Stoichiometry Reality II:
Percent Yield

 A reaction is never 100% efficient. In other words:
Not all of the reactants in a chemical reaction are converted into the predicted products. The products either don’t form or other products form in unpredictable “side reactions”.

 When a chemist wants to analyze the efficiency of a reaction, they calculate the percent yield .
 Under different conditions of temperature, pressure, etc. a reaction efficiency may change .
 This calculation allows a chemist to compare a reaction under different conditions to determine what conditions give maximum efficiency for a reaction.

 The calculation requires two bits of information:
1. Actual Yield The Sweat
The mass of the product measured in an experiment.
2. Theoretical Yield The Stoichiometry
The mass of the product calculated by stoichiometry – the ideal amount.

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 To calculate the Percent Yield, use this equation:
% Yield =
Actual Yield (g or mol)
Theoretical Yield (g or mol)
 100%

Expected Percent Yield
The percent yield is expected to be less than
100% , and there are two reasons for this.
a.
Most of the reactants in a reaction don’t have enough energy to react.
b.
Some of the reactants may react to form other products in “side reactions” and these products may not be measured in the Actual Yield.

Example Percent Yield Question
Two industrious Chem students mixed
15.52 g of Aluminum with 46.75 g of chlorine gas. When the reaction was complete, they found the mass of the product, aluminum chloride, to be 40.58 g.
What was the percent yield?

Example Percent Yield Question
Two industrious Chem students mixed
15.52 g of Aluminum with 46.75 g of chlorine gas. When the reaction was complete, they found the mass of the product, aluminum chloride, to be 40.58 g .
In a stoichiometry question, the mass of the product that is given is the Actual Yield .

Example Percent Yield Question
 How would you find the actual yield of a reaction?
You measure the mass of a product in an experiment.
OR
The Actual Yield is given in the question .

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Example Percent Yield Question
 Determine the theoretical yield of a reaction.
1) Balanced reaction equation:
2 Al
(s)
+ 3 Cl
2(g)
2) Moles of each reactant:
→ 2 AlCl
3(s) mol Al = mass Al 
1 mol
MM
= 15.52g 
1 mol
26.98g
= 0.5752 mol Al mol Cl
2
= mass Cl
2

1 mol
MM
= 46.75g 
1 mol
70.90g
= 0.6594 mol Cl
2



Example Percent Yield Question
 Determine the theoretical yield of a reaction.
3) Limiting reactant: mol AlCl
3
= mol Al 
2 AlCl
3
2 Al mol AlCl
3
= mol Cl
2

2 AlCl
3
3 Cl
2
= 0.5752 mol Al 
2 AlCl
3
2 Al
= 0.6594 mol Cl
2

2 AlCl
3 Cl
2
3
= 0.5752 mol AlCl
3
= 0.4396 mol AlCl
3
Cl
2 is the limiting reactant, we predict a theoretical yield of
0.4396 mol of AlCl
3 which translates into: mass AlCl
3
= mol AlCl
3

MM
1 mol
= 0.4396 mol 
133.33g
1 mol
= 58.61g


Example Percent Yield Question
 Determine the Percent Yield of this reaction.
% Yield =
Actual Yield
Theoretica l Yield
 100%
=
40.58g
58.61g
 100%
= 69.24%

Percent Yield
 How do you interpret a Percent Yield greater than
100%?
A percent yield greater than 100% means that the
Actual Yield is falsely high.
This is most likely due to impurities in the product.
In this example, the impurities could be excess Al or unknown products of side reactions that were not separated out before the final mass was measured.