1. Empirical and Molecular Formulas
As part of Dalton’s atomic theory, he stated that atoms will combine with one another in simple whole number ratios to form compounds.
An example of a molecular formula, actual formula, is C6H6. This is benzene.
An example of a simple whole number ratio, known as the empirical formula, is CH.

2. Comparing Molecular and Empirical Formulas
Name of compound
Molecular Formula
Empirical Formula
Hydrogen peroxide
H2O2 HO
Glucose
C6H12O6
CH2O
Benzene
C6H6 CH
Ethyne
C2H2
Aniline
C6H7N
Water
H2O

3. Determining Empirical Formula
You already know how to calculate percentage composition. In this section you are going to use the percent composition to find the formula of the compound.
There are three steps to determining the empirical formula.
Determine the mass of the elements present
Determine the moles of the elements present
Moles of element/divided by the smallest mole will give you a ratio and that will help you determine the formula.

4. Empirical Formula Example
What is the empirical formula of a compound with 79.9 g of Cu and 20.1 g of S?
Mass is given, so no need to determine.
Moles
79.9 g x 1mol = 1.24 mol Cu g
20.1 g x 1mol = mol S g
Mole Ratio
Cu  1.24/0.627 = 1.98 ~ 2
S  0.627/0.627 = 1
Formula
Cu2S

5. Another Example!
Calculate the empirical formula of a compound that has 85.6% carbon and 14.4% hydrogen.
Assume that the percentage composition is based on 100 g of the compound and that way you can use it in the calculation directly.

6. Mass is not given, but we are assuming 85. 6 % C is 85. 6 g
Mass is not given, but we are assuming 85.6 % C is 85.6 g. And we are assuming 14.4% H is 14.4 g.
Moles
85.6 g x 1mol = 7.13 mol C g
14.4 g x 1mol = 14.3 mol H g
Mole Ratio
C  7.13/7.13 = 1
H  14.3/7.13 = 2
Formula
CH2

7. A Tricky Example
What happens if the mole ratio step does not give you nice whole numbers?
The percentage composition of a fuel is 81.7% C and 18.3% H. What is the empirical formula?

8. Mass is not given, but we are assuming 81. 7 % C is 81. 7 g
Mass is not given, but we are assuming 81.7 % C is 81.7 g. And we are assuming 18.3% H is 18.3 g.
Moles
81.7 g x 1mol = 6.80 mol C g
18.3 g x 1mol = 18.2 mol H g
Mole Ratio
C  6.80/6.80 = 1 x 3  3
H  18.2/6.80 = x 3  8
Formula
C3H8

9. 0.2, 1.2, 2.2, 3.2……. Multiply by 5
0.25, 1.25, … …Multiply by 4
0.3, 1.3, 2.3, 3.3…… Multiply by 3
0.5, 1.5, 2.5, 3.5…… Multiply by 2