Example 3.13: Determination of the Molecular Formula for Nicotine
Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If
40.57 g of nicotine contains
0.2500 mol nicotine, what is the molecular formula?
Solution
\(f_{\mathrm{C}}\) \(= 74.02\ \mathrm{％}\)
\(f_{\mathrm{H}}\) \(= 8.710\ \mathrm{％}\)
\(f_{\mathrm{N}}\) \(= 17.27\ \mathrm{％}\)
\(m_{\mathrm{sample}}\) \(= 40.57\ \mathrm{g}\)
\(n_{\mathrm{sample}}\) \(= 0.2500\ \mathrm{mol}\)
Determining the molecular formula from the provided data will require comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula.
\(n_{\mathrm{C}}\) \(= \dfrac{m_{\mathrm{sample}} \cdot f_{\mathrm{C}}}{M_{\mathrm{\ce{C}}}}\)
\(\ \ \ =\dfrac{40.57\ \mathrm{g} \cdot 74.02\ \mathrm{％}}{12.011\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =\dfrac{30.0299\ \mathrm{g}}{12.011\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =2.5002\ \mathrm{mol}\)
\(n_{\mathrm{H}}\) \(= \dfrac{m_{\mathrm{sample}} \cdot f_{\mathrm{H}}}{M_{\mathrm{\ce{H}}}}\)
\(\ \ \ =\dfrac{40.57\ \mathrm{g} \cdot 8.710\ \mathrm{％}}{1.008\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =\dfrac{3.5336\ \mathrm{g}}{1.008\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =3.506\ \mathrm{mol}\)
\(n_{\mathrm{N}}\) \(= \dfrac{m_{\mathrm{sample}} \cdot f_{\mathrm{N}}}{M_{\mathrm{\ce{N}}}}\)
\(\ \ \ =\dfrac{40.57\ \mathrm{g} \cdot 17.27\ \mathrm{％}}{14.007\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =\dfrac{7.0064\ \mathrm{g}}{14.007\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =0.5002\ \mathrm{mol}\)
Now, divide by the chemical amount of nicotine present in the sample to obtain the coefficients of the molecular formula.
\(\mathrm{coeff}_{\mathrm{C}}\) \(= \dfrac{n_{\mathrm{C}}}{n_{\mathrm{sample}}}\)
\(\ \ \ =\dfrac{2.5002\ \mathrm{mol}}{0.2500\ \mathrm{mol}}\)
\(\ \ \ =1000.1\ \mathrm{％}\)
\(\mathrm{coeff}_{\mathrm{H}}\) \(= \dfrac{n_{\mathrm{H}}}{n_{\mathrm{sample}}}\)
\(\ \ \ =\dfrac{3.506\ \mathrm{mol}}{0.2500\ \mathrm{mol}}\)
\(\ \ \ =1402.\ \mathrm{％}\)
\(\mathrm{coeff}_{\mathrm{N}}\) \(= \dfrac{n_{\mathrm{N}}}{n_{\mathrm{sample}}}\)
\(\ \ \ =\dfrac{0.5002\ \mathrm{mol}}{0.2500\ \mathrm{mol}}\)
\(\ \ \ =200.1\ \mathrm{％}\)
The molecular formula is
\(\ce{C10H14N2}\). As a check, we can calculate the molar mass of nicotine from the experimental data and compare it to that obtained from the molecular formula
\(M_{\mathrm{nicotine}}\) \(= \dfrac{m_{\mathrm{sample}}}{n_{\mathrm{sample}}}\)
\(\ \ \ =\dfrac{40.57\ \mathrm{g}}{0.2500\ \mathrm{mol}}\)
\(\ \ \ =162.28\ \frac{\mathrm{g}}{\mathrm{mol}}\)
\(M_{\mathrm{\ce{C10H14N2}}}\) \(= \sum (10 \cdot M_{\mathrm{\ce{C}}}, 14 \cdot M_{\mathrm{\ce{H}}}, 2 \cdot M_{\mathrm{\ce{N}}})\)
\(\ \ \ =\sum (10 \cdot 12.011\ \frac{\mathrm{g}}{\mathrm{mol}}, 14 \cdot 1.008\ \frac{\mathrm{g}}{\mathrm{mol}}, 2 \cdot 14.007\ \frac{\mathrm{g}}{\mathrm{mol}})\)
\(\ \ \ =\sum (120.110\ \frac{\mathrm{g}}{\mathrm{mol}}, 14.112\ \frac{\mathrm{g}}{\mathrm{mol}}, 28.0140\ \frac{\mathrm{g}}{\mathrm{mol}})\)
\(\ \ \ =162.24\ \frac{\mathrm{g}}{\mathrm{mol}}\)