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CHEMISTRY CHAPTER 7 SECTIONS 3-4
SECTION 3. USING CHEMICAL FORMULAS
Introduction
A chemical formula indicates:
• the elements present in a
compound
• the relative number of atoms or
ions of each element present in
a compound
Chemical formulas also allow
chemists to calculate a number
of other characteristic values for
a compound:
• formula mass
• molar mass
• percentage composition
Formula Masses
• 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.
• The molar mass is the same
number, but expressed in grams
instead of AMU. It is the mass of
one mole of the compound.
Example 1 (molecular compound):
formula mass of water, H2O
average atomic mass of H: 1.01 amu
ave. atomic mass of O: 16.00 amu
1.01 amu
2 H atoms 
 2.02 amu
H atom
16.00 amu
1 O atom 
 16.00 amu
O atom
average mass of H2O molecule:
18.02 amu
• Average mass of H2O molecule =
18.02 amu
• Molar mass of H2O = 18.02 g/mol
Example 2 (ionic compound)
• calcium nitrate = Ca(NO3)2
• A formula unit has 1 Ca2+ ion and
2 NO3- ions
Formula mass:
1 Ca atom x 40.08 amu/atom = 40.08 amu
2 N atoms x 14.01 amu/atom = 28.02 amu
6 O atoms x 16.00 amu/atom = 96.00 amu
total
164.10 amu
What is the molar mass of Ca(NO3)2?
Formula mass:
1 Ca atom x 40.08 amu/atom = 40.08 amu
2 N atoms x 14.01 amu/atom = 28.02 amu
6 O atoms x 16.00 amu/atom = 96.00 amu
total
164.10 amu
What is the molar mass of Ca(NO3)2?
164.10 g/mol
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.
• This uses the method of
dimensional analysis (factor-label
method) just as we did for
elements.
Mole-Mass Calculations
Example 1: grams to moles
How many moles of CO2 in 325
g?
First you need to calculate the
molar mass of CO2:
C: 1 x 12.0 = 12.0
O: 2 x 16.0 = 32.0
44.0 g/mol
Then convert g to mol:
1 mol
325
325 g 

mol  7.39 mol
44.0 g 44.0
Example 2: moles to grams
What is the mass of 0.15 mol
MgSO4?
First you need to calculate the
molar mass of MgSO4:
Mg 1 x 24.3 = 24.3
S 1 x 32.1 = 32.1
O 4 x 16.0 = 64.0
120.4 g/mol
Then convert mol to g:
120.4 g
0.15 mol 
 18 g
1 mol
Molar Mass as a Conversion
Factor - 75159
Recall that Avogadro’s number
(=
) is the number of
particles (atoms, molecules, or
formula units) in 1 mole of a
substance.
So we can convert between grams
or moles of a compound and
numbers of molecules or formula
units.
Recall that Avogadro’s number
( = 6.02 x 1023 ) is the number of
particles (atoms, molecules, or
formula units) in 1 mole of a
substance.
So we can convert between grams
or moles of a compound and
numbers of molecules or formula
units.
Converting Between Amount in Moles
and Number of Particles
Example 1: How many formula
units and atoms are there in
25.0 g MgCl2?
Calculate molar mass:
Mg 1 x 24.3 = 24.3
Cl 2 x 35.3 = 71.0
95.3 g/mol
Then convert g to formula units:
1 mol 6.02  10 formula units
25.0 g 

95.3 g
1 mol
23
 1.58  10 formula units
23
= 1.58 x 1023 atoms of Mg
and 3.16 x 1023 atoms of Cl
Example 2: What is the mass of
8.0 x 1030 molecules of CO2?
Calculate molar mass:
C 1 x 12.0 = 12.0
O 2 x 16.0 = 32.0
44.0 g/mol
Then convert molecules to g:
1 mol
44.0 g
8.0  10 molecules 

