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.