Chapter 5. Chemical Composition

# 5.3 Percent Composition

### Learning Objectives

By the end of this section, you will be able to:

• Compute the percent composition of a compound from experimental mass measurements
• Compute the percent composition of a compound from its chemical formula

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 begin exploring how a chemist may go about determining the identity of a compound from experimental mass measurements and calculating percent composition.

## 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:

$\% \;\text{H} = \frac{\text{mass H}}{\text{mass compound}} \times 100 \%$
$\% \;\text{C} = \frac{\text{mass C}}{\text{mass compound}} \times 100 \%$

If analysis of a 10.0-g sample of this gas showed that it contains 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

$\%\;\text{H} = \frac{2.5 \;\text{g H}}{10.0 \;\text{g compound}} \times 100 \% = 25 \%$
$\%\;\text{C} = \frac{7.5 \;\text{g C}}{10.0 \;\text{g compound}} \times 100 \% = 75 \%$

### Example 1

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed that it contains 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

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:

$\%\;\text{C} = \frac{7.34 \;\text{g C}}{12.04 \;\text{g compound}} \times 100\% = 61.0\%$
$\%\;\text{H} = \frac{1.85 \;\text{g H}}{12.04 \;\text{g compound}} \times 100\% = 15.4\%$
$\%\;\text{N} = \frac{2.85 \;\text{g N}}{12.04 \;\text{g compound}} \times 100\% = 23.7\%$

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Test Yourself
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?

12.1% C, 16.1% O, 71.8% Cl

## Determining Percent Composition from The Chemical Formula

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 14.0067 amu and three H atoms weighing a total of (3 × 1.00794 amu) = 3.02382 amu. The formula mass of ammonia is therefore (14.0067 amu + 3.02382 amu) = 17.0305 amu, and its percent composition is:

$\%\;\text{N} = \frac{14.0067 \;\text{amu N}}{17.0305 \;\text{amu NH}_3} \times 100\% = 82.2448\%$
$\%\;\text{H} = \frac{3.02382 \;\text{amu N}}{17.0305 \;\text{amu NH}_3} \times 100\% = 17.7553\%$

This same approach may be taken for 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 in 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 2

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 (180.160 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

$\begin{array}{r @{{}={}} l} \%\text{C} & \frac{9 \;\text{mol C} \;\times\; \text{molar mass C}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{9 \times 12.011 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{108.099 \;\text{g/mol}}{180.160 \;\text{g/mol}} \times 100 \\[1em] \%\text{C} & 60.002\%\;\text{C} \end{array}$

$\begin{array}{r @{{}={}} l} \%\text{H} & \frac{8 \;\text{mol H} \;\times\; \text{molar mass H}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{8 \times 1.00794 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{8.06352 \;\text{g/mol}}{180.160 \;\text{g/mol}} \times 100 \\[1em] \%\text{H} & 4.47575\%\;\text{H} \end{array}$

$\begin{array}{r @{{}={}} l} \%\text{O} & \frac{4 \;\text{mol O} \;\times\; \text{molar mass O}}{\text{molar mass} \;\text{C}_9\text{H}_{18}\text{O}_4} \times 100 = \frac{4 \times 15.994 \;\text{g/mol}}{180.159 \text{g/mol}} \times 100 = \frac{63.9976 \;\text{g/mol}}{180.160 \;\text{g/mol}} \times 100 \\[1em] \%\text{O} & 35.5226\%\;\text{O} \end{array}$

Note that these percentages sum to equal 100.00% when appropriately rounded.

Test Yourself
To three significant digits, what is the mass percentage of iron in the compound Fe2O3?

69.9% Fe

### Example 3

What is the percent by mass of H in 5.00g of H2O

Solution

We are given the total mass of the compound (5.00g). To solve for percent composition, we need to determine the mass of H in the sample. Remember that ONLY the chemical formula gives us the ability to go from H2O to H, and it only refers to particles or moles. Thus, our pathway:

grams H2O $\longrightarrow$ mol H2O $\longrightarrow$ mol H $\longrightarrow$ grams H

Then we will apply the percent composition formula.

$5.00\; \rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g of H}_2\text{O}\; \times \frac{1\; \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}_2\text{O}} {18.0153\; \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g H}_2\text{O}} \times \frac{2\; \rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}_2\text{O}} \times \frac{1.01\; \text{g H}}{1\;\rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol H}_2\text{O}}$ $= 0.561\; \text{g of H}$

$\%\;\text{H} = \frac{0.561 \;\text{g H}}{5.00 \;\text{g water}} \times 100 \% = 11.2 \%$

Test Yourself

What is the percent by mass of O in 10.0 g of Cu(NO3)2?

51.18%

### Example 4

Determine the percent composition of H in propanol (C3H7OH)

Solution

Considering one mol of C3H7OH, % of H in C3H7OH  =

$\%\;\text{H} = \frac{\text{mass of 8 mol H}}{\text{mass of 1 mol propanol}} \times 100 \% =$ $\frac{8.06352\; \text{g H}}{60.096\; \text{g propanol}} \times 100 \% = 13.418 \%$

Test Yourself

What is the percent composition of each element in H2SO4?

H = 2.06%, S = 32.69%, O = 65.25%

### Example 5

Hemoglobin contains 0.33% Fe by mass. The molar mass of hemoglobin is 6.8 x 104g/mol. How many grams of Fe are in 0.020 mol of hemoglobin?

Solution

The percent composition essentially refers to the mass of element in 100 g of compound. Thus 0.33% = 0.33g of Fe in 100 g of hemoglobin. We can use this as a conversion factor (but don’t forget that it’s based on 100 g of compound!)

Mol hemoglobin $\longrightarrow$ mass hemoglobin $\longrightarrow$ mass of Fe

$0.020\; \rule[0.5ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol of hemoglobin}\; \times \frac{6.8 \times 10^4\; \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g hemoglobin}} {1\; \rule[0.25ex]{1.25em}{0.1ex}\hspace{-1.25em}\text{mol hemoglobin}} \times \frac{0.33\; \text{g Fe}}{100\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{g hemoglobin}}$ $= 4.5\; \text{g of Fe}$

Test Yourself

How many grams of Ba3Pwould contain 4.23 g of Ba, if Ba3Pcontains 86.9% Ba by mass?

4.87 g Ba3P2

## Key Concepts and Summary

The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound.

## Key Equations

• $\%\text{X} = \frac{\text{mass X}}{\text{mass commpound}} \times 100\%$

### Exercises

1. Calculate the following to four significant figures:

a) the percent composition of ammonia, NH3

b) the percent composition of photographic “hypo,” Na2S2O3

c) the percent of calcium ion in Ca3(PO4)2

2. Determine the percent ammonia, NH3, in Co(NH3)6Cl3, to three significant figures.

1. a) % N = 82.24%, % H = 17.76%

b) % Na = 29.08%, % S = 40.56%, % O = 30.36%
c) % Ca2+ = 38.76%

2. % NH3 = 38.2%

## Glossary

percent composition: percentage by mass of the various elements in a compound