What Could It Be?
Empirical Formulas
The empirical formula is the simplest whole
number ratio of the atoms of each element in a
compound.
Note: it is not necessarily the true
formula of the compound.
Example: The molecular formula for
glucose is C6H12O6, but its empirical
formula is CH2O
It’s kind of like a
fraction reduced to its
lowest terms.
Molecular Formulas
The molecular formula gives the actual
numbers of each element, and thereby
represents the true formula of the
compound.
Calculating the Empirical
Formula
Example 1: A compound is found to contain
the following…
2.199 g Copper
0.277 g Oxygen
Calculate it’s empirical formula.
Calculating the Empirical
Formula
Step 1: Convert the masses to moles.
Copper:
1 m oleCu
2.199g Cu
 0.03460 m olesCu
63.5 g Cu
Oxygen:
1 m ole O
0.277g O
 0.01731m oles O
16.0 g O
Calculating the Empirical
Formula
Step 2: Divide all the moles by the smallest
value. This gives the “mole ratio”
0.03460 mol Cu
 1.999 Cu
0.0173mol
0.0173 molO
 1.000O
0.0173 mol
Calculating the Empirical
Formula
Step 3: Round off these numbers, they
become the subscripts for the elements.
Cu2O
Example 2: A material is found to be composed of 38.7%
Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it
is known that the molecular weight is 62.0 g/mol Calculate the
empirical and molecular formula for the compound.
If you assume a sample weight
of 100grams, then the percents
are really grams.
Example 2: A material is found to be composed of 38.7%
Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it
is known that the molecular weight is 62.0 Calculate the
empirical and molecular formula for the compound.
1m ole
 3.23m ol
Carbon: 38.7 gram s
12.0 gram s
Oxygen:
Hydrogen:
1 m ole
51.6 gram s
 3.23m ol
16.0 gram s
1 m ole
9.7 gram s
 9.7m ol
1.0 gram s
Example 2: A material is found to be composed of 38.7%
Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it
is known that the molecular weight is 62.0 Calculate the
empirical and molecular formula for the compound.
1m ole
 3.23m ol
Carbon: 38.7 gram s
1.00
12.0 gram s
3.23
Now, divide all the moles
by the smallest one, 3.23
Oxygen:
1 m ole
51.6 gram s
 3.23m ol
1.00
16.0 gram s
3.23
Hydrogen:
1 m ole
9.7 gram s
 9.7m ol  3.00
1.0 gram s 3.23
Example 2: A material is found to be composed of 38.7%
Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it
is known that the molecular weight is 62.0 Calculate the
empirical and molecular formula for the compound.
So, the empirical formula must be
CH3O
The molecular weight of the empirical formula
is….
C
12 x 1
=
12 g/mol
H
1x3
=
3 g/mol
O
16 x 1
=
16 g/mol
31 g/mol
Remember, the
empirical formula is not
necessarily the molecular
formula!
MW of the empirical formula = 31
MW of the molecular formula = 62
Molecular
Multiplying Factor
Em pirical
62

2
31
2x
Empirical
Formula
( CH3O )
= C2H6O2
Molecular
Formula
Remember, the molecular
formula represents the
actual formula.
What if the mole ratios
don’t come out even?
Example 3: A compound is analyzed and found to
contain 2.42g aluminum and 2.15g oxygen. Calculate
its empirical formula.
1m ol
Aluminum: 2.42g
 .0896m ol  1.0 X 2 = 2
27g
.0896
Oxygen:
1m ol
2.15g
 .134m ol  1.495 X 2 = 3
16g .0896
Al2O3