Percent Yield Examples in Chemistry

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Percent Yield.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Chemistry.

Concept Recap

The ratio of the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry, expressed as a percentage.

How much of the possible product you actually got — 100% is perfect, real reactions are always less.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Percent Yield starts with the given amount, names the substance, and chooses the conversion factor that cancels the old unit.

Common stuck point: Students often know a formula related to percent yield but skip the recognition step: Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts? That leads to a correct-looking substitution attached to the wrong chemical model.

Sense of Study hint: Ask: Am I using a mole bridge, molar mass, formula ratio, or balanced-equation ratio to connect measured amounts?

Worked Examples

Example 1

easy
A reaction has a theoretical yield of 50.050.0 g but only 42.042.0 g of product is obtained. Calculate the percent yield.

Answer

84.0%84.0\%

First step

1
Start with the percent-yield formula: %yield=actual yieldtheoretical yield×100%\%\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%.

Full solution

  1. 2
    Substitute the given masses: %yield=42.050.0×100%\%\text{yield} = \frac{42.0}{50.0} \times 100\%.
  2. 3
    Compute the ratio and convert to a percent: %yield=84.0%\%\text{yield} = 84.0\%.
Percent yield measures the efficiency of a reaction. It is always ≤100%\leq 100\% because side reactions, incomplete reactions, and losses during purification reduce the actual yield.

Example 2

hard
In the reaction 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3, 10.010.0 g of Al reacts with excess Cl2\text{Cl}_2 to produce 40.040.0 g of AlCl3\text{AlCl}_3. Calculate the percent yield.

Example 3

medium
In Mg+Cl2→MgCl2\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2, 24 g Mg (M=24) reacts with excess Cl2\text{Cl}_2 to give 76 g MgCl2\text{MgCl}_2 (M=95). Find the percent yield.

Example 4

medium
A student divides theoretical by actual yield and gets 125%. What mistake did they make and what's the correct percent yield?

Example 5

hard
In C3H8+5O2→3CO2+4H2O\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}, burning 44 g propane (M=44) with excess O2\text{O}_2 gives 105.6 g CO2\text{CO}_2 (M=44). Percent yield?

Example 6

challenge
In N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3, you start with 28 g N2\text{N}_2 (M=28) and 4 g H2\text{H}_2 (M=2). Identify the limiting reactant, then compute percent yield if 10.2 g NH3\text{NH}_3 (M=17) is collected.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
If the theoretical yield is 25.025.0 g and the percent yield is 90.0%90.0\%, what is the actual yield?

Example 2

easy
A reaction has a theoretical yield of 18.518.5 g and a percent yield of 92.0%92.0\%. What actual mass of product was collected?

Example 3

easy
Actual yield 40 g, theoretical yield 50 g. What is the percent yield?

Example 4

easy
Actual yield 9 g, theoretical 18 g. Percent yield?

Example 5

easy
Actual 25 g, theoretical 25 g. Percent yield?

Example 6

easy
Which goes in the numerator of percent yield: actual or theoretical?

Example 7

easy
Actual 3 g, theoretical 12 g. Percent yield?

Example 8

easy
Can percent yield exceed 100%? What does it usually mean?

Example 9

easy
Theoretical yield 200 g, actual 150 g. Percent yield?

Example 10

easy
Percent yield is 90% and theoretical is 100 g. What is the actual yield?

Example 11

medium
Theoretical yield is 2 mol of a product (M=18M=18). Actual mass recovered is 27 g. Percent yield?

Example 12

medium
In C+O2→CO2C + O_2 \rightarrow CO_2, 12 g C (M=12M=12) gives 33 g CO2CO_2 (M=44M=44). Percent yield?

Example 13

medium
In 2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O, 2 mol H2H_2 (excess O2O_2) gives 27 g water (M=18M=18). Percent yield?

Example 14

medium
A reaction has 85% yield and a theoretical yield of 200 g. What actual mass is expected?

