Question

Outline the steps needed to solve the following problem, then do the calculations. Ether,\({\left( {{C_2}{H_5}} \right)_2}O\), which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.\(2{C_2}{H_5}OH + {H_2}S{O_4} \to \left( {{C_2}{H_5}} \right)2O + {H_2}S{O_4} \cdot {H_2}O\)What is the percent yield of ether if 1.17 L (d = 0.7134 g/mL) is isolated from the reaction of 1.500 L of C₂H₅OH (d = 0.7894 g/mL)?

Short answer

The percent yield of ether is 87.5%.

Step-by-step solution

Determine molar mass

1 mol\({C_2}{H_5}OH = 2\left( {12.011} \right) + 6\left( {1.008} \right) + 15.999 = 46.069g\)

1 mol \({\left( {{C_2}{H_5}} \right)_2}O = 4\left( {12.011} \right) + 10\left( {1.008} \right) + \left( {15.999} \right) = 74.123g\)

Determine number of moles of C2H5OH

Find the mole of C₂H₅OH present.

\(1.500\,L\,{C_2}{H_5}OH\left( {\frac{{1000\,mL}}{{1\,L}}} \right)\left( {0.7894\frac{g}{{mL}}} \right)\left( {\frac{{1\,mol}}{{46.069\,g}}} \right) = 25.703\,mol\,{C_2}{H_5}OH\)

Determine actual yield

Find the actual yield of\({\left( {{C_2}{H_5}} \right)_2}O\)

\(1.17L\left( {\frac{{1000mL}}{{1L}}} \right)\left( {0.7134\frac{g}{{mL}}} \right) = 834.68g{\left( {{C_2}{H_5}} \right)_2}O\)

Find the theoretical yield of\({\left( {{C_2}{H_5}} \right)_2}O\)

\(25.703\,mol\,{C_2}{H_5}OH\left( {\frac{{1mol\,\,{{\left( {{C_2}{H_5}} \right)}_2}O}}{{2mol\,\,\,{{\left( {{C_2}{H_5}} \right)}_2}O}}} \right)\left( {\frac{{74.123g{{\left( {{C_2}{H_5}} \right)}_2}O}}{{1mol{{\left( {{C_2}{H_5}} \right)}_2}O}}} \right) = 952.59g\,\,{\left( {{C_2}{H_5}} \right)_2}O\)

Determine percentage yield

Calculate the percent yield.

%yield =\(\frac{{{\rm{actual yield}}}}{{{\rm{theoretical yield}}}} \times 100\% \)

\(\begin{aligned}{} &= \frac{{834.68g}}{{952.59g}} \times 100\\ &= 87.6\% \end{aligned}\)