Percentage yield

What you'll learn

Why the mass of product you collect is often less than the calculated maximum.
What actual yield, theoretical yield and percentage yield mean.
How to calculate percentage yield using masses.
Higher Tier: how to calculate the theoretical mass of product from a balanced equation.

Why yield matters

In calculations, we often imagine a reaction working perfectly: all the reactants react exactly as shown in the balanced equation, and all the product is collected.

Real practical chemistry is messier. Even though atoms are not created or destroyed in a chemical reaction, you may not collect the full amount of product predicted by the calculation.

Schematic showing theoretical yield, actual yield, and reasons actual yield can be lower

Key Idea

No atoms lost, but product can be lost

A low yield does not mean atoms have disappeared. It means the atoms may be in unreacted reactants, unwanted side products, or product that was not successfully collected.

Key terms

Definition

Yield

The yield is the amount of product obtained from a chemical reaction. It is usually measured as a mass, such as grams (g).

Definition

Actual yield and theoretical yield

The actual yield is the mass of product actually collected in the experiment. The theoretical yield is the maximum mass of product that could be made, calculated from the balanced equation.

If the reaction and collection were perfect, the actual yield would equal the theoretical yield. In reality, actual yield is usually smaller.

Why actual yield is often less than theoretical yield

There are three GCSE reasons you need to know.

1. The reaction may be reversible

A reversible reaction can go both forwards and backwards. It uses the symbol ⇌.

For example:

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

If a reaction is reversible, not all reactants are converted into products. Some product can turn back into reactants, so the reaction may not go to completion.

2. Product may be lost during separation

After making a product, you often need to separate it from the reaction mixture. For example, you might filter, crystallise, dry or transfer it between containers.

During these steps, some product may:

stick to filter paper
remain dissolved in solution
be left behind on glassware
be spilled or lost during transfer

3. Reactants may form unwanted products

Sometimes reactants react in a different way from the expected reaction. This is called a side reaction.

A side reaction uses up some reactants, so less of the desired product is made.

Common Mistake

Thinking low yield breaks conservation of mass

Conservation of mass still applies. The “missing” mass is not destroyed — it is usually in other substances, left in the apparatus, or not converted into the desired product.

Percentage yield

Percentage yield compares the actual yield with the theoretical yield.

Definition

Percentage yield

The percentage yield is the actual yield as a percentage of the theoretical yield.

product×100\text{percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical mass of product}} \times 100

A percentage yield of 100% would mean you collected exactly the maximum possible mass. Most real reactions have a percentage yield below 100%.

Tip

Sanity check

For GCSE questions, percentage yield should normally be between 0% and 100%. If you get more than 100%, check whether you swapped actual and theoretical masses.

Calculating percentage yield from actual and theoretical masses

This is the most direct type of question. You are given:

the actual mass collected
the theoretical mass predicted

Then you substitute into the formula.

Example

Calculating percentage yield

A reaction has a theoretical yield of 12.5 g. In the experiment, 9.8 g of product is collected. Calculate the percentage yield.

Identify the two masses needed by the formula: actual mass = 9.8 g and theoretical mass = 12.5 g.
Substitute into the percentage yield equation:
$yield=9.812.5×100\text{percentage yield} = \frac{9.8}{12.5} \times 100$
Calculate the fraction first, then convert to a percentage:
$yield=78.4%\text{percentage yield} = 78.4\%$
Give the answer to a sensible number of significant figures: 78.4%, or 78% if rounding to two significant figures.

Rearranging the percentage yield formula

Sometimes you are given the percentage yield and one mass, then asked to find the other mass.

Start with:

mass×100\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100

If you need the actual mass:

mass\text{actual mass} = \frac{\text{percentage yield}}{100} \times \text{theoretical mass}

If you need the theoretical mass:

yield×100\text{theoretical mass} = \frac{\text{actual mass}}{\text{percentage yield}} \times 100

Example

Finding the actual mass produced

The theoretical yield of a product is 40.0 g. The percentage yield is 65%. Calculate the actual mass obtained.

Choose the rearranged formula because the actual mass is unknown:
$mass\text{actual mass} = \frac{\text{percentage yield}}{100} \times \text{theoretical mass}$
Substitute the values:
$mass=65100×40.0\text{actual mass} = \frac{65}{100} \times 40.0$
Calculate:
$g\text{actual mass} = 26.0\ \text{g}$

So the actual mass obtained is 26.0 g.

Common Mistake

Using 65 instead of 0.65

A percentage must be divided by 100 before it is used as a multiplier. A yield of 65% means 0.65 of the theoretical mass, not 65 times the theoretical mass.

Higher Tier: calculating theoretical mass from a reactant

This part is Higher Tier in GCSE Chemistry.

