Q52E - Chemistry (OpenStax)

Question

Yields in organic reactions are sometimes low. What is the percent yield of a process that produces 13.0 g of ethyl acetate from \(10.0\;g\) of\(C{H_3}C{O_2}H\) ?

Step-by-Step Solution

Answer

The yield is 93 %.

Step 1: To find the acetic acid and ethyl acetate:

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

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

Step 2: To find the acetic acid:

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

It's molar mass is:

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

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

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

Step 3: The ethyl acetate

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

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

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

Step 4: Yield

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

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