The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly ubraintv-jp.comicals. 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.

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## Percent Yield

ubraintv-jp.comical 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 thanpredicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. ubraintv-jp.comists 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 can 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 ubraintv-jp.comical production. When complex ubraintv-jp.comicals 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 ubraintv-jp.comist synthesizes a desired ubraintv-jp.comical, 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 $$14.9 \: \text{g}$$.

What is the theoretical yield of oxygen gas? 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:

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 ubraintv-jp.comical 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}\>

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

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## Summary

Theoretical yield is calculated based on the stoichiometry of the ubraintv-jp.comical equation. The actual yield is experimentally determined. The percent yield is determined by calculating the ratio of actual yield to theoretical yield.