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Wize High School Grade 11 Chemistry Textbook > Stoichiometry

Percent Yield

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Percent Yield

  • Often, during chemical reactions in a laboratory, we cannot recover 100% of the product expected. This could be because:
  • Reactants or products are lost when they are being transferred
  • You may have undesirable side reactions
  • Your reaction may be incomplete
  • The amount of product that we expect from a stoichiometric calculation is known as the theoretical yield
  • We refer to the product amount that is weighed and recovered as the actual yield
  • We get the percent yield of a reaction by comparing the actual yield to the theoretical yield of a reaction
% yield= actual yieldtheoretical yield×100%\boxed{\%\ \text{yield}=\ \frac{\text{actual yield}}{\text{theoretical yield}}\times100\%}




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Example: Calculating Percent Yield

When 49.00g of a hydrocarbon fuel with formula C7H10O2 is reacted with excess oxygen, a total of 21.56g of water is collected. What was the percent yield of the reaction?

2C7H10O2+17O2→10H2O+14CO22C_7H_{10}O_2 + 17O_2 \to 10H_2O + 14CO_2

actual yield=21.56g water

m of C7H10O2 = 49g
MM of C7H10O2 =126 g/mol

nC7H10O2=mMM=49g126g/mol=0.38889mol C7H10O2n_{C_7H_{10}O_2} = \dfrac{m}{MM} = \dfrac{ 49 g} { 126 g/mol} = 0.38889mol \ C_7H_{10}O_2

Now we need to convert that into theoretical moles of water:
0.38889mol C7H10O2×10mol H2O2mol C7H10O2=1.945mol H2O0.38889 mol\ C_7H_{10}O_2\times\dfrac{10mol \ H_2O}{2mol \ C_7H_{10}O_2} = 1.945 mol \ H_2O

MM of H2O = 18 g/mol

Now we need to calculate theoretical mass:

mH2O=n×MM=1.945mol×18gmol=35.01gm_{H_2O} = n\times MM = 1.945 mol \times \dfrac{18 g}{mol}= 35.01 g

In the question it states we produced 21.56 g of water (actual mass).

Now we can solve for % yield:

%yield=actual yieldtheoretical yield×100=21.56g35.01g×100=62%\% \text{yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}}\times100 = \dfrac{21.56 g}{35.01 g} \times 100 = 62\%

% yield = actual yield/theoretical yield x 100 % yield = 21.56 g / 35.01 g x 100 % yield = 62%


The balanced equation for the complete combustion of butane is as follows:

4C4H10+13O2→8CO2+10H2O4C_4H_10 + 13 O_2 → 8 CO_2 + 10 H_2O
In an experiment, 12.37 g of carbon dioxide was produced when 14.25 g was predicted. What is the percentage yield? Round your answer to the nearest integer; do not include the % symbol.

Consider the following reaction:

PCℓ3+Cℓ2→PCℓ5PC\ell_3 + C\ell_2 \to PC\ell_5

If the yield of the reaction is 76.5 %, what is the mass of PCl5 , in grams, obtained from the reaction of 27.0 g of PCl3 with excess Cl2? Give your answer to one decimal place; do not include units in your answer.
In the balanced reaction below, a student reacts 1.25g of copper with 5.0mL of 12.0mol/L HCl.

Cu(s)+2HCℓ(aq)→CuCℓ2(aq)+H2(g)Cu(s)+ 2HC \ell(aq)\to CuC\ell_2(aq) +H_2(g)

Calculate the theoretical yield of hydrogen gas produced in grams. Give your answer to four decimal points; do not include units.
Extra Practice
Chemical Reactions: Actual Yield
Consider the following reaction:
PCℓ3+Cℓ2→PCℓ5PC\ell_3 + C\ell_2 \to PC\ell_5
If the yield of the reaction is 76.5 %, what is the mass of PCl5 , in grams, obtained from the reaction of 27.0 g of PCl3 with excess Cl2? Give your answer to one decimal place; do not include units in your answer.
Stoichiometry: Percent yield
In the balanced reaction below, a student reacts 1.25g of copper with 5.0mL of 12.0mol/L HCl.
Cu(s)+2HCℓ(aq)→CuCℓ2(aq)+H2(g)Cu(s)+ 2HC \ell(aq)\to CuC\ell_2(aq) +H_2(g)
Chemical Reaction: Percent Yield
The balanced equation for the complete combustion of butane is as follows:
4C4H10+13O2→8CO2+10H2O4C_4H_10 + 13 O_2 → 8 CO_2 + 10 H_2O
In an experiment, 12.37 g of carbon dioxide was produced when 14.25 g was predicted. What is the percentage yield? Round your answer to the nearest integer; do not include the % symbol.