Percentage yield

The percentage yield of a chemical reaction is an important consideration in industrial chemistry.

It can be calculated to compare the yield (quantity) of product actually obtained with what could have been obtained in theory, if all of the reactants were converted with no loss or waste.

Obviously total conversion under ideal circumstances will be 100 per cent, but in reality, that will not happen.

The formula for the percentage yield calculation as found in the data booklet is:

\[\%\,yield = \frac{{Actual\,yield}}{{Theoretical\,yield}} \times 100\%\]

The higher the percentage yield is, the more efficient the reaction. Esterification and other reversible reactions can never result in 100 per cent conversion of reactants into products.

Example

5 g of methanol (CH3OH, formula mass = 32 g) reacts with excess ethanoic acid (CH3COOH) to produce 9.6 g of methyl ethanoate (CH3OOCCH3, formula mass = 74 g). Calculate the percentage yield.

The actual yield (9.6 g) is given in the question. You must first calculate the theoretical yield.

The balanced equation for the reaction shows that one mole of methanol can produce one mole of methyl ethanoate.

\[CH_{3}OH + CH_{3}COOH \rightleftharpoons CH_{3}OOCCH_{3} + H_{2}O\]

\[1\,mole\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,mole\]

Replacing the numbers of moles with their equivalent formula masses shows that:

\[32g\,\,C{H_3}OH = 74g\,\,C{H_3}OOCC{H_3}\\]

\[5g\,\,C{H_3}OH = \frac{5}{{32}} \times 74\\]

\[= 11.56g\,ester\\]

\[Theoretical\,yield = 11.56g\\]

\[Actual\,yield = 9.6g\\]

%\(\,\,yield = \frac{{Actual\,yield}}{{Theoretical\,yield}} \times 100\% \\)

\[ = \frac{{9.6}}{{11.56}} \times 100\\]

\[ = 83\% \\]