8.7: Theoretical Yield and Percent Yield


⚙️ Learning Objectives

• Calculate percentage or actual yields from known amounts of reactants.

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.

Percent Yield

Chemical reactions in the real world don't always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than predicted. Besides spills and other experimental errors, there are usually 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. As we have previously learned, this is called the theoretical yield, the maximum amount of product that can 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. Once again,

$$\mathrm{percent}\;\mathrm{yield}\;=\;\dfrac{\mathrm{actual}\;\mathrm{yield}}{\mathrm{theoretical}\;\mathrm{yield}}\times100\%$$

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.

The percent yield of a desired product may never exceed 100% for the reasons indicated earlier. Should a percent yield ever be greater than 100%, this would be an indication of some sort of experimental error or that the reaction contains impurities that cause its mass to be greater than if the product was pure. When chemists synthesize a desired chemical, they are always careful to purify the products of the reaction. Example $$\PageIndex{1}$$ illustrates the steps for determining percent yield.

✅ Example $$\PageIndex{1}$$: Theoretical Yield and Percent Yield

Potassium chlorate decomposes upon heating, according to the reaction below:

$$2\;{\mathrm{KClO}}_3\;(s)\;\xrightarrow\triangle\;2\;\mathrm{KCl}\;(s)\;+\;3\;{\mathrm O}_2\;(g)$$

If 40.0 g KClO3 is heated until it completely decomposes, resulting in the collection of 14.3 g of oxygen gas,

1. What is the theoretical yield of oxygen gas?
2. What is the percent yield of oxygen gas?

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: 40.0 g KClO3 reacted; 14.3 g O2 produced

Find: theoretical yield O2; percent yield O2

List other known quantities.
1 mol KClO3 = 122.55 g KClO3

1 mol O2 = 32.00 g O2

2 mol KClO3: 3 mol O2

Prepare concept maps using the proper conversion factor(s).
$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\;\mathrm g\;{\mathrm{KClO}}_3\;\;}}\xrightarrow[{122.55\;\mathrm g\;{\mathrm{KClO}}_3}]{1\;\mathrm{mol}\;{\mathrm{KClO}}_3}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm{KClO}}_3\;}}\xrightarrow[{2\;\mathrm{mol}\;{\mathrm{KClO}}_3}]{3\;\mathrm{mol}\;{\mathrm O}_2}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\;\;\;\mathrm{mol}\;{\mathrm O}_2\;\;\;}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm O}_2}]{32.00\;\mathrm g\;{\mathrm O}_2}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\;\;\;\;\mathrm g\;{\mathrm O}_2\;\;\;\;\;}}$$

Calculate the theoretical yield.
$$40.0\:\cancel{\mathrm g\;{\mathrm{KClO}}_3}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm{KClO}}_3}}{122.55\:\cancel{\mathrm g\;{\mathrm{KClO}}_3}}\times\dfrac{3\;\cancel{\mathrm{mol}\;{\mathrm O}_2}}{2\:\cancel{\mathrm{mol}\;{\mathrm{KClO}}_3}}\times\dfrac{32.00\;\mathrm g\;{\mathrm O}_2}{1\;\cancel{\mathrm{mol}\;{\mathrm O}_2}}=\boxed{15.7\;\mathrm g\;{\mathrm O}_2}$$

Calculate the percent yield. $$\mathrm{percent}\;\mathrm{yield}=\dfrac{\mathrm{actual}\;\mathrm{yield}}{\mathrm{theoretical}\;\mathrm{yield}}\times100\%=\dfrac{14.3\;\cancel{\mathrm g\;{\mathrm O}_2}}{15.7\;\cancel{\mathrm g\;{\mathrm O}_2}}\times100\%=\boxed{91.1\%\;\mathrm{yield}}$$

The percent yield is less than 100%, but still quite high (over 90%).
This seems like a reasonable yield for a chemical reaction.

