What is an example of a percent yield?

In this tutorial, you will learn what percent yield is and how to calculate it. In addition, you will walk through an example calculation.

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What is Percent Yield?

When performing an experiment, there is a maximum yield you can obtain if there are perfect reaction conditions; this is the theoretical yield. However, even if you follow an experiment correctly, it is likely that you will not have a perfect yield of the product; the amount of product you end up with is your actual yield.

The percentage of the theoretical yield you obtained in your experiment is the percent yield. Let’s learn how to calculate it below!

How to Calculate Percent Yield

You can use the equation below to calculate your percent yield from an experiment:

Calculating Theoretical Yield

First, you should calculate the theoretical yield of your experiment; usually, this will involve stoichiometric calculations. By first looking at the chemical equation and information given, you can get an idea of what is reacting and how the product is forming.

Since no more product can be formed once the limiting reagent runs out, the next step is to identify the limiting reagent.

You can then use dimensional analysis to see how much product can be formed based on the amount of limiting reagent that is given. This is the theoretical yield of your experiment.

Calculating Actual Yield

If you are physically doing an experiment, your actual yield will be the amount of product you weigh out on your balance. If you are doing a word problem, the actual yield may be given within the problem.

Percent Yield Equation

The last step, once you have both theoretical and actual yield, is plugging the numbers into the equation. Dividing the actual by the theoretical gives you the fraction of product you made. Multiplying that by 100 gives you the percent yield.

Calculating Percent Yield Example

Now that we know the steps to calculate percent yield, let’s walk through an example:

Use the balanced chemical reaction below. If 40.00 g of Acetylene (C2H2) and 65.00 g of Oxygen are used, and 25.00 g of water are produced, what is the percent yield?

First step is to find limiting reagent & theoretical yield of water:

Using dimensional analysis on both reagents, acetylene is found to produce a lower amount of product than oxygen; because of this acetylene is our limiting reagent.

27.67g is our theoretical yield. The last step is to plug our numbers into the percent yield equation.

Our percent yield is 90.35%.

For more example questions to try, click here!

Further Reading

• Common Polyatomic Ions
• Kinetic Molecular Theory
• E1 Reaction
• E2 Reaction

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.

Chemical reactions in the real world do not 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 often 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:

\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\nonumber \]

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 previously indicated. 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 \(\PageIndex{1}\): Calculating the Theoretical Yield and the Percent Yield

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. What is the theoretical yield of oxygen gas? The experiment is performed, the oxygen gas is collected, and its mass is found to be \(14.9 \: \text{g}\). What is the percent yield for the reaction?

Solution

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

Step 1: List the known quantities and plan the problem.

Known

• Given: Mass of \(\ce{KClO_3} = 40.0 \: \text{g}\)
• Molar mass \(\ce{KClO_3} = 122.55 \: \text{g/mol}\)
• Molar mass \(\ce{O_2} = 32.00 \: \text{g/mol}\)

Unknown

• theoretical yield O2 = ? g

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

\[\text{g} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{O_2} \rightarrow \text{g} \: \ce{O_2} \nonumber\nonumber \]

Step 2: Solve.

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

The theoretical yield of \(\ce{O_2}\) is \(15.7 \: \text{g}\).

Step 3: Think about your result.

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

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

Step 1: List the known quantities and plan the problem.

Known

• Actual yield \(= 14.9 \: \text{g}\)
• Theoretical yield \(= 15.7 \: \text{g}\)

Unknown

\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \nonumber\nonumber \]

Use the percent yield equation above.

Step 2: Solve.

\[\text{Percent Yield} = \frac{14.9 \: \text{g}}{15.7 \: \text{g}} \times 100\% = 94.9\% \nonumber\nonumber \]

Step 3: Think about your result.

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