&
Percent Composition
Indicates relative % of each element in
a compound
Total % of the components ~ 100%
Examples
1. Determine the % of children in a group
consisting of 10 men, 8 women, and 9
children.
10 + 8 + 9 = 27 total people
9 children x 100 = 33.3%
27 people
Examples
2. Find the % of aluminum in aluminum
oxide.
3. Find the % of nitrogen in ammonium
nitrate, a compound used in fertilizers.
4. Copper(II) sulfate pentahydrate is a blue
compound used to make colored
pigments, insecticides, and electric
batteries. Determine the % water.
Empirical Formulas
Empirical formulas are the lowest possible
ratio of components in the compound.
Ionic formulas are almost always empirical
in nature.
Covalent compounds require more
information
…..is the EF= MF?
Steps to determining the EF
1. Find the percent composition of the
substance.
2. Assume you have a 100 gram sample
assuming 100g, you can remove the %
sign and replace it with grams
3. Convert the grams found in step 2 into
moles.
Steps to Determine the EF
4. Ratio the moles you found in step 3
5. Divide all moles by the smallest value
this determines the relative molar
amount of each of the atoms—
~ whole number
6. If the numbers are not whole numbers
multiply by a factor that will yield a
whole number
Example: 1: 1.3 --- x 3 = 3:4
Steps to Determine the EF
Use molar ratios obtained as subscripts in the
formula.
Example 1
What is the EF of a compound that contains
53.73% Fe and 46.27% S?
( Fe2S3)
Example 2
What is the EF of a compound that contains
90.7% Pb and 9.33% O?
( Pb3O4)
Molecular Formulas
Molecular formulas can be found from the
EF
1. Find the empirical formula and mass
2. Find the relationship between the EF
mass and the molecular mass
n = molecular formula mass
empirical formula mass
3. (Empirical formula)n = Molecular formula
EF = CH4 n = 2 C2H8
Example:
• A compound composed of hydrogen and
oxygen is analyzed and a sample of the
compound yields 0.59 g of hydrogen and
9.40 g of oxygen. The molecular mass of
the compound is 34 g/mol. Find the
empirical formula and molecular formula
for this compound.
Step 1: Find the percent composition
H = 0.59 g
O = 9.40 g
9.99 g Total Sample
0.59 g
100  5.9%H
9.99 g
100% - 5.9% = 94.1% oxygen
Step 2: Assume a 100 g Sample
• Drop the percent sign and replace it with a
“g” for grams.
Step 3: Convert grams to moles
5.9 g H 1 mole = 5.9 moles H
1.0 g
94.1 g
1 mole = 5.9 moles O
16 g
Step 4: Ratio the moles and
reduce
Moles of H:Moles of O
5.9 moles : 5.9 moles
5.9 moles 5.9 moles
1:1
Step 5: Use mole ratio for
empirical formula subscripts
• 1 hydrogen : 1 oxygen
Empirical Formula: HO
Step 6: Find the empirical
formula mass
• Empirical Formula: HO
1 g + 16 g = 17 g/mol
Step 7: Divide Molecular Mass
by the Empirical Formula Mass
Molecular mass (sometimes called molar
mass or mass of the molecule) = 34.0 g/mol
Divide the molecular mass by the empirical
formula mass to find “n”, a multiplier.
n = (molecular mass
) = 34 g/mol
(empirical formula mass) 17 g/mol
n= 2
Step 8: Multiply the subscripts
in the empirical formula by “n”
Molecular Formula = (empirical formula)n
Molecular Formula = (HO)2
*distribute the 2 through the parentheses
Molecular Formula = H2O2