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.

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

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.

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 64.546 g 20.1 g x 1mol = 0.627 mol S 32.065 g Mole Ratio Cu  1.24/0.627 = 1.98 ~ 2 S  0.627/0.627 = 1 Formula Cu2S

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.

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 12.011 g 14.4 g x 1mol = 14.3 mol H 1.00794 g Mole Ratio C  7.13/7.13 = 1 H  14.3/7.13 = 2 Formula CH2

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?

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 12.011 g 18.3 g x 1mol = 18.2 mol H 1.00794 g Mole Ratio C  6.80/6.80 = 1 x 3  3 H  18.2/6.80 = 2.68 x 3  8 Formula C3H8

0.2, 1.2, 2.2, 3.2……. Multiply by 5 0.25, 1.25, 2.25 … …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