APPLICATIONS
OF THE MOLE
Molar Mass of a Compound
• The molar mass of a compound is the
mass of a mole of the representative
particles of the compound.
• Because each representative particle is
composed of two or more atoms, the
molar mass of the compound is found by
adding the molar masses of all of the
atoms in the representative particle.
Molar Mass of an Element


To determine the molar mass of an
element, find the element’s symbol on the
periodic table and round the mass so there
is one digit beyond the decimal.
The molar mass of carbon (C) is
12.0 g/mol, of chlorine (Cl) is 35.5 g/mol
and of iron (Fe) is 55.8 g/mol.
Molar Mass of a Compound
• In the case of NH3, the molar mass
equals the mass of one mole of nitrogen
atoms plus the mass of three moles of
hydrogen atoms.
Molar Mass of a Compound
Molar mass of NH3 = molar mass of N +
3(molar mass of H)
Molar mass of NH3 = 14.0 g +
3(1.0 g)
= 17.0 g/mol
Molar Mass of a Compound
• You can use the molar mass of a
compound to convert between mass and
moles, just as you used the molar mass
of elements to make these conversions.
Molar Mass Conversion

How many moles of magnesium in
56.3 g of Mg?
56.3 g Mg
__
1 mole Mg
__
24.3 g Mg
(2.32)
Molar Mass Conversion
• How many moles are in 146 grams of
NH3?
(Answer: 8.59 mol)
Molar Mass Conversion
• How many moles are in 295 grams of
Cr(OH)3?
(Answer: 2.86 mol)
Molar Mass Conversion
• How many moles are in 22.5 grams of
HCl?
(Answer: 0.616 mol)
Molar Mass Conversion

How many grams of sodium chloride in
3.45 moles of NaCl?
3.45 moles NaCl
__ g NaCl
58.5
__
1 mole NaCl
(202)
Molar Mass Conversion
• How many grams are in 0.120 moles of
AlF3?
(Answer: 10.1 g)
Molar Mass Conversion
• How many grams are in 13.0 moles of
H2SO4?
(Answer: 1280 g)
Molar Mass Conversion
• How many grams are in 1.6 moles of
K2CrO4?
(Answer: 310 g)
Percent Composition
• Recall that every chemical compound
has a definite composition - a
composition that is always the same
wherever that compound is found.
• The composition of a compound is
usually stated as the percent by mass of
each element in the compound.
Percent Composition
• The percent of an element (X) in a
compound can be found in the following
way.
(molar mass of X) (# X ‘s)
%X=
molar mass of compound
x 100
Percent Composition
Example
• Determine the percent composition of
chlorine in calcium chloride (CaCl2).
• First, analyze the information available
from the formula.
• A mole of calcium chloride consists of
one mole of calcium ions and two
moles of chloride ions.
Calculating Percent Composition
• Next, gather molar mass information
from the atomic masses on the periodic
table.
Calculating Percent Composition
• To the mass of one mole of CaCl2, a
mole of calcium ions contributes 40.1 g,
and two moles of chloride ions
contribute 2 x 35.5 g = 71.0 g for a
total molar mass of 111.1 g/mol for
CaCl2.
Calculating Percent Composition
• Finally, use the data to set up a
calculation to determine the percent by
mass of an element in the compound.
• The percent by mass of chlorine in CaCl2
can be calculated as follows.
Calculating Percent Composition
CaCl2
(molar mass of X) (# X ions)
%X=
x 100
molar mass of compound
( 35.5 g) ( 2 Cl ions)
% Cl =
X 100
111.1 g
% Cl in CaCl2 = 63.9%
Example
• Determine the percent composition of
carbon in sodium acetate (NaC2H3O2).
(molar mass of X) (# X ions)
%X=
molar mass of compound
x 100
Calculating Percent Composition
NaC2H3O2
( 12.0 g) ( 2 C’s)
%C=
X 100
82.0 g
% C = 29.3%
Problem
Calculate the percent composition
aluminum of aluminum oxide (Al2O3).
