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14.4: Percent Yield

Created by: CK-12

Lesson Objectives

The student will:

  • explain what is meant by the terms “actual yield” and “theoretical yield.”
  • calculate the theoretical yield for a chemical reaction when given the amount(s) of reactant(s) available.
  • determine the percent yield for a chemical reaction.

Vocabulary

  • actual yield
  • percent yield
  • theoretical yield

Introduction

The amount of product that should be formed when the limiting reactant is completely consumed is called the theoretical yield. This is the maximum amount of the product that could form from the quantities of reactants used. In actual practice, however, this theoretical yield is seldom obtained due to side reactions, failure of the reaction to go to completion, and other complications. The actual amount of product produced in a laboratory or industrial reaction is called the actual yield. The actual yield is almost always less than the theoretical yield and is often expressed as a percentage of the theoretical yield. This is also referred to as the percent yield.

Calculating the Theoretical Yield

When we calculate the amount of product that can be produced from limiting reactants, we are determining the maximum theoretical amount of product we could obtain from the reaction. In other words, the theoretical yield is the maximum amount obtained when all of the limiting reactant has reacted in the balanced chemical equation. You have already been calculating the theoretical yields in the previous section, but let us consider one more example.

Example:

Kerry-Sue is studying the reaction between silver nitrate and potassium sulfide. The unbalanced equation for the reaction is:

\text{AgNO}_{3(aq)}\ + \ \text{K}_2\text{S}_{(aq)} \rightarrow \text{Ag}_2\text{S}_{(s)}\ + \ \text{KNO}_{3(aq)}

When Kerry-Sue added 5.00 \ \text{g} of the silver nitrate with 2.50 \ \text{g} of potassium sulfide, what was her theoretical yield for silver sulfide?

Notice that this is a limiting reagent problem because we are given two reactant amounts, and we have to determine which of these reactants will run out first.

Step 1: Write the balanced equation.

2 \ \text{AgNO}_{3(aq)} + \ \text{K}_2\text{S}_{(aq)} \rightarrow \text{Ag}_2\text{S}_{(s)} + 2 \ \text{KNO}_{3(aq)}

Step 2: Convert given amounts to moles.

\text{mol AgNO}_3\text{:} \ \frac {5.00 \ \text{g}} {170. \ \text{g/mol}} = 0.0294 \ \text{mol AgNO}_3
\text{mol K}_2\text{S:} \ \frac {2.50 \ \text{g}} {110. \ \text{g/mol}} = 0.0227 \ \text{mol K}_2\text{S}

Step 3: Divide each amount by the coefficients from the balanced equation.

\text{AgNO}_3\text{:} \ \frac {0.0294 \ \text{mol}} {2} = 0.0147 \ \text{mol}
\text{K}_2\text{S:} \ \frac {0.0227 \ \text{mol}} {1} = 0.0227 \ \text{mol}

\text{AgNO}_3 has the lowest final value, so it is the limiting reagent. This means \text{K}_2\text{S} is in excess. We use the values for \text{AgNO}_3 for the rest of our theoretical yield calculations.

Step 4: Convert moles of the limiting reagent to mass of the desired product.

\text{mol Ag}_2\text{S:} \ (0.0294 \ \text{mol AgNO}_3) \cdot \frac {1 \ \text{mol Ag}_2\text{S}} {2 \ \text{mol AgNO}_3} = 0.0147 \ \text{Ag}_2\text{S}
\text{mass Ag}_2\text{S:} \ (0.0147 \ \text{Ag}_2\text{S}) \cdot (248 \ \text{g/mol}) = 3.64 \ \text{g Ag}_2\text{S}

Therefore, when Kerry-Sue did her experiment in the lab, her theoretical yield would have been 3.64 \ \text{g of Ag}_2\text{S}_{(s)}.

Alternatively, if we use dimensional analysis the calculation would be:

5.00 \ \text{g AgNO}_3 \cdot \frac {1 \ \text{mol AgNO}_3} {170.\ \text{g AgNO}_3} \cdot \frac {1 \ \text{mol Ag}_2\text{S}} {2 \ \text{mol AgNO}_3} \cdot \frac {248 \ \text{g Ag}_2\text{S}} {1 \ \text{mol Ag}_2\text{S}} = 3.64 \ \text{g Ag}_2\text{S}

Calculating Percent Yield

Theoretical yields are the optimum yields if conditions allow all 100% of the reactant to react. If, on the other hand, anything were to happen to jeopardize this, the actual yield will differ from the theoretical yield. Remember that the actual yield is measured, while the theoretical yield is calculated.

The actual yield is almost always less than the theoretical yield, and we often calculate how close this actual yield is to the theoretical yield. The percentage of the theoretical yield that is actually produced (actual yield) is known as the percent yield. The efficiency of a chemical reaction is determined by the percent yield. The percent yield is found using the following formula.

