Q67E - Chemistry (OpenStax)

Question

Toluene, C₆H₅CH₃, is oxidized by air under carefully controlled conditions to benzoic acid, C₆H₅CO₂H, which is used to prepare the food preservative sodium benzoate, C₆H₅CO₂Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid? \(2{C_6}{H_5}C{H_3} + 3{O_2} \to 2{C_6}{H_5}C{O_2}H + 2{H_2}O\).

Step-by-Step Solution

Answer

The percent yield is  \(91.3\% \).

Step 1: Determine molar mass

1 mol \({C_6}{H_5}C{H_3} = 7\left( {12.011} \right) + 8\left( {1.008} \right) = 92.141g\)

1 mol \({C_6}{H_5}C{O_2}H = 7\left( {12.011} \right) + 6\left( {1.008} \right) + 2\left( {15.999} \right) = 122.123g\)

\(\begin{aligned}{\underline{\phantom{xx}}}1.000kg{C_6}{H_5}C{H_3}\left( {\frac{{1mol{C_6}{H_5}C{H_3}}}{{342.297g{C_6}{H_5}C{H_3}}}} \right)\left( {\frac{{2mol{C_6}{H_5}C{O_2}H}}{{1mol{C_6}{H_5}C{H_3}}}} \right)\left( {\frac{{122.123g{C_6}{H_5}C{O_2}H}}{{1mol{C_6}{H_5}C{O_2}H}}} \right)\\ = 1.325g{C_6}{H_5}C{O_2}H\end{aligned}\)

Step 2: Determine percentage yield

Calculate the percent yield.

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

\( = \frac{{1.21}}{{1.325}} = 91.3\% \)