Example 3.16: Calculating Molar Concentrations from the Mass of Solute

Distilled white vinegar ( Figure 15) is a solution of acetic acid, \(\ce{CH3CO2H}\), in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of mol/L or M?
Figure 15: Distilled white vinegar is a solution of acetic acid in water.

Solution

\(V_{\mathrm{solution}}\) \(= 0.500\ \mathrm{L}\)


\(m_{\mathrm{\ce{CH3CO2H}}}\) \(= 25.2\ \mathrm{g}\)


\(c_{\mathrm{\ce{CH3CO2H}}}\) = ?


As in previous examples, the definition of concentration is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the chemical amount of solute:

\(n_{\mathrm{\ce{CH3CO2H}}}\) \(= \dfrac{m_{\mathrm{\ce{CH3CO2H}}}}{M_{\mathrm{\ce{CH3CO2H}}}}\)

\(\ \ \ =\dfrac{25.2\ \mathrm{g}}{60.052\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =0.420\ \mathrm{mol}\)


\(c_{\mathrm{\ce{CH3CO2H}}}\) \(= \dfrac{n_{\mathrm{\ce{CH3CO2H}}}}{V_{\mathrm{solution}}}\)

\(\ \ \ =\dfrac{0.420\ \mathrm{mol}}{0.500\ \mathrm{L}}\)

\(\ \ \ =0.839\ \frac{\mathrm{mol}}{\mathrm{L}}\)


\(c_{\mathrm{\ce{CH3CO2H}}}\) \(= 0.839\ \frac{\mathrm{mol}}{\mathrm{L}}\)

\(\ \ \ =0.839\ \mathrm{M}\)