52.9% Al
Problem
Determine the percent composition of
oxygen in magnesium nitrate, which has
the formula Mg(NO3)2.
64.7% O
Problem
Determine the percent composition of
sulfur in aluminum sulfate, which has the
formula Al2(SO4)3.
28.1% S
Problem
Determine the percent composition of
oxygen in zinc nitrite, which has the
formula Zn(NO2)2.
40.7% O
Percent Water in a Hydrate
• Hydrates are compounds that
incorporate water molecules into their
fundamental solid structure. In a hydrate
(which usually has a specific crystalline
form), a defined number of water
molecules are associated with each
formula unit of the primary material.
Percent Water in a Hydrate
• Gypsum is a hydrate with two water
molecules present for every formula unit
of CaSO4. The chemical formula for
gypsum is CaSO4 • 2 H2O and the
chemical name is calcium sulfate
dihydrate. Note that the dot in the
formula (or multiplication sign) indicates
that the waters are there.
Percent Water in a Hydrate
• Other examples of hydrates are:



lithium perchlorate trihydrate LiClO4 • 3 H2O;
magnesium carbonate pentahydrate MgCO3 • 5 H2O;
and copper (II) sulfate pentahydrate CuSO4 • 5 H2O.
Percent Water in a Hydrate
• The water in the hydrate (referred to as
"water of hydration") can be removed by
heating the hydrate. When all hydrating
water is removed, the material is said to
be anhydrous and is referred to as an
anhydrate.
Percent Water in a Hydrate
• Experimentally measuring the percent
water in a hydrate involves first heating a
known mass of the hydrate to remove
the waters of hydration and then
measuring the mass of the anhydrate
remaining. The difference between the
two masses is the mass of water lost.
Percent Water in a Hydrate
• Dividing the mass of the water lost by the
original mass of hydrate used is equal to
the fraction of water in the compound.
Multiplying this fraction by 100 gives the
percent water.
Example
• Determine the percent water in
CuSO4 • 5 H2O (s).
Calculating Percent Composition
• Mass of CuSO4 =
63.6 + 32.1 + 16.0 (4) = 159.7
• Mass of 5 H2O =
1.0 (2) + 16.0 = 18.0 x 5 = 90.0
• Mass of CuSO4 . 5 H2O =
159.7 + 90.0 = 249.7
Calculating Percent Composition
% H2O =
( 18.0 g) ( 5 H2O’s)
X 100
249.7 g
% H2O = 36.0%
Problem
• Determine the percent water in
MgCO3 • 5 H2O (s).
51.6% H2O
Problem
• Determine the percent water in
LiClO4 • 3 H2O (s).
33.7% H2O
Empirical Formula
• You can use percent composition data to
help identify an unknown compound by
determining its empirical formula.
• The empirical formula is the simplest
whole-number ratio of atoms of elements
in the compound. In many cases, the
empirical formula is the actual formula
for the compound.
Empirical Formula
• For example, the simplest ratio of atoms
of sodium to atoms of chlorine in sodium
chloride is 1 atom Na : 1 atom Cl.
• So, the empirical formula of sodium
chloride is Na1Cl1, or NaCl, which is the
true formula for the compound.
Empirical Formula
• The formula for glucose is C6H12O6.
• The coefficients in glucose are all
divisible by 6.
• The empirical formula of glucose is
CH2O.
Problem
• Determine the empirical formula for
Tl2C4H4O6.
TlC2H2O3
Problem
• Determine the empirical formula for
N2O4.
NO2
Empirical Formula
from Percent Composition
• The percent composition of an unknown
compound is found to be 38.43% Mn,
16.80% C, and 44.77% O. Determine
the compound’s empirical formula.
• Because percent means “parts per
hundred parts,” assume that you have
100 g of the compound.
Empirical Formula
from Percent Composition
• Then calculate the number of moles of
each element in the 100 g of compound.