To calculate theoretical yield from a balanced equation, you use moles.

Definition

Mole

A mole is an amount of substance. In GCSE calculations, the number of moles is found using:

moles=massMr\text{moles} = \frac{\text{mass}}{M_r}

where $M_r$ is the relative formula mass.

The balanced equation gives the mole ratio between reactants and products.

For example:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This equation tells you that 1 mole of magnesium makes 1 mole of hydrogen gas. The big numbers in front of formulae are the important ratio numbers.

Example

Calculating theoretical mass from a balanced equation

Copper oxide reacts with hydrogen to form copper and water:

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

Calculate the theoretical mass of copper made from 8.0 g of copper oxide, CuO. Use relative atomic masses: Cu = 64, O = 16.

Calculate the relative formula mass of copper oxide:
$Mr(CuO)=64+16=80M_r(\text{CuO}) = 64 + 16 = 80$
Convert the mass of copper oxide into moles:
$mol\text{moles of CuO} = \frac{8.0}{80} = 0.10\ \text{mol}$
Use the balanced equation ratio. CuO and Cu are in a 1:1 ratio, so 0.10 mol of CuO makes 0.10 mol of Cu.
Convert moles of copper into mass:
$g\text{mass of Cu} = 0.10 \times 64 = 6.4\ \text{g}$

The theoretical mass of copper is 6.4 g.

Combining theoretical yield with percentage yield

Some questions ask you to do both parts:

Use the balanced equation to calculate the theoretical mass.
Use the actual mass to calculate the percentage yield.

Example

Calculating percentage yield after finding theoretical mass

Calcium carbonate thermally decomposes:

CaCO₃(s) → CaO(s) + CO₂(g)

A student heats 10.0 g of calcium carbonate and collects 4.20 g of calcium oxide. Calculate the percentage yield of calcium oxide. Use relative formula masses: CaCO₃ = 100, CaO = 56.

Convert the mass of calcium carbonate into moles:
$mol\text{moles of CaCO}_3 = \frac{10.0}{100} = 0.100\ \text{mol}$
Use the balanced equation ratio. CaCO₃ and CaO are in a 1:1 ratio, so 0.100 mol of CaCO₃ makes 0.100 mol of CaO.
Calculate the theoretical mass of calcium oxide:
$g\text{mass of CaO} = 0.100 \times 56 = 5.60\ \text{g}$
Use the actual mass and theoretical mass in the percentage yield formula:
$yield=4.205.60×100=75.0%\text{percentage yield} = \frac{4.20}{5.60} \times 100 = 75.0\%$

The percentage yield is 75.0%.

Percentage yield vs atom economy

Percentage yield is not the same as atom economy.

Percentage yield is about how much product you actually collect compared with the maximum possible amount.

Atom economy is about how much of the reactant mass ends up in the desired product according to the balanced equation.

Key Idea

Different questions, different formulae

Use percentage yield when the question compares actual mass with theoretical mass. Use atom economy when the question compares the mass of the desired product with the total mass of products in the equation.

Presenting your answer

In yield calculations, always include:

the correct formula or rearranged formula
substitution of values
units for masses, usually g
a percentage sign for percentage yield
sensible rounding, usually 2 or 3 significant figures unless told otherwise

Common Mistake

Percentage yield over 100%

In real chemistry, a calculated yield above 100% usually suggests an error, such as wet product, impurities in the product, or an incorrect mass measurement. In GCSE calculations, it usually means you have used the formula the wrong way round.

Exam technique

In the exam

Check whether the question gives you the theoretical mass directly, or whether you must calculate it from the balanced equation first.
Put the actual mass on top in the percentage yield formula: actual divided by theoretical, then multiplied by 100.
If the answer is bigger than 100%, pause and check your substitution, mole ratio and units.

Self review

Check yourself

Why might a reaction produce less product than the maximum theoretical amount?
What is the formula for percentage yield?
Higher Tier: how does a balanced equation help you calculate theoretical yield?

Percentage yield

What you'll learn

Why the mass of product you collect is often less than the calculated maximum.
What actual yield, theoretical yield and percentage yield mean.
How to calculate percentage yield using masses.
Higher Tier: how to calculate the theoretical mass of product from a balanced equation.

Why yield matters

In calculations, we often imagine a reaction working perfectly: all the reactants react exactly as shown in the balanced equation, and all the product is collected.

Real practical chemistry is messier. Even though atoms are not created or destroyed in a chemical reaction, you may not collect the full amount of product predicted by the calculation.

Schematic showing theoretical yield, actual yield, and reasons actual yield can be lower

Key Idea

No atoms lost, but product can be lost

A low yield does not mean atoms have disappeared. It means the atoms may be in unreacted reactants, unwanted side products, or product that was not successfully collected.

Key terms

Definition

Yield

The yield is the amount of product obtained from a chemical reaction. It is usually measured as a mass, such as grams (g).

Definition

Actual yield and theoretical yield

The actual yield is the mass of product actually collected in the experiment. The theoretical yield is the maximum mass of product that could be made, calculated from the balanced equation.

If the reaction and collection were perfect, the actual yield would equal the theoretical yield. In reality, actual yield is usually smaller.

Why actual yield is often less than theoretical yield

There are three GCSE reasons you need to know.

1. The reaction may be reversible

A reversible reaction can go both forwards and backwards. It uses the symbol ⇌.

For example:

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

If a reaction is reversible, not all reactants are converted into products. Some product can turn back into reactants, so the reaction may not go to completion.

2. Product may be lost during separation

After making a product, you often need to separate it from the reaction mixture. For example, you might filter, crystallise, dry or transfer it between containers.

During these steps, some product may:

stick to filter paper
remain dissolved in solution
be left behind on glassware
be spilled or lost during transfer

3. Reactants may form unwanted products

Sometimes reactants react in a different way from the expected reaction. This is called a side reaction.

A side reaction uses up some reactants, so less of the desired product is made.

Common Mistake

Thinking low yield breaks conservation of mass

Conservation of mass still applies. The “missing” mass is not destroyed — it is usually in other substances, left in the apparatus, or not converted into the desired product.

Percentage yield

Percentage yield compares the actual yield with the theoretical yield.

Definition

Percentage yield

The percentage yield is the actual yield as a percentage of the theoretical yield.

product×100\text{percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical mass of product}} \times 100

A percentage yield of 100% would mean you collected exactly the maximum possible mass. Most real reactions have a percentage yield below 100%.

Tip

Sanity check

For GCSE questions, percentage yield should normally be between 0% and 100%. If you get more than 100%, check whether you swapped actual and theoretical masses.

Calculating percentage yield from actual and theoretical masses

This is the most direct type of question. You are given:

the actual mass collected
the theoretical mass predicted

Then you substitute into the formula.

Example

Calculating percentage yield

A reaction has a theoretical yield of 12.5 g. In the experiment, 9.8 g of product is collected. Calculate the percentage yield.

Identify the two masses needed by the formula: actual mass = 9.8 g and theoretical mass = 12.5 g.
Substitute into the percentage yield equation:
$yield=9.812.5×100\text{percentage yield} = \frac{9.8}{12.5} \times 100$
Calculate the fraction first, then convert to a percentage:
$yield=78.4%\text{percentage yield} = 78.4\%$
Give the answer to a sensible number of significant figures: 78.4%, or 78% if rounding to two significant figures.

Rearranging the percentage yield formula

Sometimes you are given the percentage yield and one mass, then asked to find the other mass.

Start with:

mass×100\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100

If you need the actual mass:

mass\text{actual mass} = \frac{\text{percentage yield}}{100} \times \text{theoretical mass}

If you need the theoretical mass:

yield×100\text{theoretical mass} = \frac{\text{actual mass}}{\text{percentage yield}} \times 100

Example

Finding the actual mass produced

The theoretical yield of a product is 40.0 g. The percentage yield is 65%. Calculate the actual mass obtained.

Choose the rearranged formula because the actual mass is unknown:
$mass\text{actual mass} = \frac{\text{percentage yield}}{100} \times \text{theoretical mass}$
Substitute the values:
$mass=65100×40.0\text{actual mass} = \frac{65}{100} \times 40.0$
Calculate:
$g\text{actual mass} = 26.0\ \text{g}$

So the actual mass obtained is 26.0 g.

Common Mistake

Using 65 instead of 0.65

A percentage must be divided by 100 before it is used as a multiplier. A yield of 65% means 0.65 of the theoretical mass, not 65 times the theoretical mass.

Higher Tier: calculating theoretical mass from a reactant

This part is Higher Tier in GCSE Chemistry.

To calculate theoretical yield from a balanced equation, you use moles.

Definition

Mole

A mole is an amount of substance. In GCSE calculations, the number of moles is found using:

moles=massMr\text{moles} = \frac{\text{mass}}{M_r}

where $M_r$ is the relative formula mass.

The balanced equation gives the mole ratio between reactants and products.

For example:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This equation tells you that 1 mole of magnesium makes 1 mole of hydrogen gas. The big numbers in front of formulae are the important ratio numbers.

Example

Calculating theoretical mass from a balanced equation

Copper oxide reacts with hydrogen to form copper and water:

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

Calculate the theoretical mass of copper made from 8.0 g of copper oxide, CuO. Use relative atomic masses: Cu = 64, O = 16.

