Percent Yield Vs Percent Recovery

12.9: Theoretical Yield and Percent Yield

1. Final updated

2. Salvage as PDF

Folio ID

53797

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 we relieve some money?

The world of pharmaceutical production is an expensive ane. Many drugs accept several steps in their synthesis and use costly chemicals. A groovy deal of research takes identify to develop ameliorate means to make drugs faster and more than efficiently. Studying how much of a compound is produced in any given reaction is an important office of cost control.

Percent Yield

Chemical reactions in the real earth do non always go exactly as planned on newspaper. In the class of an experiment, many things will contribute to the formation of less production than would be predicted. Besides spills and other experimental errors, in that location 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 percentage yield.

To compute the per centum yield, it is first necessary to decide how much of the product should exist formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could exist 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 bodily yield to the theoretical yield, expressed as a pct:

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

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 unlike reactions, one step with a low pct yield can apace crusade a large waste of reactants and unnecessary expense.

Typically, per centum yields are understandably less than \(100\%\) because of the reasons previously indicated. Yet, pct yields greater than \(100\%\) are possible if the measured product of the reaction contains impurities that cause its mass to exist greater than information technology actually would be if the product was pure. When a chemist synthesizes a desired chemic, he or she is always careful to purify the products of the reaction.

Example \(\PageIndex{1}\): Computing the Theoretical Yield and the Percent Yield

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

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

In a sure 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 nerveless, and its mass is found to be \(xiv.nine \: \text{chiliad}\). What is the percent yield for the reaction?

Solution

Starting time, we will calculate the theoretical yield based on the stoichiometry.

Stride 1: List the known quantities and program the problem.

Known

• Given: Mass of \(\ce{KClO_3} = twoscore.0 \: \text{g}\)
• Molar mass \(\ce{KClO_3} = 122.55 \: \text{g/mol}\)
• Molar mass \(\ce{O_2} = 32.00 \: \text{1000/mol}\)

Unknown

• theoretical yield O₂ = ? g

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

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

Footstep 2: Solve.

\[xl.0 \: \text{g} \: \ce{KClO_3} \times \frac{ane \: \text{mol} \: \ce{KClO_3}}{122.55 \: \text{chiliad} \: \ce{KClO_3}} \times \frac{three \: \text{mol} \: \ce{O_2}}{2 \: \text{mol} \: \ce{KClO_3}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{one \: \text{mol} \: \ce{O_2}} = fifteen.seven \: \text{g} \: \ce{O_2} \nonumber\nonumber \]

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

Step iii: Recall about your result.

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

Now we will employ the actual yield and the theoretical yield to summate the percent yield.

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

Known

• Actual yield \(= fourteen.9 \: \text{grand}\)
• Theoretical yield \(= 15.7 \: \text{grand}\)

Unknown

• Pct yield = ? %

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

Employ the pct yield equation to a higher place.

Step 2: Solve.

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

Step 3: Recall most your result.

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

Summary

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

Review

1. What practise we demand in guild to calculate theoretical yield?
2. If I spill some of the product before I weigh it, how will that bear on the actual yield?
3. How volition spilling some of the product affect the pct yield?
4. I make a product and weigh it earlier information technology is dry out. How volition that affect the actual yield?

Percent Yield Vs Percent Recovery,

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

Posted by: jamesyourneart64.blogspot.com

Share this post