Empirical and Molecular Formulas Section 11.4 Chemistry

Objectives Explain what is meant by the percent composition of a compound. Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.

Key Terms Percent composition Empirical formula Molecular formula

Percent Composition Percent by mass of each element in a compound. Mass of Element x 100 Mass of Compound

Example Calculate percent composition of H in H2O Molar mass of water: 18.02 g/mol Determine mass of H in 1 mol of H2O

Mass of H 1.01 x 2 = 2.02 g H in 1 mol water Atomic mass of H from periodic table Number of H in H2O

Percent Composition of H 2.02 g of H x 100 = 11.2% H 18.02 g of H2O

Empirical Formula Formula for a compound with the smallest whole-number ratio of elements. Percent composition can be used to find the chemical formula.

Empirical Formula When given percent composition, assume: The total mass of the compound is 100 g The percent composition of the element is equal to the mass in grams of the element.

Example A compound has a percent composition of 40.05% S and 59.95% O. So in 100 g of the compound, 40.05 g are S and 59.95 g are O. Find the amount of mol for each element.

Empirical Formula 40.05g S x 1 mol S = 1.249 mol S 32.07g S 59.95g O x 1 mol O = 3.747 mol O 16.00g O

Empirical Formula The element with the smallest number of mol gets the subscript 1.

Empirical Formula S has a subscript 1 Then divide the mol O by the mol S 3.747 mol O/ 1.249 mol S = 3 mol O Then write your empirical formula using your subscripts: SO3

Practice Problems Pg. 333

Molecular Formula Formula that specifies the actual number of atoms of each element in one molecule or formula unit of a substance. n = molecular formula mass empirical formula mass

Example Compound is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.

Example 40.68 g C x 1 mol C = 3.387 mol C 12.01 g C 5.08 g H x 1 mol H = 5.04 mol H 1.008 g H 54.24 g O x 1 mol O = 3.390 mol O 16.00 g O

Example 3.387 mol C/ 3.387 = 1 mol C 5.040 mol H/ 3.387 = 1.49 = 1.5 mol H 3.390 mol O/ 3.387 = 1.001 = 1 mol O Ratio of C : H : O = 1 : 1.5 : 1

Example Empirical Formula: C2H3O2

Example n = molecular formula mass empirical formula mass To find molecular formula, calculate n. n = molecular formula mass empirical formula mass Molecular mass is in the problem! Calculate molar mass of empirical

Example n = 118.1 = 2 59.04 Multiply the subscripts of the empirical by n to find the molecular formula. Molecular Formula: C4H6O4

Practice Problems Pg 335

Homework Section 11.4 Problems 27-29 on page 877