Transcript Section 11.4

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
Mass of Compound
100
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
 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