Chemistry

TABLE 4 Binary Compounds of Nitrogen and Oxygen

Formula Prefix-system name

N2O dinitrogen monoxide
NO nitrogen monoxide

NO2 nitrogen dioxide
N2O3 dinitrogen trioxide
N2O4 dinitrogen tetroxide
N2O5 dinitrogen pentoxide

The prefix system is illustrated further in Table 4, which lists the

names of the six oxides of nitrogen. Note the application of rule 1, for

example, in the name nitrogen dioxide for NO2. No prefix is needed with
nitrogen because only one atom of nitrogen is present in a molecule of

NO2. On the other hand, according to rule 2, the prefix di- in dioxide is
needed to indicate the presence of two atoms of oxygen. Take a moment

to review the prefixes in the other names in Table 4.

SAMPLE PROBLEM D

a. Give the name for As2O5.
b. Write the formula for oxygen difluoride.

SOLUTION a. A molecule of the compound contains two arsenic atoms, so the first word in the name is
diarsenic. The five oxygen atoms are indicated by adding the prefix pent- to the word
oxide. The complete name is diarsenic pentoxide.

b. The first symbol in the formula is that for oxygen. Oxygen is first in the name because
it is less electronegative than fluorine. Since there is no prefix, there must be only one
oxygen atom. The prefix di- in difluoride shows that there are two fluorine atoms in
the molecule. The formula is OF2.

PRACTICE Answers in Appendix E

1. Name the following binary molecular compounds: Go to go.hrw.com for
a. SO3 more practice problems
b. ICl3 that ask you to write
c. PBr5 names and formulas for
binary molecular
2. Write formulas for the following compounds: compounds.
a. carbon tetraiodide
b. phosphorus trichloride Keyword: HC6FRMX
c. dinitrogen trioxide

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 229

Covalent-Network Compounds

As you read in Chapter 6, some covalent compounds do not consist
of individual molecules. Instead, each atom is joined to all its neighbors
in a covalently bonded, three-dimensional network. There are no dis-
tinct units in these compounds, just as there are no such units in ionic
compounds. The subscripts in a formula for a covalent-network com-
pound indicate the smallest whole-number ratio of the atoms in the
compound. Naming such compounds is similar to naming molecular
compounds. Some common examples are given below.

SiC SiO2 Si3N4
silicon carbide silicon dioxide trisilicon tetranitride

Acids and Salts

www.scilinks.org An acid is a distinct type of molecular compound about which you will
Topic: Acids read in detail in Chapter 14. Most acids used in the laboratory can
Code: HC60012 be classified as either binary acids or oxyacids. Binary acids are acids
that consist of two elements, usually hydrogen and one of the halogens—
fluorine, chlorine, bromine, iodine. Oxyacids are acids that contain
hydrogen, oxygen, and a third element (usually a nonmetal).

Acids were first recognized as a specific class of compounds based on
their properties in solutions of water. Consequently, in chemical nomen-
clature, the term acid usually refers to a solution in water of one of these
special compounds rather than to the compound itself. For example,
hydrochloric acid refers to a water solution of the molecular compound
hydrogen chloride, HCl. Some common binary and oxyacids are listed
in Table 5.

Many polyatomic ions are produced by the loss of hydrogen ions
from oxyacids. A few examples of the relationship between oxyacids
and oxyanions are shown below.

sulfuric acid H2SO4 sulfate SO24−
nitric acid HNO3 nitrate NO−3
phosphoric acid H3PO4 phosphate PO34−

TABLE 5 Common Binary Acids and Oxyacids

HF hydrofluoric acid HNO2 nitrous acid HClO hypochlorous acid
HCl hydrochloric acid HNO3 nitric acid chlorous acid
HBr hydrobromic acid H2SO3 sulfurous acid HClO2 chloric acid
HI hydriodic acid H2SO4 sulfuric acid HClO3 perchloric acid
H3PO4 phosphoric acid CH3COOH acetic acid HClO4 carbonic acid
H2CO3

230 C H A P T E R 7

FIGURE 2 Some common labora-
tory acids. Acids should always be
handled with care and according to
instructions. They can burn the skin,
and they burn holes in clothing.

An ionic compound composed of a cation and the anion from an acid
is often referred to as a salt. Table salt, NaCl, contains the anion from
hydrochloric acid. Calcium sulfate, CaSO4, is a salt containing an anion
from sulfuric acid. Some salts contain anions in which one or more
hydrogen atoms from the acid are retained. Such anions are named by
adding the word hydrogen or the prefix bi- to the anion name. The best
known such anion comes from carbonic acid, H2CO3.

HCO−3
hydrogen carbonate ion

bicarbonate ion

SECTION REVIEW 3. Name the following compounds by using the
Stock system:
1. What is the significance of a chemical formula?
a. NaI c. CaO e. CuBr
2. Write formulas for the compounds formed
between the following: b. MgS d. K2S f. FeCl2

a. aluminum and bromine 4. Write formulas for each of the following compounds:

b. sodium and oxygen a. sodium hydroxide e. carbon diselenide

c. magnesium and iodine b. lead(II) nitrate f. acetic acid
d. Pb2+ and O2−
e. Sn2+ and I− c. iron(II) sulfate g. chloric acid
f. Fe3+ and S2−
g. Cu2+ and NO3− d. diphosphorus trioxide h. sulfurous acid
h. NH+4 and SO24−
Critical Thinking

5. RELATING IDEAS Draw the Lewis structure, give
the name, and predict VSEPR geometry of SCl2.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 231

SECTION 2 Oxidation Numbers

OBJECTIVES T he charges on the ions composing an ionic compound reflect the

List the rules for assigning electron distribution of the compound. In order to indicate the general
oxidation numbers. distribution of electrons among the bonded atoms in a molecular com-
Give the oxidation number for pound or a polyatomic ion, oxidation numbers, also called oxidation
each element in the formula of states, are assigned to the atoms composing the compound or ion. Unlike
a chemical compound. ionic charges, oxidation numbers do not have an exact physical meaning.
Name binary molecular In fact, in some cases they are quite arbitrary. However, oxidation num-
compounds using oxidation bers are useful in naming compounds, in writing formulas, and in bal-
numbers and the Stock ancing chemical equations. And, as will be discussed in Chapter 19, they
system. are helpful in studying certain types of chemical reactions.

232 C H A P T E R 7 Assigning Oxidation Numbers

As a general rule in assigning oxidation numbers, shared electrons are
assumed to belong to the more electronegative atom in each bond.
More specific rules for determining oxidation numbers are provided by
the following guidelines.
1. The atoms in a pure element have an oxidation number of zero. For

example, the atoms in pure sodium, Na, oxygen, O2, phosphorus, P4,
and sulfur, S8, all have oxidation numbers of zero.
2. The more-electronegative element in a binary molecular compound
is assigned the number equal to the negative charge it would have as
an anion. The less-electronegative atom is assigned the number equal
to the positive charge it would have as a cation.
3. Fluorine has an oxidation number of −1 in all of its compounds
because it is the most electronegative element.
4. Oxygen has an oxidation number of −2 in almost all compounds.
Exceptions include when it is in peroxides, such as H2O2, in which its
oxidation number is −1, and when it is in compounds with fluorine,
such as OF2, in which its oxidation number is +2.
5. Hydrogen has an oxidation number of +1 in all compounds contain-
ing elements that are more electronegative than it; it has an oxida-
tion number of −1 in compounds with metals.
6. The algebraic sum of the oxidation numbers of all atoms in a neutral
compound is equal to zero.
7. The algebraic sum of the oxidation numbers of all atoms in a poly-
atomic ion is equal to the charge of the ion.
8. Although rules 1 through 7 apply to covalently bonded atoms, oxi-
dation numbers can also be assigned to atoms in ionic compounds.

9. A monatomic ion has an oxidation number equal to the charge of the
ion. For example, the ions Na+, Ca2+, and Cl− have oxidation numbers
of +1, +2, and −1, respectively.

Let’s examine the assignment of oxidation numbers to the atoms in
two molecular compounds, hydrogen fluoride, HF, and water, H2O. Both
compounds have polar-covalent bonds. In HF, the fluorine atom should
have an oxidation number of −1 (see Rule 3 on the previous page).
Rule 5 tells us that hydrogen should have an oxidation number of +1.
This makes sense because fluorine is more electronegative than hydro-
gen, and in a polar-covalent bond, shared electrons are assumed to
belong to the more-electronegative element. For water, Rules 4 and 5
tell us that the oxidation number of oxygen should be −2 and the oxi-
dation number of each hydrogen atom should be +1. Again, oxygen is
more electronegative than hydrogen, so the shared electrons are
assumed to belong to oxygen.

Because the sum of the oxidation numbers of the atoms in a com-
pound must satisfy Rule 6 or 7 of the guidelines on the previous page, it
is often possible to assign oxidation numbers when they are not known.
This is illustrated in Sample Problem E.

SAMPLE PROBLEM E

Assign oxidation numbers to each atom in the following compounds or ions:
a. UF6
b. H2SO4
c. ClO–3

SOLUTION a. Start by placing known oxidation numbers above the appropriate elements. From the
guidelines, we know that fluorine always has an oxidation number of −1.

−1

UF6

Multiply known oxidation numbers by the appropriate number of atoms and place the
totals underneath the corresponding elements. There are six fluorine atoms, 6 × −1 = −6.

−1

UF6

−6

The compound UF6 is molecular. According to the guidelines, the sum of the oxidation
numbers must equal zero. The total of positive oxidation numbers is therefore +6.

−1

UF6

+6 −6

Divide the total calculated oxidation number by the appropriate number of atoms. There
is only one uranium atom in the molecule, so it must have an oxidation number of +6.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 233

+6 −1

UF6

+6 −6

b. Oxygen and sulfur are each more electronegative than hydrogen, so hydrogen has an oxi-
dation number of +1. Oxygen is not combined with a halogen, nor is H2SO4 a peroxide.
Therefore, the oxidation number of oxygen is −2. Place these known oxidation numbers
above the appropriate symbols. Place the total of the oxidation numbers underneath.

+1 −2

H2SO4

+2 −8

The sum of the oxidation numbers must equal zero, and there is only one sulfur atom in
each molecule of H2SO4. Because (+2) + (−8) = −6, the oxidation number of each sulfur
atom must be +6.

c. To assign oxidation numbers to the elements in ClO−3, proceed as in parts (a) and (b).
Remember, however, that the total of the oxidation numbers should equal the overall
charge of the anion, 1−. The oxidation number of a single oxygen atom in the ion is −2.
The total oxidation number due to the three oxygen atoms is −6. For the chlorate ion to
have a 1− charge, chlorine must be assigned an oxidation number of +5.

+5 −2

ClO−3

+5 −6

PRACTICE Answers in Appendix E

1. Assign oxidation numbers to each atom in the following com-

pounds or ions:

a. HCl e. HNO3 h. HClO3 Go to go.hrw.com for
f. KH i. N2O5 more practice problems
b. CF4 j. GeCl2 that ask you to assign
c. PCl3 g. P4O10 oxidation numbers.
d. SO2
Keyword: HC6FRMX

Using Oxidation Numbers
for Formulas and Names

As shown in Table 6, many nonmetals can have more than one oxidation
number. (A more extensive list of oxidation numbers is given in
Appendix Table A-15.) These numbers can sometimes be used in the
same manner as ionic charges to determine formulas. Suppose, for exam-
ple, you want to know the formula of a binary compound formed
between sulfur and oxygen. From the common +4 and +6 oxidation states
of sulfur, you could expect that sulfur might form SO2 or SO3. Both are
known compounds. Of course, a formula must represent facts. Oxidation
numbers alone cannot be used to prove the existence of a compound.

234 C H A P T E R 7

TABLE 6 Common Oxidation Numbers of Some
Nonmetals That Have Variable Oxidation
States*

Group 14 carbon −4, +2, +4

Group 15 nitrogen −3, +1, +2, +3, +4, +5
phosphorus −3, +3, +5
Group 16
Group 17 sulfur −2, +4, +6

chlorine −1, +1, +3, +5, +7
bromine −1, +1, +3, +5, +7
iodine −1, +1, +3, +5, +7

*In addition to the values shown, atoms of each element in its pure state
are assigned an oxidation number of zero.

In Section 1 we introduced the use of Roman numerals to denote
ionic charges in the Stock system of naming ionic compounds. The Stock
system is actually based on oxidation numbers, and it can be used as an
alternative to the prefix system for naming binary molecular com-
pounds. In the prefix system, for example, SO2 and SO3 are named sul-
fur dioxide and sulfur trioxide, respectively. Their names according to
the Stock system are sulfur(IV) oxide and sulfur(VI) oxide. The inter-
national body that governs nomenclature has endorsed the Stock sys-
tem, which is more practical for complicated compounds. Prefix-based
names and Stock-system names are still used interchangeably for many
simple compounds, however. A few additional examples of names in
both systems are given below.