23
6.02 10 molecules 1 mol
30
 5.8  10 g  5.8  10 kg
8
5
Percentage Composition
( = Percent by Mass)
= the percent of the mass of a
compound contributed by each
element.
Use the following equation:
mass of element in sample of compound
 100 
mass of sample of compound
% element in compound
• The mass percentage of an element
in a compound is the same
regardless of the sample’s size.
• If you know the molar mass (or
calculate it from the formula), it is
convenient to use one mole as the
sample size.
• The sum of the percents for all
elements in the compound must add
up to 100%.
Percentage Composition of Iron
Oxides
Example 1: Calculate the percentage
composition of a sample that is 1.67
g Ce and 4.54 g I.
Total mass = 1.67 g + 4.54 g = 6.21 g
1.67 g
Ce 
100%  0.269 100%  26.9%
6.21 g
4.54 g
I
100%  0.731100%  73.1%
6.21 g
Example 2: Calculate the percent
by mass of each element in
Pb(NO3)4.
First calculate molar mass:
Pb 1 x 207.2 = 207.2
N
O
Example 2: Calculate the percent
by mass of each element in
Pb(NO3)4.
First calculate molar mass:
Pb 1 x 207.2 = 207.2
N 4 x 14.0 = 56.0
O 12 x 16.0 = 192.0
455.2 g/mol
We can then consider the sample size
to be 1 mol, and use the fraction that
each element contributed to the total:
207.2 g
Pb :
100%  45.5%
455.2 g
56.0 g
N:
100%  12.3%
455.2 g
192.0 g
O:
100%  42.2%
455.2 g
Check: 45.5%+12.3%+42.2% = 100%
Section 4 Determining Chemical Formulas
Objectives
• Define empirical formula, and explain how
the term applies to ionic and molecular
compounds.
• Determine an empirical formula from either a
percentage or a mass composition.
• Explain the relationship between the
empirical formula and the molecular formula
of a given compound.
• Determine a molecular formula from an
empirical formula.
• Determine the formula of a hydrated crystal.
• An empirical formula gives the
relative numbers of atoms in a
compound, using the smallest
whole number ratios.
• For an ionic compound, it is
usually the same as the formula
unit.
• For a molecular compound, the
molecular formula gives the
actual number of atoms in a
molecule.
• It may be the same as the
empirical formula, or it may be a
whole number multiple of it.
Examples:
molecular formula
CO2
Pb2O4
Hg2I2
glucose C6H12O6
empirical formula
Examples:
molecular formula
CO2
Pb2O4
Hg2I2
glucose C6H12O6
empirical formula
CO2
PbO2
HgI
CH2O
Calculation of Empirical Formulas
1. Convert grams to moles for
each element (if using
percentage composition,
assume a total mass of 100 g).
2. Divide by the smallest number
of moles to get subscripts.
Change to whole numbers if
necessary.
3. Write the formula using the
subscripts.
Example 1: What is the empirical
formula of a compound containing
28.3 g Ca and 14.6 g P?
1 mol
Ca : 28.3 g 
 0.706 mol
40.1 g
1 mol
P : 14.6 g 
 0.471 mol
31.0 g
Divide by smaller number (0.471
mol):
0.706 mol
0.471 mol
Ca :
 1.50 P :
1
0.471 mol
0.471 mol
Change to whole number ratio:
1.5:1 = 3:2. Formula: Ca3P2
(Name =
)
Divide by smaller number (0.471
mol):
0.706 mol
0.471 mol
Ca :
 1.50 P :
1
0.471 mol
0.471 mol
Change to whole number ratio:
1.5:1 = 3:2. Formula: Ca3P2
(Name = calcium phosphide )
Example 2: A compound has a
percent composition of 40.0%C,
6.71% H, and 53.3% O. What is
its empirical formula?
For a total mass of 100 g, there
are 40.0 g C, 6.71 g H, and 53.3
g O.
1 mol
C : 40.0 g 
 3.33 mol
12.0 g
1 mol
H : 6.71 g 
 6.71 mol
1.0 g
1 mol
O : 53.3 g 
 3.33 mol
16.0 g
Divide by smallest number (3.33 mol):
3.33 mol
6.71 mol
3.33 mol
C:
1 H :
 2.02 O :
1
3.33 mol
3.33 mol
3.33 mol
Change to whole number ratio: 1:2:1.
Empirical formula: CH2O
Calculation of Molecular Formulas
• The empirical formula contains the
smallest possible whole numbers
that describe the atomic ratio.
• The molecular formula is the actual
formula of a molecular compound.
• An empirical formula may or may
not be a correct molecular formula.
• The relationship between a
compound’s empirical formula
and its molecular formula can
be written as follows:
n(empirical formula) =
molecular formula
• The formula masses have a
similar relationship:
n(empirical formula mass) =
molecular formula mass
Comparing Empirical and Molecular
Formulas
• To determine the molecular
formula of a compound, you
must know the compound’s
molecular formula mass (molar
mass) [usually given in a
problem].
• First determine the empirical
formula and the empirical
formula mass.
Then determine n from the ratio:
molecular formula mass
n
empirical formula mass
Finally, multiply the empirical
formula by n to get the
molecular formula.
Example 1: A sample contains 0.076
mol C and 0.0765 mol H, and has a
molecular mass of 78 AMU. What are
the empirical and molecular formulas?
First determine the empirical formula:
mol H 0.0765

