Percent Composition and Empirical Formulas Notes
Percent Composition
The percent composition shows the relative percent (by _______________) of each element in a compound. The percent
composition is determined by dividing the _______________ of the individual elements in a compound by the
_______________ formula mass of the compound.
Percent Composition =
Mass of individual
element (g)
Formula Mass of
Compound (g)
× 100 %
= % of element
For example, when the correct percent composition for HF is determined, the process is as follows:
1.
Find the _______________ formula molar mass (from the Periodic Table):
1 mol H =
1.0079 g/mol
1 mol F =
+ 18.9884 g/mol
total =
19.9963 g/mol
2.
Take the _______________ molar mass of each element and divide by the total formula mass, and turn it into a
percent:
for H
1.0079 g/mol H
19.9963 g/mol HF
× 100 %
= ______________ % H
for F
18.9884 g/mol F
19.9963 g/mol HF
× 100 %
= _____________ % F
A good way to quickly check the answers is to sum the percentages, which should equal _______________ % (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%.
Note: some calculators have a “%” button. Please do not use this button. In a calculator simply multiply by 100, and then
write the percent sign as the label at the end.
As another example, consider sulfuric acid:
H2SO4 :
2 mol H
1 mol S
4 mol O
× 1.008 g/mol
× 32.066 g/mol
× 15.999 g/mol
total
= 2.016 g/mol
= 32.066 g/mol
= 63.996 g/mol
= 98.078 g/mol
for H
g/mol H
g/mol H2SO4
for S
g/mol S
g/mol H2SO4
× 100%
= 32.69% S
for O
g/mol O
g/mol H2SO4
× 100%
= 65.25% O
× 100%
= 2.06% H
Empirical Formulas
Once the percent composition of a compound is known, the _______________ _______________ of the compound can be
determined. An empirical formula shows the lowest _______________ -number ratio of the elements in a compound
1.
Turn the percent composition information into _______________. This is made simple by assuming a theoretical
amount of _______________ 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 _______________ for each element that would contain the amount of mass from step 1.
This involves dividing the mass from step 1 by the _______________ mass shown for the element on the periodic
table.
3.
The simplest _______________ -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 2 by the _______________ amount of moles. This will
guarantee at least one whole number to start with (a “1” amount).
a. If the other molar amounts are within 0.15 of a whole number, it is usually safe to _______________ up or down to
that whole number.
b. If the other molar amounts cannot be rounded, it will be necessary to _______________ 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?
Turning percent composition into moles:
75.0% Au →
75.0 g Au
1 mole Au
g
= 0.3807 moles Au
10.0% Pd →
10.0 g Pd
1 mole Pd
g
= 0.09398 moles Pd
10.0% Ni →
10.0 g Ni
1 mole Ni
g
= 0.1704 moles Ni
5.00% Zn →
5.00 g Zn
1 mole Zn
g
= 0.07646 moles Zn
Dividing by the lowest amount of moles from above (0.07646 mol):
0.3807 moles Au
0.07646 moles
= 4.979 Au ≈ 5 Au
0.09398 moles Pd
0.07646 moles
= 1.229 Pd
0.1704 moles Ni
0.07646 moles
= 2.229 Ni
0.07646 moles Zn
0.07646 moles
= 1 Zn
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 ___________ Au, ___________ Pd, ___________ Ni, ___________ Zn.
The empirical formula is ______________________
Practice Exercise:
Find the empirical formula for purple gold, Purple Gold = 80% Au, 20% Al
for Au
mass
for Al
mass
1 mol
P.T. mass
=
answer
lowest #
=
answer
×
factor
=
whole #
1 mol
P.T. mass
=
answer
lowest #
=
answer
×
factor
=
whole #
Empirical Formula = _______________