How to find the actual yield and the percentage yield
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
Step Two: Find the theoretical yield.
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x 100%
Step Two: Find the theoretical yield.
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x 100%
3 worked out examples of how to find the actual/percentage yield
1) For the balanced equation shown below, if the reaction of 19.2 grams of O2 produced 6.76 grams of H2O, what is the percent yield?
C7H16 + 11O2 --> 7CO2 + 8H2O
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
The limiting reagent is O2
Step Two: Find the theoretical yield.
O2 moles = 19.2 grams/32 GFW --> O2 moles = 0.6 moles
ratio between O2 and H2O --> 11:8 or 1.375:1
moles of H2O = (0.6 moles)/(1.375) = O.436 moles
grams of H2O = (0.436 moles)x(18 GFW) = 7.848 grams of H2O
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x 100%
Percentage Yield = (6.76 grams)/(7.848 grams) x 100%
Percentage Yield = 86.14%
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2) For the balanced equation shown below, if the reaction of 83.3 grams of H2O produced a 31.7% yield, how many grams of O2 would be produced?
2F2 + 2H2O --> 4HF + O2
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
The limiting reagent is H2O.
Step Two: Find the theoretical yield.
H2O moles = 88.3 grams/18 GFW --> H2O moles = 4.906 moles
ratio between H2O and O2 --> 2:1
moles of O2 = (4.906 moles)/(2) = 2.453 moles
grams of O2 = (2.453 moles)x(32 GFW) = 78.496 grams of O2
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x100%
31.7% = (actual yield)/(78.496 grams) x 100%
Actual Yield = 24.883 grams
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3) For the balanced equation shown below, if the reaction of 57.8 grams of O2 produced a 35.8% yield, how many grams of CO2 would be produced? C4H8O2 + 5O2 --> 4CO2 + 4H2O
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
The limiting reagent is O2.
Step Two: Find the theoretical yield.
O2 moles = 57.8 grams/32 GFW --> O2 moles = 1.806 moles
ratio between O2 and CO2 --> 5:4 OR 1.25:1
moles of CO2 = (1.806 moles)/(1.25) = 1.445 moles
grams of CO2 = (1.445 moles)x(44 GFW) = 63.571 grams of CO2
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x100%
35.8% = (actual yield)/(63.571 grams) x 100%
Actual Yield = 22.758 grams
C7H16 + 11O2 --> 7CO2 + 8H2O
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
The limiting reagent is O2
Step Two: Find the theoretical yield.
O2 moles = 19.2 grams/32 GFW --> O2 moles = 0.6 moles
ratio between O2 and H2O --> 11:8 or 1.375:1
moles of H2O = (0.6 moles)/(1.375) = O.436 moles
grams of H2O = (0.436 moles)x(18 GFW) = 7.848 grams of H2O
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x 100%
Percentage Yield = (6.76 grams)/(7.848 grams) x 100%
Percentage Yield = 86.14%
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2) For the balanced equation shown below, if the reaction of 83.3 grams of H2O produced a 31.7% yield, how many grams of O2 would be produced?
2F2 + 2H2O --> 4HF + O2
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
The limiting reagent is H2O.
Step Two: Find the theoretical yield.
H2O moles = 88.3 grams/18 GFW --> H2O moles = 4.906 moles
ratio between H2O and O2 --> 2:1
moles of O2 = (4.906 moles)/(2) = 2.453 moles
grams of O2 = (2.453 moles)x(32 GFW) = 78.496 grams of O2
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x100%
31.7% = (actual yield)/(78.496 grams) x 100%
Actual Yield = 24.883 grams
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3) For the balanced equation shown below, if the reaction of 57.8 grams of O2 produced a 35.8% yield, how many grams of CO2 would be produced? C4H8O2 + 5O2 --> 4CO2 + 4H2O
Step One: Identify the limiting reagent. (the question only gives you one of the reactants, therefore, making that compound the limiting reagent)
The limiting reagent is O2.
Step Two: Find the theoretical yield.
O2 moles = 57.8 grams/32 GFW --> O2 moles = 1.806 moles
ratio between O2 and CO2 --> 5:4 OR 1.25:1
moles of CO2 = (1.806 moles)/(1.25) = 1.445 moles
grams of CO2 = (1.445 moles)x(44 GFW) = 63.571 grams of CO2
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x100%
35.8% = (actual yield)/(63.571 grams) x 100%
Actual Yield = 22.758 grams