Achemist ran a reaction and obtained 5.50 g of ethyl butyrate. what was the percent yield?

a. The amount of ethyl butyrate that would be synthesized from this chemical reaction is 9.361 grams.

b. The percentage yield of 5.50 grams of ethyl butyrate is 58.75%.

Given the following data:

First of all, we would write a properly balanced chemical equation for this chemical reaction between butanoic acid and ethanol.

Since, we are assuming a complete 100% yield, all of the 7.10 grams of butanoic acid are used up in the chemical reaction.

So, we would determine the number of moles of butanoic acid used:

Number of moles = 0.0807 moles

Next, we would determine the amount of ethyl butyrate that would be synthesized from this chemical reaction:

1 mole of butanoic acid = 1 mole of ethyl butyrate

0.0807 mole of butanoic acid = 0.0807 mole of ethyl butyrate

Mass = 9.361 grams

Part B:

To find the percentage yield of 5.50 grams of ethyl butyrate:

Percent yield = 58.75%

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Complete Question:

A. Given 7.10 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.

B. A chemist ran the reaction and obtained 5.50 g of ethyl butyrate. What was the percent yield? Express your answer as a percent to three significant figures

9.359 g of ethyl butyrate

% yield = 58.766 %

Explanation:

The reaction is:

C4H8O2 +  C2H5OH -> C6H12O2 + H2O

Molecular weight of butanoic acid: 88 g/mole

Molecular weight of ethyl butyrate: 116 g/mole

Assuming 100% yield, all 7.10 g of butanoic acid reacts. In moles are:

(7.10 g) / (88 g/mole) = 0.0806 moles of butanoic acid

From the balanced equation we know that 1 mole of butanoic acid produce 1 mole of ethyl butyrate, then 0.0806 moles of ethyl butyrate are produced. In grams are:

0.0806 moles * 116 g/mole = 9.359 g of ethyl butyrate (this represents the theoretical yield)

If 5.50 g of ethyl butyrate are produced, then the percent yield is:

% yield = (actual yield /theoretical yield) * 100

% yield = (5.50 g/9.359 g) * 100

% yield = 58.766 %

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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