PCl3 Prefix system Stock system
PCl5
N2O phosphorus trichloride phosphorus(III) chloride
NO phosphorus pentachloride phosphorus(V) chloride
dinitrogen monoxide nitrogen(I) oxide
PbO2 nitrogen monoxide nitrogen(II) oxide
Mo2O3 lead dioxide lead(IV) oxide
dimolybdenum trioxide molybdenum(III) oxide

SECTION REVIEW 2. Name each of the following binary molecular
compounds according to the Stock system:
1. Assign oxidation numbers to each atom in the
following compounds or ions: a. CI4 c. As2S3
b. SO3 d. NCl3
a. HF e. CS2 i. SO42−

b. CI4 f. Na2O2 Critical Thinking

c. H2O g. H2CO3 3. DRAWING CONCLUSIONS Determine the oxidation
d. PI3 numbers for iron oxide, Fe3O4. (Recall that oxida-
h. NO − tion numbers are integers.)
2

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 235

Mass Spectrometry: Identifying Molecules

Tests for locating oil deposits in the molecular mass of the molecular ion. mass spectrometry, and Koichi Tanaka
ground and detecting dioxins in our By analyzing the peaks, scientists (Shimadzu Corporation, Japan)
food supply are commonly per- can determine the identity and struc- developed matrix-assisted laser
formed today. These tests can be ture of a compound. Computers are desorption ionization (MALDI) mass
performed by using a technique used to help interpret the spectrum spectrometry. In 2002, both scientists
known as mass spectrometry. Mass and identify the molecule by using received the Nobel Prize in chemistry
spectrometry is now used in many online spectral database libraries. for their work. Their methods opened
fields, such as medicine, chemistry, the field to the study of large mol-
forensic science, and astronomy. Mass spectrometry has been an ecules, allowing scientists to use mass
essential tool for scientists since its spectrometry to study the structure of
What is mass spectrometry? It is invention in the early 1900s. But its macromolecules such as nucleic acids
the most accurate technique avail- use was limited to small molecules and steroids and to identify the
able to measure the mass of an indi- from which ion creation was easy. sequence of proteins.
vidual molecule or atom. Knowing Large biological molecules could not
the molecular mass is an essential be studied because they would Questions
part of identifying an unknown com- break down or decompose during
pound and determining the structure conventional ion-formation tech- 1. Why is it necessary to convert
of a molecule of the compound. As niques. In the late 1980s, two groups the sample into ions in the mass
the diagram of a mass spectrometer developed ion-formation methods spectrometer?
shows, the molecules in a gaseous that are used today in commercial
sample are converted into ions. The mass spectrometers. John Fenn 2. How have recent developments
ions then are separated and sorted (Virginia Commonwealth University) in mass spectrometry con-
according to their mass-to-charge developed electrospray ionization tributed to the field of medicine?
ratio by a combination of electric
and magnetic fields. The fields cause Scientists use mass spectrometers to
the ions’ trajectories to change
based on the ions’ masses and identify and study the structure of Ion separation
charges. Then, the sorted ions are molecules. and sorting
detected, and a mass spectrum is
obtained. The mass spectrum is a Ion detection
graph of relative intensity (related to
the number of ions detected) versus Ion creation
mass-to-charge ratio. Mass spec-
trometry uses a very small sample heated
size (10−12 g) to obtain results.
filament (–)
The resulting spectrum is like a
puzzle. It contains numerous peaks detector
that correspond to fragments of the
initial molecule. The largest peak (–) negative magnets
(parent peak) corresponds to the (+) grid

gaseous to vacuum mass spectrum
sample pump

236 C H A P T E R 7

Using Chemical SECTION 3
Formulas
OBJECTIVES

Calculate the formula mass
or molar mass of any given
compound.

A s you have seen, a chemical formula indicates the elements as well as Use molar mass to convert
between mass in grams
the relative number of atoms or ions of each element present in a com- and amount in moles of
pound. Chemical formulas also allow chemists to calculate a number of a chemical compound.
characteristic values for a given compound. In this section, you will learn
how to use chemical formulas to calculate the formula mass, the molar Calculate the number of
mass, and the percentage composition by mass of a compound. molecules, formula units, or
ions in a given molar amount
Formula Masses of a chemical compound.

In Chapter 3, we saw that hydrogen atoms have an average atomic mass Calculate the percentage
of 1.007 94 amu and that oxygen atoms have an average atomic mass of composition of a given
15.9994 amu. Like individual atoms, molecules, formula units, or ions chemical compound.
have characteristic average masses. For example, we know from the
chemical formula H2O that a single water molecule is composed of
exactly two hydrogen atoms and one oxygen atom. The mass of a water
molecule is found by adding the masses of the three atoms in the
molecule. (In the calculation, the average atomic masses have been
rounded to two decimal places.)

average atomic mass of H: 1.01 amu
average atomic mass of O: 16.00 amu

2 H atoms × 1.01 amu = 2.02 amu

H atom

1 O atom × 16.00 amu = 16.00 amu

O atom

average mass of H2O molecule = 18.02 amu

The mass of a water molecule can be correctly referred to as a www.scilinks.org
molecular mass. The mass of one NaCl formula unit, on the other hand, Topic: Photochemical Reaction
is not a molecular mass because NaCl is an ionic compound. The mass Code: HC61137
of any unit represented by a chemical formula, whether the unit is a
molecule, a formula unit, or an ion, is known as the formula mass. The
formula mass of any molecule, formula unit, or ion is the sum of the
average atomic masses of all atoms represented in its formula.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 237

The procedure illustrated for calculating the formula mass of a water
molecule can be used to calculate the mass of any unit represented by a
chemical formula. In each of the problems that follow, the atomic
masses from the periodic table in the back of the book have been
rounded to two decimal places.

SAMPLE PROBLEM F

Find the formula mass of potassium chlorate, KClO3.

SOLUTION The mass of a formula unit of KClO3 is found by summing the masses of one K atom,
one Cl atom, and three O atoms. The required atomic masses can be found in the

periodic table in the back of the book. In the calculation, each atomic mass has been

rounded to two decimal places.

1 K atom × 39.10 amu = 39.10 amu

K atom

1 Cl atom × 35.45 amu = 35.45 amu

Cl atom

3 O atoms × 16.00 amu = 48.00 amu

O atom

formula mass of KClO3 = 122.55 amu

PRACTICE Answers in Appendix E

1. Find the formula mass of each of the following: Go to go.hrw.com for
more practice problems
a. H2SO4 that ask you to calculate
b. Ca(NO3)2 formula mass.
c. PO43−
d. MgCl2 Keyword: HC6FRMX

Molar Masses

In Chapter 3 you learned that the molar mass of a substance is equal to
the mass in grams of one mole, or approximately 6.022 × 1023 particles,
of the substance. For example, the molar mass of pure calcium, Ca, is
40.08 g/mol because one mole of calcium atoms has a mass of 40.08 g.

The molar mass of a compound is calculated by summing the
masses of the elements present in a mole of the molecules or formula
units that make up the compound. For example, one mole of water
molecules contains exactly two moles of H atoms and one mole of O
atoms. Rounded to two decimal places, a mole of hydrogen atoms has a

238 C H A P T E R 7

mass of 1.01 g, and a mole of oxygen atoms has a mass of 16.00 g. The
molar mass of water is calculated as follows.

2 mol H × 1.01 g H = 2.02 g H

mol H

1 mol O × 16.00 g O = 16.00 g O

mol O

molar mass of H2O = 18.02 g/mol

Figure 3 shows a mole of water as well as a mole of several other FIGURE 3 Every compound has a
substances. characteristic molar mass. Shown
here are one mole each of nitrogen
You may have noticed that a compound’s molar mass is numeri- (in balloon), water (in graduated
cally equal to its formula mass. For instance, in Sample Problem F the cylinder), cadmium sulfide, CdS (yel-
formula mass of KClO3 was found to be 122.55 amu. Therefore, because low substance), and sodium chloride,
molar mass is numerically equal to formula mass, we know that the NaCl (white substance).
molar mass of KClO3 is 122.55 g/mol.

SAMPLE PROBLEM G

What is the molar mass of barium nitrate, Ba(NO3)2?

SOLUTION One mole of barium nitrate contains exactly one mole of Ba2+ ions and two moles of NO−3
ions. The two moles of NO−3 ions contain two moles of N atoms and six moles of O atoms.
Therefore, the molar mass of Ba(NO3)2 is calculated as follows.

1 mol Ba × 137.33 g Ba = 137.33 g Ba

mol Ba

2 mol N × 14.01 g N = 28.02 g N

mol N

6 mol O × 16.00 g O = 96.00 g O

mol O

molar mass of Ba(NO3)2 = 261.35 g/mol

PRACTICE Answers in Appendix E

1. How many moles of atoms of each element are there in one mole Go to go.hrw.com for
of the following compounds? more practice problems
a. Al2S3 that ask you to calculate
b. NaNO3 molar mass.
c. Ba(OH)2
Keyword: HC6FRMX
2. Find the molar mass of each of the compounds listed in item 1.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 239

Mass of ؋ 1‫؍‬ Amount of ؋ 6.022 ؋ 1023 ‫؍‬ Number of
compound molar mass of compound compound 1 molecules or
‫ ؍‬molar mass of compound ؋ formula units
(grams) (moles) ‫ ؍‬6.022 ؋ 1023 ؋ of compound

(a)

1 Amount of ‫؍‬؋
؋ molar mass of element ‫؍‬ element 6.022 ؋ 1023
1 Number
Mass of element in compound of atoms
in compound (moles) ‫ ؍‬6.022 ؋ 1023 ؋ of element
(grams) in compound
‫ ؍‬molar mass of element ؋

(b)

FIGURE 4 (a) The diagram shows the relationships between mass in grams,
amount in moles, and number of molecules or atoms for a given compound.
(b) Similar relationships exist for an element within a compound.

Molar Mass as a Conversion Factor

The molar mass of a compound can be used as a conversion factor to
relate an amount in moles to a mass in grams for a given substance.
Recall that molar mass usually has the units of grams per mole. To con-
vert a known amount of a compound in moles to a mass in grams, mul-
tiply the amount in moles by the molar mass.

amount in moles × molar mass (g/mol) = mass in grams

Conversions of this type for elements and compounds are summarized
above in Figure 4.

SAMPLE PROBLEM H
What is the mass in grams of 2.50 mol of oxygen gas?

SOLUTION

1 ANALYZE Given: 2.50 mol O2
Unknown: mass of O2 in grams

2 PLAN moles O2 ⎯→ grams O2
To convert amount of O2 in moles to mass of O2 in grams, multiply by the molar mass of O2.

amount of O2 (mol) × molar mass of O2 (g/mol) = mass of O2 (g)

240 C H A P T E R 7

3 COMPUTE First the molar mass of O2 must be calculated.
4 EVALUATE 16.00 g O

2 mol O × mol O = 32.00 g (mass of one mole of O2)

The molar mass of O2 is therefore 32.00 g/mol. Now do the calculation shown in step 2.

2.50 mol O2 × 32.00 g O2 = 80.0 g O2
mol O2

The answer is correctly given to three significant figures and is close to an estimated value
of 75 g (2.50 mol × 30 g/mol).

To convert a known mass of a compound in grams to an amount in
moles, the mass must be divided by the molar mass. Or you can invert
the molar mass and multiply so that units are easily canceled.

mass in grams × 1 = amount in moles

molar mass (g/mol)

SAMPLE PROBLEM I

Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers.
Its molar mass is 206.31 g/mol.
a. If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of

ibuprofen are in the bottle?
b. How many molecules of ibuprofen are in the bottle?
c. What is the total mass in grams of carbon in 33 g of ibuprofen?

SOLUTION

1 ANALYZE Given: 33 g of C13H18O2, molar mass 206.31 g/mol
Unknown: a. moles C13H18O2

b. molecules C13H18O2
c. total mass of C

2 PLAN a. grams ⎯→ moles

To convert mass of ibuprofen in grams to amount of ibuprofen in moles, multiply by the
inverted molar mass of C13H18O2.

g C13H18O2 × 1 mol C13H18O2 = mol C13H18O2
206.31 g C13H18O2

b. moles ⎯→ molecules
To find the number of molecules of ibuprofen, multiply amount of C13H18O2 in moles by
Avogadro’s number.

6.022 × 1023 molecules
mol C13H18O2 × mol = molecules C13H18O2

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 241

c. moles C13H18O2 ⎯→ moles C ⎯→ grams C
To find the mass of carbon present in the ibuprofen, the two conversion factors needed

are the amount of carbon in moles per mole of C13H18O2 and the molar mass of carbon.

13 mol C 12.01 g C
mol C13H18O2 × mol C13H18O2 × mol C = g C

3 COMPUTE a. 33 g C13H18O2 × 1 mol C13H18O2 = 0.16 mol C13H18O2
206.31 g C13H18O2

b. 0.16 mol C13H18O2 × 6.022 × 1023 molecules = 9.6 × 1022 molecules C13H18O2
mol

13 mol C 12.01 g C
c. 0.16 mol C13H18O2 × mol C13H18O2 × mol C = 25 g C

The bottle contains 0.16 mol of ibuprofen, which is 9.6 × 1022 molecules of ibuprofen.
The sample of ibuprofen contains 25 g of carbon.

4 EVALUATE Checking each step shows that the arithmetic is correct, significant figures have been used
correctly, and units have canceled as desired.