✅ Example $$\PageIndex{2}$$: Limiting Reactants, Theoretical Yield, and Percent Yield

4 V (s) + 5 O2 (g) → 2 V2O5 (s)

1. What mass of V2O5 can be made when 41.3 g V reacts with 35.0 g O2?
2. If 62.3 g V2O5 were actually produced, what is the percent yield?

Solution

Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: 41.3 g V reacted; 35.0 g O2 reacted; 62.3 g V2O5 produced

Find: theoretical yield V2O5; percent yield V2O5

List other known quantities.
1 mol V = 50.94 g V

1 mol O2 = 32.00 g O2

1 mol V2O5 = 181.88 g V2O5

4 mol V: 2 mol V2O5

5 mol O2: 5 mol V2O5

Prepare concept maps using the proper conversion factor(s).
$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\;\;\;\mathrm g\;\mathrm V\;\;\;\;}}\xrightarrow[{50.94\;\mathrm g\;\mathrm V}]{1\;\mathrm{mol}\;\mathrm V}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;\mathrm V\;\;}}\xrightarrow[{4\;\mathrm{mol}\;\mathrm V}]{2\;\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}]{181.88\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5\;\;}}$$

$${\color[rgb]{0.5, 0.0, 0.5}\boxed{\;\;\;\mathrm g\;{\mathrm O}_2\;\;\;}}\xrightarrow[{32.00\;\mathrm g\;{\mathrm O}_2}]{1\;\mathrm{mol}\;{\mathrm O}_2}{\color[rgb]{0.8, 0.0, 0.0}\boxed{\;\mathrm{mol}\;{\mathrm O}_2\;}}\xrightarrow[{5\;\mathrm{mol}\;{\mathrm O}_2}]{2\;\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}{\color[rgb]{0.0, 0.0, 1.0}\boxed{\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}}\xrightarrow[{1\;\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}]{181.88\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5}{\color[rgb]{0.0, 0.5, 0.0}\boxed{\;\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5\;\;}}$$

Calculate the theoretical yield.

$$41.3\:\cancel{\mathrm g\;\mathrm V}\times\dfrac{1\;\cancel{\mathrm{mol}\;\mathrm V}}{50.94\:\cancel{\mathrm g\;\mathrm V}}\times\dfrac{2\;\cancel{\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}}{4\;\cancel{\mathrm{mol}\;\mathrm V}}\times\dfrac{181.88\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5}{1\;\cancel{\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}}=\boxed{73.7\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5}$$

$$35.0\:\cancel{\mathrm g\;{\mathrm O}_2}\times\dfrac{1\;\cancel{\mathrm{mol}\;{\mathrm O}_2}}{32.00\:\cancel{\mathrm g\;{\mathrm O}_2}}\times\dfrac{2\;\cancel{\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}}{5\;\cancel{\mathrm{mol}\;{\mathrm O}_2}}\times\dfrac{181.88\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5}{1\;\cancel{\mathrm{mol}\;{\mathrm V}_2{\mathrm O}_5}}=\xcancel{79.6\;\mathrm g\;{\mathrm V}_2{\mathrm O}_5}$$

Calculate the percent yield. $$\mathrm{percent}\;\mathrm{yield}=\dfrac{\mathrm{actual}\;\mathrm{yield}}{\mathrm{theoretical}\;\mathrm{yield}}\times100\%=\dfrac{62.3\;\cancel{\mathrm g\;{\mathrm V}_2{\mathrm O}_5}}{73.7\;\cancel{\mathrm g\;{\mathrm V}_2{\mathrm O}_5}}\times100\%=\boxed{84.5\%\;\mathrm{yield}}$$

The percent yield is less than 100%, but still quite high (almost 85%).
This seems like a reasonable yield for a chemical reaction.

✏️ Exercise $$\PageIndex{1}$$

What is the percent yield of a reaction that produces 12.5 g of the Freon CF2Cl2 from 32.9 g of CCl4 and an excess of HF?

CCl4 + 2 HF → CF2Cl2 + 2 HCl

48.3% yield

Summary

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