Empirical Formulas Molecular Formulas

Since the formula of a compound indicates the number of atoms in the compound, CH4 = 4H atoms and 1C atom we must first convert the masses of the elements to number of atoms g  moles  atoms (ratio)

Step 1: g  moles Given: 38.67% Carbon by mass 38.67g Carbon per 100g of the compound Given: 16.22% Hydrogen by mass 16.22g Hydrogen per 100g of the compound Given: 45.11% Nitrogen by mass 45.11g Nitrogen per 100g of the compound

Step 1: g  moles Given: 38.67% Carbon by mass 38.67g Carbon per 100g of the compound 38.67g C 1 mol C = 3.220 mol C 12.01g C

Step 1: g  moles Given: 16.22% Hydrogen by mass 16.22g Hydrogen per 100g of the compound 16.22g H 1 mol H = 16.09 mol H 1.008 H

Step 1: g  moles Given: 45.11% Nitrogen by mass 45.11g Nitrogen per 100g of the compound 45.11g N 1 mol N = 3.219 mol N 14.01g N

Step 2: Which number is the smallest? 38.67g C 1 mol C = 3.220 mol C 12.01g C 16.22g H 1 mol H = 16.09 mol H 1.008 H 45.11g N 1 mol N = 3.219 mol N 14.01g N

Step 3: Whole Number Ratio The smallest number from step two becomes the denominator when solving for atoms C: 3.220 = 1.000 = 1 3.220 H: 16.09 = 4.997 = 5 N: 3.219 = 1.000 = 1 CH5N Can find molar mass C – 1 x 12.01 H – 5 x 1.008 N – 1 x 14.01 Compare to given and divide 31.06 by 31.06 = 1, 1 = multiplier

Molecular Formula Molar Mass (given) = Molecular Formula Multiplier Empirical Formula Mass Given: ClCH2 has a molar mass of 98.96 g/mol and an empirical formula mass of 49.48 g/mol MM = 98.69 g/mol = 2 EFM 49.48 g/mol We now multiple (ClCH2) by 2 = (ClCH2) 2  Cl2C2H4

Step 3: Whole Number Ratio Example from Empirical Formula Notes Can find molar mass C – 1 x 12.01 H – 5 x 1.008 N – 1 x 14.01 Compare to given (31.06 g/mol) and divide 31.06 (found molar mass) = 1, 1 = multiplier CH5N Can find molar mass C – 1 x 12.01 H – 5 x 1.008 N – 1 x 14.01 Compare to given and divide 31.06 by 31.06 = 1, 1 = multiplier