A chemist ran a reaction and obtained 5.50 g of ethyl butyrate. What was the percent yield?

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 %

58.8%

Explanation:

n = Experimental/Theoretical ·100

n = 5.50 / (your answer to part A) 9.36·100 = 58.8%