Percent Actual And Theoretical Yield

12.9: Theoretical Yield and Percent Yield

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Figure \(\PageIndex{1}\) (Credit: Courtesy of Linda Bartlett and the National Cancer Institute; Source: http://commons.wikimedia.org/wiki/File:Drug_synthesis_%25282%2529.jpg(opens in new window); License: Public Domain)

Tin can nosotros save some money?

The world of pharmaceutical production is an expensive i. Many drugs have several steps in their synthesis and use plush chemicals. A keen 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 do non always become exactly equally planned on paper. In the class of an experiment, many things will contribute to the germination of less product than would be predicted. Likewise spills and other experimental errors, there are often 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 pct yield, it is get-go necessary to determine how much of the product should exist formed based on stoichiometry. This is called the theoretical yield, the maximum corporeality of product that could be formed from the given amounts of reactants. The bodily yield is the amount of production that is actually formed when the reaction is carried out in the laboratory. The per centum yield is the ratio of the actual yield to the theoretical yield, expressed as a per centum:

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

Percent yield is very important in the manufacture of products. Much time and coin is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many dissimilar reactions, 1 step with a low percent yield tin can speedily crusade a big waste product of reactants and unnecessary expense.

Typically, percent yields are understandably less than \(100\%\) because of the reasons previously indicated. However, percentage yields greater than \(100\%\) are possible if the measured product of the reaction contains impurities that crusade 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.

Instance \(\PageIndex{ane}\): 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.

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

In a certain experiment, \(forty.0 \: \text{yard} \: \ce{KClO_3}\) 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 \(xiv.9 \: \text{thousand}\). 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 programme the problem.

Known

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

Unknown

• theoretical yield O_two = ? m

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

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

Step ii: Solve.

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

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

Step 3: Think about your outcome.

The mass of oxygen gas must be less than the \(twoscore.0 \: \text{g}\) of potassium chlorate that was decomposed.

Now we will utilise 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 \(= xiv.9 \: \text{g}\)
• Theoretical yield \(= fifteen.vii \: \text{thousand}\)

Unknown

• Pct yield = ? %

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

Utilise the pct yield equation higher up.

Step 2: Solve.

\[\text{Percent Yield} = \frac{14.9 \: \text{one thousand}}{15.vii \: \text{g}} \times 100\% = 94.nine\% \nonumber\nonumber \]

Pace 3: Remember about your effect.

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

Summary

• Theoretical yield is calculated based on the stoichiometry of the chemic equation.
• The actual yield is experimentally determined.
• The percent yield is adamant by computing the ratio of actual yield/theoretical yield.

Review

1. What do we need in order to calculate theoretical yield?
2. If I spill some of the production before I counterbalance it, how volition that affect the bodily yield?
3. How will spilling some of the product affect the percentage yield?
4. I make a production and weigh information technology earlier information technology is dry out. How volition that bear on the actual yield?

Percent Actual And Theoretical Yield,

Source: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/12%3A_Stoichiometry/12.09%3A_Theoretical_Yield_and_Percent_Yield

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