Example 15

medium
Percent yield is 60% and actual yield is 30 g. What was the theoretical yield?

Example 16

medium
In CaCO3→CaO+CO2CaCO_3 \rightarrow CaO + CO_2, 100 g CaCO3CaCO_3 (M=100M=100) gives 42 g CaOCaO (M=56M=56). Percent yield?

Example 17

medium
Two steps each have 80% yield. What is the overall percent yield?

Example 18

medium
In N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3, 1 mol N2N_2 (excess H2H_2) gives 28.9 g NH3NH_3 (M=17M=17). Percent yield?

Example 19

medium
Percent yield is 50% and theoretical yield is 80 g. What is the actual yield?

Example 20

challenge
In 2Al+3Cl2→2AlCl32Al + 3Cl_2 \rightarrow 2AlCl_3, 54 g Al (M=27M=27, excess Cl2Cl_2) gives 200 g AlCl3AlCl_3 (M=133.5M=133.5). Percent yield?

Example 21

challenge
In C+O2→CO2C + O_2 \rightarrow CO_2 with 6 g C (M=12M=12) and 8 g O2O_2 (M=32M=32), 8.8 g CO2CO_2 forms (M=44M=44). Percent yield?

Example 22

challenge
A two-step synthesis gives an overall yield of 49%. If both steps have equal yield, what is each step's yield?

Example 23

easy
Actual yield 18 g, theoretical yield 20 g. What is the percent yield?

Example 24

easy
Actual yield 6 g, theoretical 15 g. Percent yield?

Example 25

easy
A reaction has theoretical yield 60 g and percent yield 50%. What is the actual yield?

Example 26

easy
Actual yield 12 g, theoretical 16 g. Percent yield?

Example 27

easy
A reaction's percent yield is 80% and actual yield is 16 g. What was the theoretical yield?

Example 28

medium
In S+O2→SO2\text{S} + \text{O}_2 \rightarrow \text{SO}_2, 32 g S (M=32) with excess O2\text{O}_2 gives 51.2 g SO2\text{SO}_2 (M=64). Percent yield?

Example 29

medium
In 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}, 4 g H2\text{H}_2 (M=2) with excess O2\text{O}_2 gives 32.4 g water (M=18). Percent yield?

Example 30

medium
In N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3, 14 g N2\text{N}_2 (M=28) with excess H2\text{H}_2 gives 13.6 g NH3\text{NH}_3 (M=17). Percent yield?

Example 31

medium
A reaction is reported with 65% yield. Theoretical yield is 80 g. Actual yield?

Example 32

medium
Three reaction steps each have 90% yield. Overall percent yield?

Example 33

medium
In CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2, 200 g CaCO3\text{CaCO}_3 (M=100) produces 89.6 g CaO\text{CaO} (M=56). Percent yield?

Example 34

medium
In Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2, 6.5 g Zn (M=65) with excess HCl gives 0.18 g H2\text{H}_2 (M=2). Percent yield?

Example 35

hard
A reaction reports percent yield of 120%. List two reasonable explanations.

Example 36

hard
In Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2, 160 g Fe2O3\text{Fe}_2\text{O}_3 (M=160) yields 89.6 g Fe (M=56). Percent yield?

Example 37

hard
In 2KClO3→2KCl+3O22\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2, 24.5 g KClO3\text{KClO}_3 (M=122.5) gives 8.64 g O2\text{O}_2 (M=32). Percent yield?

Example 38

hard
A four-step synthesis has yields 80%, 75%, 90%, and 50%. Overall percent yield?

Example 39

hard
In 2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}, 8 g NaOH (M=40) with excess acid gives 12.78 g Na2SO4\text{Na}_2\text{SO}_4 (M=142). Percent yield?

Example 40

challenge
A chemist wants an overall yield of at least 50% over 3 equal-yield steps. What is the minimum yield each step must achieve?
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Related Concepts

Theoretical Yield

Background Knowledge

These ideas may be useful before you work through the harder examples.

theoretical yield