# Percent Composition Problems

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```					                           Percent Composition
It was observed experimentally that the percent of a certain element in a compound didn’t
change even if the quantity of the compound increased or decreased. Therefore, you could
describe a new material as being 60% carbon and 40% hydrogen by mass. Using the periodic
table we can convert mass of an element into moles of that element or vice versa. You will see 2
types of problems involving percent composition: 1. You are given information so that you can
determine the mass of each of the compound and can determine the percent mass of each element
in the compound; or 2. You are given a percent of each of the elements in a compound and you
have to determine the empirical and possibly molecular formula of the compound.

Examples:
i.     What is the percent of carbon and oxygen in CO2?
grams of carbon
x 100% = % of carbon in CO2. (repeat for oxygen)
total grams of carbon dioxide

12.01 g C
x 100% = 27.29% carbon
44.01 g CO2

32.00 g O
x 100% = 72.71% oxygen
44.01 g CO2

ii.    What is the molecular formula for a material that is 92.2% C and 7.80% H by mass? The
material has a molar mass of 52.1.

1.     Assume that we have 100g of this material, therefore we have 92.2g of carbon and
7.80g of hydrogen in this compound.

2.     Determine the number of moles of all the elements in the compound.

1mol C
92.2g C x            = 7.68 moles C
12.01g C

1mol H
7.80g H x           = 7.72 moles H
1.01g H

3.     Divide all the moles by the lowest number of moles.

7.68 moles C                         7.66 moles H
= 1.00 moles C                       = 1.01 moles H
7.68 moles                           7.68 moles

1
4.     Determine the empirical formula. In this case the empirical formula is C1H1.

5.     To determine the molecular formula we need to use information about the molar
mass of the compound.

The molar mass of the empirical formula is 12.01 + 1.01 = 13.02g/mole.
The molar mass of the unknown compound is 52.1g/mole.

Molar mass of the compound          =      the whole number that you multiply the subscripts
Molar mass of the empirical formula        by to get the molecular formula.

52.1 = 4.00           (CH)4 → C4H4
13.02

2
Problems:
1.   What is the empirical formula of aspirin? It has 4.48 % H, 60.00 % C, and   35.52 % O
by mass.

Solution

2    An unknown compound is found in tree sap. It has been shown to be composed of 40.0%
carbon, 6.7% hydrogen and 53.3% oxygen by mass. It was also discovered that 5.00
moles of the material has a mass of 900 grams. What is the molecular formula of this
compound?

Solution

3
Solutions:
1.   What is the empirical formula of aspirin? It has 4.48 % H, 60.00 % C, and 35.52 % O by
mass.

1. Assume 100g of material.

1 mole H
2. 4.48 g H x                = 4.44 moles H
1.01g H

1 mole C
60.00 g C x            = 5.00 mole C
12.01g C

1mole O
35.52 g O x             = 2.22 mole O
16.00 g O

4.44 mole H
3.                   = 2.00 mole H
2.22 mole

5.00 mole C
= 2.25 mole C
2.22 mole

2.22 mole O
= 1.00 mole O
2.22 mole

4.       Need all to be whole numbers, and multiply each by 4.

2.00 moles H x 4 = 8.00 moles H
2.25 moles C x 4 = 9.01 moles C
1.00 mole O x 4 = 4.00 moles O

Empirical Formula: C9H8O4

4
2.    An unknown compound is found in tree sap. It has been shown to be composed of 40.0%
carbon, 6.7% hydrogen and 53.3% oxygen by mass. It was also discovered that 5.00
moles of the material has a mass of 900 grams. What is the molecular formula of this
compound?

1. Assume 100 grams of material

1 mole C
2. 40.0 g C x               = 3.33 mole C
12.01g C

1 mole H
6.7 g H x             = 6.63 mole H
1.01g H

1 mole O
53.3 g O x              =3.33 mole O
16.00 g O

3.33 mole C
3.               = 1.00 moles C
3.33 mole

6.63 mole H
= 1.99 moles H
3.33 mole

3.33 mole O
= 1.00 moles O
3.33 mole

4. Empirical Formula CH2O

Empirical molar mass

CH2O ⇒            12.01
+ 2.02
16.00
30.03 g/mole

5. Molar Mass = 900 g            = 180 g/mol
5.00 mole

5
Molar mass          180.0 g
=             mol = 6   ⇒ Molecular Formula: C6H12O6
Empirical mass       30.0 g
mol

6

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