Empirical Formula: Smallest ratio of atoms of all elements in a compound Molecular Formula: Actual numbers of atoms of each element in a compound Determined experimentally The empirical and molecular formulas may or my not be the same.

Empirical and Molecular: H2O CO2 Empirical CH2O Molecular C6H12O6

Using % Comp to Find Empirical Formulas Steps: Convert mass of each element into moles. If given a % element, that percentage is equal to grams (assume 100g) Divide moles of each element by the smallest number of moles available to generate mole ratios If ratios are not whole numbers, multiply to make them whole. Write the empirical formula

Using % Comp to Find Empirical Formulas A sample of a compound containing sulfur and oxygen contains 59.95% O and 40.05% S. What is the empirical formula? 55.95g O x 1mol O = 3.747mol O/ 1.249 = 3 15.99g O 40.05g S x 1mol S = 1.249mol S/1.249 = 1 32.07g S Empirical formula is SO3

Using % Comp to Find Empirical Formulas Determine the empirical formula for methyl acetate, which has the following analysis: 48.64% C, 8.16% H, 43.20% O. 43.20g O x 1mol O = 2.700 mol O/ 2.700 = 1x2 = 2 15.99g O 48.64g C x 1mol C = 4.050 mol C/2.700 = 1.5 x 2 =3 12.01g C 8.16g H x 1mol H = 8.10 mol H/2.700 = 3 x 2 = 6 1.008g H Empirical formula is C3H6O2

Molecular Formula Determine the molar mass of the empirical formula Find the molecular molar mass in the problem and underline it. Divide the molecular molar mass by the empirical molar mass Multiply all the subscripts in empirical formula by the number obtained from step #3.

Practice Acetylene (molar mass 26.04g/mol) and benzene (molar mass 78.12g/mol) both have an empirical formula of CH. The mass of the empirical formula is 13.02g/mol. What are the molecular formulas for acetylene and benzene? Acetylene = 26.04 g/mol = 2 C2H2 13.02 g/mol Benzene = 78.12 g/mol = 6 C6H6

%Comp, Empirical, and Molecular Formulas Succinic acid is composed of 40.68%C, 5.08%H, and 54.24% O, and has a molar mass of 118.1 g/mol. What are the empirical and molecular formulas? 54.24g O x 1mol O = 3.390 mol O/ 3.387 = 1x2 = 2 15.99g O 40.68g C x 1mol C = 3.387 mol C/3.387 = 1 x 2 =2 12.01g C 5.08g H x 1mol H = 5.04 mol H/3.387 = 1.5 x 2 = 3 1.008g H Empirical Formula = C2H3O2

Problem Continued…… Molar Mass of empirical formula: 2mol C x 12.01g/mol = 24.02g C 3mol H x 1.008g/mol = 3.024g H 2mol O x 15.99g/mol = 32.00g O 59.04g/mol succinic acid To produce a molecular formula, multiply empirical by: 118.1 g/mol = 2 C4H6O4 59.04 g/mol