Question

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0 % yield. How many grams would be produced from 7.45 g of butanoic acid and excess ethanol?
Express your answer in grams to three significant figures.

Answer 1

7.66g of ethyl butyrate is produced.

The reaction is

CH₃CH₂CH₂COOH + CH₃CH₂OH ----> CH₃CH₂CH₂COOCH₂CH₃

• The molar mass of butanoic acid is 88.11g/mol
• We have 7.45g of butanoic acid
• The moles of butanoic acid we have is 7.45/88.11 = 0.0845 mol
• If the yield is 100%,  1 mole of butanoic acid gives 1 mole of ethyl butyrate
• But the reaction yield is 78%
• 1 mole of butanoic acid gives 0.78 mole of ethyl butyrate

From 0.0845 mol of butanoic acid we get 0.78 x 0.0845 = 0.66 mol of ethyl butyrate.

The molar mass of ethyl butyrate is 116.16g/mol

So 0.66 x 116.16 = 7.66g

7.66g of ethyl butyrate is produced.

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