what would cause the percent yield of a reaction to be less than 100%?

The world of pharmaceutical production is an expensive 1. Many drugs accept several steps in their synthesis and employ costly chemicals. A bang-up bargain of enquiry takes place to develop ameliorate ways to brand drugs faster and more efficiently. Studying how much of a chemical compound is produced in whatsoever given reaction is an important role of toll control.

Percent Yield

Chemical reactions in the real world don't always become exactly every bit planned on paper. In the grade of an experiment, many things will contribute to the formation of less product than predicted. Besides spills and other experimental errors, there are unremarkably 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 per centum yield.

To compute the percent yield, information technology is get-go necessary to determine how much of the product should be formed based on stoichiometry. This is chosen the theoretical yield, the maximum corporeality of production that can exist formed from the given amounts of reactants. The actual yield is the corporeality 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 pct.

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

Pct yield is very of import in the manufacture of products. Much time and money is spent improving the percent yield for chemic product. When complex chemicals are synthesized by many unlike reactions, one step with a depression percent yield can rapidly cause a large waste matter of reactants and unnecessary expense.

Typically, percent yields are understandably less than \(100\%\) considering of the reasons indicated before. However, percent yields greater than \(100\%\) are possible if the measured product of the reaction contains impurities that crusade its mass to be greater than information technology actually would exist if the production was pure. When a pharmacist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Example \(\PageIndex{1}\) illustrates the steps for determining percentage yield.

Example \(\PageIndex{1}\): Decomposition of Potassium Chlorate

Potassium chlorate decomposes upon slight heating in the presence of a catalyst, co-ordinate to the reaction below:

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

In a certain experiment, \(xl.0 \: \text{g} \: \ce{KClO_3}\) is heated until it completely decomposes. The experiment is performed and the oxygen gas is collected and its mass is found to be \(14.nine \: \text{chiliad}\).

1. What is the theoretical yield of oxygen gas?
2. What is the percentage yield for the reaction?

Solution

a. Calculation of theoretical yield

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

Step one: Place the "given" information and what the problem is asking you to "notice".

Given: Mass of \(\ce{KClO_3} = 40.0 \: \text{thou}\)

Mass of O₂ collected = 14.9g

Observe: Theoretical yield, 1000 O₂

Step 2: List other known quantities and plan the trouble.

1 mol KClO_iii = 122.55 1000/mol

1 mol O₂ - 32.00 m/mol

Step 3: Apply stoichiometry to catechumen from the mass of a reactant to the mass of a production:

Stride four: Solve.

\[40.0 \: \cancel{\text{g} \: \ce{KClO_3}} \times \frac{1 \: \abolish{\text{mol} \: \ce{KClO_3}}}{122.55 \: \cancel{\text{g} \: \ce{KClO_3}}} \times \frac{three \: \cancel{\text{mol} \: \ce{O_2}}}{2 \: \cancel{\text{mol} \: \ce{KClO_3}}} \times \frac{32.00 \: \text{thou} \: \ce{O_2}}{one \: \abolish{\text{mol} \: \ce{O_2}}} = 15.7 \: \text{g} \: \ce{O_2}\nonumber\]

The theoretical yield of \(\ce{O_2}\) is \(15.vii \: \text{thou}\), 15.67 g unrounded.

Step v: Think about your result.

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

b. Adding of percent yield

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

Stride 1: Identify the "given" data and what the problem is asking you to "find".

Given: Theoretical yield =xv.half-dozenvii 1000, use the united nations-rounded number for the adding.

Actual yield = 14.9g

Observe: Percent yield, % Yield

Step 2: List other known quantities and plan the problem.

No other quantities needed.

Step 3: Utilise the percent yield equation below.

\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)

         Step iv: Solve.

\(\text{Percentage Yield} = \frac{14.ix \: \text{chiliad}}{fifteen.\underline{vi}7 \: \text{g}} \times 100\% = 94.9\%\)

Step 5: Think about your result.

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

Example \(\PageIndex{2}\): Oxidation of Zinc

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 yard copper metal was obtained according to the equation:

\(\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(southward)+\ce{ZnSO4}(aq)\)

What is the percentage yield?

Solution

Steps for Problem Solving-The Product Method	Example \(\PageIndex{ane}\)
Identify the "given" information and what the trouble is request you to "notice."	Given: i.274 g CuSO4 Actual yield = 0.392 thou Cu Notice: Percent yield
Listing other known quantities.	1 mol CuSO4= 159.62 yard/mol 1 mol Cu = 63.55 m/mol Since the amount of product in grams is not required, only the tooth mass of the reactants is needed.
Balance the equation.	The chemical equation is already balanced. The balanced equation provides the human relationship of ane 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 equally the limiting reactant, and and then the theoretical yield (g Cu) is found past performing mass-mass calculation based on the initial amount of CuSO4.
Abolish units and summate.	\[\mathrm{ane.274\:\cancel{chiliad\:Cu_SO_4}\times \dfrac{1\:\abolish{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{ane\:\cancel{mol\: Cu}}{ane\:\cancel{mol\:CuSO_4}}\times \dfrac{63.55\:one thousand\: Cu}{i\:\cancel{mol\: Cu}}=0.5072\: g\: Cu}\nonumber\] Using this theoretical yield and the provided value for actual yield, the percentage yield is calculated to be: \[\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100}\] \[\brainstorm{marshal} \mathrm{percent\: yield}&=\mathrm{\left(\dfrac{0.392\: g\: Cu}{0.5072\: chiliad\: Cu}\right)\times 100} \\ &=77.3\% \terminate{align}\nonumber\]
Think about your result.	Since the actual yield is slightly less than the theoretical yield, the percent yield is just under \(100\%\).

Exercise \(\PageIndex{i}\)

What is the percent yield of a reaction that produces 12.5 1000 of the Freon CF₂Cl₂ from 32.9 thousand of CCl_iv and excess HF?

\[\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber\]

Reply

48.3%

Summary

Theoretical yield is calculated based on the stoichiometry of the chemical equation. The actual yield is experimentally determined. The percent yield is adamant past calculating the ratio of actual yield to theoretical yield.

Contributions & Attributions

This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts evolution team to meet platform style, presentation, and quality:

• Marisa Alviar-Agnew (Sacramento Metropolis College)

• Henry Agnew (UC Davis)

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Source: https://chem.libretexts.org/Courses/Sacramento_City_College/SCC:_CHEM_300_-_Beginning_Chemistry/SCC:_CHEM_300_-_Beginning_Chemistry_%28Faculty%29/08:_Quantities_in_Chemical_Reactions/8.06:_Limiting_Reactant,_Theoretical_Yield,_and_Percent_Yield_from_Initial_Masses_of_Reactants

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