12.3: Percent Composition
Lesson Objectives
The student will:
 calculate the percent composition by mass given the masses of elements in a compound.
 calculate the percent composition by mass given the formula or name of a compound.
Vocabulary
 percent composition
Introduction
Metals useful to man are typically extracted from ore. The ore is removed from the mine and partially purified by washing away dirt and other materials not chemically bound to the metal of interest. This partially purified ore is then treated chemically in smelters or other purifying processes to separate pure metal from the other elements. The value of the original ore is very dependent on how much pure metal can eventually be separated from it. Highgrade ore and lowgrade ore command significantly different prices. The ore can be evaluated before it is mined or smelted to determine what percent of the ore can eventually become pure metal. This process involves determining what percentage of the ore is metal compounds and what percentage of the metal compounds is pure metal.
Percent Composition from Masses
Compounds are made up of two or more elements. The law of definite proportions tells us that the proportion, by mass, of elements in a compound is always the same. Water, for example, is always \begin{align*}11\%\end{align*}
Percentage composition can be determined experimentally. To do this, a known quantity of a compound is decomposed in the laboratory. The mass of each element is measured and then divided by the total mass of the original compound. This tells us what fraction of the compound is made up of that element. The fraction can then be multiplied by 100% to convert it into a percent.
Example:
Laboratory procedures show that \begin{align*}50.0 \ \mathrm{grams}\end{align*}
Solution:



\begin{align*}\% \ \text{nitrogen} = \left (\frac {41.0\ \text{grams}} {50.0\ \text{grams}}\right ) \cdot (100\%) = 82\%\end{align*}
% nitrogen=(41.0 grams50.0 grams)⋅(100%)=82%

\begin{align*}\% \ \text{nitrogen} = \left (\frac {41.0\ \text{grams}} {50.0\ \text{grams}}\right ) \cdot (100\%) = 82\%\end{align*}




\begin{align*}\% \ \text{hydrogen} = \left (\frac {9.00\ \text{grams}} {50.0\ \text{grams}}\right ) \cdot (100\%) = 18\%\end{align*}
% hydrogen=(9.00 grams50.0 grams)⋅(100%)=18%

\begin{align*}\% \ \text{hydrogen} = \left (\frac {9.00\ \text{grams}} {50.0\ \text{grams}}\right ) \cdot (100\%) = 18\%\end{align*}

Example:
The decomposition of \begin{align*}25.0 \ \mathrm{grams}\end{align*}
Solution:



\begin{align*}\% \ \text{calcium} = \left (\frac {13.5\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 54.0\%\end{align*}
% calcium=(13.5 grams25.0 grams)⋅(100%)=54.0%

\begin{align*}\% \ \text{calcium} = \left (\frac {13.5\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 54.0\%\end{align*}




\begin{align*}\% \ \text{oxygen} = \left(\frac {10.8\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 43.2\%\end{align*}
% oxygen=(10.8 grams25.0 grams)⋅(100%)=43.2%

\begin{align*}\% \ \text{oxygen} = \left(\frac {10.8\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 43.2\%\end{align*}




\begin{align*}\% \ \text{hydrogen} = \left (\frac {0.68\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 2.8\%\end{align*}
% hydrogen=(0.68 grams25.0 grams)⋅(100%)=2.8%

\begin{align*}\% \ \text{hydrogen} = \left (\frac {0.68\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 2.8\%\end{align*}

You should note that the sum of the percentages always adds to \begin{align*}100 \%\end{align*}
Percent Composition from the Formula
Percent composition can also be calculated from the formula of a compound. Consider the formula for the compound iron(III) oxide, \begin{align*}\text{Fe}_2\text{O}_3\end{align*}
Example:
What is the percent composition of iron(III) oxide, \begin{align*}\text{Fe}_2\text{O}_3\end{align*}
Solution:



