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Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 5
Ionic and Covalent
Compounds
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
5
Ionic and Covalent Compounds
5.1 Compounds
5.2 Lewis Dot Symbols
5.3 Ionic Compounds and Bonding
5.4 Naming Ions and Ionic Compounds
Formulas of Ionic Compounds
Naming Ionic Compounds
5.5 Covalent Bonding and Molecules
Molecules
Molecular Formulas
Empirical Formulas
5.6 Naming Molecular Compounds
Specifying Numbers of Atoms
Compounds Containing Hydrogen
Organic Compounds
5
Ionic and Covalent Compounds
5.7 Covalent Bonding in Ionic Species
Polyatomic Ions
Oxoacids
Hydrates
Familiar Inorganic Compounds
5.8 Molecular and Formula Masses
5.9 Percent Composition of Compounds
5.10 Molar Mass
Interconverting Mass, Moles, and Number of Particles
Determination of Empirical Formula and Molecular Formula
from Percent Composition
5.1
Compounds
A compound is a substance composed of two or more elements
combined in a specific ratio and held together by chemical bonds.
Familiar examples of compounds are water and salt (sodium
chloride).
5.2
Lewis Dot Symbols
When atoms form compounds, it is their valence electrons that
actually interact.
A Lewis dot symbol consists of the element’s symbol surround by
dots.
Each dot represents a valence electron.
•
•B•
Boron 1s22s22p1
3 valence electrons
•
B•
•
Lewis dot symbol for boron
•B•
•
•
•B
•
other reasonable Lewis dot
symbols for boron
Lewis Dot Symbols
Atoms combine in order to achieve a more stable electron
configuration.
Maximum stability results when a chemical species is isoelectronic
with a noble gas.
Na: 1s22s22p63s1
Na+: 1s22s22p6
10 electrons total,
isoelectronic with Ne
Cl: 1s22s22p63s23p5
Cl‒: 1s22s22p63s23p6
18 electrons total,
isoelectronic with Ar
Lewis Dot Symbols
Lewis dot symbols of the main group elements.
Lewis Dot Symbols
Dots are not paired until absolutely necessary.
•
•B•
1s22s22p1
•
•C•
•
1s22s22p2
••
•N•
•
1s22s22p3
5 valence electrons; first pair formed
in the Lewis dot symbol
Na •
For main group metals such as Na, the number of dots is
the number of electrons that are lost.
••
•O•
••
For nonmetals in the second period, the number of
unpaired dots is the number of bonds the atom can form.
Lewis Dot Symbols
Ions may also be represented by Lewis dot symbols.
Remember the charge
Na•
Na 1s22s22p63s1
Valence electron lost in the formation
of the Na+ ion.
••
•O•
••
O 1s22s22p4
Na+
Na+ 1s22s22p6
Core electrons not represented in the
Lewis dot symbol
••
•• O ••
••
2‒
O2‒ 1s22s22p6
Worked Example 5.1
Write Lewis dot symbols for (a) fluoride ion (F-), (b) potassium ion (K+), and
(c) sulfide ion (S2-).
Strategy Starting with the Lewis dot symbols for each element, add dots (for
anions) or remove dots (for cations) as needed to achieve the correct charge on
each ion. Don’t forget to include the appropriate charge on the Lewis dot symbol.
Solution (a)
(b) K+
(c)
Think About It For ions that are isoelectronic with noble gases, cations should
have no dots remaining around the element symbol, whereas anions should have
eight dots around the element symbol. Note, too, that for anions, we put square
brackets around the Lewis dot symbol and place the negative charge outside the
brackets. Because the symbol for a common cation such as the potassium ion
has no remaining dots, square brackets are not necessary.
5.3
Ionic Compounds and Bonding
Ionic bonding refers to the electrostatic attraction that holds
oppositely charged ions together in an ionic compound.
Na•
••
••Cl
• + e−
••
••
••Cl
• + Na•
••
Na+ + e−
−
••
••Cl••
••
Na+ +
•• •
••Cl
•
••
−
The attraction between the cation and
anion draws them together to form NaCl
Ionic Compounds and Bonding
The resulting electrically neutral compound, sodium chloride, is
represented with the chemical formula NaCl.
