Part A Given 7.70 g of butanoic acid and excess ethanol, how many grams of ethyl

Question

Part A

Given 7.70 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 5.70 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 7.70 g of butanoic acid and excess ethanol?

Express your answer in grams to three significant figures.

Ethyl buty rate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such favors 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+): CH3CH2CH CO2H() +CH2CH3OH H, CH3CH2CH2CO2CH2CH3 (1) H2o(I)

Explanation / Answer

molar mass of butanoic acid = 88g/mol

molar mass of ethylbutyrate = 116g/mol

According to the equation 1 mole of acid produces one mole of ester product.

Thus 7.7g of acid = 7.7/88 =0.0875 moles produces equal moles = 0.0875 moles of product ester.

Thus the theoretical yield of ester = 0.0875molx116 g/mol

= 10.15 g

Part B

When we obtain only 5.7 g of ester from this reaction

% yield = experimental yield x 100/ theoretical yield

= 5.7g x 100 /10.15g

= 56.16 %

Part C

When the yild was 78% the amount of ester formed is calculated as follows:

  

% yield = experimental yield x 100/ theoretical yield

78 = mass x 100/ 10.15

Thus mass of ester produced = 7.917 g

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