Calculate the relative formula mass of copper oxide:
$Mr(CuO)=64+16=80M_r(\text{CuO}) = 64 + 16 = 80$
Convert the mass of copper oxide into moles:
$mol\text{moles of CuO} = \frac{8.0}{80} = 0.10\ \text{mol}$
Use the balanced equation ratio. CuO and Cu are in a 1:1 ratio, so 0.10 mol of CuO makes 0.10 mol of Cu.
Convert moles of copper into mass:
$g\text{mass of Cu} = 0.10 \times 64 = 6.4\ \text{g}$

The theoretical mass of copper is 6.4 g.

Combining theoretical yield with percentage yield

Some questions ask you to do both parts:

Use the balanced equation to calculate the theoretical mass.
Use the actual mass to calculate the percentage yield.

Example

Calculating percentage yield after finding theoretical mass

Calcium carbonate thermally decomposes:

CaCO₃(s) → CaO(s) + CO₂(g)

A student heats 10.0 g of calcium carbonate and collects 4.20 g of calcium oxide. Calculate the percentage yield of calcium oxide. Use relative formula masses: CaCO₃ = 100, CaO = 56.

Convert the mass of calcium carbonate into moles:
$mol\text{moles of CaCO}_3 = \frac{10.0}{100} = 0.100\ \text{mol}$
Use the balanced equation ratio. CaCO₃ and CaO are in a 1:1 ratio, so 0.100 mol of CaCO₃ makes 0.100 mol of CaO.
Calculate the theoretical mass of calcium oxide:
$g\text{mass of CaO} = 0.100 \times 56 = 5.60\ \text{g}$
Use the actual mass and theoretical mass in the percentage yield formula:
$yield=4.205.60×100=75.0%\text{percentage yield} = \frac{4.20}{5.60} \times 100 = 75.0\%$

The percentage yield is 75.0%.

Percentage yield vs atom economy

Percentage yield is not the same as atom economy.

Percentage yield is about how much product you actually collect compared with the maximum possible amount.

Atom economy is about how much of the reactant mass ends up in the desired product according to the balanced equation.

Key Idea

Different questions, different formulae

Presenting your answer

In yield calculations, always include:

the correct formula or rearranged formula
substitution of values
units for masses, usually g
a percentage sign for percentage yield
sensible rounding, usually 2 or 3 significant figures unless told otherwise

Common Mistake

Percentage yield over 100%

Exam technique

In the exam

Check whether the question gives you the theoretical mass directly, or whether you must calculate it from the balanced equation first.
Put the actual mass on top in the percentage yield formula: actual divided by theoretical, then multiplied by 100.
If the answer is bigger than 100%, pause and check your substitution, mole ratio and units.

Self review

Check yourself

Why might a reaction produce less product than the maximum theoretical amount?
What is the formula for percentage yield?
Higher Tier: how does a balanced equation help you calculate theoretical yield?

Percentage yield

What you'll learn

Why yield matters

Key terms

Why actual yield is often less than theoretical yield

1. The reaction may be reversible

2. Product may be lost during separation

3. Reactants may form unwanted products

Percentage yield

Calculating percentage yield from actual and theoretical masses

Rearranging the percentage yield formula

Higher Tier: calculating theoretical mass from a reactant

Combining theoretical yield with percentage yield

Percentage yield vs atom economy

Presenting your answer

You've reached the end

Percentage yield

What you'll learn

Why yield matters

Key terms

Why actual yield is often less than theoretical yield

1. The reaction may be reversible

2. Product may be lost during separation

3. Reactants may form unwanted products

Percentage yield

Calculating percentage yield from actual and theoretical masses

Rearranging the percentage yield formula

Higher Tier: calculating theoretical mass from a reactant

Combining theoretical yield with percentage yield

Percentage yield vs atom economy

Presenting your answer

You've reached the end

Percentage yield

What you'll learn

Why yield matters

Key terms

Why actual yield is often less than theoretical yield

1. The reaction may be reversible

2. Product may be lost during separation

3. Reactants may form unwanted products

Percentage yield

Calculating percentage yield from actual and theoretical masses

Rearranging the percentage yield formula

Higher Tier: calculating theoretical mass from a reactant

Combining theoretical yield with percentage yield

Percentage yield vs atom economy

Presenting your answer

You've reached the end

Percentage yield Revision Guide

Percentage yield

What you'll learn

Why yield matters

Key terms

Why actual yield is often less than theoretical yield

1. The reaction may be reversible

2. Product may be lost during separation

3. Reactants may form unwanted products

Percentage yield

Calculating percentage yield from actual and theoretical masses

Rearranging the percentage yield formula

Higher Tier: calculating theoretical mass from a reactant

Combining theoretical yield with percentage yield

Percentage yield vs atom economy

Presenting your answer

You've reached the end

Percentage yield Revision Guide