Example 4.15: Gravimetric Analysis

Gravimetric Analysis

A 0.4550 g solid mixture containing \(\ce{MgSO4}\) is dissolved in water and treated with an excess of \(\ce{Ba(NO3)2}\), resulting in the precipitation of 0.6168 g of \(\ce{BaSO4}\).

\(\ce{MgSO4(aq)}\)\(\ce{ + }\)\(\ce{Ba(NO3)2(aq)}\)\(\ce{->}\)\(\ce{BaSO4(s)}\)\(\ce{ + }\)\(\ce{Mg(NO3)2(aq)}\)\(\ce{ }\)

What is the mass ratio (percent) of \(\ce{MgSO4}\) in the mixture?

Solution

\(m_{\mathrm{mix}}\) \(= 0.4550\ \mathrm{g}\)


\(m_{\mathrm{\ce{BaSO4}}}\) \(= 0.6168\ \mathrm{g}\)


The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of \(\ce{BaSO4}\) and \(\ce{MgSO4}\) through their stoichiometric factor. Once the mass of \(\ce{MgSO4}\) is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.

\(n_{\mathrm{\ce{BaSO4}}}\) \(= \dfrac{m_{\mathrm{\ce{BaSO4}}}}{M_{\mathrm{\ce{BaSO4}}}}\)

\(\ \ \ =\dfrac{0.6168\ \mathrm{g}}{233.38\ \frac{\mathrm{g}}{\mathrm{mol}}}\)

\(\ \ \ =2.6429\times 10^{-3}\ \mathrm{mol}\)


\(\dfrac{n_{\mathrm{\ce{BaSO4}}}}{n_{\mathrm{\ce{MgSO4}}}} = \dfrac{1}{1}\)     

\(n_{\mathrm{\ce{MgSO4}}}\) \(= n_{\mathrm{\ce{BaSO4}}}\)

\(\ \ \ =2.6429\times 10^{-3}\ \mathrm{mol}\)

\(\ \ \ =2.6429\times 10^{-3}\ \mathrm{mol}\)


The mass of \(\ce{MgSO4}\) that would yield the provided precipitate mass is

\(m_{\mathrm{\ce{MgSO4}}}\) \(= n_{\mathrm{\ce{MgSO4}}} \cdot M_{\mathrm{\ce{MgSO4}}}\)

\(\ \ \ =2.6429\times 10^{-3}\ \mathrm{mol} \cdot 120.36\ \frac{\mathrm{g}}{\mathrm{mol}}\)

\(\ \ \ =0.31810\ \mathrm{g}\)


The concentration of \(\ce{MgSO4}\) in the sample mixture is then calculated to be

\(\mathrm{ratio}_{\mathrm{mass\ \ce{MgSO4}}}\) \(= \dfrac{m_{\mathrm{\ce{MgSO4}}}}{m_{\mathrm{mix}}} \cdot 100\ \mathrm{%}\)

\(\ \ \ =\dfrac{0.31810\ \mathrm{g}}{0.4550\ \mathrm{g}} \cdot 100\ \mathrm{%}\)

\(\ \ \ =0.69912 \cdot 100\ \mathrm{%}\)

\(\ \ \ =69.91\ \mathrm{%}\)