Percent Composition and
Empirical Formulas
Percent Composition
Percent Composition
The percent composition shows the relative percent (by
mass) of each element in a compound.
Percent Composition
The percent composition shows the relative percent (by
mass) of each element in a compound.
The percent composition is determined by dividing the
mass of the individual elements in a compound by the
entire formula mass of the compound.
Mass of individual
element (g)
× 100 % = % of element
Percent Composition =
Formula Mass of
Compound (g)
For example, when the correct percent composition for HF
is determined, the process is as follows:
For example, when the correct percent composition for HF
is determined, the process is as follows:
Find the total formula molar mass:
1 mol H =
1.0079 g/mol
1 mol F = + 18.9884 g/mol
total =
19.9963 g/mol
Take the individual molar mass of each element and divide
by the total formula mass, and turn it into a percent:
Take the individual molar mass of each element and divide
by the total formula mass, and turn it into a percent:
1.0079 g/mol H
for H
× 100 %
= 5.04% H
× 100 %
= 94.96% F
19.9963 g/mol HF
for F
18.9884 g/mol H
19.9963 g/mol HF
A good way to quickly check the answers is to sum the
percentages, which should equal 100% (or 1). There will
be cases where the percentages might not equal exactly
100% because of rounding, but the total should always
be VERY close to 100%.
As another example, consider sulfuric acid:
H2SO4 :
As another example, consider sulfuric acid:
H2SO4 :
2 mol H × 1.008 g/mol
1 mol S × 32.066 g/mol
= 2.016 g/mol
= 32.066 g/mol
4 mol O × 15.999 g/mol
= 63.996 g/mol
total = 98.078 g/mol
As another example, consider sulfuric acid:
H2SO4 :
2.016 g/mol H
× 100%
for H
= 2.06% H
98.078 g/mol H2SO4
for S
32.066 g/mol S
98.078 g/mol H2SO4
for O
63.996 g/mol O
98.078 g/mol H2SO4
× 100%
= 32.69% S
× 100%
= 65.25% O
Empirical Formula
Empirical Formula
Once the percent composition of a compound is known, the
empirical formula of the compound can be
determined. An empirical formula shows the lowest
whole-number ratio of the elements in a compound
1. Turn the percent composition information into
mass. This is made simple by assuming a theoretical
amount of 100 grams.
Thus 50% composition is turned into 50 grams, and
36.8% composition is turned into 36.8 grams, etc.
2. Calculate the number of moles for each element that
would contain the amount of mass from step 2.
This involves dividing the mass from step 2 by the molar
mass shown for the element on the periodic table.
3. The simplest whole-number ratio of each element needs
to be found.
One of the ways to get a good start on this is to divide
each number of moles from step 3 by the smallest
amount of moles.
This will guarantee at least one whole number to start
with (a “1” amount).
3. a. If the other molar amounts are within 0.15 of a whole
number, it is usually safe to round up or down to that
whole number.
3. b. If the other molar amounts cannot be rounded, it will
be necessary to multiply ALL the molar amounts by a
whole number to obtain a whole number
(or a number close to a whole number.) Thus, if a molar
amount had the decimal value of 0.20, it would be
necessary to multiply by 5. If the decimal value is 0.25,it
would be necessary to multiply by 4, and it would if the
decimal value is 0.33, it would be necessary to multiply
by 3, etc.
Example:
White gold is 75.0% gold, 10.0% palladium, 10.0% nickel,
and 5.00% zinc. What would be the empirical formula
of white gold?
75.0% Au →
75.0 g Au
1 mole Au
= 0.3807 moles Au
197.0 g
10.0% Pd →
10.0 g Pd
1 mole Pd
= 0.09398 moles Pd
106.4 g
10.0% Ni →
10.0 g Ni
1 mole Ni
= 0.1704 moles Ni
58.69 g
5.00% Zn →
5.00 g Zn
1 mole Zn
= 0.07646 moles Zn
65.39 g
Dividing by the lowest amount of moles from above (0.07646 mol):
0.3807 moles Au
0.07646
0.09398 moles Pd
0.07646
0.1704 moles Ni
0.07646
= 4.979 moles Au ≈ 5 moles Au
= 1.229 moles Pd
= 2.229 moles Ni
0.07646 moles Zn
= 1 moles Zn
0.07646
The gold and zinc are already expressed in a whole
number, but to express the palladium and nickel as a
whole number, it will be necessary to multiply
everything
by 4. This would make the palladium and nickel 4.916
moles and 8.916 moles (respectively), which are now
close enough to round. Do not forget to multiply
everything, even the ones that are already whole
numbers!
The gold and zinc are already expressed in a whole
number, but to express the palladium and nickel as a
whole number, it will be necessary to multiply
everything
by 4. This would make the palladium and nickel 4.916
moles and 8.916 moles (respectively), which are now
close enough to round. Do not forget to multiply
everything, even the ones that are already whole
numbers!
Thus the final relative amount of moles is 20 Au, 5 Pd, 9
Ni, 4 Zn. The empirical formula is Au20Pd5Ni9Zn4.
Practice Exercise:
Find the empirical formula for purple gold, Purple Gold =
80% Au, 20% Al
mass
1 mol
answer
=
P.T. mass
mass
×
answer
factor
=
whole #
lowest #
1 mol
P.T. mass
=
answer
=
lowest #
=
answer
×
factor
=
whole #
80 g
1 mol
196.97 g
20 g
0.4062 mol
=
1 mol
26.98 g
lowest #
0.7413 mol
=
lowest #
answe
=
r
answe
=
r
×
×
fact
o
r
fact
o
r
=
=
whole
#
whole
#
80 g
1 mol
196.97 g
20 g
0.4062 mol
=
1 mol
26.98 g
0.4062 mol
=
1
×
0.7413 mol
=
0.4062 mol
= 1.825
×
fact
o
r
fact
o
r
=
=
whole
#
whole
#
80 g
1 mol
196.97 g
20 g
0.4062 mol
=
1 mol
26.98 g
0.4062 mol
=
1
×
5
=
5
0.7413 mol
=
0.4062 mol
= 1.825
×
5
=
9
80 g
1 mol
196.97 g
20 g
0.4062 mol
=
1 mol
26.98 g
0.4062 mol
=
1
×
5
=
5
0.7413 mol
=
0.4062 mol
Empirical Formula = Au5Al9
= 1.825
×
5
=
9