A student made ethyl ethanoate by heating 6.00 g of ethanoic acid with an excess of ethanol and a little concentrated sulfuric acid.
CH3COOH + CH3CH2OH ⇌ CH3COOCH2CH3 + H2O
The 5.80 g of liquid collected was analysed by gas chromatography.
| Retention time of peak / min | 1.9 | 3.1 | 4.4 |
|---|---|---|---|
| Peak area / arbitrary units | 24 | 540 | 36 |
Under the same conditions, standards gave these retention times: ethanol 1.9 min, ethyl ethanoate 3.1 min, ethanoic acid 4.4 min.
Use Mr values: ethanoic acid = 60.0, ethyl ethanoate = 88.0
Identify the two impurities in the liquid collected.
[1]
Assume that the percentage by mass of each component is proportional to its peak area. Calculate the mass of ethyl ethanoate in the 5.80 g of liquid.
[2]
Calculate the percentage yield of ethyl ethanoate.
[3]
Suggest why the student's yield is less than 100% and why the result in part (c) is more reliable than a yield calculated from the 5.80 g of liquid.
[2]