91Ó°ÊÓ

One molecule of acetic acid and one molecule of ethanol produce one molecule of ethyl acetate and a water molecule.

This means that \(n{\;_{\left( {acetic{\rm{ }}acid} \right)}}\; = {n_{\left( {ethyl{\rm{ }}acetate} \right)}}\)

There is \(10.0\;g\) of acetic acid.

It's molar mass is:

\(\begin{aligned}{}{M_{(aceticacid)}} &= 2(C) + 4(H) + 2(O)\\ &= 64.05gmo{l^ - }1\end{aligned}\)

Then, \(n\) of acetic acid is:

\(\begin{aligned}{}{n_{\left( {C{H_3}COOH} \right)}} &= \frac{{{m_{\left( {C{H_3}COOH} \right)}}}}{{{M_{\left( {C{H_3}COOH} \right)}}}}\\ &= \frac{{(10.0\;g)}}{{\left( {64.05gmo{l^{ - 1}}} \right)}}\\ &= 0.16\;mol\end{aligned}\)

The expected mass of ethyl acetate is the product of number of moles(n), and the molar mass (M) is:

\(\begin{aligned}{}{M_{\left( {C{H_3}COOC{H_2}C{H_3}} \right)}} &= 4(C) + 2(O) + 8(H)\\ &= 88.1gmo{l^{ - 1}}\end{aligned}\)

\(\begin{aligned}{}{m_{C{H_3}COOC{H_2}C{H_3}}} &= {n_{\left( {C{H_3}COOCHC{H_3}} \right)}} \times {M_{\left( {C{H_3}COOC{H_2}C{H_3}} \right)}}\\ &= (0.16\;mol) \times \left( {88.1gmo{l^{ - 1}}} \right)\\ &= 14.096\;g\end{aligned}\)

The ratio of experimental mass and theoretical mass multiplied by 100 results the yield.

\(\begin{aligned}{}Yield &= \frac{{m( experimental )}}{{m( theoretical )}} \times 100\\ &= \frac{{(13\;g)}}{{(14.096\;g)}} \times 100\\ &= 93\% \end{aligned}\)