The chemical formula, or simply formula, of an ionic compound
denotes the constituent elements and the ratio in which they
combine.
Ionic Compounds and Bonding
A three-dimensional array of oppositely-charged ions is called a
lattice. Lattice energy is the amount of energy required to convert a
mole of ionic solid to its constituent ions in the gas phase.
−
+
+
−
NaCl(s)  Na+(g) + Cl−(g)
−
−
+
+
−
+
−
−
+
Hlattice = +788 kJ/mol
+
Ionic Compounds and Bonding
The magnitude of lattice energy is a measure of an ionic
compound’s stability.
Lattice energy depends on the magnitudes of the charge and on the
distance between them.
Q1 Q2
F 2
d
Q1
Q2

d
amount of charge
distance of separation
Ionic Compounds and Bonding
Ionic Compounds and Bonding
The magnitude of lattice energy is a measure of an ionic
compound’s stability.
Lattice energy depends on the magnitudes of the charge and on the
distance between them.
Ionic Compounds and Bonding
Worked Example 5.2
Arrange MgO, CaO, and SrO in order of increasing lattice energy.
Strategy Consider the charges on the ions and the distances between them.
Apply Coulomb’s law to determine the relative lattice energies. All three
compounds contain O2- and all three cations are +2. Recalling that lattice energy
increases as the distance between ions decreases, we need only consider the radii
of the cations as all three contain the same anion. From Figure 4.13, the ionic
radii are 0.72 Å (Mg2+), 1.00 Å (Ca2+), and 1.18 Å (Sr2+).
Solution MgO has the smallest distance between ions, whereas SrO has the
largest distance between ions. Therefore, in order of increase lattice energy: SrO <
CaO < MgO.
Think About It Mg, Ca, and Sr are all Group 2A metals, so we could have
predicted this result without knowing their radii. Recall that ionic radii increase
as we move down a column in the periodic table, and charges that are farther
apart are more easily separated (meaning the lattice energy will be smaller.) The
lattice energies of SrO, CaO, and MgO are 3217, 3414, and 3890 kJ/mol,
respectively.
5.4
Naming Ions and Ionic Compounds
A monatomic ion is named by changing the ending of the element’s
name to –ide.
Cl– is chloride
O2– is oxide
Some metals can form cations of more than one possible charge.
Fe2+ : ferrous ion [Fe(II)]
Fe3+ : ferric ion [Fe(III)]
Mn2+ : manganese(II) ion
Mn3+ : manganese(III) ion
Mn4+ : manganese(IV) ion
Naming Ions and Ionic Compounds
Naming Ions and Ionic Compounds
Formulas for ionic compounds are generally empirical formulas.
Ionic compounds are electronically neutral.
Formulas of Ionic Compounds
In order for ionic compounds to be electronically neutral, the sum
of the charges on the cation and anion in each formula must be
zero.
Aluminum oxide:
Al3+
O2–
Al2O3
Sum of charges:
2(+3) + 3(–2) = 0
Formulas of Ionic Compounds
Naming Ions and Ionic Compounds
To name ionic compounds:
1) Name the cation
omit the word ion
use a Roman numeral if the cation can have more than one charge
2) Name the anion
omit the word ion
Examples:
NaCN sodium cyanide
FeCl2 iron(II) chloride
FeCl3 iron(III) chloride
Worked Example 5.3
Name the following ionic compounds: (a) CaO, (b) Mg3N2, and (c) Fe2S3.
Strategy Begin by identifying the cation and anion in each compound, and then
combine the names for each, eliminating the word ion.
Solution (a) CaO is calcium oxide.
(b) Mg3N2 is magnesium nitride.
(c) Fe2S3 is iron(III) sulfide.
Think About It Be careful not to confuse the subscript in the formula with the
charge in the metal ion. In part (c), for example, the subscript on Fe is 2, but this
is an iron(III) compound.
