1. Molecular Formula

2. FormaldehydeFormaldehyde Acetic Acid Glucose

3. CompoundEmpirical FormulaMolecular FormulaE.F.WM.F.W Formaldehyde CH 2 O Acetic Acid C2H4O2C2H4O2 Glucose C 6 H 12 O 6 CH 2 O 30.03 g/mol (30.03 x 6) = 180.18 g/mol (30.03 x 2) = 60.06 g/mol (30.03 x 1) = 30.03 g/mol

4. Molecular Formula: formula that specifies the actual number of atoms of each element in one molecule or formula unit of the substance. It is a multiple of the empirical formula Definition

5. Steps 1.Find the empirical formula 2.Calculate the molar mass of the empirical formula 3.Divide the molar mass given, by the molar mass of the empirical formula to get the multiplier 4.Multiply empirical formula subscripts by the multiplier to get the molecular formula

6. A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen has a molar mass of 118.1 g/mol. Determine the empirical and molecular formula. Example 1 40.68 g C 1 mol C 12.01 g C = 3.387 mol C 3.387 = 1.00 mol C ×2 = 2 5.08 g H 1 mol H 1.01 g H = 5.03 mol H 3.387 = 1.49 mol H ≈ 1.5 ×2 = 3 54.24 g O 1 mol O 16.00 g O = 3.390 mol O 3.387 = 1.00 mol O ×2 = 2 E.F. = C 2 H 3 O 2 Molar Mass: C: 12.01 ×2 H: 1.01 ×3 + O: 16.00 ×2 59.05g 118.1 g 59.05g = 2 M.F. = C 4 H 6 O 4

7. A compound has a molar mass of 462.8 g/mol and contains 77.87% C, 11.76% H, and 10.37% O. Determine the empirical and molecular formulas. Example 2 77.87 g C 1 mol C 12.01 g C = 6.484 mol C 0.6481 = 10 mol C 11.76 g H 1 mol H 1.01 g H = 11.64 mol H 0.6481 = 17.99 mol H ≈ 18 10.37 g O 1 mol O 16.00 g O = 0.6481 mol O 0.6481 = 1.00 mol O E.F. = C 10 H 18 O Molar Mass: C: 12.01 ×10 H: 1.01 ×18 + O: 16.00 154.28 g 462.8 g 154.28 g = 3 M.F. = C 30 H 54 O 3