Example 4.13: Calculation of Percent Yield
Calculation of Percent Yield
Upon reaction of
1.274 g of copper sulfate with excess zinc metal,
0.392 g copper metal was obtained according to the equation:
\(\ce{CuSO4(aq)}\)\(\ce{ + }\)\(\ce{Zn(s)}\)\(\ce{->}\)\(\ce{Cu(s)}\)\(\ce{ + }\)\(\ce{ZnSO4(aq)}\)\(\ce{ }\)
What is the relative yield expressed in percent?
Solution
\(m_{\mathrm{\ce{CuSO4}}}\) \(= 1.274\ \mathrm{g}\)
\(m_{\mathrm{\ce{Cu(s)}}}\) \(= 0.392\ \mathrm{g}\)
\(\dfrac{n_{\mathrm{\ce{Cu(s)},theo}}}{n_{\mathrm{\ce{CuSO4}}}} = 1\)
The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:
\(n_{\mathrm{\ce{CuSO4}}}\) \(= \dfrac{m_{\mathrm{\ce{CuSO4}}}}{M_{\mathrm{\ce{CuSO4}}}}\)
\(\ \ \ =\dfrac{1.274\ \mathrm{g}}{159.60\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =7.982\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{Cu(s)},actual}}\) \(= \dfrac{m_{\mathrm{\ce{Cu(s)}}}}{M_{\mathrm{\ce{Cu}}}}\)
\(\ \ \ =\dfrac{0.392\ \mathrm{g}}{63.546\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =6.17\times 10^{-3}\ \mathrm{mol}\)
\(n_{\mathrm{\ce{Cu(s)},theo}}\) \(= 1 \cdot n_{\mathrm{\ce{CuSO4}}}\)
\(\ \ \ =1 \cdot 7.982\times 10^{-3}\ \mathrm{mol}\)
\(\ \ \ =7.982\times 10^{-3}\ \mathrm{mol}\)
Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be
\(\mathrm{yield}_{\mathrm{rel}}\) \(= \dfrac{n_{\mathrm{\ce{Cu(s)},actual}}}{n_{\mathrm{\ce{Cu(s)},theo}}} \cdot 100\ \mathrm{％}\)
\(\ \ \ =\dfrac{6.17\times 10^{-3}\ \mathrm{mol}}{7.982\times 10^{-3}\ \mathrm{mol}} \cdot 100\ \mathrm{％}\)
\(\ \ \ =0.7728 \cdot 100\ \mathrm{％}\)
\(\ \ \ =77.3\ \mathrm{％}\)