PRACTICE Answers in Appendix E

1. How many moles of compound are there in the following? Go to go.hrw.com for
a. 6.60 g (NH4)2SO4 more practice problems
b. 4.5 kg Ca(OH)2 that ask you to use
molar mass as a
2. How many molecules are there in the following? conversion factor.
a. 25.0 g H2SO4
b. 125 g of sugar, C12H22O11 Keyword: HC6FRMX

3. What is the mass in grams of 6.25 mol of copper(II) nitrate?

Percentage Composition

It is often useful to know the percentage by mass of a particular element
in a chemical compound. For example, suppose the compound potas-
sium chlorate, KClO3, were to be used as a source of oxygen. It would
be helpful to know the percentage of oxygen in the compound. To find
the mass percentage of an element in a compound, one can divide the
mass of the element in a sample of the compound by the total mass of
the sample, then multiply this value by 100.

mass of element in sample of compound × 100 = % element in
mass of sample of compound compound

242 C H A P T E R 7

The mass percentage of an element in a compound is the same regard-
less of the sample’s size. Therefore, a simpler way to calculate the per-
centage of an element in a compound is to determine how many grams
of the element are present in one mole of the compound. Then divide
this value by the molar mass of the compound and multiply by 100.

mass of element in 1 mol of compound × 100 = % element in
molar mass of compound compound

The percentage by mass of each element in a compound is known as the
percentage composition of the compound.

SAMPLE PROBLEM J For more help, go to the Math Tutor at the end of this chapter.

Find the percentage composition of copper(I) sulfide, Cu2S.

SOLUTION

1 ANALYZE Given: formula, Cu2S
Unknown: percentage composition of Cu2S

2 PLAN formula ⎯→ molar mass ⎯→ mass percentage of each element

The molar mass of the compound must be found. Then the mass of each element present
in one mole of the compound is used to calculate the mass percentage of each element.

3 COMPUTE 2 mol Cu × 63.55 g Cu = 127.1 g Cu
4 EVALUATE
mol Cu

1 mol S × 32.07 g S = 32.07 g S

mol S

molar mass of Cu2S = 159.2 g
127.1 g Cu

× 100 = 79.85% Cu
159.2 g Cu2S

32.07 g S
× 100 = 20.15% S

159.2 g Cu2S
A good check is to see if the results add up to about 100%. (Because of rounding, the total
may not always be exactly 100%.)

SAMPLE PROBLEM K For more help, go to the Math Tutor at the end of this chapter.

As some salts crystallize from a water solution, they bind water molecules in their
crystal structure. Sodium carbonate forms such a hydrate, in which 10 water molecules
are present for every formula unit of sodium carbonate. Find the mass percentage of
water in sodium carbonate decahydrate, Na2CO3•10H2O, which has a molar mass of
286.19 g/mol.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 243

SOLUTION

1 ANALYZE Given: chemical formula, Na2CO3•10H2O
molar mass of Na2CO3•10H2O

Unknown: mass percentage of H2O

2 PLAN chemical formula ⎯→ mass H2O per mole of Na2CO3•10H2O ⎯→ % water

3 COMPUTE The mass of water per mole of sodium carbonate decahydrate must first be found.
This value is then divided by the mass of one mole of Na2CO3•10H2O.

One mole of Na2CO3•10H2O contains 10 mol H2O. Recall from page 239 that the molar
mass of H2O is 18.02 g/mol. The mass of 10 mol H2O is calculated as follows.

18.02 g H2O
10 mol H2O × mol H2O = 180.2 g H2O

mass of H2O per mole of Na2CO3•10H2O = 180.2 g

The molar mass of Na2CO3•10H2O is 286.19 g/mol, so we know that 1 mol of the hydrate
has a mass of 286.19g. The mass percentage of 10 mol H2O in 1 mol Na2CO3•10H2O can
now be calculated.

mass percentage of H2O in Na2CO3•10H2O = 180.2 g H2O × 100 = 62.97% H2O
286.19 g Na2CO3•10H2O

4 EVALUATE Checking shows that the arithmetic is correct and that units cancel as desired.

PRACTICE Answers in Appendix E

1. Find the percentage compositions of the following:

a. PbCl2 b. Ba(NO3)2

2. Find the mass percentage of water in ZnSO4•7H2O. Go to go.hrw.com for
more practice problems
3. Magnesium hydroxide is 54.87% oxygen by mass. How many that ask you to calculate
grams of oxygen are in 175 g of the compound? How many moles percentage
of oxygen is this? composition.

Keyword: HC6FRMX

SECTION REVIEW 5. Calculate the percentage composition of (NH4)2CO3.

1. Determine both the formula mass and molar mass Critical Thinking
of ammonium carbonate, (NH4)2CO3.
6. RELATING IDEAS A sample of hydrated copper(II)
2. How many moles of atoms of each element are sulfate (CuSO4 •nH2O) is heated to 150°C and pro-
there in one mole of (NH4)2CO3? duces 103.74 g anhydrous copper(II) sulfate and
58.55 g water. How many moles of water mol-
3. What is the mass in grams of 3.25 mol Fe2(SO4)3? ecules are present in 1.0 mol of hydrated
4. How many molecules of aspirin, C9H8O4, are there copper(II) sulfate?

in a 100.0 mg tablet of aspirin?

244 C H A P T E R 7

Determining SECTION 4
Chemical Formulas
OBJECTIVES
W hen a new substance is synthesized or is discovered, it is analyzed
Define empirical formula,
quantitatively to reveal its percentage composition. From these data, and explain how the term
the empirical formula is then determined. An empirical formula consists applies to ionic and molecular
of the symbols for the elements combined in a compound, with subscripts compounds.
showing the smallest whole-number mole ratio of the different atoms in
the compound. For an ionic compound, the formula unit is usually the Determine an empirical formula
compound’s empirical formula. For a molecular compound, however, from either a percentage or a
the empirical formula does not necessarily indicate the actual numbers mass composition.
of atoms present in each molecule. For example, the empirical formula
of the gas diborane is BH3, but the molecular formula is B2H6. In this Explain the relationship between
case, the number of atoms given by the molecular formula corre- the empirical formula and the
sponds to the empirical ratio multiplied by two. molecular formula of a given
compound.

Determine a molecular formula
from an empirical formula.

Calculation of Empirical Formulas

To determine a compound’s empirical formula from its percentage com-
position, begin by converting percentage composition to a mass compo-
sition. Assume that you have a 100.0 g sample of the compound. Then
calculate the amount of each element in the sample. For example, the
percentage composition of diborane is 78.1% B and 21.9% H.
Therefore, 100.0 g of diborane contains 78.1 g of B and 21.9 g of H.

Next, the mass composition of each element is converted to a com-
position in moles by dividing by the appropriate molar mass.

78.1 g B × 1 mol B = 7.22 mol B

10.81 g B

21.9 g H × 1 mol H = 21.7 mol H

1.01 g H

These values give a mole ratio of 7.22 mol B to 21.7 mol H. However,
this is not a ratio of smallest whole numbers. To find such a ratio, divide
each number of moles by the smallest number in the existing ratio.

7.22 mol B 21.7 mol H
: = 1 mol B: 3.01 mol H

7.22 7.22

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 245

Because of rounding or experimental error, a compound’s mole ratio
sometimes consists of numbers close to whole numbers instead of exact
whole numbers. In this case, the differences from whole numbers may be
ignored and the nearest whole number taken. Thus, diborane contains
atoms in the ratio 1 B:3 H. The compound’s empirical formula is BH3.

Sometimes mass composition is known instead of percentage
composition. To determine the empirical formula in this case, convert
mass composition to composition in moles. Then calculate the smallest
whole-number mole ratio of atoms. This process is shown in Sample
Problem M.

SAMPLE PROBLEM L For more help, go to the Math Tutor at the end of Chapter 22.

Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen.
Find the empirical formula of this compound.

SOLUTION

1 ANALYZE Given: percentage composition: 32.38% Na, 22.65% S, and 44.99% O
Unknown: empirical formula

2 PLAN percentage composition ⎯→ mass composition ⎯→ composition in moles
⎯→ smallest whole-number mole ratio of atoms

3 COMPUTE Mass composition (mass of each element in 100.0 g sample): 32.38 g Na, 22.65 g S, 44.99 g O

Composition in moles: 32.38 g Na × 1 mol Na = 1.408 mol Na

22.99 g Na

22.65 g S × 1 mol S = 0.7063 mol S

32.07 g S

44.99 g O × 1 mol O = 2.812 mol O

16.00 g O

Smallest whole-number mole ratio of atoms:
The compound contains atoms in the ratio 1.408 mol Na:0.7063 mol S:2.812 mol O. To find the
smallest whole-number mole ratio, divide each value by the smallest number in the ratio.

1.408 mol Na 0.7063 mol S 2.812 mol O
: : = 1.993 mol Na:1 mol S:3.981 mol O
0.7063 0.7063 0.7063

Rounding each number in the ratio to the nearest whole number yields a mole ratio
of 2 mol Na:1 mol S:4 mol O. The empirical formula of the compound is Na2SO4.

4 EVALUATE Calculating the percentage composition of the compound based on the empirical formula
determined in the problem reveals a percentage composition of 32.37% Na, 22.58% S, and
45.05% O. These values agree reasonably well with the given percentage composition.

246 C H A P T E R 7

SAMPLE PROBLEM M For more help, go to the Math Tutor at the end of Chapter 22.

Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a
phosphorus content of 4.433 g. What is the empirical formula of this compound?

SOLUTION

1 ANALYZE Given: sample mass = 10.150 g
phosphorus mass = 4.433 g

Unknown: empirical formula

2 PLAN Mass composition ⎯→ composition in moles ⎯→ smallest whole-number ratio of atoms

3 COMPUTE The mass of oxygen is found by subtracting the phosphorus mass from the sample mass.
sample mass − phosphorus mass = 10.150 g − 4.433 g = 5.717 g

Mass composition: 4.433 g P, 5.717 g O

Composition in moles:

4.433 g P × 1 mol P = 0.1431 mol P

30.97 g P

5.717 g O × 1 mol O = 0.3573 mol O

16.00 g O

Smallest whole-number mole ratio of atoms:

0.1431 mol P 0.3573 mol O
:
0.1431 0.1431

1 mol P : 2.497 mol O

The number of O atoms is not close to a whole number. But if we multiply each number
in the ratio by 2, then the number of O atoms becomes 4.994 mol, which is close to 5 mol.
The simplest whole-number mole ratio of P atoms to O atoms is 2:5. The compound’s
empirical formula is P2O5.

4 EVALUATE The arithmetic is correct, significant figures have been used correctly, and units cancel as
desired. The formula is reasonable because +5 is a common oxidation state of phosphorus.

PRACTICE Answers in Appendix E

1. A compound is found to contain 63.52% iron and 36.48% sulfur. Go to go.hrw.com
Find its empirical formula. for more practice
problems that ask you
2. Find the empirical formula of a compound found to contain to determine empirical
26.56% potassium, 35.41% chromium, and the remainder oxygen. formulas.

3. Analysis of 20.0 g of a compound containing only calcium and Keyword: HC6FRMX
bromine indicates that 4.00 g of calcium are present. What is the
empirical formula of the compound formed?

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 247

CROSS-DISCIPLINARY Calculation of Molecular Formulas

Go to go.hrw.com for a full-length Remember that the empirical formula contains the smallest possible
article on the importance of molecu- whole numbers that describe the atomic ratio. The molecular formula is
lar shape in odors. the actual formula of a molecular compound. An empirical formula may
or may not be a correct molecular formula. For example, diborane’s
Keyword: HC6FRMX empirical formula is BH3. Any multiple of BH3, such as B2H6, B3H9,
B4H12, and so on, represents the same ratio of B atoms to H atoms. The
molecular compounds ethene, C2H4, and cyclopropane, C3H6, also share
an identical atomic ratio (2 H:1 C), yet they are very different sub-
stances. How is the correct formula of a molecular compound found
from an empirical formula?

The relationship between a compound’s empirical formula and its
molecular formula can be written as follows.

x(empirical formula) = molecular formula

The number represented by x is a whole-number multiple indicating the
factor by which the subscripts in the empirical formula must be multi-
plied to obtain the molecular formula. (The value of x is sometimes 1.)
The formula masses have a similar relationship.

x(empirical formula mass) = molecular formula mass

To determine the molecular formula of a compound, you must know
the compound’s formula mass. For example, experimentation shows the
formula mass of diborane to be 27.67 amu. The formula mass for the
empirical formula, BH3, is 13.84 amu. Dividing the experimental formula
mass by the empirical formula mass gives the value of x for diborane.

27.67 amu
x = = 2.000

13.84 amu

The molecular formula of diborane is therefore B2H6.

2(BH3) = B2H6

Recall that a compound’s molecular formula mass is numerically
equal to its molar mass, so a compound’s molecular formula can also be
found given the compound’s empirical formula and its molar mass.

SAMPLE PROBLEM N For more help, go to the Math Tutor at the end of Chapter 22.

In Sample Problem M, the empirical formula of a compound of phosphorus and oxygen was found to be

P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the com-
pound’s molecular formula?