Worked Example 5.4
Deduce the formulas of the following ionic compounds: (a) mercury(II) chloride,
(b) lead(II) bromide, and (c) potassium nitride.
Strategy Identify the ions in each compound, and determine their ratios of
combination using the charges on the cation and anion in each.
Solution (a) Mercury(II) chloride is a combination of Hg2+ and Cl-. To produce
a neutral compound, these two ions must combine in a 1:2 ratio – HgCl2.
(b) Lead(II) bromide is a combination of Pb2+ and Cl-. These ions combine in a
1:2 ratio to give PbBr2.
(c) Potassium nitride is a combination of K+ and N3-. These ions combine in a 3:1
ratio to give K3N.
Think About It Make sure that the charges sum to zero in each compound
formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b),
(+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
5.5
Covalent Bonding and Molecules
When compounds form between elements with similar properties,
electrons are not transferred from one element to another but instead
are shared in order to give each atom a noble gas configuration.
This approach is known as the Lewis theory of bonding, named for it’s
proponent, Gilbert Lewis.
Lewis theory depicts bond formation in H2 as
H∙ + ∙H → H:H
This type of arrangement, where two atoms share a pair of electrons, is
known as covalent bonding, and the shared pair of electrons
constitutes a covalent bond.
Covalent Bonding and Molecules
A molecule is a combination of at least two atoms in a specific
arrangement held together by chemical forces (chemical bonds).
A molecule may be an element or a compound.
Different samples of a given compound always contain the same
elements in the same ratio. This is known as the law of definite
proportions.
Sample
Mass of O (g)
Mass of C (g)
Ratio (g O:g C)
123 g carbon dioxide
89.4
33.6
2.66:1
50.5 g carbon dioxide
36.7
13.8
2.66:1
88.6 g carbon dioxide
64.4
24.2
2.66:2
Covalent Bonding and Molecules
If two elements can for two or more different compounds, the law of
multiple proportions tells us that the ratio of masses of one element
that combine with a fixed mass of the other element can be expressed
in small whole numbers.
In addition to carbon dioxide, carbon also combines with oxygen to
form carbon monoxide.
Sample
Mass of O (g)
Mass of C (g)
Ratio (g O:g C)
16.3 g carbon
monoxide
9.31
6.99
1.33:1
25.9 g carbon
monoxide
14.8
11.1
1.33:1
88.4 g carbon
monoxide
50.5
37.9
1.33:2
Covalent Bonding and Molecules
The mass ratio of oxygen to carbon in carbon dioxide is 2.66:1, and
the ratio of oxygen to carbon in carbon monoxide is 1.33:1.
The ratio of two such mass ratios can be expressed as small whole
numbers.
mass ratio of O to C in carbon dioxide
2.66
= 2:1
=
mass ratio of O to C in carbon monoxide
1.33
Covalent Bonding and Molecules
Diatomic molecules contain two atoms and may be either
heteronuclear or homonuclear.
Polyatomic molecules contain more than two atoms.
Covalent Bonding and Molecules
Covalent Bonding and Molecules
A chemical formula denotes the
composition of the substance.
A molecular formula shows the exact
number of atoms of each element in a
molecule.
Some elements have two or more distinct
forms known as allotropes.
For example, oxygen (O2) and ozone
(O3) are allotropes of oxygen.
A structural formula shows not only the
elemental composition, but also the
general arrangements.
Worked Example 5.5
Write the molecular formula of ethanol based on its ball-and-stick model, shown
here.
Strategy Refer to the labels on the atoms (or see Table 5.3). There are two
carbon atoms, six hydrogen atoms, and one oxygen atom, so the subscript on C
will be 2 and the subscript on H will be 6, and there will be no subscript on O.
Solution C2H6O
Think About It Often the molecular formula for a compound such as ethanol
(consisting of carbon, hydrogen, and oxygen) is written so that the formula more
closely resembles the actual arrangement of atoms in the molecule. Thus, the
molecular formula for ethanol is commonly written as C2H5OH.
Covalent Bonding and Molecules
Molecular substances can also be represented using empirical
formulas, the whole-number ratio of elements.
