Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used

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 7.05 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? Express your answer in grams to three significant figures.

Part B A chemist ran the reaction and obtained 5.55 g  of ethyl butyrate. What was the percent yield? Express your answer as a percent to three significant figures.

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 7.05 g of butanoic acid and excess ethanol? Express your answer in grams to three significant figures.

Explanation / Answer

A)

MW of butanoic acid = 88.11 g/mol

mol of butanoic acid = mass/MW = 7.05/88.11 = 0.08 mol of But. Acid

1 mol of butanoic acid + 1 mol of ethanol --> 1 mol of ethyl butyrate

Therefore, ratio is 1:1 for 0.08 mol of Butanoic acid we produce 0.08 mol of ethyl butryate

Now, MW of Ethyl butyrate = 116.16 g/mol

change moles to mass

mass =mol*MW = 0.08*116.16 = 9.29 g of Ethyl Bytyrate

B)

Find the % yield:

% yield = real/theoretical = 5.55/9.29 *100% = 59.7%

C)

Yield = 78%

how many grams could be produced from those 7.05 g of butanoic acid?

We already proved in A that we could create up to 9.29 grams of product

but only 78% will be converted so,

0.78*9.29 = 7.249 g is the product yield

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