SOLUTION

1 ANALYZE Given: empirical formula
Unknown: molecular formula

248 C H A P T E R 7

2 PLAN x(empirical formula) = molecular formula
3 COMPUTE
molecular formula mass
x=

empirical formula mass

Molecular formula mass is numerically equal to molar mass. Thus, changing the g/mol unit
of the compound’s molar mass to amu yields the compound’s molecular formula mass.

molecular molar mass = 283.89 g/mol
molecular formula mass = 283.89 amu

The empirical formula mass is found by adding the masses of each of the atoms indicated in
the empirical formula.

mass of phosphorus atom = 30.97 amu
mass of oxygen atom = 16.00 amu

empirical formula mass of P2O5 = 2 × 30.97 amu + 5 × 16.00 amu = 141.94 amu

Dividing the experimental formula mass by the empirical formula mass gives the value of x.
The formula mass is numerically equal to the molar mass.

283.89 amu
x = = 2.0001

141.94 amu
The compound’s molecular formula is therefore P4O10.

2 × (P2O5) = P4O10

4 EVALUATE Checking the arithmetic shows that it is correct.

PRACTICE Answers in Appendix E

1. Determine the molecular formula of the compound with an Go to go.hrw.com
empirical formula of CH and a formula mass of 78.110 amu. for more practice
problems that ask you
2. A sample of a compound with a formula mass of 34.00 amu is to determine molecular
found to consist of 0.44 g H and 6.92 g O. Find its molecular formulas.
formula.
Keyword: HC6FRMX

SECTION REVIEW 4. If 4.04 g of N combine with 11.46 g O to produce
a compound with a formula mass of 108.0 amu,
1. A compound contains 36.48% Na, 25.41% S, and what is the molecular formula of this compound?
38.11% O. Find its empirical formula.
Critical Thinking
2. Find the empirical formula of a compound that
contains 53.70% iron and 46.30% sulfur. 5. RELATING IDEAS A compound containing sodium,
chlorine, and oxygen is 25.42% sodium by mass. A
3. Analysis of a compound indicates that it contains 3.25 g sample gives 4.33 × 1022 atoms of oxygen.
1.04 g K, 0.70 g Cr, and 0.86 g O. Find its empiri- What is the empirical formula?
cal formula.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 249

CHAPTER HIGHLIGHTS

Chemical Names and Formulas

Vocabulary • A positive monatomic ion is identified simply by the name of
monatomic ion
binary compound the appropriate element. A negative monatomic ion is named
nomenclature
oxyanion by dropping parts of the ending of the element’s name and
salt
adding -ide to the root.

• The charge of each ion in an ionic compound may be used to

determine the simplest chemical formula for the compound.

• Binary compounds are composed of two elements.
• Binary ionic compounds are named by combining the names

of the positive and negative ions.

• The old system of naming binary molecular compounds uses

prefixes. The new system, known as the Stock system, uses oxi-

dation numbers.

Oxidation Numbers • Oxidation numbers are useful in naming compounds, in writing

Vocabulary formulas, and in balancing chemical equations.
oxidation number
oxidation state • Compounds containing elements that have more than one oxi-

dation state are named by using the Stock system.

• Stock-system names and prefix-system names are used inter-

changeably for many molecular compounds.

• Oxidation numbers of each element in a compound may be

used to determine the compound’s simplest chemical formula.

• By knowing oxidation numbers, we can name compounds with-

out knowing whether they are ionic or molecular.

Using Chemical Formulas • Formula mass, molar mass, and percentage composition can be

Vocabulary calculated from the chemical formula for a compound.
formula mass
percentage composition • The percentage composition of a compound is the percentage

by mass of each element in the compound.

• Molar mass is used as a conversion factor between amount in

moles and mass in grams of a given compound or element.

Determining Chemical Formulas

Vocabulary • An empirical formula shows the simplest whole-number ratio
empirical formula
of atoms in a given compound.

• Empirical formulas indicate how many atoms of each element

are combined in the simplest unit of a chemical compound.

• A molecular formula can be found from the empirical formula

if the molar mass is measured.

250 C H A P T E R 7

CHAPTER REVIEW

For more practice, go to the Problem Bank in Appendix D. 9. What determines the order in which the compo-

Chemical Names nent elements of binary molecular compounds
and Formulas
are written?

SECTION 1 REVIEW 10. Name the following binary molecular com-

1. a. What are monatomic ions? pounds according to the prefix system. (Hint:

b. Give three examples of monatomic ions. See Sample Problem D.)

2. How does the chemical formula for the nitrite a. CO2 d. SeF6

ion differ from the chemical formula for the b. CCl4 e. As2O5

nitrate ion? c. PCl5

3. Using only the periodic table, write the symbol 11. Write formulas for each of the following binary

of the ion most typically formed by each of the molecular compounds. (Hint: See Sample

following elements: Problem D.)

a. K d. Cl a. carbon tetrabromide

b. Ca e. Ba b. silicon dioxide

c. S f. Br c. tetraphosphorus decoxide

4. Write the formula for and indicate the charge d. diarsenic trisulfide

on each of the following ions: 12. Distinguish between binary acids and oxyacids,

a. sodium ion and give two examples of each.

b. aluminum ion 13. a. What is a salt?

c. chloride ion b. Give two examples of salts.

d. nitride ion 14. Name each of the following acids:

e. iron(II) ion a. HF d. H2SO4

f. iron(III) ion b. HBr e. H3PO4

5. Name each of the following monatomic ions: c. HNO3

a. K+ d. Cl− 15. Give the molecular formula for each of the fol-

b. Mg2+ e. O2− lowing acids:

c. Al3+ f. Ca2+ a. sulfurous acid

6. Write formulas for the binary ionic compounds b. chloric acid

formed between the following elements. (Hint: c. hydrochloric acid

See Sample Problem A.) d. hypochlorous acid

a. sodium and iodine e. perchloric acid

b. calcium and sulfur f. carbonic acid

c. zinc and chlorine g. acetic acid

d. barium and fluorine

e. lithium and oxygen PRACTICE PROBLEMS

7. Give the name of each of the following binary 16. Write formulas for each of the following
compounds:
ionic compounds. (Hint: See Sample Problem B.) a. sodium fluoride
b. calcium oxide
a. KCl c. Li2O c. potassium sulfide
d. magnesium chloride
b. CaBr2 d. MgCl2 e. aluminum bromide
f. lithium nitride
8. Write the formulas for and give the names of g. iron(II) oxide

the compounds formed by the following ions:
a. Cr2+ and F−
b. Ni2+ and O2−
c. Fe3+ and O2−

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 251

CHAPTER REVIEW

17. Name each of the following ions: 24. Assign oxidation numbers to each atom in the
following compounds. (Hint: See Sample
a. NH4+ f. CO23− Problem E.)
b. ClO3− g. PO43− a. HI
c. OH− h. CH3COO− b. PBr3
d. SO42− i. HCO3− c. GeS2
e. NO3− j. CrO42− d. KH
e. As2O5
18. Write the formula and charge for each of the f. H3PO4

following ions: 25. Assign oxidation numbers to each atom in the
following ions. (Hint: See Sample Problem E.)
a. ammonium ion g. copper(II) ion a. NO−3
b. ClO−4
b. acetate ion h. tin(II) ion c. PO43−
d. Cr2O72−
c. hydroxide ion i. iron(III) ion e. CO32−

d. carbonate ion j. copper(I) ion Using Chemical Formulas

e. sulfate ion k. mercury(I) ion SECTION 3 REVIEW

f. phosphate ion l. mercury(II) ion 26. a. Define formula mass.
b. In what unit is formula mass expressed?
Oxidation Numbers
27. What is meant by the molar mass of a
SECTION 2 REVIEW compound?

19. Name each of the following ions according to PRACTICE PROBLEMS

the Stock system: d. Pb4+ 28. Determine the formula mass of each of the fol-
a. Fe2+ e. Sn2+ lowing compounds or ions. (Hint: See Sample
b. Fe3+ f. Sn4+ Problem F.)
c. Pb2+ a. glucose, C6H12O6
b. calcium acetate, Ca(CH3COO)2
20. Name each of the binary molecular compounds c. the ammonium ion, NH4+
d. the chlorate ion, ClO−3
in item 11 by using the Stock system.
29. Determine the number of moles of each type of
21. Write formulas for each of the following monatomic or polyatomic ion in one mole of
the following compounds. For each polyatomic
compounds: ion, determine the number of moles of each
atom present in one mole of the ion.
a. phosphorus(III) iodide a. KNO3
b. Na2SO4
b. sulfur(II) chloride c. Ca(OH)2
d. (NH4)2SO3
c. carbon(IV) sulfide e. Ca3(PO4)2
f. Al2(CrO4)3
d. nitrogen(V) oxide

22. a. What are oxidation numbers?

b. What useful functions do oxidation numbers

serve?

PRACTICE PROBLEMS

23. Name each of the following ionic compounds by
using the Stock system:
a. NaCl
b. KF
c. CaS
d. Co(NO3)2
e. FePO4
f. Hg2SO4
g. Hg3(PO4)2

252 C H A P T E R 7

CHAPTER REVIEW

30. Determine the molar mass of each compound MIXED REVIEW
listed in item 29. (Hint: See Sample Problem G.)
40. Chemical analysis shows that citric acid contains
31. Determine the number of moles of compound
in each of the following samples. (Hint: See 37.51% C, 4.20% H, and 58.29% O. What is the
Sample Problem I.)
a. 4.50 g H2O empirical formula for citric acid?
b. 471.6 g Ba(OH)2
c. 129.68 g Fe3(PO4)2 41. Name each of the following compounds by

32. Determine the percentage composition of each using the Stock system:
of the following compounds. (Hint: See Sample
Problem J.) a. LiBr f. Fe2O3
a. NaCl g. AgNO3
b. AgNO3 b. Sn(NO3)2 h. Fe(OH)2
c. Mg(OH)2 c. FeCl2 i. CrF2
d. MgO
33. Determine the percentage by mass of water in
the hydrate CuSO4•5H2O. (Hint: See Sample e. KOH
Problem K.)
42. What is the mass in grams of each of the follow-
Determining Chemical
Formulas ing samples?

SECTION 4 REVIEW a. 1.000 mol NaCl

34. What three types of information are used to b. 2.000 mol H2O
find an empirical formula from percentage c. 3.500 mol Ca(OH)2
composition data? d. 0.625 mol Ba(NO3)2
43. Determine the formula mass and molar mass of
35. What is the relationship between the empirical
formula and the molecular formula of a each of the following compounds:
compound?
a. XeF4
PRACTICE PROBLEMS b. C12H24O6
c. Hg2I2
36. Determine the empirical formula of a com- d. CuCN
pound containing 63.50% silver, 8.25% nitro-
gen, and 28.25% oxygen. (Hint: See Sample 44. Write the chemical formulas for the following
Problem L.)
compounds:
37. Determine the empirical formula of a com-
pound found to contain 52.11% carbon, 13.14% a. aluminum fluoride
hydrogen, and 34.75% oxygen.
b. magnesium oxide
38. What is the molecular formula of the molecule
that has an empirical formula of CH2O and a c. vanadium(V) oxide
molar mass of 120.12 g/mol?
d. cobalt(II) sulfide
39. A compound with a formula mass of 42.08 amu
is found to be 85.64% carbon and 14.36% e. strontium bromide
hydrogen by mass. Find its molecular formula.
f. sulfur trioxide

45. How many atoms of each element are con-

tained in a single formula unit of iron(III) for-

mate, Fe(CHO2)3•H2O? What percentage by
mass of the compound is water?

46. Name each of the following acids, and assign

oxidation numbers to the atoms in each:

a. HNO2 c. H2CO3

b. H2SO3 d. HI

47. Determine the percentage composition of the

following compounds:

a. NaClO

b. H2SO3
c. C2H5COOH
d. BeCl2

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 253

CHAPTER REVIEW

48. Name each of the following binary compounds: USING THE HANDBOOK

a. MgI2 e. SO2 53. Review the common reactions of Group 1
metals in the Elements Handbook, and answer
b. NaF f. PBr3 the following questions:
a. Some of the Group 1 metals react with oxy-
c. CS2 g. CaCl2 gen to form superoxides. Write the formulas
for these compounds.
d. N2O4 h. AgI b. What is the charge on each cation for the
formulas that you wrote in (a)?
49. Assign oxidation numbers to each atom in the c. How does the charge on the anion vary for
oxides, peroxides, and superoxides?
following molecules and ions:
54. Review the common reactions of Group 2
a. CO2 e. H2O2 metals in the Elements Handbook, and answer
b. NH+4 the following questions:
c. MnO4− f. P4O10 a. Some of the Group 2 metals react with oxy-
gen to form oxides. Write the formulas for
d. S2O32− g. OF2 these compounds.
b. Some of the Group 2 metals react with oxy-
50. A 175.0 g sample of a compound contains gen to form peroxides. Write the formulas for
these compounds.
56.15 g C, 9.43 g H, 74.81 g O, 13.11 g N, and c. Some of the Group 2 metals react with nitro-
gen to form nitrides. Write the formulas for
21.49 g Na. What is the compound’s empirical these compounds.
d. Most Group 2 elements form hydrides.
formula? What is hydrogen’s oxidation state in these
compounds?
CRITICAL THINKING
55. Review the analytical tests for transition metals
51. Analyzing Information Sulfur trioxide is pro- in the Elements Handbook, and answer the
duced in the atmosphere through a reaction of following questions:
sulfur dioxide and oxygen. Sulfur dioxide is a a. Determine the oxidation state of each metal
primary air pollutant. Analyze the formula for in the precipitates shown for cadmium, zinc,
sulfur trioxide. Then, list all of the chemical and lead.
information from the analysis that you can. b. Determine the oxidation state of each metal
in the complex ions shown for iron, man-
52. Analyzing Data In the laboratory, a sample of ganese, and cobalt.
pure nickel was placed in a clean, dry, weighed c. The copper compound shown is called a
crucible. The crucible was heated so that the coordination compound. The ammonia
nickel would react with the oxygen in the air. shown in the formula exists as molecules that
After the reaction appeared complete, the cru- do not have a charge. Determine copper’s
cible was allowed to cool and the mass was oxidation state in this compound.
determined. The crucible was reheated and
allowed to cool. Its mass was then determined
again to be certain that the reaction was com-
plete. The following data were collected:

Mass of crucible = 30.02 g
Mass of nickel and crucible = 31.07 g
Mass of nickel oxide and crucible = 31.36 g

Determine the following information based on
the data given above:

Mass of nickel =
Mass of nickel oxide =
Mass of oxygen =

Based on your calculations, what is the empiri-
cal formula for the nickel oxide?

