Example 3.6: Deriving Moles from Grams for a Compound

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula \(\ce{C2H5O2N}\). How many moles of glycine molecules are contained in 28.35 g of glycine?

Solution

\(m_{\mathrm{glycine}}\) \(= 28.35\ \mathrm{g}\)


\(n_{\mathrm{glycine}}\) = ?


We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 2:

The molar mass of glycine is required for this calculation. One mole of glycine, \(\ce{C2H5O2N}\), contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

\(M_{\mathrm{\ce{C2H5O2N}}}\) \(= \sum (2 \cdot M_{\mathrm{\ce{C}}}, 5 \cdot M_{\mathrm{\ce{H}}}, 2 \cdot M_{\mathrm{\ce{O}}}, 1 \cdot M_{\mathrm{\ce{N}}})\)

\(\ \ \ =\sum (2 \cdot 12.011\ \frac{\mathrm{g}}{\mathrm{mol}}, 5 \cdot 1.008\ \frac{\mathrm{g}}{\mathrm{mol}}, 2 \cdot 15.999\ \frac{\mathrm{g}}{\mathrm{mol}}, 1 \cdot 14.007\ \frac{\mathrm{g}}{\mathrm{mol}})\)

\(\ \ \ =\sum (24.0220\ \frac{\mathrm{g}}{\mathrm{mol}}, 5.0400\ \frac{\mathrm{g}}{\mathrm{mol}}, 31.9980\ \frac{\mathrm{g}}{\mathrm{mol}}, 14.0070\ \frac{\mathrm{g}}{\mathrm{mol}})\)

\(\ \ \ =75.067\ \frac{\mathrm{g}}{\mathrm{mol}}\)



The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

\(n_{\mathrm{glycine}}\) \(= \dfrac{m_{\mathrm{glycine}}}{M_{\mathrm{\ce{C2H5O2N}}}}\)

\(\ \ \ =\dfrac{28.35\ \mathrm{g}}{75.067\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.3777\ \mathrm{mol}\)


This result is consistent with our rough estimate.