 1.0066  1
mol C 0.076
Empirical formula =
Example 1: A sample contains 0.076
mol C and 0.0765 mol H, and has a
molecular mass of 78 AMU. What are
the empirical and molecular formulas?
First determine the empirical formula:
mol H 0.0765

 1.0066  1
mol C 0.076
Empirical formula =
CH
Next determine the empirical formula
mass (work in AMU since molecular
mass was given in AMU):
C:
1 x 12.0 = 12.0
H:
1 x 1.0 = 1.0
13.0 AMU
molecular formula ma ss 78 AMU
n

6
empirical formula mass 13 AMU
Molecular formula = n(empirical
formula) =
Next determine the empirical formula
mass (work in AMU since molecular
mass was given in AMU):
C:
1 x 12.0 = 12.0
H:
1 x 1.0 = 1.0
13.0 AMU
molecular formula ma ss 78 AMU
n

6
empirical formula mass 13 AMU
Molecular formula = n(empirical
formula) = 6(CH) = C6H6
Example 2: (p. 249, #4):
4.04 g of N combine with 11.46 g
of O to produce a compound with
a formula mass of 108.0 amu.
What is the molecular formula?
Empirical formula:
1 mol
N : 4.04 g 
 0.289 mol
14.0 g
1 mol
O : 11.46 g 
 0.716 mol
16.0 g
mol O 0.716

 2.48  2.5
mol N 0.289
Whole number ratio: 5:2.
Empirical formula =
Empirical formula:
1 mol
N : 4.04 g 
 0.289 mol
14.0 g
1 mol
O : 11.46 g 
 0.716 mol
16.0 g
mol O 0.716

 2.48  2.5
mol N 0.289
Whole number ratio: 5:2.
Empirical formula = N2O5
Next determine the empirical formula
mass:
N:
2 x 14.0 = 28.0
O:
5 x 16.0 = 80.0
108.0 AMU
molecular formula mass 108.0 AMU
n

1
empirical formula mass 108.0 AMU
Molecular formula =
n(empirical formula) = N2O5
Hydrated Crystals
Some compounds crystallizing from
water solution incorporate water
molecules into the crystal. These
hydrates contain a specific ratio of
water to compound:
(formula for compound)•xH2O
Example: CuSO4•2H2O
Formulas for hydrates are
determined by determining the
mass of the hydrated compound,
heating to remove all water, and
then determining the mass of the
anhydrous (= without water)
compound.
From this the moles of water per
mole of anhydrous compound can
be calculated.
Example: a 10.407 g sample of
hydrated barium iodide (BaI2) is
heated to drive off water. The dry
sample has a mass of 9.520 g.
What is the formula for the hydrate?
First determine the mass of
water:
hydrated (= BaI2 + H2O) 10.407 g
anhydrous (= BaI2)
-9.520 g
water alone
0.887 g
Then determine moles of each
compound and their ratio:
BaI2:
Ba
I
1 x 137.3 = 137.3
2 x 126.9 = 253.8
391.1 g/mol
1 mol
9.520 g 
 0.0243 mol
391.1 g
H2O:
H
O
2 x 1.0 = 2 .0
1 x 16.0 = 16.0
18.0 g/mol
1 mol
0.887 g 
 0.0493 mol
18.0 g
Then calculate the ratio:
mol H 2 O 0.0493

 2.03  2
mol BaI 2 0.0243
Formula =
H2O:
H
O
2 x 1.0 = 2 .0
1 x 16.0 = 16.0
18.0 g/mol
1 mol
0.887 g 
 0.0493 mol
18.0 g
Then calculate the ratio:
mol H 2 O 0.0493

 2.03  2
mol BaI 2 0.0243
Formula = BaI2•2H2O