254 C H A P T E R 7

CHAPTER REVIEW

56. Review the common reactions of Group 15 ele- ALTERNATIVE ASSESSMENT
ments in the Elements Handbook, and answer
the following questions: 59. Performance Assessment Your teacher will
a. Write formulas for each of the oxides listed supply you with a note card that has one of the
for the Group 15 elements. following formulas on it: NaCH3COO•3H2O,
b. Determine nitrogen’s oxidation state in the MgCl2•6H2O, LiC2H3O2•2H2O, or MgSO4•7H2O.
oxides listed in (a). Design an experiment to determine the percent-
age of water by mass in the hydrated salt
RESEARCH & WRITING assigned to you. Be sure to explain what steps
you will take to ensure that the salt is com-
57. Nomenclature Biologists who name newly dis- pletely dry. If your teacher approves your
covered organisms use a system that is struc- design, obtain the salt and perform the experi-
tured very much like the one used by chemists ment. What percentage of water does the salt
in naming compounds. The system used by bi- contain?
ologists is called the Linnaeus system, after its
creator, Carolus Linnaeus. Research this system 60. Both ammonia, NH3, and ammonium nitrate,
in a biology textbook, and then note similarities NH4NO3, are used in fertilizers as a source of
and differences between the Linnaeus system nitrogen. Which compound has the higher per-
and chemical nomenclature. centage of nitrogen? Research the physical
properties of both compounds, and find out how
58. Common Chemicals Find out the systematic each compound is manufactured and used.
Explain why each compound has its own partic-
chemical name and write the chemical formula ular application. (Consider factors such as the
cost of raw ingredients, the ease of manufacture,
for each of the following common compounds: and shipping costs.)

a. baking soda d. limestone

b. milk of magnesia e. lye

c. Epsom salts f. wood alcohol

Graphing Calculator Calculating Molar
Mass
Go to go.hrw.com for a graphing calculator
exercise that asks you to calculate the molar
mass of a compound.

Keyword: HC6FRMX

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 255

Math Tutor CALCULATING PERCENTAGE COMPOSITION

Chemists can analyze an unknown substance by determining its percentage composi-
tion by mass. Percentage composition is determined by finding the mass of each ele-
ment in a sample of the substance as a percentage of the mass of the whole sample.
The results of this analysis can then be compared with the percentage composition of
known compounds to determine the probable identity of the unknown substance.
Once you know a compound’s formula, you can determine its percentage composition
by mass.

SAMPLE 1 SAMPLE 2

Determine the percentage composition of potas- Determine the percentage of nitrogen in ammo-
sium chlorate, KClO3.
nium sulfate, (NH4)2SO4.
First, calculate the molar mass of KClO3. The Even though you want to find the percentage of
formula shows you that one mole of KClO3 con-
sists of 1 mol K atoms, 1 mol Cl atoms, and 3 mol only one element, you must calculate the molar
O atoms. Thus, the molar mass of KClO3 is molar
mass K + molar mass Cl + 3(molar mass O) = mass of (NH4)2SO4. To do that, examine the for-
39.10 g K + 35.45 g Cl + 3(16.00 g O). mula to find the number of moles of each element

molar mass KClO3 = 122.55 g in the compound. The two ammonium groups, indi-

The percentage composition of KClO3 is deter- cated by (NH4)2, contain 2 mol N and 8 mol H per
mined by calculating the percentage of the total mole of (NH4)2SO4. The sulfate group, SO4, con-
molar mass contributed by each element. tains 1 mol S and 4 mol O per mole of (NH4)2SO4.

2 mol N = 2 × 14.01 g = 28.02 g

8 mol H = 8 × 1.01 g = 8.08 g

1 mol S = 1 × 32.07 = 32.07 g

⎯mass of⎯element⎯in 1 mo⎯l of com⎯pound × 100 4 mol O = 4 × 16.00 = 64.00 g
molar mass of compound
molar mass (NH4)2SO4 = 132.17 g

= % element in compound Now, you can determine the percentage of

⎯39.10⎯g K nitrogen in the compound as follows.
122.55 g KClO3
% K in KClO3 = × 100 = 31.91% % N in (NH4)2SO4 = ⎯28⎯.02 g N⎯
132.17 g (NH4)2SO4

% Cl in KClO3 = ⎯35.45⎯g Cl × 100 = 28.93% × 100 = 21.20%
122.55 g KClO3

% O in KClO3 = ⎯48.00⎯g O × 100 = 39.17%
122.55 g KClO3

PRACTICE PROBLEMS 2. What is the percentage of iodine in zinc iodate,
Zn(IO3)2?
1. What is the percentage composition of sodium
carbonate, Na2CO3?

256 C H A P T E R 7

Standardized Test Prep

Answer the following items on a separate piece of paper. 7.Which of the following shows the percentage
composition of H2SO4?
MULTIPLE CHOICE A. 2.5% H, 39.1% S, 58.5% O
B. 2.1% H, 32.7% S, 65.2% O
1.Which of the following compounds does not C. 28.6% H, 14.3% S, 57.1% O
contain a polyatomic ion? D. 33.3% H, 16.7% S, 50% O
A. sodium carbonate
B. sodium sulfate 8.Which of the following compounds has the
C. sodium sulfite highest percentage of oxygen?
D. sodium sulfide A. CH4O
B. CO2
2.The correct formula for ammonium phosphate C. H2O
is D. Na2CO3
A. (NH4)3PO4.
B. (NH4)2PO4. 9.The empirical formula for a compound that is
C. NH4PO4. 1.2% H, 42.0% Cl, and 56.8% O is
D. NH4(PO4)2. A. HClO.
B. HClO2.
3.When writing the formula for a compound that C. HClO3.
contains a polyatomic ion, D. HClO4.
A. write the anion’s formula first.
B. use superscripts to show the number of poly- SHORT ANSWER
atomic ions present.
C. use parentheses if the number of polyatomic 10.When a new substance is synthesized or is dis-
ions is greater than 1. covered experimentally, the substance is ana-
D. always place the polyatomic ion in lyzed quantitatively. What information is
parentheses. obtained from this typical analysis, and how is
this information used?
4. The correct name for NH4CH3COO is
A. ammonium carbonate. 11.An oxide of selenium is 28.8% O. Find the
B. ammonium hydroxide. empirical formula. Assuming that the empirical
C. ammonium acetate. formula is also the molecular formula, name the
D. ammonium nitrate. oxide.

5.Which of the following is the correct formula EXTENDED RESPONSE
for iron(III) sulfate?
A. Fe3SO4 12.What is an empirical formula, and how does it
B. Fe3(SO4)2 differ from a molecular formula?
C. Fe2(SO4)3
D. 3FeSO4 13.What are Stock system names based on?

6.The molecular formula for acetylene is C2H2. Whenever possible, highlight or
The molecular formula for benzene is C6H6. underline numbers or words critical to answering a
The empirical formula for both is question correctly.
A. CH.
B. C2H2.
C. C6H6.
D. (CH)2.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 257

CHAPTER LAB SEE PRE-LAB

GRAVIMETRIC ANALYSIS

Determining the Empirical
Formula of Magnesium Oxide

OBJECTIVES BACKGROUND

• Measure the mass of magnesium oxide. This gravimetric analysis involves the combustion
of magnesium metal in air to synthesize magnesium
• Perform a synthesis reaction by using oxide. The mass of the product is greater than the
gravimetric techniques. mass of magnesium used because oxygen reacts with
the magnesium metal. As in all gravimetric analyses,
• Determine the empirical formula of success depends on attaining a product yield near
magnesium oxide. 100%. Therefore, the product will be heated and
cooled and have its mass measured until two of these
• Calculate the class average and standard mass measurements are within 0.02% of one another.
deviation for moles of oxygen used. When the masses of the reactant and product have
been carefully measured, the amount of oxygen used
MATERIALS • crucible and lid, in the reaction can be calculated. The ratio of oxygen
metal or ceramic to magnesium can then be established, and the empir-
• 15 cm magnesium ical formula of magnesium oxide can be determined.
ribbon, 2 • crucible tongs
SAFETY
• 25 mL beaker • distilled water

• Bunsen burner • eyedropper or
assembly micropipet

• clay triangle • ring stand

For review of safety, please see Safety in the
Chemistry Laboratory in the front of your book.

PREPARATION

1. Copy the following data table in your lab
notebook.

DATA TABLE

Trial 1 Trial 2

1. Mass of crucible, lid, and metal (g)
2. Mass of crucible, lid, and product (g)
3. Mass of crucible and lid (g)

FIGURE A

258 C H A P T E R 7

PROCEDURE 10. Turn off the burner. Cool the crucible, lid, and
1. Construct a setup for heating a crucible as contents to room temperature. Measure the mass
shown in Figure A and as demonstrated in the of the crucible, lid, and product. Record the
Pre-Laboratory Procedure “Gravimetric measurement in the margin of your data table.
Analysis.”
11. Replace the crucible, lid, and contents on the
2. Heat the crucible and lid for 5 min to burn off clay triangle, and reheat for another 2 min.
any impurities. Cool to room temperature, and remeasure the
mass of the crucible, lid, and contents. Compare
3. Cool the crucible and lid to room temperature. this mass measurement with the measurement
Measure their combined mass, and record the obtained in step 10. If the new mass is ±0.02%
measurement on line 3 of your data table. of the mass in step 10, record the new mass on
line 2 of your data table and go on to step 12. If
NOTE: Handle the crucible and lid with crucible not, your reaction is still incomplete, and you
tongs at all times during this experiment. Such should repeat step 11.
handling prevents burns and the transfer of dirt
and oil from your hands to the crucible and lid. 12. Clean the crucible, and repeat steps 2–11 with a
second strip of magnesium ribbon. Record your
4. Polish a 15 cm strip of magnesium with steel measurements under Trial 2 in your data table.
wool. The magnesium should be shiny. Cut the
strip into small pieces to make the reaction pro- CLEANUP AND DISPOSAL
ceed faster, and place the pieces in the crucible. 13. Put the solid magnesium oxide in the

5. Cover the crucible with the lid, and measure designated waste container. Return
the mass of the crucible, lid, and metal. Record any unused magnesium ribbon to your
the measurement on line 1 of your data table. teacher. Clean your equipment and lab station.
Thoroughly wash your hands after completing
6. Use tongs to replace the crucible on the clay the lab session and cleanup.
triangle. Heat the covered crucible gently. Lift
the lid occasionally to allow air in. ANALYSIS AND INTERPRETATION
1. Applying Ideas: Calculate the mass of the
CAUTION: Do not look directly at the burning magnesium metal and the mass of the product.
magnesium metal. The brightness of the light
can blind you. 2. Evaluating Data: Determine the mass of the
oxygen consumed.
7. When the magnesium appears to be fully re-
acted, partially remove the crucible lid and 3. Applying Ideas: Calculate the number of
continue heating for 1 min. moles of magnesium and the number of moles
of oxygen in the product.
8. Remove the burner from under the crucible.
After the crucible has cooled, use an eyedropper CONCLUSIONS
to carefully add a few drops of water to decom- 1. Inferring Relationships: Determine the
pose any nitrides that may have formed. empirical formula for magnesium oxide, MgxOy.

CAUTION: Use care when adding water. Using
too much water can cause the crucible to crack.

9. Cover the crucible completely. Replace the
burner under the crucible, and continue heating
for about 30 to 60 s.

C H E M I C A L F O R M U L A S A N D C H E M I C A L C O M P O U N D S 259

CHAPTER 8

Chemical Equations
and Reactions

The evolution of energy as light and heat is an indication
that a chemical reaction is taking place.

