Stoichiometry Percent Yield Worksheet Answers

12.nine: Theoretical Yield and Percent Yield

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The world of pharmaceutical product is an expensive i. Many drugs have several steps in their synthesis and apply plush chemicals. A great bargain of research takes place to develop better ways to make drugs faster and more than efficiently. Studying how much of a compound is produced in whatever given reaction is an important part of toll command.

Per centum Yield

Chemical reactions in the real globe practise 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 ofttimes losses due to an incomplete reaction, undesirable side reactions, etc. Chemists demand a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.

To compute the percentage yield, it is first necessary to determine how much of the product should exist 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 really 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 per centum:

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

Percent yield is very of import in the manufacture of products. Much time and coin is spent improving the pct yield for chemical product. When complex chemicals are synthesized by many different reactions, one footstep with a depression percent yield can quickly crusade a large waste product of reactants and unnecessary expense.

Typically, per centum yields are understandably less than \(100\%\) considering of the reasons previously indicated. Still, pct yields greater than \(100\%\) are possible if the measured product of the reaction contains impurities that crusade its mass to be greater than it really would exist 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 \(\PageIndex{i}\): Calculating the Theoretical Yield and the Per centum Yield

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

\[2 \ce{KClO_3} \left( s \right) \rightarrow two \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( k \correct)\nonumber \]

In a certain experiment, \(forty.0 \: \text{g} \: \ce{KClO_3}\) is heated until information technology 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 \(14.9 \: \text{one thousand}\). What is the per centum yield for the reaction?

Solution

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

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

Known

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

Unknown

theoretical yield O_two = ? g

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

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

Step 2: Solve.

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

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

Step three: Remember about your upshot.

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

Now we will utilise the bodily yield and the theoretical yield to calculate the percent yield.

Stride ane: List the known quantities and plan the trouble.

Known

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

Unknown

Percent yield = ? %

\[\text{Pct Yield} = \frac{\text{Bodily Yield}}{\text{Theoretical Yield}} \times 100\% \nonumber\nonumber \]

Use the percent yield equation in a higher place.

Pace 2: Solve.

\[\text{Per centum Yield} = \frac{14.9 \: \text{thousand}}{15.seven \: \text{k}} \times 100\% = 94.9\% \nonumber\nonumber \]

Step iii: Think about your result.

Since the actual yield is slightly less than the theoretical yield, the per centum yield is but nether \(100\%\).

Summary

Theoretical yield is calculated based on the stoichiometry of the chemical equation.
The actual yield is experimentally adamant.
The per centum yield is determined by calculating the ratio of bodily yield/theoretical yield.

Review

What do we need in order to calculate theoretical yield?
If I spill some of the product before I weigh it, how will that touch the actual yield?
How will spilling some of the product affect the percent yield?
I make a product and weigh it before it is dry. How volition that touch on the actual yield?