Empirical Formula
from Percent Composition
• The number of moles of manganese
may be calculated as follows.
Given: 38.43% Mn
38.43 g Mn 1 mol Mn
54.9 g Mn
= 0.7000 mol Mn
Empirical Formula
from Percent Composition
• The number of moles of carbon may be
calculated as follows.
Given: 16.80% C
16.80 g C
1 mol C
12.0 g C
= 1.400 mol C
Empirical Formula
from Percent Composition
• The number of moles of oxygen may be
calculated as follows.
Given: 44.77% O
44.77 g O
1 mol O
16.0 g O
= 2.798 mol O
Empirical Formula
from Percent Composition
• The results show the following
relationship.
mol Mn : mol C : mol O
0.7000 : 1.400 : 2.798
Empirical Formula
from Percent Composition
• To obtain the simplest whole-number
ratio of moles, divide each number of
moles by the smallest number of moles.
0.7000 mol Mn
= 1 mol Mn
0.7000
Empirical Formula
from Percent Composition
1.400 mol C
= 2 mol C
0.7000
2.798 mol O
= 4 mol O
0.7000
Empirical Formula
from Percent Composition
• The mole ratio for the compound is:
1 Mn : 2 C : 4 O
• The empirical formula for the compound
is MnC2O4.
Example
• Determine the empirical formula of the
following compound.
31.9 g Mg, 27.1 g P
Example
• The number of moles of magnesium
may be calculated as follows.
Given: 31.9 g Mg
31.9 g Mg
1 mol Mg
24.3 g Mg
= 1.31 mol Mg
Example
• The number of moles phosphorous may
be calculated as follows.
Given: 27.1 g P
27.1 g P
1 mol P
31.0 g P
= 0.874 mol P
Empirical Formula
from Percent Composition
• The results show the following
relationship.
mol Mg : mol P
1.31 : 0.874
Empirical Formula
from Percent Composition
• To obtain the simplest whole-number
ratio of moles, divide each number of
moles by the smallest number of moles.
0.874 mol P
= 1 mol P
0.874
Empirical Formula
from Percent Composition
1.31 mol Mg
= 1.5 mol Mg
0.874
Empirical Formula
from Percent Composition
• The mole ratio for the compound is:
1.5 Mg : 1 P
• You cannot have a fraction, so multiply
both numbers by 2.
3 Mg : 2 P
Empirical Formula
from Percent Composition
• The empirical formula for the compound
is Mg3P2.
Problem
The composition of an unknown acid is
40.00% carbon, 6.71% hydrogen, and
53.29% oxygen. Calculate the empirical
formula for the acid.
CH20
Problem
The composition of an unknown ionic
compound is 60.7% nickel and 39.3%
fluorine. Calculate the empirical formula
for the ionic compound.
NiF2
Problem
The composition of a compound is 6.27 g
calcium and 1.46 g nitrogen. Calculate
the empirical formula for the compound.
Ca3N2
Problem
Find the empirical formula for a compound
consisting of 63.0% Mn and 37.0% O.
MnO2
Molecular Formulas
• For many compounds, the empirical
formula is not the true formula.
• A molecular formula tells the exact
number of atoms of each element in a
molecule or formula unit of a compound.
Molecular Formulas
• The molecular formula for a compound is
either the same as the empirical formula
or a whole-number multiple of the
empirical formula.
Molecular Formulas
• In order to determine the molecular
formula for an unknown compound, you
must know the molar mass of the
compound in addition to its empirical
formula.
Molecular Formulas
• Then you can compare the molar mass
of the compound with the molar mass
represented by the empirical formula as
shown in the following example
problem.
Example
• The molecular mass of benzene is
78 g/mol and its empirical formula is CH.
What is the molecular formula for
benzene?
Determining a Molecular Formula
• Calculate the molar mass represented by
the formula CH.
• Here, the molar mass is the sum of the
masses of one mole of each element.