Fireworks

Describing Chemical SECTION 1
Reactions
OBJECTIVES
A chemical reaction is the process by which one or more substances
List three observations that
are changed into one or more different substances. In any chemical suggest that a chemical
reaction, the original substances are known as the reactants and the reaction has taken place.
resulting substances are known as the products. According to the law of
conservation of mass, the total mass of reactants must equal the total List three requirements for
mass of products for any given chemical reaction. a correctly written chemical
equation.
Chemical reactions are described by chemical equations. A chemical
equation represents, with symbols and formulas, the identities and Write a word equation and a
relative molecular or molar amounts of the reactants and products in a formula equation for a given
chemical reaction. For example, the following chemical equation shows chemical reaction.
that the reactant ammonium dichromate yields the products nitrogen,
chromium(III) oxide, and water. Balance a formula equation
by inspection.
(NH4)2Cr2O7(s) ⎯→ N2(g) + Cr2O3(s) + 4H2O(g)

This strongly exothermic reaction is shown in Figure 1.

Indications of a Chemical Reaction

To know for certain that a chemical reaction has taken place requires FIGURE 1 The decomposition
evidence that one or more substances have undergone a change in iden- of ammonium dichromate proceeds
tity. Absolute proof of such a change can be provided only by chemical rapidly, releasing energy in the form
analysis of the products. However, certain easily observed changes usu- of light and heat.
ally indicate that a chemical reaction has occurred.
1. Evolution of energy as heat and light. A change in matter that re-

leases energy as both heat and light is strong evidence that a chemi-
cal reaction has taken place. For example, you can see in Figure 1
that the decomposition of ammonium dichromate is accompanied by
the evolution of energy as heat and light. And you can see evidence
that a chemical reaction occurs between natural gas and oxygen if
you burn gas for cooking in your house. Some reactions involve only
heat or only light. But heat or light by itself is not necessarily a sign
of chemical change, because many physical changes also involve
either heat or light.

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 261

FIGURE 2 (a) The reaction of
vinegar and baking soda is evidenced
by the production of bubbles of car-
bon dioxide gas. (b) When water
solutions of ammonium sulfide
and cadmium nitrate are combined,
the yellow precipitate cadmium
sulfide forms.

www.scilinks.org (a) (b)
Topic: Chemical Reactions
Code: HC60274 2. Production of a gas. The evolution of gas bubbles when two sub-
stances are mixed is often evidence of a chemical reaction. For exam-
262 C H A P T E R 8 ple, bubbles of carbon dioxide gas form immediately when baking
soda is mixed with vinegar, in the vigorous reaction that is shown in
Figure 2a.

3. Formation of a precipitate. Many chemical reactions take place
between substances that are dissolved in liquids. If a solid appears
after two solutions are mixed, a reaction has likely occurred. A solid
that is produced as a result of a chemical reaction in solution and that
separates from the solution is known as a precipitate. A precipitate-
forming reaction is shown in Figure 2b.

4. Color change. A change in color is often an indication of a chemical
reaction.

Characteristics of Chemical Equations

A properly written chemical equation can summarize any chemical
change. The following requirements will aid you in writing and reading
chemical equations correctly.
1. The equation must represent known facts. All reactants and products

must be identified, either through chemical analysis in the labora-
tory or from sources that give the results of experiments.
2. The equation must contain the correct formulas for the reactants and
products. Remember what you learned in Chapter 7 about symbols
and formulas. Knowledge of the common oxidation states of the ele-
ments and of methods of writing formulas will enable you to supply
formulas for reactants and products if they are not available. Recall
that the elements listed in Table 1 exist primarily as diatomic mol-
ecules, such as H2 and O2. Each of these elements is represented in an
equation by its molecular formula. Other elements in the elemental
state are usually represented simply by their atomic symbols. For
example, iron is represented as Fe and carbon is represented as C.The
symbols are not given any subscripts because the elements do not

TABLE 1 Elements That Normally Exist as Diatomic
Molecules

Physical state at
Element Symbol Molecular formula room temperature

Hydrogen H H2 gas
Nitrogen N N2 gas
Oxygen O O2 gas
Fluorine F F2 gas
Chlorine Cl Cl2 gas
Bromine Br Br2 liquid
Iodine I I2 solid

form definite molecular structures. Two exceptions to this rule are Chemistry in Action
sulfur, which is usually written S8, and phosphorus, which is usually
written P4. In these cases, the formulas reflect each element’s unique Go to go.hrw.com for a full-length
atomic arrangement in its natural state. article on water treatment and our
3. The law of conservation of mass must be satisfied. Atoms are neither public water supply.
created nor destroyed in ordinary chemical reactions. Therefore, the
same number of atoms of each element must appear on each side of Keyword: HC6RXNX
a correct chemical equation. To balance numbers of atoms, add coef-
ficients where necessary. A coefficient is a small whole number that
appears in front of a formula in a chemical equation. Placing a coef-
ficient in front of a formula specifies the relative number of moles of
the substance; if no coefficient is written, the coefficient is assumed
to be 1. For example, the coefficient 4 in the equation on page 261
indicates that 4 mol of water are produced for each mole of nitrogen
and chromium(III) oxide that is produced.

Word and Formula Equations

The first step in writing a chemical equation is to identify the facts to be
represented. It is often helpful to write a word equation, an equation in
which the reactants and products in a chemical reaction are represented
by words. A word equation has only qualitative (descriptive) meaning.
It does not give the whole story because it does not give the quantities
of reactants used or products formed.

Consider the reaction of methane, the principal component of natu-
ral gas, with oxygen. When methane burns in air, it combines with oxy-
gen to produce carbon dioxide and water vapor. In the reaction,
methane and oxygen are the reactants, and carbon dioxide and water
are the products. The word equation for the reaction of methane and
oxygen is written as follows.

methane + oxygen ⎯→ carbon dioxide + water

The arrow, ⎯→, is read as react to yield or yield (also produce or form).
So the equation above is read, “methane and oxygen react to yield

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 263

carbon dioxide and water,” or simply, “methane and oxygen yield car-
bon dioxide and water.”

The next step in writing a correct chemical equation is to replace the
names of the reactants and products with appropriate symbols and for-
mulas. Methane is a molecular compound composed of one carbon
atom and four hydrogen atoms. Its chemical formula is CH4. Recall that
oxygen exists in nature as diatomic molecules; it is therefore represented
as O2. The correct formulas for carbon dioxide and water are CO2 and
H2O, respectively.

A formula equation represents the reactants and products of a chemi-
cal reaction by their symbols or formulas. The formula equation for the
reaction of methane and oxygen is written as follows.

CH4(g) + O2(g) ⎯→ CO2(g) + H2O(g) (not balanced)

The g in parentheses after each formula indicates that the correspond-
ing substance is in the gaseous state. Like a word equation, a formula
equation is a qualitative statement. It gives no information about the
amounts of reactants or products.

A formula equation meets two of the three requirements for a cor-
rect chemical equation. It represents the facts and shows the correct
symbols and formulas for the reactants and products. To complete the
process of writing a correct equation, the law of conservation of mass
must be taken into account. The relative amounts of reactants and prod-
ucts represented in the equation must be adjusted so that the numbers
and types of atoms are the same on both sides of the equation. This
process is called balancing an equation and is carried out by inserting
coefficients. Once it is balanced, a formula equation is a correctly writ-
ten chemical equation.

Look again at the formula equation for the reaction of methane and
oxygen.

CH4(g) + O2(g) ⎯→ CO2(g) + H2O(g) (not balanced)

To balance the equation, begin by counting atoms of elements that are
combined with atoms of other elements and that appear only once on
each side of the equation. In this case, we could begin by counting either
carbon or hydrogen atoms. Usually, the elements hydrogen and oxygen
are balanced only after balancing all other elements in an equation.
(You will read more about the rules of balancing equations later in the
chapter.) Thus, we begin by counting carbon atoms.

Inspecting the formula equation reveals that there is one carbon atom
on each side of the arrow. Therefore, carbon is already balanced in the
equation. Counting hydrogen atoms reveals that there are four hydrogen
atoms in the reactants but only two in the products.Two additional hydro-
gen atoms are needed on the right side of the equation. They can be
added by placing the coefficient 2 in front of the chemical formula H2O.

CH4(g) + O2(g) ⎯→ CO2(g) + 2H2O(g) (partially balanced)

264 C H A P T E R 8

A coefficient multiplies the number of atoms of each element indi- +
cated in a chemical formula. Thus, 2H2O represents four H atoms and
two O atoms. To add two more hydrogen atoms to the right side of the
equation, one may be tempted to change the subscript in the formula of
water so that H2O becomes H4O. However, this would be a mistake
because changing the subscripts of a chemical formula changes the
identity of the compound. H4O is not a product in the combustion of
methane. In fact, there is no such compound. One must use only coeffi-
cients to change the relative number of atoms in a chemical equation
because coefficients change the numbers of atoms without changing the
identities of the reactants or products.

Now consider the number of oxygen atoms. There are four oxygen
atoms on the right side of the arrow in the partially balanced equation.
Yet there are only two oxygen atoms on the left side of the arrow. One
can increase the number of oxygen atoms on the left side to four by
placing the coefficient 2 in front of the molecular formula for oxygen.
This results in a correct chemical equation, or balanced formula equa-
tion, for the burning of methane in oxygen.

CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(g)

This reaction is further illustrated in Figure 3.

Additional Symbols Used in Chemical Equations

Table 2 on the next page summarizes the symbols commonly used in
chemical equations. Sometimes a gaseous product is indicated by an
arrow pointing upward, ↑, instead of (g), as shown in the table. A down-
ward arrow, ↓, is often used to show the formation of a precipitate dur-
ing a reaction in solution.

The conditions under which a reaction takes place are often indicated
by placing information above or below the reaction arrow.The word heat,

(a)
(b)

+

Molecules CH4 + 2O2 CO2 + 2H2O
+ (4O) (1C, 2O) + (4H, 2O)
{
{

Atoms (1C, 4H)

FIGURE 3 (a) In a Bunsen burner, methane combines with oxygen in the air to form
carbon dioxide and water vapor. (b) The reaction is represented by both a molecular
model and a balanced equation. Each shows that the number of atoms of each element in
the reactants equals the number of atoms of each element in the products.

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 265

TABLE 2 Symbols Used in Chemical Equations

Symbol Explanation

⎯→ “Yields”; indicates result of reaction

←⎯⎯→ Used in place of a single arrow to indicate a
reversible reaction

(s) A reactant or product in the solid state; also
used to indicate a precipitate

Alternative to (s), but used only to indicate a
↓ precipitate

(l) A reactant or product in the liquid state

(aq) A reactant or product in an aqueous solution
(dissolved in water)

(g) A reactant or product in the gaseous state

Alternative to (g), but used only to indicate a
↑ gaseous product

⎯⎯Δ → or ⎯h⎯eat→ Reactants are heated
⎯2 ⎯atm→
Pressure at which reaction is carried out, in
this case 2 atm

⎯pr⎯ess⎯ure→ Pressure at which reaction is carried out
exceeds normal atmospheric pressure

⎯0⎯°C→ Temperature at which reaction is carried out,
in this case 0°C

⎯M⎯nO→2 Formula of catalyst, in this case manganese
dioxide, used to alter the rate of the reaction

symbolized by a Greek capital delta, Δ, indicates that the reactants must
be heated. The specific temperature at which a reaction occurs may also
be written over the arrow. For some reactions, it is important to specify
the pressure at which the reaction occurs or to specify that the pressure
must be above normal. Many reactions are speeded up and can take
place at lower temperatures in the presence of a catalyst. A catalyst is a
substance that changes the rate of a chemical reaction but can be recov-
ered unchanged. To show that a catalyst is present, the formula for the
catalyst or the word catalyst is written over the reaction arrow.

In many reactions, as soon as the products begin to form, they imme-
diately begin to react with each other and re-form the reactants. In
other words, the reverse reaction also occurs. The reverse reaction may
occur to a greater or lesser degree than the original reaction, depend-
ing on the specific reaction and the conditions. A reversible
reaction is a chemical reaction in which the products re-form the original

266 C H A P T E R 8

reactants. The reversibility of a reaction is indicated by writing two
arrows pointing in opposite directions. For example, the reversible reac-
tion between iron and water vapor is written as follows.

3Fe(s) + 4H2O(g) ←⎯⎯→ Fe3O4(s) + 4H2(g)

With an understanding of all the symbols and formulas used, it is pos-
sible to translate a chemical equation into a sentence. Consider the fol-
lowing equation.

2HgO(s) ⎯⎯Δ → 2Hg(l) + O2(g)

Translated into a sentence, this equation reads, “When heated, solid
mercury(II) oxide yields liquid mercury and gaseous oxygen.”

It is also possible to write a chemical equation from a sentence describ-
ing a reaction. Consider the sentence, “Under pressure and in the pres-
ence of a platinum catalyst, gaseous ethene and hydrogen form gaseous
ethane.” This sentence can be translated into the following equation.

C2H4(g) + H2(g) ⎯pre⎯s⎯sur⎯e, P→t C2H6(g)

Throughout this chapter we will often include the symbols for physi-
cal states (s, l, g, and aq) in balanced formula equations. You should be
able to interpret these symbols when they are used and to supply them
when the necessary information is available.

SAMPLE PROBLEM A For more help, go to the Math Tutor at the end of this chapter.

