Section 7.3 - Percent
Composition and
Chemical Formulas
Calculating the Percent
Composition of a
compound
The relative amounts of
each element (Part) in a
compound (Whole) are
expressed as the Percent
Composition or the Percent
by Mass of each element in
the compound.
Part
%
x 100
Whole
The percent
composition of a
compound has as many
percent values as there
are different elements
in the compound.
The percent composition
of CaCO3 is:
Ca = 40.1 %
C = 12.0 %
O = 47.9 %
To calculate the percent
composition of a
compound, use the
chemical formula to
determine which elements
and how many moles of
each are in the compound.
Divide the total mass of
one of the elements in the
compound by the molar
mass of the compound
then multiply by 100 to
convert to a percent.
Remember that the total
mass of the element is
the molar mass of the
element multiplied by its
subscript in the formula.
Example: What is the
percent composition of
FeSO4?
55.8g
Fe = 151.9g x 100 = 36.7 %
32.1g
S = 151.9g x 100 = 21.1 %
64.0g
O = 151.9g x 100 = 42.2 %
The Percent By Mass in
a compound is the
number of grams of the
element divided by the
number of grams of the
compound multiplied by
100.
Sample Problem 7 - 10 p 189
An 8.20 g piece of magnesium
combines with 5.40 g of
oxygen to form a compound.
What is the Percent
Composition (Percent By
Mass) of this compound?
% Mg =
% Mg
8.20 g
=13.60 g x 100  60.3 %
%O=
%O
mass of Mg
x 100
mass of compound
mass of O
x 100
mass of compound
5.40 g
= 13.6 g x 100  39.7 %
Using Percent
Composition as a
Conversion Factor
You can use the percent
composition to
determine the mass of
each element in a
sample of a compound.
Example:
Aluminum makes up
52.9 % of Al2O3.
Calculate the mass of
Aluminum in a 50.0 g
sample of Al2O3.
We first assume that we
have a 100.0g sample of
Al2O3. 52.9 % of 100.0 g
is 52.9 g. We can use
52.9 g divided by 100.0 g
sample as a conversion
factor.
50.0g 52.9g
x
 26.5g Al
1
100.0g
Calculating Empirical
Formula
Empirical Formula
EMPIRICAL FORMULA
An empirical
formula is the
smallest whole
number ratio of
atoms in a formula.
EMPIRICAL FORMULA
If the atoms are in a
small whole number
ratio, then the moles of
atoms of each element
in the substance will
also be in a small
whole number ratio.
Two Sources of Data From
Which to Calculate the
Empirical Formula
Experimental Analysis - What
is the empirical formula of a
compound if a 2.50 gram
sample contains 0.900 g of
calcium and 1.60 g of chlorine?
Two Sources of Data From
Which to Calculate the
Empirical Formula
Percentage Composition - A
compound has a percentage
composition of 40.0% C, 6.71%
H, and 53.3% O. What is the
empirical Formula?
The following is an
Example of the Empirical
Formula Calculation from
Experimental Analysis.
What is the empirical
formula for a compound
if a 2.50 g sample
contains 0.900 g of
calcium and 1.60 g of
chlorine?
Determine the number of
moles of each element in
the compound.
0.900 g Ca 1 mole Ca
x

1
40.1 g Ca
0. 0224 moles Ca
1.60 g Cl
1 mole Cl
x

1
35.5 g Cl
0.0450 moles Cl
The next step is to determine
the smallest whole number
ratio.
To obtain the smallest whole
number ratio, divide both
numbers of moles by the
smaller number of moles.
0.0224 moles Ca
1
0.0224 moles
0.0450 moles Cl
2
0.0224 moles
The mole ratio is 1 mole
of Calcium to 2 moles of
Chlorine.
The Empirical formula is
therefore CaCl2.
Example of Empirical
Formula Calculation From
Percentage Composition:
A compound has a
percentage composition of
40.0% C, 6.71% H, and
53.3% O. What is the
empirical formula?
Example of Empirical
Formula Calculation From
Percentage Composition
 To calculate the ratio
of moles of these
elements, we assume a
100 g sample . 40.0%
of 100 g is 40.0 g.
Example of Empirical
Formula Calculation From
Percentage Composition
 Next we calculate the
number of moles of
each element present.
Example of Empirical
Formula Calculation From
Percentage Composition
Mol C
Mol H
Mol O
40.0gC 1molC
x

1
12.0gC
3.33
6.71gH 1molH
x
 6.67
1
1.01gH
53.3gO 1molO
x
 3.33
1
16.0gO
4 Determine the smallest
Whole Number Ratio.
The number of moles of each
element calculated in step 2 were:
3.33 mol C, 6.64 mol H, and 3.33
mol O. Dividing by the smallest
value we obtain the smallest whole
number ratio of moles.
3.33 mol
1
For C
3.33 mol
6.64 mol
2
For H
3.33 mol
3.33 mol
1
For O
3.33 mol
With a ratio of 1-C to
2-H to 1-O
The empirical formula
is therefore --- CH2O.
What is the empirical
formula for a compound
if a 40.0 g sample
contains 29.7 g of sodium
and 10.3 g of oxygen?
Solution:
Step 1 Determine the
number of moles of each
element.
29.7 g Na 1 mole Na
x
 1.29 moles Na
1
23.0 g Na
10.3 g O 1 mole O
x
 .644 moles O
1
16.0 g O
Step 2 Divide each
number of moles by the
smallest.
This will give the
smallest whole number
ratio.
1.29
2
0.644
0.644
1
0.644
This ratio shows that
there are 2 moles of
sodium for one mole of
oxygen.
The empirical formula is
therefore Na2O.
Calculating Molecular
Formula
The Empirical Formula
tells us the ratio of each
kind of atom present in a
compound.
The Molecular Formula
tells us the actual number
of each kind of atom in a
molecule of a compound.
If the empirical formula
and the molar mass are
known, the molecular
formula can be
calculated.
Example:
A certain gas, having a
molar mass of 78 g, has
an empirical formula of
CH. What is the
molecular formula?
If we multiply the mass
of the empirical formula
by an integer, the
product must equal the
molar mass.
(CH)x = 78 g
Where x represents an
integer.
(13)x = 78 g
X=6
By multiplying each
subscript in the
empirical formula by the
integer (6), we obtain the
molecular formula.
(CH)6 gives C6H6
Problem:
The empirical formula of
a compound is CHOCl
and the molar mass is
129 g. What is the
molecular formula?