Molar mass CH = 12.0 g + 1.0 g
Molar mass CH = 13.0 g/mol
Example
• As stated in the problem, the molecular
mass of benzene is known to be
78.0 g/mol.
Determining a Molecular Formula
• To determine the molecular formula for
benzene, calculate the whole number
multiple, n, to apply to its empirical
formula.
78.0 g/mol
n=
=6
13.0 g/mol
Determining a Molecular Formula
• This calculation shows that the molar
mass of benzene is six times the molar
mass of its empirical formula CH.
Determining a Molecular Formula
• Therefore, the molecular formula must
have six times as many atoms of each
element as the empirical formula.
• Thus, the molecular formula is
(CH)6 = C6H6
Example
• The simplest formula for butane is C2H5
and its molecular mass is about
60.0 g/mol. What is the molecular
formula of butane?
Determining a Molecular Formula
• To determine the molecular formula for
butane, calculate the whole number
multiple, n, to apply to its empirical
formula.
60.0 g/mol
n=
=2
29.0 g/mol
Determining a Molecular Formula
• Therefore, the molecular formula must
have two times as many atoms of each
element as the empirical formula.
• Thus, the molecular formula is
(C2H5)2 = C4H10
Problem
What is its molecular formula of cyanuric
chloride, if the empirical formula is CClN
and the molecular mass is 184.5 g/mol?
C3Cl3N3
Problem
The simplest formula for vitamin C is
C3H4O3. Experimental data indicates that
the molecular mass of vitamin C is about
180. What is the molecular formula of
vitamin C?
C6H8O6
Example
• Maleic acid is a compound that is widely
used in the plastics and textiles
industries.
• The composition of maleic acid is 41.39%
carbon, 3.47% hydrogen, and 55.14%
oxygen.
• Its molar mass is 116.1 g/mol.
Calculate the molecular formula for
maleic acid.
Determining a Molecular Formula
• Start by determining the empirical
formula for the compound.
Example
• The number of moles of carbon may be
calculated as follows.
Given: 41.39% C
41.39 g C
1 mol C
12.0 g C
= 3.449 mol C
Example
• The number of moles of hydrogen may
be calculated as follows.
Given: 3.47% H
3.47 g H
1 mol H
1.0 g H
= 3.47 mol H
Example
• The number of moles of oxygen may be
calculated as follows.
Given: 55.14% O
55.14 g O
1 mol O
16.0 g O
= 3.446 mol O
Determining a Molecular Formula
• The numbers of moles of C, H, and O
are nearly equal, so it is not necessary to
divide through by the smallest value.
• You can see by inspection that the
smallest whole-number ratio is
1C : 1H : 1O, and the empirical formula
is CHO.
Determining a Molecular Formula
• Next, calculate the molar mass
represented by the formula CHO.
• Here, the molar mass is the sum of the
masses of one mole of each element.
Molar mass CHO = 12.0 g + 1.0 g + 16.0 g
Molar mass CHO = 29.0 g/mol
Example
• As stated in the problem, the molar
mass of maleic acid is known to be
116.1 g/mol.
Determining a Molecular Formula
• To determine the molecular formula for
maleic acid, calculate the whole number
multiple, n, to apply to its empirical
formula.
116.1 g/mol
n=
=4
29.0 g/mol
Determining a Molecular Formula
• This calculation shows that the molar
mass of maleic acid is four times the
molar mass of its empirical formula
CHO.
Determining a Molecular Formula
• Therefore, the molecular formula must
have four times as many atoms of each
element as the empirical formula.
• Thus, the molecular formula is
(CHO)4 = C4H4O4
Determining a Molecular Formula
Determining a Molecular Formula
Problem
The composition of silver oxalate is
71.02% silver, 7.91% carbon, and
21.07% oxygen. If the molar mass of
silver oxalate is 303.8 g/mol, what is its
molecular formula?
Ag2C2O4
Problem
The composition of a compound is 85.6%
carbon and 14.4% hydrogen. If the molar
mass of the compound is 42.1 g/mol,
what is its molecular formula?
C3H6