Write word and formula equations for the chemical reaction that occurs when solid sodium oxide is added to
water at room temperature and forms sodium hydroxide (dissolved in the water). Include symbols for physi-
cal states in the formula equation. Then balance the formula equation to give a balanced chemical equation.

SOLUTION The word equation must show the reactants, sodium oxide and water, to the left of the
arrow. The product, sodium hydroxide, must appear to the right of the arrow.

sodium oxide + water ⎯→ sodium hydroxide

The word equation is converted to a formula equation by replacing the name of each com-
pound with the appropriate chemical formula. To do this requires knowing that sodium has
an oxidation state of +1, that oxygen usually has an oxidation state of −2, and that a hydrox-
ide ion has a charge of 1−.

Na2O + H2O ⎯→ NaOH (not balanced)

Adding symbols for the physical states of the reactants and products and the coefficient 2 in
front of NaOH produces a balanced chemical equation.

Na2O(s) + H2O(l) ⎯→ 2NaOH(aq) 267

CHEMICAL EQUATIONS AND REACTIONS

SAMPLE PROBLEM B
Translate the following chemical equation into a sentence:

BaCl2(aq) + Na2CrO4(aq) ⎯→ BaCrO4(s) + 2NaCl(aq)

SOLUTION Each reactant is an ionic compound and is named according to the rules for such com-
pounds. Both reactants are in aqueous solution. One product is a precipitate and the other
remains in solution. The equation is translated as follows: Aqueous solutions of barium
chloride and sodium chromate react to produce a precipitate of barium chromate plus sodi-
um chloride in aqueous solution.

PRACTICE Answers in Appendix E

1. Write word and balanced chemical equations for the following
reactions. Include symbols for physical states when indicated.
a. Solid calcium reacts with solid sulfur to produce solid calcium
sulfide.

b. Hydrogen gas reacts with fluorine gas to produce hydrogen flu-
oride gas. (Hint: See Table 1.)

c. Solid aluminum metal reacts with aqueous zinc chloride to pro-
duce solid zinc metal and aqueous aluminum chloride.

2. Translate the following chemical equations into sentences: Go to go.hrw.com for
a. CS2(l) + 3O2(g) ⎯→ CO2(g) + 2SO2(g) more practice problems
that ask you to write
b. NaCl(aq) + AgNO3(aq) ⎯→ NaNO3(aq) + AgCl(s) balanced chemical
equations.
3. Hydrazine, N2H4, is used as rocket fuel. Hydrazine reacts violently
with oxygen to produce gaseous nitrogen and water. Write the Keyword: HC6RXNX
balanced chemical equation.

Significance of a Chemical Equation

Chemical equations are very useful in doing quantitative chemical
work. The arrow in a balanced chemical equation is like an equal sign.
And the chemical equation as a whole is similar to an algebraic equa-
tion in that it expresses an equality. Let’s examine some of the quanti-
tative information revealed by a chemical equation.
1. The coefficients of a chemical reaction indicate relative, not absolute,

amounts of reactants and products. A chemical equation usually
shows the smallest numbers of atoms, molecules, or ions that will sat-
isfy the law of conservation of mass in a given chemical reaction.

268 C H A P T E R 8

Consider the equation for the formation of hydrogen chloride from
hydrogen and chlorine.

H2(g) + Cl2(g) ⎯→ 2HCl(g)

The equation indicates that 1 molecule of hydrogen reacts with 1
molecule of chlorine to produce 2 molecules of hydrogen chloride,
giving the following molecular ratio of reactants and products.

1 molecule H2 : 1 molecule Cl2 : 2 molecules HCl

This ratio shows the smallest possible relative amounts of the reac-
tion’s reactants and products. To obtain larger relative amounts, we
simply multiply each coefficient by the same number. Thus, 20 mol-
ecules of hydrogen would react with 20 molecules of chlorine to yield
40 molecules of hydrogen chloride. The reaction can also be consid-
ered in terms of amounts in moles: 1 mol of hydrogen molecules
reacts with 1 mol of chlorine molecules to yield 2 mol of hydrogen
chloride molecules.
2. The relative masses of the reactants and products of a chemical reac-
tion can be determined from the reaction’s coefficients. Recall from
Figure 4 in Chapter 7 that an amount of an element or compound in
moles can be converted to a mass in grams by multiplying by the
appropriate molar mass. We know that 1 mol of hydrogen reacts with
1 mol of chlorine to yield 2 mol of hydrogen chloride. The relative
masses of the reactants and products are calculated as follows.

2.02 g H2
1 mol H2 × mol H2 = 2.02 g H2

1 mol Cl2 × 70.90 g Cl2 = 70.90 g Cl2
mol Cl2

36.46 g HCl FIGURE 4 This representation
2 mol HCl × mol HCl = 72.92 g HCl of the reaction of hydrogen and
chlorine to yield hydrogen chloride
The chemical equation shows that 2.02 g of hydrogen will react with shows several ways to interpret the
70.90 g of chlorine to yield 72.92 g of hydrogen chloride. quantitative information of a chemi-
cal reaction.
H2 + Cl2
2 HCl

++

1 molecule H2 1 molecule Cl2 2 molecules HCl
1 mol H2 1 mol Cl2 2 mol HCl
2.02 g H2 70.90 g Cl2
2 ϫ 36.46 g ϭ 72.92 g HCl

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 269

3. The reverse reaction for a chemical equation has the same relative
amounts of substances as the forward reaction. Because a chemical
equation is like an algebraic equation, the equality can be read in
either direction. Reading the hydrogen chloride formation equation
on the previous page from right to left, we can see that 2 molecules of
hydrogen chloride break down to form 1 molecule of hydrogen plus
1 molecule of chlorine. Similarly, 2 mol (72.92 g) of hydrogen chlo-
ride yield 1 mol (2.02 g) of hydrogen and 1 mol (70.90 g) of chlorine.

We have seen that a chemical equation provides useful quantitative
information about a chemical reaction. However, there is also impor-
tant information that is not provided by a chemical equation. For
instance, an equation gives no indication of whether a reaction will actu-
ally occur. A chemical equation can be written for a reaction that may
not even take place. Some guidelines about the types of simple reac-
tions that can be expected to occur are given in Sections 2 and 3. And
later chapters provide additional guidelines for other types of reactions.
In all these guidelines, it is important to remember that experimenta-
tion forms the basis for confirming that a particular chemical reaction
will occur.

In addition, chemical equations give no information about the speed
at which reactions occur or about how the bonding between atoms or
ions changes during the reaction. These aspects of chemical reactions
are discussed in Chapter 17.

FIGURE 5 When an electric cur- Balancing Chemical Equations
rent is passed through water that has
been made slightly conductive, the Most of the equations in the remainder of this chapter can be balanced
water molecules break down to yield by inspection. The following procedure demonstrates how to master
hydrogen (in tube at right) and oxy- balancing equations by inspection using a step-by-step approach. The
gen (in tube at left). Bubbles equation for the decomposition of water (see Figure 5) will be used as
of each gas are evidence of the an example.
reaction. Note that twice as much 1. Identify the names of the reactants and the products, and write a word
hydrogen as oxygen is produced.
equation. The word equation for the reaction shown in Figure 5 is
written as follows.

water ⎯→ hydrogen + oxygen

2. Write a formula equation by substituting correct formulas for the
names of the reactants and the products. We know that the formula
for water is H2O. And recall that both hydrogen and oxygen exist as
diatomic molecules. Therefore, their correct formulas are H2 and O2,
respectively.

H2O(l) ⎯→ H2(g) + O2(g) (not balanced)

270 C H A P T E R 8

3. Balance the formula equation according to the law of conservation
of mass. This last step is done by trial and error. Coefficients are
changed and the numbers of atoms are counted on both sides of the
equation. When the numbers of each type of atom are the same for
both the products and the reactants, the equation is balanced. The
trial-and-error method of balancing equations is made easier by the
use of the following guidelines.

• Balance the different types of atoms one at a time.
• First balance the atoms of elements that are combined and that

appear only once on each side of the equation.
• Balance polyatomic ions that appear on both sides of the equa-

tion as single units.
• Balance H atoms and O atoms after atoms of all other elements

have been balanced.

The formula equation in our example shows that there are two oxygen
atoms on the right and only one on the left. To balance oxygen atoms,
the number of H2O molecules must be increased. Placing the coefficient
2 before H2O gives the necessary two oxygen atoms on the left.

2H2O(l) ⎯→ H2(g) + O2(g) (partially balanced)

The coefficient 2 in front of H2O has upset the balance of hydrogen atoms.
Placing the coefficient 2 in front of hydrogen, H2, on the right, gives an
equal number of hydrogen atoms (4) on both sides of the equation.

2H2O(l) ⎯→ 2H2(g) + O2(g)

4. Count atoms to be sure that the equation is balanced. Make sure that
equal numbers of atoms of each element appear on both sides of the
arrow.

2H2O(l) ⎯→ 2H2(g) + O2(g)
(4H + 2O) = (4H) + (2O)

Occasionally at this point, the coefficients do not represent the smallest
possible whole-number ratio of reactants and products.When this hap-
pens, the coefficients should be divided by their greatest common fac-
tor in order to obtain the smallest possible whole-number coefficients.

Balancing chemical equations by inspection becomes easier as
you gain experience. Learn to avoid the most common mistakes:
(1) writing incorrect chemical formulas for reactants or products and
(2) trying to balance an equation by changing subscripts. Remember
that subscripts cannot be added, deleted, or changed. Eventually, you
will probably be able to skip writing the word equation and each sep-
arate step. However, do not leave out the final step of counting atoms
to be sure the equation is balanced.

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 271

SAMPLE PROBLEM C For more help, go to the Math Tutor at the end of this chapter.

The reaction of zinc with aqueous hydrochloric acid produces a solution of zinc
chloride and hydrogen gas. This reaction is shown at right in Figure 6. Write a
balanced chemical equation for the reaction.

SOLUTION
1 ANALYZE Write the word equation.

zinc + hydrochloric acid ⎯→ zinc chloride + hydrogen

2 PLAN Write the formula equation.

3 COMPUTE Zn(s) + HCl(aq) ⎯→ ZnCl2(aq) + H2(g) (not balanced)

Adjust the coefficients. Note that chlorine and hydrogen each FIGURE 6 Solid zinc
appear only once on each side of the equation. We balance reacts with hydrochloric
chlorine first because it is combined on both sides of the equa- acid to form aqueous zinc
tion. Also, recall from the guidelines on the previous page that chloride and hydrogen gas.
hydrogen and oxygen are balanced only after all other elements
in the reaction are balanced. To balance chlorine, we place the
coefficient 2 before HCl. Two molecules of hydrogen chloride
also yield the required two hydrogen atoms on the right. Finally,
note that there is one zinc atom on each side in the formula
equation. Therefore, no further coefficients are needed.

4 EVALUATE Zn(s) + 2HCl(aq) ⎯→ ZnCl2(aq) + H2(g)
Count atoms to check balance.

Zn(s) + 2HCl(aq) ⎯→ ZnCl2(aq) + H2(g)
(1Zn) + (2H + 2Cl) = (1Zn + 2Cl) + (2H)

The equation is balanced.

PRACTICE Answers in Appendix E

1. Write word, formula, and balanced chemical equations for each of Go to go.hrw.com for
the following reactions: more practice problems
a. Solid magnesium and aqueous hydrochloric acid react to pro- that ask you to write
duce aqueous magnesium chloride and hydrogen gas. balanced chemical
equations.
b. Aqueous nitric acid reacts with solid magnesium hydroxide to
produce aqueous magnesium nitrate and water. Keyword: HC6RXNX

2. Solid calcium metal reacts with water to form aqueous calcium
hydroxide and hydrogen gas. Write a balanced chemical equation
for this reaction.

272 C H A P T E R 8

SAMPLE PROBLEM D For more help, go to the Math Tutor at the end of this chapter.

Solid aluminum carbide, Al4C3, reacts with water to produce methane gas and solid aluminum hydroxide.
Write a balanced chemical equation for this reaction.

SOLUTION The reactants are aluminum carbide and water. The products are methane and aluminum
hydroxide. The formula equation is written as follows.

Al4C3(s) + H2O(l) ⎯→ CH4(g) + Al(OH)3(s) (not balanced)

Begin balancing the formula equation by counting either aluminum atoms or carbon atoms.
(Remember that hydrogen and oxygen atoms are balanced last.) There are four Al atoms
on the left. To balance Al atoms, place the coefficient 4 before Al(OH)3 on the right.

Al4C3(s) + H2O(l) ⎯→ CH4(g) + 4Al(OH)3(s) (partially balanced)

Now balance the carbon atoms. With three C atoms on the left, the coefficient 3 must be
placed before CH4 on the right.

Al4C3(s) + H2O(l) ⎯→ 3CH4(g) + 4Al(OH)3(s) (partially balanced)

Balance oxygen atoms next because oxygen, unlike hydrogen, appears only once on each
side of the equation. There is one O atom on the left and 12 O atoms in the four Al(OH)3
formula units on the right. Placing the coefficient 12 before H2O balances the O atoms.

Al4C3(s) + 12H2O(l) ⎯→ 3CH4(g) + 4Al(OH)3(s)

This leaves the hydrogen atoms to be balanced. There are 24 H atoms on the left. On the
right, there are 12 H atoms in the methane molecules and 12 in the aluminum hydroxide
formula units, totaling 24 H atoms. The H atoms are balanced.