While, the molecular formulas tell us the actual number of atoms (the
true formula), the empirical formula gives the simplest formula.
Molecular formula: N2H4
Empirical formula: NH2
The molecular and empirical formulas are often the same.
Covalent Bonding and Molecules
Worked Example 5.6
Write the empirical formulas for the following molecules: (a) glucose (C6H12O6),
a substance known as blood sugar; (b) adenine (C5H5N5), also known as vitamin
B4; and (c) nitrous oxide (N2O), a gas that is used as an anesthetic (“laughing
gas”) and as an aerosol propellant for whipped cream.
Strategy To write the empirical formula, the subscripts in the molecular formula
must be reduced to the smallest possible whole numbers (without altering the
relative numbers of atoms).
Worked Example 5.6 (cont.)
Solution (a) Dividing each of the subscripts in the molecular formula for
glucose by 6, we obtain the empirical formula CH2O. If we had divided the
subscripts by 2 or 3, we would have obtained the formulas C3H6O3 and C2H4O2,
respectively. Although the ratio of carbon to hydrogen to oxygen atoms in each of
these formulas is correct (1:2:1), neither is the simplest formula because the
subscripts are not in the smallest possible whole-number ratio.
(b) Dividing each subscript by 5, we get the empirical formula CHN.
(c) Because the subscripts in the formula for nitrous oxide are already the smallest
possible whole numbers, its empirical formula is the same as its molecular
formula N2O.
Think About It Make sure that the ratio in each empirical formula is the same
as that in the corresponding molecular formula and that the subscripts are the
smallest possible whole numbers. In part (a), for example, the ratio of C:H:O in
the molecular formula is 6:12:6, which is equal to 1:2:1, the ratio expressed in
the empirical formula.
5.6
Naming Molecular Compounds
Remember that binary molecular compounds are substances that
consist of just two different elements.
Nomenclature:
1) Name the first element that appears in the formula.
2) Name the second element that appears in the formula, changing
its ending to –ide.
Examples:
HCl hydrogen chloride
HI
hydrogen iodide
Naming Molecular Compounds
Greek prefixes are used to denote the number of atoms of each
element present.
Naming Molecular Compounds
The prefix mono- is generally omitted for the first element.
For ease of pronunciation, we usually eliminate the last letter of a
prefix that ends in “o” or “a” when naming an oxide.
Example: N2O5 is dinitrogen pentoxide not dinitrogen
pentaoxide
Worked Example 5.7
Name the following binary molecular compounds: (a) NF3 and (b) N2O4.
Strategy Each compound will be named using the systematic nomenclature
including, where necessary, appropriate Greek prefixes.
Solution (a) nitrogen trifluoride
(b) dinitrogen tetroxide
Think About It Make sure that the prefixes match the subscripts in the
molecular formulas and that the word oxide is not preceded immediately by an
“a” or an “o”.
Worked Example 5.8
Write the chemical formulas for the following binary molecular compounds:
sulfur tetrafluoride and (b) tetraphosphorus decasulfide.
(a)
Strategy The formula for each compound will be deduced using the systematic
nomenclature guidelines.
Solution (a) SF4
(b) P4S10
Think About It Double-check that the subscripts in the formulas match the
prefixes in the compound names: (a) 4 = tetra and (b) 4 = tetra and 10 = deca.
Compounds Containing Hydrogen
The names of molecular compounds containing hydrogen do not
usually conform to the systematic nomenclature guidelines.
Many are called by the common, nonsystematic names or by names
that do not indicate explicitly the number of H atoms present.
Examples:
B2H6
Diborane
SiH4
Silane
NH3
Ammonia
PH3
Phosphine
H2O
Water
H2S
Hydrogen sulfide
Compounds Containing Hydrogen
One definition of an acid is a substance that produces hydrogen
ions (H+) when dissolved in water.
HCl is an example of a binary compound that is an acid when
dissolved in water.
