Percent yield

Actual Yield

The measured amount of product actually collected from a reaction.

Actual Yield is Lower Than the Theoretical Yield

There are many reasons why the actual yield of a reaction is lower than the theoretical yield. Sometimes conditions do not allow a reaction to run to completion, while other times the purification of the product results in a loss of some of the product.

Percent Yield

the ratio of the actual yield to the theoretical yield, multiplied by 100 to make it a percentage.

Equation to Remember

percent yield = actual yield / theoretical yield × 100%

the actual yield is a measured value while the theoretical yield is a calculated value.

Steps for Calculating Percent Yield

1. Write the balanced equation for the reaction.

2. Identify all important information provided in the word problems or data table.

3. Solve for the theoretical yield of the reaction, following all the steps of a stoichiometry calculation.

4. Divide the actual yield (the measured amount of product produced) by the theoretical yield (the calculated amount of product from the stoichiometry calculation) and multiply by 100 to get the percent yield of the reaction.

Example 1:

9.80 grams of magnesium metal react with excess hydrochloric acid, and 29.8 grams of magnesium chloride are produced by the reaction. What is the percent yield of this reaction?

Mg + HCl → MgCl2 + H2

1) The first thing you need to always make sure you do is balance the chemical equation.

Mg +2 HCl → MgCl2 + H2

2) Identify all important information provided in the word problems or data table. 9.80 grams of magnesium metal react with excess hydrochloric acid, and 29.8 grams of magnesium chloride are produced by the reaction. What is the percent yield of this reaction?

9.80 grams of magnesium metal is the given amount of one reactant, and the other reactant (HCl) is in excess. This means that it is not a limiting reactant problem, you will only need to do one stoichiometry calculation to determine theoretical yield.

29.8 grams of magnesium chloride is the actual yield that was collected and measured in the lab.

3) Solve for the theoretical yield of the reaction, following all the steps of a stoichiometry calculation.

9.80 g Mg x 1 mol Mg / 24.305 g Mg x 1 mol MgCl2 / 1 mol Mg x 95.21 MgCl2 / 1 mol MgCl2 = 38.4 g MgCl2 (theoretical yield)

4) Divide the actual yield (the measured amount of product produced) by the theoretical yield (the calculated amount of product from the stoichiometry calculation) and multiply by 100 to get the percent yield of the reaction.

percent yield = 29.8 g MgCl2 (actual) / 38.4 g MgCl2 (theoretical) x 100%

percent yield = 77.6% yield

Example 2:

When Mandy allows 38.5 grams of oxygen to react with 4.05 grams of hydrogen gas, she is able to collect 35.3 grams of water. What is the theoretical yield of this reaction, and what is the percent yield?

O2 + H2 → H2O

1) The first thing you need to always make sure you do is balance the chemical equation.

O2 +2 H2 →2 H2O

2) Identify the given amounts provided in the word problem, as well as what you are being asked to solve for.

When Mandy allows 38.5 grams of oxygen to react with 4.05 grams of hydrogen gas, she is able to collect 35.3 grams of water.

3) Solve for the theoretical yield of the reaction, following all the steps of a stoichiometry calculation.

38.5 g O2 x 1 mol O2 / 32.0 g O2 x 2 mol H2O / 1 mol O2 x 18.0 g H2O / 1 mol H2O = 43.3 g H2O

4.05 g H2 x 1 mol H2 / 2.02 g H2 x 2 mol H2O / 2 mol H2 x 18.0 g H2O / 1 mol H2O = 36.1 H2O

4) Divide the actual yield (the measured amount of product produced) by the theoretical yield (the calculated amount of product from the stoichiometry calculation) and multiply by 100 to get the percent yield of the reaction.

percent yield = 35.3 g H2O (actual) / 36.1 g H2O (theoretical) x 100%

percent yield = 97.8% yield

Example 3:

When 70.0 grams of MnO2 reacted with 128.0 grams of HCl, the reaction resulted in a 62.7 percent yield of chlorine gas. What is the actual yield of chlorine gas in grams?

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

1) The first thing you need to always make sure you do is balance the chemical equation.

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

2) Identify all important information provided in the word problems or data table.

When 70.0 grams of MnO2 reacted with 128.0 grams of HCl, the reaction resulted in a 62.7 percent yield of chlorine gas. What is the actual yield of chlorine gas in grams?

There is a given, measured, amount for both reactants (70.0 g MnO2 and 128.0 g HCl) so this will be a limiting reactant problem. This means that you will solve two separate stoichiometry calculations to determine the theoretical yield.

The percent yield of Cl2 is 62.7 percent. You can use the percent yield and the theoretical yield of chlorine to solve for the actual yield of chlorine.

3) Solve for the theoretical yield of chlorine gas.

70.0 g MnO2 x 1 mol MnO2 / 86.9 g MnO2 x 1 mol Cl2 / 1 mol MnO2 x 71.0 Cl2 / 1 mol Cl2 = 57.2 g Cl2

128.0 g HCl x 1 mol HCl / 36.5 g HCl x 1 mol Cl2 / 4 mol HCl x 71.0 Cl2 / 1 mol Cl2 = 62.2 g Cl2

4) Now that you have the theoretical yield (57.2 g Cl2) and the percent yield (62.7 percent) of Cl2, you can solve for the actual yield of Cl2.

percent yield x theoretical yield / 100% = actual yield

62.7% x 57.2 g Cl2 / 100% = actual yield

actual yield = 35.9 g Cl2

What is the percent yield of a reaction in which 3.8 moles of iron metal react with excess sulfur to produce 230 grams of iron (II) sulfide?

Fe + S --> FeS

69% yield FeS

3.8 mol Fe x 1 mol FeS / 1 mol Fe x 87.9 g FeS / 1 mol FeS = 334 g FeS (theoretical)

230 g FeS (acutal) / 334 g FeS (theoretical) x 100% = 69% yield

108 grams of potassium chloride are produced when 305 grams of potassium chlorate are heated. What is the theoretical yield and the percent yield of this reaction?

KClO3 --> KCl + O2

58.4% yield of KCl

305 g KClO3 x 1 mol KClO3 / 122.5 g KClO3 x 2 mol KCl / 2 mol KClO3 x 74.5 g KCl / 1 mol KCl = 185 g KCl (theoretical)

108 g KCl (actual) / 185 g KCl (theoretical) x 100% = 58.4% yield

Aluminum reacts with excess copper (II) sulfate according to the equation below. If 12.8 grams of aluminum metal reacts and the percent yield of copper metal is 72.5%, what mass of copper metal is actually produced?