Al4C3(s) + 12H2O(l) ⎯→ 3CH4(g) + 4Al(OH)3(s)
(4Al + 3C) + (24H + 12O) = (3C + 12H) + (4Al + 12H + 12O)

The equation is balanced.

SAMPLE PROBLEM E For more help, go to the Math Tutor at the end of this chapter.

Aluminum sulfate and calcium hydroxide are used in a water-purification process. When added to water, they
dissolve and react to produce two insoluble products, aluminum hydroxide and calcium sulfate. These products
settle out, taking suspended solid impurities with them. Write a balanced chemical equation for the reaction.

SOLUTION Each of the reactants and products is an ionic compound. Recall from Chapter 7 that the
formulas of ionic compounds are determined by the charges of the ions composing each
compound. The formula reaction is thus written as follows.

Al2(SO4)3 + Ca(OH)2 ⎯→ Al(OH)3 + CaSO4 (not balanced)

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 273

There is one Ca atom on each side of the equation, so the calcium atoms are already balanced.
There are two Al atoms on the left and one Al atom on the right. Placing the coefficient 2 in
front of Al(OH)3 produces the same number of Al atoms on each side of the equation.

Al2(SO4)3 + Ca(OH)2 ⎯→ 2Al(OH)3 + CaSO4 (partially balanced)

Next, checking SO24− ions shows that there are three SO24− ions on the left side of the equa-
tion and only one on the right side. Placing the coefficient 3 before CaSO4 gives an equal
number of SO24− ions on each side.

Al2(SO4)3 + Ca(OH)2 ⎯→ 2Al(OH)3 + 3CaSO4 (partially balanced)

There are now three Ca atoms on the right, however. By placing the coefficient 3 in front of
Ca(OH)2, we once again have an equal number of Ca atoms on each side. This last step also
gives six OH− ions on both sides of the equation.

Al2(SO4)3(aq) + 3Ca(OH)2(aq) ⎯→ 2Al(OH)3(s) + 3CaSO4(s)
(2Al + 3SO24−) + (3Ca + 6OH−) = (2Al + 6OH−) + (3Ca + 3SO24−)

The equation is balanced.

PRACTICE Answers in Appendix E

1. Write balanced chemical equations for each of the following Go to go.hrw.com for
reactions: more practice problems
a. Solid sodium combines with chlorine gas to produce solid sodi- that ask you to write
um chloride. balanced chemical
equations.
b. When solid copper reacts with aqueous silver nitrate, the prod-
ucts are aqueous copper(II) nitrate and solid silver. Keyword: HC6RXNX

c. In a blast furnace, the reaction between solid iron(III) oxide
and carbon monoxide gas produces solid iron and carbon diox-
ide gas.

SECTION REVIEW 4. Write the word, formula, and chemical equations
for the reaction between hydrogen sulfide gas and
1. Describe the differences between word equations, oxygen gas that produces sulfur dioxide gas and
formula equations, and chemical equations. water vapor.

2. Write word and formula equations for the reaction Critical Thinking
in which aqueous solutions of sulfuric acid and
sodium hydroxide react to form aqueous sodium 5. INTEGRATING CONCEPTS The reaction of vanadi-
sulfate and water. um(II) oxide with iron(III) oxide results in the for-
mation of vanadium(V) oxide and iron(II) oxide.
3. Translate the following chemical equations into Write the balanced chemical equation.
sentences:

a. 2K(s) + 2H2O(l ) ⎯→ 2KOH(aq) + H2(g)
b. 2Fe(s) + 3Cl2(g) ⎯→ 2FeCl3(s)

274 C H A P T E R 8

Carbon Monoxide Catalyst

Colorless, odorless, and deadly— NASA’s space-based carbon dioxide converters in cars oxidize carbon
carbon monoxide, “the silent killer,” laser needed to be fed a continuous monoxide and unburned hydrocar-
causes the deaths of hundreds of supply of CO2. This was necessary bons to minimize pollution. Many
Americans every year. When fuel because as a byproduct of its opera- substances are oxidized into new
does not burn completely in a com- tion, the laser degraded some of the materials for manufacturing purposes.
bustion process, carbon monoxide is CO2 into carbon monoxide and oxy- But both of these types of catalytic
produced. Often this occurs in a mal- gen. To address this problem, NASA reactions occur at very high tempera-
functioning heater, furnace, or fire- scientists developed a catalyst made tures. NASA’s catalyst is special,
place. When the carbon monoxide is of tin oxide and platinum that oxi- because it’s able to eliminate carbon
inhaled, it bonds to the hemoglobin dized the waste carbon monoxide monoxide at room temperature.
in the blood, leaving the body oxy- back into carbon dioxide. The NASA
gen starved. Before people realize a scientists then realized that this cat- According to David Schryer, low-
combustion device is malfunctioning, alyst had the potential to be used in temperature catalysts constitute a
it’s often too late. many applications here on Earth, whole new class of catalysts with
including removing carbon monoxide abundant applications for the future.
O2Hb ؉ CO B COHb ؉ O2 from houses and other buildings.
Questions
Carbon monoxide, CO, has almost Typically, a malfunctioning heater
200 times the affinity to bind with circulates the carbon monoxide it 1. How did NASA’s research on the
the hemoglobin, Hb, in the blood as produces through its air intake sys- space-based carbon dioxide laser
oxygen. This means that hemoglobin tem back into a dwelling space. result in a benefit for consumers?
will bind to carbon monoxide rather Installing the catalyst in the air
than oxygen in the body. If enough intake would oxidize any carbon 2. According to the chemical reac-
carbon monoxide is present in the monoxide to nontoxic carbon dioxide tion, if there are 4.5 mol of oxy-
blood, it can be fatal. before it reentered the room. genated hemoglobin present in
an excess of carbon monoxide,
Carbon monoxide poisoning can “The form of our catalyst is a very how many moles of hemoglobin
be prevented by installing filters that thin coating on some sort of a sup- would release oxygen and bind
absorb the gas. After a time, howev- port, or substrate as we call it,” says to carbon monoxide? Explain
er, filters become saturated, and then NASA chemist David Schryer. “And your answer.
carbon monoxide can pass freely that support, or substrate, can be
into the air. The best way to prevent any one of a number of things. The www.scilinks.org
carbon monoxide poisoning is not great thing about a catalyst is that Topic: Carbon Monoxide
just to filter out the gas, but to elimi- the only thing that matters about it Code: HC60219
nate it completely. is its surface. So a catalyst can be
incredibly thin and still be very
The solution came to research effective.”
chemists at NASA who were working
on a problem with a space-based The idea of using catalysts to oxi-
laser. In order to operate properly, dize gases is not a new one. Catalytic

C H E M I C A L E Q U A T I O N S A N D R E A C T I O N S 275

SECTION 2 Types of Chemical
Reactions
OBJECTIVES
T housands of known chemical reactions occur in living systems, in
Define and give general equa-
tions for synthesis, decomposi- industrial processes, and in chemical laboratories. Often it is necessary
tion, single-displacement, and to predict the products formed in one of these reactions. Memorizing
double-displacement the equations for so many chemical reactions would be a difficult task.
reactions. It is therefore more useful and realistic to classify reactions according to
Classify a reaction as a various similarities and regularities. This general information about
synthesis, decomposition, reaction types can then be used to predict the products of specific
single-displacement, double- reactions.
displacement, or combustion
reaction. There are several ways to classify chemical reactions, and none are
List three kinds of synthesis entirely satisfactory. The classification scheme described in this section
reactions and six kinds of provides an introduction to five basic types of reactions: synthesis,
decomposition reactions. decomposition, single-displacement, double-displacement, and combus-
List four kinds of single- tion reactions. In later chapters, you will be introduced to categories
displacement reactions and that are useful in classifying other types of chemical reactions.
three kinds of double-
displacement reactions. Synthesis Reactions
Predict the products of simple
reactions given the reactants. In a synthesis reaction, also known as a composition reaction, two or
more substances combine to form a new compound. This type of reac-
276 C H A P T E R 8 tion is represented by the following general equation.

A + X ⎯→ AX

A and X can be elements or compounds. AX is a compound. The fol-
lowing examples illustrate several kinds of synthesis reactions.

Reactions of Elements with Oxygen and Sulfur

One simple type of synthesis reaction is the combination of an element
with oxygen to produce an oxide of the element. Almost all metals react
with oxygen to form oxides. For example, when a thin strip of magne-
sium metal is placed in an open flame, it burns with bright white light.
When the metal strip is completely burned, only a fine white powder of
magnesium oxide is left. This chemical reaction, shown in Figure 7 on
the next page, is represented by the following equation.

2Mg(s) + O2(g) ⎯→ 2MgO(s)

The other Group 2 elements react in a similar manner, forming oxides (a)
with the formula MO, where M represents the metal. The Group 1 met-
als form oxides with the formula M2O, for example, Li2O. The Group 1 (b)
and Group 2 elements react similarly with sulfur, forming sulfides with FIGURE 7 Magnesium, Mg,
the formulas M2S and MS, respectively. Examples of these types of syn- pictured in (a), undergoes a synthe-
thesis reactions are shown below. sis reaction with oxygen, O2, in the
air to produce magnesium oxide,
16Rb(s) + S8(s) ⎯→ 8Rb2S(s) MgO, as shown in (b).
8Ba(s) + S8(s) ⎯→ 8BaS(s)

Some metals, such as iron, combine with oxygen to produce two dif-
ferent oxides.

2Fe(s) + O2(g) ⎯→ 2FeO(s)
4Fe(s) + 3O2(g) ⎯→ 2Fe2O3(s)

In the product of the first reaction, iron is in an oxidation state of +2. In
the product of the second reaction, iron is in an oxidation state of +3.
The particular oxide formed depends on the conditions surrounding the
reactants. Both oxides are shown below in Figure 8.

Nonmetals also undergo synthesis reactions with oxygen to form
oxides. Sulfur, for example, reacts with oxygen to form sulfur dioxide.
And when carbon is burned in air, carbon dioxide is produced.

S8(s) + 8O2(g) ⎯→ 8SO2(g)
C(s) + O2(g) ⎯→ CO2(g)

In a limited supply of oxygen, carbon monoxide is formed.

2C(s) + O2(g) ⎯→ 2CO(g)

Hydrogen reacts with oxygen to form dihydrogen monoxide, better
known as water.

2H2(g) + O2(g) ⎯→ 2H2O(g)

(a) (b) 277
FIGURE 8 Iron, Fe, and oxygen, O2, combine to form two different oxides:
(a) iron(II) oxide, FeO, and (b) iron(III) oxide, Fe2O3.

CHEMICAL EQUATIONS AND REACTIONS

FIGURE 9 Calcium hydroxide, Reactions of Metals with Halogens
a base, can be used to neutralize
hydrochloric acid in your stomach. Most metals react with the Group 17 elements, the halogens, to form
You will read more about acids, bases, either ionic or covalent compounds. For example, Group 1 metals react
and neutralization in Chapter 14. with halogens to form ionic compounds with the formula MX, where M
is the metal and X is the halogen. Examples of this type of synthesis
278 C H A P T E R 8 reaction include the reactions of sodium with chlorine and potassium
with iodine.

2Na(s) + Cl2(g) ⎯→ 2NaCl(s)
2K(s) + I2(g) ⎯→ 2KI(s)

Group 2 metals react with the halogens to form ionic compounds with
the formula MX2.

Mg(s) + F2(g) ⎯→ MgF2(s)
Sr(s) + Br2(l) ⎯→ SrBr2(s)

The halogens undergo synthesis reactions with many different met-
als. Fluorine in particular is so reactive that it combines with almost all
metals. For example, fluorine reacts with sodium to produce sodium fluo-
ride. Similarly, it reacts with cobalt to form cobalt(III) fluoride and with
uranium to form uranium(VI) fluoride.

2Na(s) + F2(g) ⎯→ 2NaF(s)
2Co(s) + 3F2(g) ⎯→ 2CoF3(s)

U(s) + 3F2(g) ⎯→ UF6(g)

Sodium fluoride, NaF, is added to municipal water supplies in trace
amounts to provide fluoride ions, which help to prevent tooth decay in
the people who drink the water. Cobalt(III) fluoride, CoF3, is a strong
fluorinating agent. And natural uranium is converted to uranium(VI)
fluoride, UF6, as the first step in the production of uranium for use in
nuclear power plants.

Synthesis Reactions with Oxides

Active metals are highly reactive metals. Oxides of active metals react
with water to produce metal hydroxides. For example, calcium oxide
reacts with water to form calcium hydroxide, an ingredient in some
stomach antacids, as shown in Figure 9.

CaO(s) + H2O(l) ⎯→ Ca(OH)2(s)

Calcium oxide, CaO, also known as lime or quicklime, is manufactured
in large quantities. The addition of water to lime to produce Ca(OH)2,
which is also known as slaked lime, is a crucial step in the setting of
cement.

Many oxides of nonmetals in the upper right portion of the periodic
table react with water to produce oxyacids. For example, sulfur dioxide,
SO2, reacts with water to produce sulfurous acid.