Calculating Percentage Yield
On this screen you will see how to calculate
the percentage yield of a product.
the percentage yield of a product.
In the reaction:
CH3COOH + C2H5OH ---------» «--------- CH3COOC2H5 + H20
40 grams of Ethanoic Acid reacts with 30 grams of Ethanol, and 45 grams of Ethyl Ethanoate are produced. What is the percentage yield of Ethyl Ethanoate, to one decimal place?
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Percentage Yield = Actual Yield x 100 %
Theoretical Yield
You need to find the theoretical yield of the reaction.
Theoretical Yield
a Find which of Ethanoic Acid and Ethanol is the limiting reactant.
Moles of Ethanoic Acid = Mass/RMM = 40/60.0 = 0.6667 moles
Moles of Ethanol = Mass/RMM = 30/46.0 = 0.6522 moles
Moles of Ethanol = Mass/RMM = 30/46.0 = 0.6522 moles
From the balanced equation you can see that:
1 mole of Ethanoic Acid would produce 1 mole of Ethyl Ethanoate. Thus:
0.667 moles of Ethanoic Acid would produce 0.667 moles of Ethyl Ethanoate.
1 mole of Ethanol would produce 1 mole of Ethyl Ethanoate. Thus:
0.6522 moles of Ethanol would produce 0.6522 moles of Ethyl Ethanoate.
0.667 moles of Ethanoic Acid would produce 0.667 moles of Ethyl Ethanoate.
1 mole of Ethanol would produce 1 mole of Ethyl Ethanoate. Thus:
0.6522 moles of Ethanol would produce 0.6522 moles of Ethyl Ethanoate.
b Find the mass of Ethyl Ethanoate produced.
0.6522 moles of Ethanol produces 0.6522 moles of Ethyl Ethanoate
Mass of Ethyl Ethanoate = Moles x RMM = 0.6522 x 88.0 g = 57.3936 grams
Mass of Ethyl Ethanoate = Moles x RMM = 0.6522 x 88.0 g = 57.3936 grams
You can now find the percentage yield of Ethyl Ethanoate.
Percentage Yield = Actual Yield/Theoretical Yield x 100 %
Percentage Yield = 45 / 57.3936 x 100 % = 78.40 % = 78.4 % to 1 dec pl
Percentage Yield = 45 / 57.3936 x 100 % = 78.40 % = 78.4 % to 1 dec pl