Determining Chemical
Formulas Experimentally
% composition, empirical and molecular
formula
Percentage Composition
The mass of each element in a compound
compared to the entire mass of the compound
times 100 is the percent composition of that
element.
Example: What is the % composition of H and O in 1
mole of H2O?
%H = 1.01 g H x 2 x 100 = 11%
18.02 g H2O
=100%
%O = 16.00 g O x 100 = 89%
18.02 g H2O
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Percentage Composition
Example:
A sample of an unknown compound with amass of
0.2370 g contains 0.0948 g of carbon, 0.126 g of
oxygen, 0.016 g of hydrogen. What is the %
composition of the compound?
%C = 0.0948g X 100 = 40%
0.2370 g
%O = 0.126 g x 100 = 53 %
0.2370 g
%H = 0.016 g x 100 = 7%
0.2370
Empirical Formula
The simplest whole number ratio of the
elements in a compound is the empirical
formula.
Steps:
1.
Find moles of each element in the compound.
2.
Divide each mole value by the smallest mole
value to find the mole ratio.
3.
Write the formula using the mole ratios
determined in step 2.

A Poem
Percent to grams
Grams to moles
Divide by smallest
Multiply till whole
Empirical Formula
Example: A sample is analyzed and
determined to contain 80 g C and 20 g
H. What is the empirical formula?
1. 80 g C x 1 mole = 6.7 mol C
12.0 g
20 g H x 1 mole = 20 mol H
1.01 g
Empirical Formula
2.
3.

Mole ratios:
C: 6.7 = 1
H: 20 = 2.98 ~ 3
6.7
6.7
CH3 is the empirical formula
You may round the mole ratio if it is
within 0.05 of a whole number. If it is
not a whole number you must multiply
all the mole ratios by a factor to get a
whole number.
Empirical Formula
Example: What is the empirical formula of a
compound that is 70% Fe and 30%O?
1. 70% Fe = 70 g Fe x 1 mole = 1.25 mol
55.85 g
30% O = 30 g O x 1 mole = 1.88 mol
16.00 g
2. Fe = 1.25/1.25 = 1
O = 1.88/ 1.25 = 1.5
Empirical Formula
3.
Empirical formula must be in whole
number ratios.
Fe = 1 and O = 1.5, multiply both ratios by 2
Fe = 2 and O = 3, so the formula is:
Fe2O3
Molecular Formula
The formula of the actual molecular
compound is the molecular formula.
Steps:
1. Calculate the empirical mass.
2. Divide the molecular mass by the
empirical mass.
3. Multiply each subscript in the empirical
formula; by the number found in step 2.

Molecular Formula
Example: The molecular mass of a
compound is found to be 180g/mol. If
the empirical formula is CH2O,
determine the molecular formula.
1. CH2O = 12.01g/mol + (1.01g/mol x 2)
+ 16.00 g/mol = 30.02 g/mol
2. I80g/mole
=6
30.02 g/mole
3. CH2O x 6 = C6H12O6
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