To name these types of acids:
1) remove the –gen ending from hydrogen
2) change the –ide ending on the second element to –ic.
hydrogen chloride → hydrochloric acid
Compounds Containing Hydrogen
A compound must contain at least one ionizable hydrogen atom to
be an acid upon dissolving.
Organic Compounds
Our nomenclature discussion so far has focused on inorganic
compounds, generally defined as those without carbon.
Organic compounds contain carbon and hydrogen, sometimes in
combination with other atoms.
Hydrocarbons contain only carbon and hydrogen.
The simplest hydrocarbons are called alkanes.
Organic Compounds
Organic Compounds
Organic Compounds
Many organic compounds contain groups of atoms known as
functional groups, which often determine a molecule’s reactivity.
5.7
Covalent Bonding in Ionic Species
Polyatomic ions consist of a combination of two or more atoms.
Formulas are determined following the same rule as for ionic
compounds containing only monatomic ions: ions must combine in
a ratio that give a neutral formula overall.
Calcium phosphate:
Ca2+
PO43–
Ca3(PO4)2
Sum of charges:
3(+2) + 2(–3) = 0
Covalent Bonding in Ionic Species
Covalent Bonding in Ionic Species
Worked Example 5.9
Name the following ionic compounds: (a) Fe2(SO4)3, (b) Al(OH)3, and (c) Hg2O.
Strategy Begin by identifying the cation and anion in each compound, and then
combine the names for each, eliminating the word ion.
Solution (a) Fe2(SO4)3 is iron(III) sulfate.
(b) Al(OH)3 is aluminum hydroxide.
(c) Hg2O is mercury(I) oxide.
Think About It Be careful not to confuse the subscript in the formula with the
charge in the metal ion. In part (a), for example, the subscript on Fe is 2, but this
is an iron(III) compound.
Covalent Bonding in Ionic Species
Oxoanions are polyatomic anions that contain one or more oxygen
atoms and one atom (the “central atom”) of another element.
Starting with the oxoanions that end in –ate, we can name these
ions as follows:
1)The ion with one more O atom than the –ate ion is called the
per…ate ion. Thus, ClO3- is the chlorate ion, so ClO4- is the
perchlorate ion.
2)The ion with one less O atom than the –ate ion is called the –ite
ion. Thus, ClO2- is the chlorite ion.
3)The ion with two fewer O atom than the –ate ion is called the
hypo…ite ion. Thus, ClO- is the hypochlorite ion.
At minimum, memorize the oxoanions that end in –ate so you can
apply these guidelines when necessary.
Covalent Bonding in Ionic Species
perchlorate
chlorate
chlorite
hypochlorite
ClO4ClO3ClO2ClO-
nitrate
nitrite
NO3NO2-
phosphate
phosphite
PO43PO33-
sulfate
sulfite
SO42SO32-
Worked Example 5.10
Name the following species: (a) BrO4-, (b) HCO3-, and (c) H2CO3.
Strategy Each species is either an oxoanion or an oxoacid. Identify the
“reference oxidation” (the one with the –ate ending) for each, and apply the rules
to determine appropriate names.
Think
It Remembering
all these
names
and formulas
Solution
(a)About
BrO4- has
one more O atom
than the
bromate
ion (BrOis3-), so BrO4greatly facilitated
is the perbromate
ion. by memorizing the common ions that end in –ate.
chlorate
ClO3nitrate
NO32IO3- ion. Becausecarbonate
CO32- hydrogen atom,
(b) COiodate
HCO3- has one ionizable
3 is the carbonate
bromate
BrO3carbonate
C2O42it is called
the hydrogen
ion.oxalate
sulfate
SO42chromate
CrO423phosphate
POhydrogen
(c) With
two ionizable
atomspermanganate
and no charge onMnO
the compound,
H2CO3 is
4
4
carbonic acid.
Think About It Make sure that the charges sum to zero in each compound
formula. In part (a), for example, Hg2+ + 2Cl- = (+2) + 2(-1) = 0; in part (b),
(+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
Worked Example 5.11
Determine the formula of sulfurous acid.
Strategy The –ous ending in the name of an acid indicates that the acid is
derived from an oxoanion ending in –ite. The oxoanion must be sulfite, SO32-, so
add enough hydrogen ions to make a neutral formula.
