Notes 4: Empirical Formulas
Name
Date
Block
A compound never before seen has just been prepared in a chemistry laboratory. What is the chemical
formula for that compound? In order to answer that question, the chemist must first determine the
percent composition of the compound. Using the % composition of the compound, the chemist can
determine the empirical formula of the compound.
The empirical formula of a compound shows the simplest, whole-number ratio of elements present in
the compound.
Percent Composition of a Compound
Percent composition of a compound tells what percent of the
total compound is made of each element in the chemical formula for the compound. (Percent is usually
written to 2 decimal places.)
Determine percent composition.
% composition = mass of element in one formula unit of compound
total mass of one formula unit of compound
x
100
NOTE: the “total mass of one formula unit of compound” is the same as the gfm of that compound.
Example: Calculate the % composition of H2O.
H 2 x 1.008 = 2.016 g
O 1 x 16.00 = 16.00 g
18.016 g = 18.02 g
% H(g) =
2.016 g H(g) x 100 = 11.187% H(g) = 11.19% H(g)
18.02 g H2O(g)
% O(g) = 16.00 g O(g) x 100 = 88.79 % O(g) = 88.79 % O(g)
18.02 g HO(g)
Percents do not always add to exactly 100.00 % because of rounding.
1
Empirical formulas give the smallest whole ratio of moles of each element in one formula unit of
the compound. We can use the percent composition values to calculate the number of moles
present of each element in 100.00 g the compound. (100.00 g is used for convenience with percent
composition values.)
Example: It has been determined that a sample of a compound contains 88.79% oxygen and
11.19% hydrogen. In 100.00 g of the compound, how many moles of each element are present?
Use the % composition values to calculate the number of moles of each element present in the
compound.
Remember that % composition means the number of parts of an element in 100.00 parts of the
compound. Thus, if a compound is known to be 88.79 % oxygen, then in a 100.00 g sample of the
sample, 88.79 g are oxygen. Since the hydrogen and oxygen are chemically combined in a compound,
we do not use the diatomic formulas for either element.
compound that contains 88.79 % O = 88.79 g O present in compound
100.00 g compound
This is usually written simply as 88.79 g O with the understanding that it is 88.79 g O per 100.00 g of
the compound.
A mass → mole conversion is calculated:
88.79 g O x
1 mole O = 88.79 mole O = 5.54937 mole O = 5.549 mole O in 100.00 g compound
16.00 g O
16.00
Since this same compound is also 11.19% hydrogen, in the 100.00 g sample of the compound 11.19 g
are hydrogen.
compound that contains 11.19% H = 11.19 g H present in compound
100.00 g compound
A mass → mole conversion is calculated:
11.19 g H x 1 mole H = 11.19 mole H = 11.101 mole H = 11.01 mole H in 100.00 g of compound
1.008 g H
1.008
2
Since an empirical formula of a compound represents the smallest whole-number ratio of elements
in the compound, we need to calculate that smallest whole-number ratio.
The smallest whole-number ratio of the elements by dividing the number of moles of each element
by the smallest number of moles present.
Since the compound has been found to contain only hydrogen and oxygen, the hydrogen is written first
since the hydrogen tends to behaves as a “metal” when combined with nonmetals. If the compound had
been found to contain hydrogen and a metal, the hydrogen would behave as a “nonmetal” and would
have been written last.
Using our example above:
# moles H : # moles O
= 11.19 moles H : 5.549mole O
Since 5.549 mole O is the smaller of the two mole values, each mole value is divided by 5.549:
11.19 mole H
5.549
:
5.549 mole O
5.549
=
2.038 mole H
: 1.000 mole O
The ratio is rounded to the nearest whole numbers.
Rounding the ratio to whole numbers gives us a ratio of H : O of 2 : 1.
The whole-number ratio gives you the SUBSCRIPTS for the empirical formula of the compound.
Therefore, the empirical formula of this compound is H2O.
Another example:
Determine the empirical formula of a compound that contains 69.5% oxygen and 30.5% nitrogen.
69.5% O means 69.5 g O in a 100.0 g sample of the compound.
30.5% N means 30.5 g N in a 100.0 g sample of the compound.
69.5 g O x 1 mole O
16.00 g O
30.5 g N x
moles N : moles O
1 mole N
14.01 g N
=
=
69.5 mole O = 4.34 mole O
16.00
=
30.5 mole N
14.01
= 2.18 mole N
2.18 moles N : 4.34 moles O
2.18 mole N : 4.34 mole O = 1 mole N : 1.99 mole O = 1 mole N : 2 mole O
2.18
2.18
The empirical formula for the compound is NO2.
3