\begin{align*}
\begin{array}{cccc}
\text{Element} &\text{Atomic Mass} &\text{Number of Atoms} &\text{Product} \\
& &\text{per Formula} & \\
\hline
\text{Fe} &55.8 \ \text{daltons} &2 &111.6 \ \text{daltons} \\
\text{O} &16.0 \ \text{daltons} &3 &\underline{\ 48.0 \ \text{daltons} \ } \\
& &\text{Formula mass =} &159.6 \ \text{daltons} \\
\end{array}
\end{align*}
ElementFeOAtomic Mass55.8 daltons16.0 daltonsNumber of Atomsper Formula23Formula mass =Product111.6 daltons 48.0 daltons −−−−−−−−−−−159.6 daltons

\begin{align*}
\begin{array}{cccc}
\text{Element} &\text{Atomic Mass} &\text{Number of Atoms} &\text{Product} \\
& &\text{per Formula} & \\
\hline
\text{Fe} &55.8 \ \text{daltons} &2 &111.6 \ \text{daltons} \\
\text{O} &16.0 \ \text{daltons} &3 &\underline{\ 48.0 \ \text{daltons} \ } \\
& &\text{Formula mass =} &159.6 \ \text{daltons} \\
\end{array}
\end{align*}




\begin{align*}\% \ \text{iron} = \left (\frac {111.6 \ \text{daltons}} {159.6 \ \text{daltons}}\right ) \cdot (100\%) = 69.9 \%\end{align*}
% iron=(111.6 daltons159.6 daltons)⋅(100%)=69.9%

\begin{align*}\% \ \text{iron} = \left (\frac {111.6 \ \text{daltons}} {159.6 \ \text{daltons}}\right ) \cdot (100\%) = 69.9 \%\end{align*}




\begin{align*}\% \ \text{oxygen} = \left (\frac {48.0 \ \text{daltons}} {159.6 \ \text{daltons}}\right ) \cdot (100\%) = 30.1 \%\end{align*}
% oxygen=(48.0 daltons159.6 daltons)⋅(100%)=30.1%

\begin{align*}\% \ \text{oxygen} = \left (\frac {48.0 \ \text{daltons}} {159.6 \ \text{daltons}}\right ) \cdot (100\%) = 30.1 \%\end{align*}

Example:
What is the percent composition of aluminum sulfate, \begin{align*}\text{Al}_2(\text{SO}_4)_3\end{align*}
Solution:
The formula mass of \begin{align*}\text{Al}_2(\text{SO}_4)_3\end{align*}



\begin{align*}\% \ \text{aluminum} = \left (\frac {54.0 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 15.8 \%\end{align*}
% aluminum=(54.0 daltons342 daltons)⋅(100%)=15.8%

\begin{align*}\% \ \text{aluminum} = \left (\frac {54.0 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 15.8 \%\end{align*}




\begin{align*}\% \ \text{sulfur} = \left (\frac {96.0 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 28.1 \%\end{align*}
% sulfur=(96.0 daltons342 daltons)⋅(100%)=28.1%

\begin{align*}\% \ \text{sulfur} = \left (\frac {96.0 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 28.1 \%\end{align*}




\begin{align*}\% \ \text{oxygen} = \left (\frac {192 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 56.1 \%\end{align*}
% oxygen=(192 daltons342 daltons)⋅(100%)=56.1%

\begin{align*}\% \ \text{oxygen} = \left (\frac {192 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 56.1 \%\end{align*}

Lesson Summary
 The percent composition of a compound is the percent of the total mass contributed by each element in the compound.
 Percent composition can be determined either from the masses of each element in the compound or from the formula of the compound.
Further Reading / Supplemental Links
This website has solved example problems for a number of topics covered in this lesson, including the calculation of percent composition by mass.
The website below reviews how to calculate percent composition.
Review Questions
Determine the percent composition of the following compounds.

\begin{align*}\text{BF}_3\end{align*}
BF3 
\begin{align*}\text{Ca(C}_2\text{H}_3\text{O}_2)_2\end{align*}
Ca(C2H3O2)2 
\begin{align*}\text{FeF}_3\end{align*}
FeF3 
\begin{align*}\text{CrCl}_3\end{align*}
CrCl3 
\begin{align*}(\text{NH}_4)_3\text{PO}_4\end{align*}
(NH4)3PO4
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