Empirical and Molecular FormulasSection 11.4Chemistry

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 compositionEmpirical formulaMolecular formula

Percent Composition

Percent by mass of each element in a compound.

Mass of Element x 100Mass of Compound

Calculate percent composition of H in H2O

Molar mass of water: 18.02 g/mol

Determine mass of H in 1 mol of H2O

Example

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% H18.02 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 S32.07g 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 1Then 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 C5.08 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

O16.00 g O

Example

3.387 mol C/ 3.387 = 1 mol C5.040 mol H/ 3.387 = 1.49 = 1.5 mol

H3.390 mol O/ 3.387 = 1.001 = 1 mol ORatio of C : H : O = 1 : 1.5 : 1

Example

Empirical Formula: C2H3O2

ExampleTo 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