8.6: Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses of Reactants

Learning Objectives

• Calculate percentage or actual yields from known amounts of reactants

The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.

Percent Yield

Chemical reactions in the real world don't always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.

To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.

Typically, percent yields are understandably less than $$100\%$$ because of the reasons indicated earlier. However, percent yields greater than $$100\%$$ are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Example $$\PageIndex{1}$$ illustrates the steps for determining percent yield.

Example $$\PageIndex{1}$$: Decomposition of Potassium chlorate

Potassium chlorate decomposes upon slight heating in the presence of a catalyst according to the reaction below:

$2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)\nonumber$

In a certain experiment, $$40.0 \: \text{g} \: \ce{KClO_3}$$ is heated until it completely decomposes. The experiment is performed and the oxygen gas is collected and its mass is found to be $$14.9 \: \text{g}$$.

1. What is the theoretical yield of oxygen gas?
2. What is the percent yield for the reaction?

Solution

a. Calculation of theoretical yield

First, we will calculate the theoretical yield based on the stoichiometry.

Step 1: Identify the "given" information and what the problem is asking you to "find".

Given: Mass of $$\ce{KClO_3} = 40.0 \: \text{g}$$

Mass of O2 collected = 14.9g

Find: Theoretical yield, g O2

Step 2: List other known quantities and plan the problem.

1 mol KClO3 = 122.55 g/mol

1 mol O2 - 32.00 g/mol

Step 3: Apply stoichiometry to convert from the mass of a reactant to the mass of a product:

Step 4: Solve.

$40.0 \: \cancel{\text{g} \: \ce{KClO_3}} \times \frac{1 \: \cancel{\text{mol} \: \ce{KClO_3}}}{122.55 \: \cancel{\text{g} \: \ce{KClO_3}}} \times \frac{3 \: \cancel{\text{mol} \: \ce{O_2}}}{2 \: \cancel{\text{mol} \: \ce{KClO_3}}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \cancel{\text{mol} \: \ce{O_2}}} = 15.7 \: \text{g} \: \ce{O_2}\nonumber$

The theoretical yield of $$\ce{O_2}$$ is $$15.7 \: \text{g}$$, 15.67 g unrounded.

The mass of oxygen gas must be less than the $$40.0 \: \text{g}$$ of potassium chlorate that was decomposed.

b. Calculation of percent yield

Now we will use the actual yield and the theoretical yield to calculate the percent yield.

Step 1: Identify the "given" information and what the problem is asking you to "find".

Given: Theoretical yield =15.67 g, use the unrounded number for the calculation.

Actual yield = 14.9g

Find: Percent yield, % Yield

Step 2: List other known quantities and plan the problem.

No other quantities needed

Step 3: Use the percent yield equation below

$$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$$

Step 4: Solve.

$$\text{Percent Yield} = \frac{14.9 \: \text{g}}{15.\underline{6}7 \: \text{g}} \times 100\% = 94.9\%$$

Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $$100\%$$.

Example $$\PageIndex{2}$$: Oxidation of Zinc

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

$$\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(s)+\ce{ZnSO4}(aq)$$

What is the percent yield?

Solution

Steps for Problem Solving-The Product Method Example $$\PageIndex{1}$$
Identify the "given"information and what the problem is asking you to "find."

Given: 1.274 g CuSO4

Actual yield = 0.392 g Cu

Find: Percent yield

List other known quantities

1 mol CuSO4= 159.62 g/mol
1 mol Cu = 63.55 g/mol

Since the amount of product in grams is not required, only the molar mass of the reactants is needed.

Balance the equation

The chemical equation is already balanced.

The balanced equation provides the relationship of 1 mol CuSO4 to 1 mol Zn to 1 mol Cu to 1 mol ZnSO4

Prepare a concept map and use the proper conversion factor.

The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield (g Cu) is found by performing mass-mass calculation based on the initial amount of CuSO4.

Cancel units and calculate.
$\mathrm{1.274\:\cancel{g\:Cu_SO_4}\times \dfrac{1\:\cancel{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{1\:\cancel{mol\: Cu}}{1\:\cancel{mol\:CuSO_4}}\times \dfrac{63.55\:g\: Cu}{1\:\cancel{mol\: Cu}}=0.5072\: g\: Cu}\nonumber$

Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be

$\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100}$

\begin{align} \mathrm{percent\: yield}&=\mathrm{\left(\dfrac{0.392\: g\: Cu}{0.5072\: g\: Cu}\right)\times 100} \\ &=77.3\% \end{align}\nonumber
Think about your result. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $$100\%$$.

Exercise $$\PageIndex{1}$$

What is the percent yield of a reaction that produces 12.5 g of the Freon CF2Cl2 from 32.9 g of CCl4 and excess HF?

$\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber$