Chapter 3
Percent Compositions
and
Empirical Formulas
AP Chemistry
1
Calculating Percent Composition of a
Compound

The following general formula is used to
determine the percent composition of
each element in a compound:

Mass of element

(in 1 mole of compound)
 Molar Mass of Compound
2
X 100
What is the percent composition
of each element in Ca(NO3)2 ?

Step 1: Find the molar mass of Ca(NO3).

Ca 40.1 grams/mole
N 2 x (14.0) grams/mole
O 6 x (16.0) grams/mole



3
= 40.1 g/mol
= 28.0g/mol
= 96.0 g/mol
164.1 g/mole
Step 2: Divide the mass of each element by the
molar mass and multiply by 100 to get the percent.

% Ca =

40.1g
164.1g
x 100% =
24.4 % Calcium
17.1% Nitrogen


%N=
28.0g
164.1 g
x 100%
%O=
96.0g
164.1g
x 100% =

=



4
58.5 % Oxygen
*Note percent composition can also be done given masses
of a compound and each element in the compound.

Calculate the percent composition of a
compound that is made of 29.0 grams of
Ag with 4.30 grams of S.
29.0 g Ag
X 100 = 87.1 % Ag
33.3 g total
Total = 100 %
4.30 g S
33.3 g total
5
X 100 = 12.9 % S
Formulas
Empirical formula: the lowest whole
number ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
• The empirical formula for C3H15N3 is
CH5N.
• you are basically dividing by the greatest
common factor , which in this case is 3,
for each element subscript.
6
Formulas (continued)
Formulas for ionic compounds are
ALWAYS empirical (the lowest whole
number ratio = cannot be reduced).
Examples:
NaCl
7
MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds
MIGHT be empirical (lowest whole
number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
H2O
CH2O
C12H22O11
(Correct formula)
Empirical:
(Lowest whole
number ratio)
8
Calculating Empirical
 We can get a ratio from the percent
composition.
1) Assume you have a 100 g sample
- the percentage become grams (75.1% = 75.1 grams)
2) Convert grams to moles.
3) Find lowest whole number ratio by
dividing each number of moles by
the smallest value.
9






Example
Calculate the empirical formula of a
compound composed of 38.67 % C, 16.22
% H, and 45.11 %N.
Step 1 : Assume 100gram sample
Step 2 : Convert grams to moles for each element
38.67 g C x 1mol C = 3.22 mole C
12.0 g C
16.22 g H x 1mol H
= 16.22 mole H
1.0 g H
45.11 g N x 1mol N
= 3.22 mole N
14.0 g N
Now divide each value by the smallest value
10
Example


The ratio is 3.22 mol C = 1 mol C
3.22 mol N
1 mol N
The ratio is 16.22 mol H = 5 mol H
3.22 mol N
1 mol N
= C1H5 N1
11
which is = CH5N
Empirical to molecular
 Since the empirical formula is the
lowest ratio, the actual molecule
would weigh more.
By a whole number multiple.
 Divide the actual molar mass by the
empirical formula mass – you get a
whole number to increase each
coefficient in the empirical formula
12
Caffeine has a molar mass of 194 g. and its
empirical formula is C4H5N2O, what is its
molecular formula?
Find the empirical formula molar mass.
 97 g/mole
 Set up ratio
194g
= 2

97g
 Multiply each subscript in the empirical
formula by 2.
 C8H10N4O2 is the molecular formula.

13
Empirical formula from combustion reactions
Example: 7.321 grams of an organic compound
containing carbon, hydrogen, and oxygen was analyzed by
combustion. The amount of carbon dioxide produced was
17.873 grams and the amount of water produced was
7.316 grams. Determine the empirical formula of the
compound.
Combustion reaction looks like the following:
CHO + O2 
14
CO2 + H2O
Assume all of the Carbon in the compound
turned into carbon dioxide, and determine the
mass percent of carbon in the compound
17.873 g CO2

1mole 1 mol C
12g C
44g CO2 1 mol CO2 1 mol
= 4.8744 grams Carbon
4.8744 g x 100
7.321 g
15
=
66.58% Carbon
Assume all of the hydrogen in the compound went to
water, and determine the mass percent of hydrogen
in the compound
7.316 g H2O 1mol H2O
18g H2O
2 mols H 1.0 g H
1mol H2O 1 mol H
= 0.8129 grams Hydrogen
16
0.8129
7.321
x 100 = 11.10% Hydrogen
With the mass percent of Carbon and
Hydrogen, you can figure out the mass
percent of oxygen in the compound.

100 - (66.58% C + 11.10% H) =

22.32 % Oxygen
Now use these percents to find the
empirical formula!
 C 4H 8O

17