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 C 6 H 6. 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 H2O2H2O2 HO GlucoseC 6 H 12 O 6 CH 2 O BenzeneC6H6C6H6 CH EthyneC2H2C2H2 CH AnilineC6H7NC6H7NC6H7NC6H7N WaterH2OH2OH2OH2O

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 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 » Cu 2 S

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. 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 » CH 2

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. 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 = 1x 3  3 » H  18.2/6.80 = 2.68 x 3  8 Formula »C3H8»C3H8

9. 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

10. Molecular Formula There are many compounds that have the empirical formula CH 2 O, but the actual formula may be some multiple of the empirical formula.