Solution The formula of sulfurous acid is H2SO3.
Think About It Remembering all these names and formulas is greatly facilitated
by memorizing the common ions that end in -ate.
Hydrates
A hydrate is a compound that has a specific number of water
molecules within its solid structure.
For example, in its normal state, copper(II) sulfate has five water
molecules associated with it.
Systematic name: copper(II) sulfate pentahydrate
Formula: Cu(SO)4 ∙ 5H2O
Some other hydrates are
BaCl2 ∙ 2H2O
LiCl ∙ H2O
MgSO4 ∙ 7H2O
Sr(NO3)2 ∙ 4H2O
Hydrates
When the water molecules are driven off by heating, the resulting
compound, Cu(SO)4, is sometimes called anhydrous copper(II)
sulfate.
Anhydrous means the compound no longer has water molecules
associated with it.
Familiar Inorganic Compounds
5.8
Molecular and Formula Mass
The molecular mass is the mass in atomic mass units (amu) of an
individual molecule.
To calculate molecular mass, multiply the atomic mass for each
element in a molecule by the number of atoms of that element and
then total the masses
Molecular mass of H2O = 2(atomic mass of H) + atomic mass of O
= 2(1.008 amu) + 16.00 amu = 18.02 amu
Because the atomic masses on the periodic table are average atomic
masses, the result of such a determination is an average molecular
mass, sometimes referred to as the molecular weight.
Molecular and Formula Mass
Although an ionic compound does not have a molecular mass, we
can use its empirical formula to determine its formula mass (the
mass of a “formula unit”), sometimes called the formula weight.
Worked Example 5.12
Calculate the molecular mass or the formula mass, as appropriate, for each of the
following corresponds: (a) propane, C3H8, (b) lithium hydroxide, LiOH, and (c)
barium acetate, Ba(C2H3O2)2.
Strategy Determine the molecular mass (for each molecular compound) or
formula mass (for each ionic compound) by summing all the atomic masses.
Solution
ForAbout
each compound,
multiply
theyou
number
atomsthe
by number
the atomic mass
Think
It Double-check
that
have of
counted
of eachofelement
and then for
sumeach
the calculated
atoms correctly
compound values.
and that you have used the
proper atomic masses from the periodic table.
(a) The molecular mass of propane is 3(12.01 amu) + 8(1.008 amu) = 44.09 amu
(b) The formula mass of lithium hydroxide is 6.941 amu + 16.00 amu + 1.008
amu = 23.95 amu.
(c) The formula mass of barium acetate is 137.3 amu + 4(12.01 amu) + 6(1.008
amu) + 4(16.00 amu) = 255.4 amu.
5.9
Percent Composition of Compounds
A list of the percent by mass of each element in a compound is
known as the compound’s percent composition by mass.
n

a
t
o
m
i
c
m
a
s
s
o
f
e
l
e
m
e
n
t
p
e
r
c
e
n
t
m
a
s
s
o
f
a
n
e
l
e
m
e
n
t
=

1
0
0
%
m
o
l
e
c
u
l
a
r
o
r
f
o
r
m
u
l
a
m
a
s
s
o
f
c
o
m
p
o
u
n
d
where n is the number of atoms of the element in a molecule or
formula unit of the compound
Percent Composition of Compounds
For a molecule of H2O2:
2

1
.
0
0
8
a
m
u
H
%
H
=

1
0
0
%
=
5
.
9
2
6
%
3
4
.
0
2
a
m
u
H
O
2
2
2

1
6
.
0
0
a
m
u
O
%
O
=

1
0
0
%
=
9
4
.
0
6
%
3
4
.
0
2
a
m
u
H
O
2
2
Percent Composition of Compounds
We could also have used the empirical formula of hydrogen peroxide
(HO) for the calculation.
In this case, we could have used the empirical formula mass, the
mass in amu of one empirical formula, in place of the molecular
formula.
The empirical formula mass of H2O2 (the mass of HO) is 17.01 amu.