% yield =  \frac { \text{actual yield}} { \text{theoretical yield}} \times 100%

Example:

A student was able to produce an actual yield of 5.12 \ \text{g} of calcium sulfate from 4.95 \ \text{g} of sulfuric acid and excess calcium hydroxide in the reaction shown below. What was her percent yield?

\text{H}_2\text{SO}_4 + \ \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2 \ \text{H}_2\text{O}

We can calculate the theoretical yield using dimensional analysis.

4.95 \ \text{g H}_2\text{SO}_4 \cdot \frac {1 \ \text{mol H}_2\text{SO}_4} {98.1 \ \text{g H}_2\text{SO}_4} \cdot \frac {1 \ \text{mol CaSO}_4} {1 \ \text{mol H}_2\text{SO}_4} \cdot \frac {136.2 \ \text{g CaSO}_4} {1 \ \text{mol CaSO}_4} = 6.88 \ \text{g CaSO}_4

Once we have found the theoretical yield, we can use the actual yield given in the problem to find the percent yield.

% yield =  \frac { \text{actual yield}} { \text{theoretical yield}} \times 100\% = \ \frac {5.12 \ \text{g}} {6.88 \ \text{g}} \times 100\% = 74.4%

Example:

Potassium hydrogen phthalate, \text{KHC}_8\text{H}_4\text{O}_4, is a compound used quite frequently in acid-base chemistry for a procedure known as standardization. Standardization has to do with the process of determining the concentration of a standard solution. In a certain experiment, 12.50 \ \text{g} of potassium hydroxide is mixed with 1.385 \ \text{g} of \text{KHP}. If 5.26 \ \text{g} of the product, \text{K}_2\text{C}_8\text{H}_4\text{O}_4, is produced, what is the percent yield?

\text{KOH} + \ \text{KHC}_8\text{H}_4\text{O}_4\ \rightarrow \text{K}_2\text{C}_8\text{H}_4\text{O}_4 + \ \text{H}_2\text{O}

Step 1: Write the balanced equation.

The equation is already balanced, so this step is done.

Step 2: Convert given amounts to moles.

\text{mol KOH}\text{:} \ \frac {1.385 \ \text{g}} {56.1 \ \text{g/mol}} = 0.0247 \ \text{mol KOH}
\text{mol KHC}_8\text{H}_4\text{O}_4\text{:} \ \frac {12.5 \ \text{g}} {204 \ \text{g/mol}} = 0.0612 \ \text{mol KHC}_8\text{H}_4\text{O}_4

Step 3: Divide each amount by the coefficients from the balanced equation.

\text{mol KOH}\text{:} \ \frac {0.0247 \ \text{mol}} {1} = 0.0247 \ \text{mol}
\text{mol KHC}_8\text{H}_4\text{O}_4\text{:} \ \frac {0.0612 \ \text{mol}} {1} = 0.0612 \ \text{mol}

\text{KOH} has the lowest final value, so it is the limiting reagent. We use the values for \text{KOH} for the rest of our theoretical yield calculations.

Step 4: Convert moles of the limiting reagent to mass of the desired product.

\text{mol K}_2\text{C}_8\text{H}_4\text{O}_4\text{:} \ (0.0247 \ \text{mol KOH}) \cdot \frac {1 \ \text{mol K}_2\text{C}_8\text{H}_4\text{O}_4} {1 \ \text{mol KOH}} = 0.0247 \ \text{mol K}_2\text{C}_8\text{H}_4\text{O}_4
\text{mass K}_2\text{C}_8\text{H}_4\text{O}_4\text{:} \ (0.0247 \ \text{mol K}_2\text{C}_8\text{H}_4\text{O}_4) \cdot (242.3 \ \text{g/mol}) = 5.98 \ \text{g K}_2\text{C}_8\text{H}_4\text{O}_4

Therefore, the theoretical yield of \text{K}_2\text{C}_8\text{H}_4\text{O}_4 is 5.98 \ \text{g}. Remember the actual yield given in the question was 5.26 \ \text{g}. Now let’s calculate the percent yield.

% yield =  \frac {\text{actual yield}} {\text{theoretical yield}} \cdot 100\% = \frac {5.26 \ \text{g}} {5.98 \ \text{g}} \cdot 100\% = 88.0%

Example:

Zeanxanthin is a compound responsible for causing the colors of the maple leaf to change in the fall. It has the formula \text{C}_{38}\text{H}_{56}\text{O}_2. In a combustion reaction of 0.95 \ \text{g} of zeanxanthin with excess oxygen, 2.2 \ \text{g} of carbon dioxide was produced. The other product was water. What is the percent yield of \text{CO}_2?

Step 1: Write the balanced equation.

