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22 August, 13:44

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

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Answers (2)
  1. 22 August, 14:18
    0
    58.8%

    Explanation:

    n = Experimental/Theoretical ·100

    n = 5.50 / (your answer to part A) 9.36·100 = 58.8%
  2. 22 August, 15:19
    0
    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 %
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