1. Calculating Empirical and Molecular Formulas

2. Calculating empirical formula A compound contains 79.80% carbon and 20.20% hydrogen. What is the empirical formula. 79.80g C 1mol C = 6.64 mol C = 1 12.01 g C 6.64 mol 20.20g H 1mol H = 20.00 mol H = 3.01 1.01 g H 6.64 mol Empirical Formula = CH 3

3. Calculating Molecular Formulas The molar mass of a compound is 92.00 g/mol. Analysis of a sample of the compound indicates that it contains 0.61g of nitrogen and 1.28 g of oxygen. Find its molecular formula. Find the empirical formula first. 0.61 g N 1mol N = 0.04 mol N = 1 14.01g N 0.04 mol 1.28 g O 1mol O = 0.08 mol O = 2 16 g O 0.04 mol Empirical formula = NO 2

4. Molecular formula continued Using the empirical formula: NO 2, find the molecular formula. Step 1: determine the molar mass of the empirical formula 1mol N x 14.01g = 14.01 g/mol 2mol O x 16.00g = 32.00 g/mol 46.01 g/mol Step 2: Divide the given molar mass by the molar mass calculated in step 1. x = 92.00 g/mol = 2 x 46.01 g/mol Step 3: Multiply the subscripts from the empirical formula by what was calculated as x in step 2: (NO 2 ) 2 = N 2 O 4 The molecular formula is N 2 O 4

5. Assignment Determine the Empirical and molecular formulas for each of the following. Box your empirical formula and your molecular formula. 1.A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molecular formulas. 2.A compound has a molar mass of 456.18 g/mol and consists of 3.05% Fe and 4.81% S. Calculate both empirical and molecular formulas 3.A compound consists of 36.48g of Na, 25.41g of S, and 38.11g of O. It has a molar mass of 252.10 g/mol. Calculate both empirical and molecular formulas.