thus, the product is dibromine

pentaoxide, Br2O5.

Think about Your Answer The whole number ratio of 5∶2 was found by realizing that

2.5 = 2 1/2 = 5/2. This problem rests on the principle of the conservation of matter.

All of the bromine used in the reaction is part of the product.

Check Your Understanding

Gallium oxide, GaxOy, forms when gallium is combined with oxygen. Suppose you allow 1.25 g of

gallium (Ga) to react with oxygen and obtain 1.68 g of GaxOy. What is the formula of the product?



EXAMPLE 2.11



Determining a Formula from Mass Data



Problem Tin metal (Sn) and purple iodine (I2) combine to form orange, solid tin iodide with an

unknown formula.

Sn metal + solid I2 n



solid SnxIy



Weighed quantities of Sn and I2 are combined, where the quantity of Sn is more than is needed

to react with all of the iodine. After SnxIy has been formed, it is isolated by filtration. The mass of

excess tin is also determined. The following data were collected:

Mass of tin (Sn) in the original mixture



1.056 g



Mass of iodine (I2) in the original mixture



1.947 g



Mass of tin (Sn) recovered after reaction



0.601 g



What is the empirical formula of the tin iodide obtained?



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