\text{C}_{38}\text{H}_{56}\text{O}_2 + 51 \ \text{O}_2 \rightarrow 38 \ \text{CO}_2 +\ 28\ H_2O

Step 2: Convert given amounts to moles.

\text{mol C}_{38}\text{H}_{56}\text{O}_2\text{:} \ \frac {0.95 \ \text{g}} {544 \ \text{g/mol}} = 0.00174 \ \text{mol C}_{38}\text{H}_{56}\text{O}_2

Step 3: Divide each amount by the coefficients from the balanced equation.

Since \text{C}_{38}\text{H}_{56}\text{O}_2 is reacted with excess oxygen, we already know that \text{C}_{38}\text{H}_{56}\text{O}_2 is the limiting reagent. We use the values for \text{C}_{38}\text{H}_{56}\text{O}_2 for the rest of our theoretical yield calculations.

Step 4: Convert moles of the limiting reagent to mass of the desired product.

\text{mol CO}_2\text{:} \ (0.00174 \ \text{mol C}_{38}\text{H}_{56}\text{O}_2) \cdot \frac {38 \ \text{mol CO}_2} {1 \ \text{mol C}_{38}\text{H}_{56}\text{O}_2} = 0.066 \ \text{mol CO}_2
\text{mass CO}_2\text{:} \ (0.066 \ \text{mol CO}_2) \cdot (44 \ \text{g/mol}) = 2.9 \ \text{g CO}_2

Therefore, the theoretical yield of \text{CO}_2 is 2.9 \ \text{g}. Remember the actual yield given in the question was 2.2 \ \text{g}. Now let’s calculate the percent yield.

% yield =  \frac {\text{actual yield}} {\text{theoretical yield}} \cdot 100\% = \frac {2.2 \ \text{g}} {2.9 \ \text{g}} \cdot 100\% = 76%

This video is an electronic blackboard presentation of the calculation of a percent yield (3f; 1b I&E): http://www.youtube.com/watch?v=1L12_TRSql8 (9:22).

Lesson Summary

  • The actual yield of a reaction is the actual amount of product that is produced in the laboratory.
  • The theoretical yield is the amount of product that is produced under ideal conditions.
  • The percentage of the theoretical yield that is actually produced (actual yield) is known as the percent yield.

Review Questions

  1. Is it possible for the actual yield to be greater than the theoretical yield?
  2. What happens when competing reactions occur while performing an experiment in the lab?
  3. If the actual yield is 4.5 \ \mathrm{g}, but the theoretical yield is 5.5 \ \mathrm{g}, what is the percent yield for this data?
  4. Solid aluminum and sulfur come together in a reaction to produce 7.5 \ \mathrm{g} of aluminum sulfide: \text{Al} + \ \text{S} \rightarrow \text{Al}_2\text{S}_3. If 5.00 \ \mathrm{g} of each solid react together, what is the percent yield? Remember to balance the reaction first.
    1. 32.0 \%
    2. 53.4 \%
    3. 96.2 \%
    4. 100.0 \%
  5. In her experiment, Gerry finds she has obtained 3.65 \ \mathrm{g} of lead(II) iodide. She knows that in her reaction, lead(II) nitrate reacted completely with potassium iodide to produce lead(II) iodide and potassium nitrate: \text{Pb(NO}_3)_{2(aq)} + \ \text{KI}_{(aq)} \rightarrow \text{PbI}_{2(s)} + \text{KNO}_{3(aq)}. The potassium iodide produced is a brilliant yellow colored precipitate. Gerry began with 5.00 \ \mathrm{g} of potassium iodide. What was her percent yield? Remember to balance your equation first.
    1. 26.3 \%
    2. 36.0 \%
    3. 52.6 \%
    4. 72.0 \%
  6. If the percentage yield in the reaction 2 \ \text{S} + 3 \ \text{O}_2 \rightarrow 2 \ \text{SO}_3 was found to be 78.3\% and the actual yield was 1.01 \ \mathrm{g}, what was the original mass of the limiting reagent, oxygen?
    1. 0.515 \ \mathrm{g}
    2. 0.773 \ \mathrm{g}
    3. 1.01 \ \mathrm{g}
    4. 1.29 \ \mathrm{g}
  7. Bromine pentafluoride can be produced from a reaction between liquid bromine and fluorine gas. If 3.25 \ \mathrm{g} of fluorine reacts with 2.74 \ \mathrm{g} of bromine to produce 4.83 \ \mathrm{g} of bromine pentafluoride, what is the percent yield of the product?
  8. Ammonia can react with oxygen in a reaction that is similar to a combustion reaction. The products, however, are nitrogen monoxide and water, rather than carbon dioxide and water. In the reaction between ammonia and oxygen, 15 \ \mathrm{g} of each reactant are placed in a container, and 10.5 \ \mathrm{g} of nitrogen monoxide was produced. What is the percent yield of the nitrogen monoxide?

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