%H =
1.008 amu H
× 100% = 5.926%
17.01 amu H2O2
%O =
16.00 amu O
× 100% = 94.06%
17.01 amu H2O2
Worked Example 5.13
Lithium carbonate, Li2CO3, was the first “mood-stabilizing” drug approved by the
FDA for the treatment of mania and manic-depressive illness, also known as
bipolar disorder. Calculate the percent composition by mass of lithium carbonate.
Strategy Use Equation 5.1 to determine the percent by mass contributed by
each element in the compound.
Think About It Make sure that the percent composition results for
Solution
For eachsum
element,
multiply the100.
number
of atoms
atomic
a compound
to approximately
(In this
case, by
thethe
results
summass,
divide to
byexactly
the formula
mass, and multiply
by 100 percent.
100 percent––18.79%
+ 16.25%
+ 64.96% = 100.00%––
but remember
that because
2×6.941
amu Li of rounding, the percentages may sum to
%Li
= slightly more or very
×100% = 18.79%
very
less.
73.89 amu Li2COslightly
3
%C =
12.01 amu C
×100% = 16.25%
73.89 amu Li2CO
3
%O =
3×16.00 amu O
×100% = 64.96%
73.89 amu Li2CO3
5.10
Molar Mass
The molar mass (M ) of a substance is the mass in grams of one
mole of the substance.
The molar mass of an element is numerically equal to its atomic
mass.
1 mol C = 12.01 g
1 C atom = 12.01 amu
The molar mass of a compound is the sum of molar masses of the
elements it contains.
1 mol H2O = 2 ×1.008 g + 16.00 g = 18.02 g
1 mol NaCl = 22.99 g + 35.45 g = 58.44 g
Molar Mass
When expressing the molar mass of elements such as oxygen and
hydrogen, we have to be careful to specify which form of the
element we mean.
For instance, the element oxygen exists predominantly as O2. But
we might also mean atomic oxygen (O). We must consider the
context to tell which form of the element, O2 or O, is intended.
Context
Oxygen means
Molar mass
How many moles of oxygen react with 2 moles of
hydrogen to produce water?
O2
32.00 g
How many moles of oxygen are there in 1 mole of
water?
O
16.00 g
Air is approximately 21 percent oxygen.
O2
32.00 g
Many organic compounds contain oxygen.
O
16.00 g
Interconverting Mass, Moles, and Numbers of Particles
Worked Example 5.14
Determine (a) the number of moles of CO2 in 10.00 g of carbon dioxide and
(b) the mass of 0.905 mole of sodium chloride.
Strategy Use molar mass to convert from mass to moles and to convert from
moles to mass. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and the
molar mass of sodium chloride (NaCl) is 58.44 g/mol.
Solution
1 mol CO2
(a) 10.00 g CO2 × 44.01 g CO = 0.2272 mol CO2
2
(b) 0.905 mol NaCl ×
58.44 g NaCl
= 52.9 g NaCl
1 mol NaCl
Think About It Always double-check unit cancellations in problems such as
these–errors are common when molar mass is used as a conversion factor. Also
make sure that the results make sense. In both cases, a mass smaller than the
molar mass corresponds to less than a mole of substance.
Worked Example 5.15
(a) Determine the number of water molecules and the numbers of H and O atoms
in 3.26 g of water.
(b) Determine the mass of 7.92×1019 carbon dioxide molecules.
Strategy Use molar mass and Avogadro’s number to convert from mass to
molecules, and vice versa. Use the molecular formula of water to determine the
numbers of H and O atoms.
Solution
1 mol H2O
6.022×1023 H2O molecules
23 H O molecules
=
1.09×10
(a) 3.26 g H2O× 18.02 g H O ×
2
1 mol H2O
2
Using the molecular formula, we can determine the number of H and O atoms in 3.26 g of
H2O as follows:
2 H atoms
23
23 H atoms
1.09×10 H2O molecules ×
=
2.18×10
1 H O molecule
2
1.09×1023 H2O molecules ×
1 O atom
23 H atoms
=
1.09×10
1 H2O molecule
Worked Example 5.15 (cont.)
