7.3 Percent Composition
Learning Objectives
By the end of this section, you will be able to:
- Compute the percent composition of a compound
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Percent Composition
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
Example 7.3a
Calculation of Percent Composition
Solution
To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
[latex]\%\;\text{H} = \frac{1.85 \;\text{g H}}{12.04 \;\text{g compound}} \times 100\% = 15.4\%[/latex]
[latex]\%\;\text{N} = \frac{2.85 \;\text{g N}}{12.04 \;\text{g compound}} \times 100\% = 23.7\%[/latex]
The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.
Exercise 7.3a
A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?
Check Your Answer[1]
Determining Percent Composition from Formula Mass
Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:
[latex]\%\;\text{H} = \frac{3.024 \;\text{amu N}}{17.03 \;\text{amu NH}_3} \times 100\% = 17.76\%[/latex]
This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 2. As long as we know the chemical formula of the substance in question, we can easily derive percent composition from the formula mass or molar mass.
Example 7.3b
Determining Percent Composition from a Molecular Formula
Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?
Solution
To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:
[latex]\begin{array}{r @{{}={}} l} \%\text{C} & \frac{9 \;\text{mol C} \;\times\; \text{molar mass C}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{9 \times 12.01 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{108.09 \;\text{g/mol}}{180.159 \;\text{g/mol}} \times 100 \\[1em] \%\text{C} & 60.00\%\;\text{C} \end{array}[/latex]
[latex]\begin{array}{r @{{}={}} l} \%\text{H} & \frac{8 \;\text{mol H} \;\times\; \text{molar mass H}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{8 \times 1.008 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{8.064 \;\text{g/mol}}{180.159 \;\text{g/mol}} \times 100 \\[1em] \%\text{H} & 4.476\%\;\text{H} \end{array}[/latex]
[latex]\begin{array}{r @{{}={}} l} \%\text{O} & \frac{4 \;\text{mol O} \;\times\; \text{molar mass O}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{4 \times 16.00 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{64.00 \;\text{g/mol}}{180.159 \;\text{g/mol}} \times 100 \\[1em] \%\text{O} & 35.52\%\;\text{O} \end{array}[/latex]
Note that these percentages sum to equal 100.00% when appropriately rounded.
Exercise 7.3b
To three significant digits, what is the mass percentage of iron in the compound Fe2O3?
Check Your Answer[2]
Key Equations
- [latex]\%\text{X} = \frac{\text{mass X}}{\text{mass commpound}} \times 100\%[/latex]
Attribution & References
Except where otherwise noted, this page is adapted by Adrienne Richards from “6.2 Determining Empirical and Molecular Formulas” In General Chemistry 1 & 2 by Rice University, a derivative of Chemistry (Open Stax) by Paul Flowers, Klaus Theopold, Richard Langley & William R. Robinson and is licensed under CC BY 4.0. Access for free at Chemistry (OpenStax)
percentage by mass of the various elements in a compound