Empirical Formula Molecular Formula Simplest whole number ratio of atoms in a compound Molecular Formula Actual formula (which is a whole number multiple of the empirical formula)

Molecular Formula Empirical Formula Examples: H2O2 HO C6H12O6 CH2O C6H6 CH

Finding the Empirical Formula Find the moles of each element Determine the mole ratio (largest/smallest) Example: Calculate the empirical formula of a compound containing 13.43 g Al and 53.18 g Cl.

13.43 g x 1 mol = 0.497 mol Al 27.0g 53.18 g x 1 mol = 1.50 mol Cl 35.5 g 1.50 = 3 0.497 1 3 mol Cl 1 mol Al AlCl3

Practice Problem Calculate the empirical formula of a compound containing 52.14 % C, 13.12% H, and 34.73 % O. 52.14 g C / 12.0 g/mol = 4.35 (2) 13.12 g H / 1.0 g/mol = 13.12 (6) 34.73 g O / 16.0 g/mol = 2.17 (1) C2H6O

Note: Sometimes the ratio is not a whole number Note: Sometimes the ratio is not a whole number. You must then convert it to a whole number ratio: 1.5/1 = 3/2 2.5/1 = 5/2 1.33 = 1 1/3 = 4/3 1.75 = 1 3/4 = 7/4

Practice Problem Calculate the empirical formula of a compound that contains 43.7 % P and 56.3% O by mass. 43.7 g P / 31.0 g/mol = 1.41 (1) 56.3 g O / 16.0 g/mol = 3.52 (2.5) P2O5

Finding the Molecular Formula from the Empirical Formula Given the molecular mass, divide by the empirical mass Example: Empirical Formula: CH2O (Empirical mass = molar mass = 30 g/mol) Molecular Mass: 180 g/mol Ratio: 180/30 = 6 Molecular Formula: 6 x (CH2O) => C6H12O6

Practice Problem Calculate the empirical and molecular formulas of a compound that contains 80.0 % C, 20.0% H, and has a molar mass (molecular mass) of 30.00 g. 80.0 g C / 12.0 g/mol = 6.67 mol (1) 20.0 g H / 1.0 g/mol = 20. mol (3) Empirical CH3 30 = 2 15 Molecular C2H6