# Percent Composition & Empirical Formulas

```CHEMISTRY 30S – MODULE 3
CHEMICAL REACTIONS
LESSON 6 
Outcomes
When you have completed this lesson, you will be able to:

Calculate the percent composition of a compound given its formula.

Calculate the empirical formula of a compound from percent composition or
mass data.

Calculate the molecular formula of a compound from percent composition or
mass data.
When you get an assignment back, it’s usually out of 10 or 20 or 30, not always
100. But we always convert it to a percentage (out of 100) so we can compare
them more easily. We take your mark, say 25 out of 30, divide what you got, by
the total and multiply by 100 to get the percentage – 25/30 = 0.833333 and
moving the decimal point two places to the right we get 83.3 %
Chemists do chemical analysis to find out the composition of a substance. They
determine the relative amounts of each element contained in the compound,
using mass, and express these amounts as a “percent composition” by mass.
The calculation is very similar to that described above for finding your marks on
an assignment or test – the mass of oxygen let’s say, in a sample of a compound
is divided by the total mass of the sample and then multiplied by 100.
Example 1. A compound contains 22.1 g of copper, 11.2 g of sulphur and 22.3 g of
oxygen. What is the percent composition of each element in this compound?
Solution:
Step 1. Determine the total mass of the sample of the compound.
Just like with the test, to determine the percent composition we must first know what the
total mass of the sample is:
22.1 g + 11.2 g + 22.3 g = 55.6 g
Step 2. Determine percent composition of each.
Copper: 22.1/55.6 = 0.3977 x 100 = 39.8%
Sulphur: 11.2/55.6 = 0.201 x 100 = 20.1%
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Oxygen: 22.3 / 55.6 = 0.401 x 100 = 40.1%
Percent Composition From Chemical Formulas
If the formula of a compound is known, the percent composition can be determined.
We can determine the amount of each element in a predetermined amount of the
compound, like a mole of that compound. We use the amount of each element in one
mole of the compound, divide by the molar mass of the compound and multiply by
100, or
% composition = mass of element in 1 mol of compound
molar mass of compond
Example 2. Determine the percent composition of each element in Na2CO3.
Solution.
Step 1. Determine the number of moles of each element in one mole of the
compound.
Recall, the number of moles of each element in one mole of the compound is equal
to the subscript for that element.
1 mole of Na2CO3 contains:
Na = 2 moles
C = 1 mole
O = 3 moles
Step 2. Determine the mass of each element in one mole of the compound.
The mass of each element is the number of moles of that element multiplied by the
molar mass of the element.
Mass of sodium = 23 x 2 = 46 g
mass of carbon = 12 x 1 = 12 g
mass of oxygen = 16 . 3 = 48 g
Step 3. Find the fraction of each and convert to a percentage.
To find the fraction, divide the mass of each element in one mole, by the mass of one
mole (molar mass) of the compound.
Na2CO3 = 106.0 g/mol
Sodium: 46 g/106.0 = 0.4339 x 100 = 43.4%
Carbon: 12 g/106 = 0.113 x 100 = 11.3 %
Oxygen: 48 g/106 = 0.453 x 100 = 45.3%
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Example 3. Determine the percent, by mass, of water in MgSO4·9H2O.
Solution.
Step 1. Determine the number of moles of water in one mole of the compound.
For every mole of magnesium sulphate, there are 9 water molecules
Step 2. Determine the mass of each in one mole of the compound.
water’s molar mass as 18, so 18 x 9 = 162 g
Step 3. Find the fraction of each and convert to a percentage.
To find the fraction, divide the mass of each element in one mole, by the molar mass of
the compound.
MgSO4·9H2O = 282.4 g/mol
water makes up 162/282.4 = 57.4% of the mass of the compound.
Empirical Formula
Once chemists have analyzed a compound and determined its percent composition,
they can use this information to determine the compound’s formula. A compound
can have two formulas, molecular and empirical formulas.
The molecular formula of a compound is the actual formula of the compound. The
molecular formula contains the actual number of atoms of each element in one unit
of that compound.
The empirical formula is the formula of a compound having the smallest whole
number ratio of the moles of each element. The subscripts of the molecular formula
are always a whole number multiple of the subscripts of the empirical formula.
Molecular
Formula
C2H4, C7H214,
C20H40
H2O2, hydrogen
peroxide
C6H12O6, glucose
Na2CO3
CHEM 30S
Empirical
Formula
CH2
HO
CH2O
Na2CO3
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Compound
Molecular
Formula
Empirical
Formula
Multiplier
Formaldehyde
CH2O
CH2O
1
Acetic acid,
vinegar
C2H4O2
CH2O
2
Glucose
C6H12O6
CH2O
6
To determine the empirical formula of a substance you need to know the elements in
the compound and the percent composition of each. Then
1. Assume that the total mass of the sample is 100 g. This means the mass of
each element in 100 g of the compound is equal to its percent
composition.
2. Determine the number of moles of each element in 100 g of compound.
3. Calculate the lowest whole number ratio by dividing each mole value by
the smallest mole value .
Example 4. Calculate the empirical formula of a compound whose percent
composition is 58.8% carbon, 9.80% hydrogen and 31.4% oxygen.
Solution.
Step 1. Assume that the total mass of the sample is 100 g.
In a 100 g sample, there will be 58.8 g carbon, 9.80 g hydrogen and 31.4 g of
oxygen.
Step 2. Find the number of moles of each element in the 100-g sample.
C: n = m/M
n = 58.8 g/12g/mol
n = 4.9 mol of C in 100 g of compound
H : n = m/M
n = 9.8 g/1.01 g/mol
n = 9.7 mol of H in 100 g of compound
O: n = m/M
n = 31.4 g/16 g/mol
n = 1.96 mol of O in 100 g of compound
Step 3. Now you take the smallest number and divide every other number of moles
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by that.
C: 4.9/1.96 = 2.5
H: 9.7/1.96 = 4.95 (close enough to 5, we round it)
O: 1.96/1.96 = 1
From this the empirical formula appears to be:
C2.5H5O1
Since this is not the lowest whole-number ratio, we must multiply each subscript by a
value that will give the lowest whole-number ratio (in this case it is 2)
2×(C2.5H5O1) = C(2.5 × 2)H(5 × 2)O(1 × 2)= C5H10O2
The empirical formula is C5H10O2.
In class we had practice questions assigned. If you missed class and want to
practice, use the following exercise from WebCT:
Exercise
Answer the following questions. Remember showing all your work is good practice.
1. Determine the percent composition of each element in the following.
a) H2SO4
b) Fe(C2H3O2)3
c) C12H22O11
d) aspartame, C14H18N2O5
2. What are the empirical formulas of the following compounds?
a) C2H4(OH)2
b) H2S2O8
c) C4H10
d) C2H5OH
e) C6H8O6
f) B2H6
3. A sample of Freon, a gas once used as a propellant in aerosol cans, was found
to contain 0.423 g of C, 2.50 g Cl, and 1.34 g F. What is the empirical
formula of this compound?
4. Determine the empirical formula for each of the following.
a) 0.104 mol potassium, 0.052 mol carbon, and 0.156 mol oxygen
b) 35.98% aluminum and 64.02% sulfur
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c) 22.9% sodium, 21.5% boron, and 55.7% oxygen
d) 21.7% carbon, 9.6% oxygen, and 68.7% fluorine
5. A compound that contains only carbon, hydrogen and oxygen is 48.64%
carbon and 8.16% hydrogen by mass. What is the empirical formula of this
substance?
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