Question

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange and pineapple. Its fragrance and taste are often associated with fresh orange juice, and thus it is most commonly used as orange flavoring.

It can be produced by the reaction of butanoic acid with ethanol in the presence of an acid catalyst (H+):

CH3CH2CH2CO2H(l)+CH2CH3OH(l)H+⟶CH3CH2CH2CO2CH2CH3(l)+H2O(l)

Part A

Given 7.75 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Part B

A chemist ran the reaction and obtained 5.15 g of ethyl butyrate. What was the percent yield?

Part C

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be produced from 7.75 g of butanoic acid and excess ethanol?

Answer 1

Part A

Molar mass of Butanoic acid = 88.11 g/mol

Molar mass of ethyl butyrate = 116.16 g/mol

1 mole of butanoic acid gives 1 mole of ethyl butyrate

88.11 g/mol of butanoic acid gives 116.16g/mol of ethyl butyrate

1g of butanoic acid gives (116.16 (g/mol)/88.11 (g/mol)) of ethyl butyrate

7.75 g of butanoic acid gives (116.16 (g/mol)/88.11 (g/mol)) x 7.75 g of ethyl butyrate

7.75 g of butanoic acid gives 10.2172 g of ethyl butyrate ( 100% yield)

.

Part B

Experimental yield by a Chemist = 5.15 g

Theoretical yield (Part A) = 10.2172

Percentage yield = (Experimental yield/ theoretical yield) x 100

= (5.15 g/10.2172 g) x 100

Percentage yield = 50.41 %

Part C

As we know that 7.75 g of butanoic acid produces 10.2172 g of ethyl butyrate

Percentage yield = 78.0%

Percentage yield = (Experimental yield/ theoretical yield) x 100

Experimental yield = (percentage yield/100) x theoretical yield

= (78/100) x 10.2172

Experimental yield = 7.9694 g