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Example 3.9: Calculation of Percent Composition

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

Solution

\(m_{\mathrm{sample}}\) \(= 12.04\ \mathrm{g}\)


\(m_{\mathrm{C}}\) \(= 7.34\ \mathrm{g}\)


\(m_{\mathrm{H}}\) \(= 1.85\ \mathrm{g}\)


\(m_{\mathrm{N}}\) \(= 2.85\ \mathrm{g}\)


We are looking for the fraction by mass of these elements in the compound.
\(f_{\mathrm{C}}\) = ?


\(f_{\mathrm{H}}\) = ?


\(f_{\mathrm{N}}\) = ?


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:

\(f_{\mathrm{C}}\) \(= \dfrac{m_{\mathrm{C}}}{m_{\mathrm{sample}}} \cdot 100\ \mathrm{％}\)

\(\ \ \ =\dfrac{7.34\ \mathrm{g}}{12.04\ \mathrm{g}} \cdot 100\ \mathrm{％}\)

\(\ \ \ =0.6096 \cdot 100\ \mathrm{％}\)

\(\ \ \ =61.0\ \mathrm{％}\)


\(f_{\mathrm{H}}\) \(= \dfrac{m_{\mathrm{H}}}{m_{\mathrm{sample}}} \cdot 100\ \mathrm{％}\)

\(\ \ \ =\dfrac{1.85\ \mathrm{g}}{12.04\ \mathrm{g}} \cdot 100\ \mathrm{％}\)

\(\ \ \ =0.15365 \cdot 100\ \mathrm{％}\)

\(\ \ \ =15.37\ \mathrm{％}\)


\(f_{\mathrm{N}}\) \(= \dfrac{m_{\mathrm{N}}}{m_{\mathrm{sample}}} \cdot 100\ \mathrm{％}\)

\(\ \ \ =\dfrac{2.85\ \mathrm{g}}{12.04\ \mathrm{g}} \cdot 100\ \mathrm{％}\)

\(\ \ \ =0.23671 \cdot 100\ \mathrm{％}\)

\(\ \ \ =23.67\ \mathrm{％}\)


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