85.7% C Percent Composition 24g C 28g CH4 mass of element x 100 x 100 The relative amounts of elements in a compound. % composition = mass of element x 100 mass of compound 24g C 85.7% C x 100 = 28g CH4

Percent Composition from Formula C2H4 2 moles of C x 12g/mol = 24 g 4 moles of H x 1 g/mol = 4 g 28 g 24g/28g x 100 = 85.7% C 4g/ 28g x 100 = 14.3% H

You try one: NaC2H3O2

Empirical Formula The lowest whole number ratio of elements in a compound

Calculate empirical formula from percent composition: 1st - Begin by pretending to have 100 g sample & convert the % to grams. 2nd - Convert grams to the moles by dividing by molar mass. 3rd - Obtain whole number ratios of moles by: 1) dividing each number by the smallest number 2) multiplying to get rid of any fractions 4th - Write the empirical formula using the number of moles as subscripts.

Example 1 Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so. . . . . 38.67 g C / 12 g/mol C = 3.22 mole C 16.22 g H / 1.07g/mol H = 16.09 mole H 45.11 g N / 14 g/mol N = 3.22 mole N Divide each by the smallest value. . . . . 3.22

Example 1 cont. 3.220 mole C / 3.22 = 1 16.09 mole H / 3.22 = 5 3.22 mole N / 3.22 = 1 The empirical formula is CH5N

Example 2 A compound is 43.64 % P and 56.36 % O. What is the empirical formula? 43.6 g P /31 g/mol = 1.4 mole P /1.4 = 1 56.4 g O /16 g/mol = 3.5 mole O /1.4 = 2.5 Divide by smallest number Multiply the result to get rid of any fractions. P1O2.5 = P2O5 2 X

Molecular Formulas Are whole number multiples of empirical formulas. Divide molar mass of compound by molar mass of empirical formula. Multiply subscripts in compound by the number from step 1. Ex. An empirical formula is CH5N. The molar mass is 93 g/mol. 93 g/mol / 31 g/mol = 3 3 x CH5N = C3H15N3