Solution
1 mol CO2
44.01 g CO2
(b) 7.92×1019 CO2 molecules × 6.022×1023 CO molecules × 1 mol CO
2
2
= 5.79×10-3 g CO2
Think About It Again, check the cancellation of units carefully and make sure
that the magnitudes of your results are reasonable.
Determination of Empirical Formula and Molecular
Formula from Percent Composition
Using the concepts of the mole and molar mass, we can now use an
experimentally determined percent composition to determine the
empirical and/or molecular formula.
The empirical formula gives only the ratio of atoms in a molecule,
so there may be multiple compounds with the same empirical
formula.
If we know the approximate molar mass of the compound, we can
determine the molecular formula by dividing the molar mass by the
empirical formula mass. We then multiply the empirical formula by
this number to obtain the molecular formula.
Worked Example 5.16
Determine the empirical formula of a compound that is 30.45 percent nitrogen
and 69.55 percent oxygen by mass. Given that the molar mass of the compound is
approximately 92 g/mol, determine the molecular formula of the compound.
Strategy Assume a 100-g sample so that the mass percentages of nitrogen and
oxygen given in the problem statement correspond to the masses of N and O in
the compound. Then, using the appropriate molar masses, convert the grams of
each element to moles. Use the resulting numbers as subscripts in the empirical
formula, reducing them to the lowest possible whole numbers for the final answer.
To calculate the molecular formula, first divide the molar mass given in the
problem statement by the empirical formula mass. Then, multiply the subscripts
in the empirical formula by the resulting number to obtain the subscripts in the
molecular formula.
The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One
hundred grams of a compound that is 30.45 percent nitrogen and 69.35 percent
oxygen by contains 30.45 g N and 69.55 g O.
Worked Example 5.16 (cont.)
Solution
30.45 g N ×
69.55 g O ×
1 mol N
= 2.173 mol N
14.01 g N
1 mol O
= 4.347 mol O
16.00 g O
The gives a formula of N2.173O4.347. Dividing both subscripts by the smaller of the
two to get the smallest possible whole numbers (2.173/2.173 = 1, 4.347/2.173 ≈
2) gives an empirical formula of NO2.
Finally, dividing the approximate molar mass (92 g/mol) by the empirical formula
mass [14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol] gives 92/46.01 ≈ 2. Then,
multiplying both subscripts in the empirical formula by 2 gives the molecular
formula, N2O4.
Think About It Use the method described in Worked Example 5.13 to calculate
the percent composition of the molecular formula N2O4 and verify that it is the
same as that given in this problem.
5
Chapter Summary: Key Points
Compounds
Lewis Dot Symbols
Ionic Compounds and Ionic Bonding
Chemical Formulas
Lattice Energy
Ionic Compound Nomenclature
Lewis Theory of Bonding
Covalent Bonding
Laws of Definite Proportions and Multiple Proportions
Molecular, Structural, and Empirical Formulas
Molecular Compound Nomenclature
Inorganic and Organic Compounds
Polyatomic Ions
Oxyanions
Oxoacids
Hydrates
Molecular Mass and Formula Mass
Empirical Formula Mass
Molar Mass
Group Quiz #6:
• Identify the type(s) of bonds (ionic or
covalent) holding ALL atoms together in
the following compounds:
• MgCl2
• NCl3
• MgCO3
• Cl2
• Na2SO4
Group Quiz #7:
• Write neutral ionic formulas using the
following pairs of ions :
– Ba and Cl
– Co+2 and N
– Fe+3 and N
– K and P
– Zn and O
80
Binary Ionic Compounds: Group Quiz #8
• Write formulas for the following names:
– aluminum bromide
– titanium(IV) chloride
– chromium(III) sulfide
• Write names for the following formulas:
– Ba3N2
– K2S
– MnF2
81
Group Quiz #9
• Write names for formulas and formulas
for names:
•
•
•
•
•
XeF6
ammonium sulfite
Cr3N2
K3PO4
hydrochloric acid
82