Using moles to balance equations (HT only)

Welcome! This is a Higher Tier (HT) only topic that flips what you already know about moles and chemical equations upside down.

Usually, you are given a beautifully balanced equation and asked to calculate an unknown mass. Here, we are going to do the reverse: you will be given the raw experimental masses, and you will use them to discover what the balanced equation looks like.

What you'll learn:

How to convert the masses of all reactants and products into moles.
How to use those moles to find the simplest whole-number ratio.
How to turn that ratio into the large balancing numbers (stoichiometric coefficients) in a symbol equation.

The basic idea

Every chemical reaction follows a recipe. The large numbers written in front of formulas in a balanced chemical equation tell us the ratio of moles involved in the reaction.

Definition

Balancing number

The large number placed in front of a chemical formula in an equation (e.g., the 2 in $2H2O2\text{H}_2\text{O}$ ). It tells you how many moles of that substance take part in the reaction. If there is no number, it means 1.

If you do an experiment and weigh all your reactants and products, those masses alone don't tell you the balancing numbers because different atoms weigh different amounts. To find the correct ratio, we have to level the playing field by converting all the masses into moles.

Key Idea

The Golden Rule of Balancing with Masses

Masses don't give you the balancing numbers directly—moles do. Once you know how many moles of each substance you have, you just find their simplest whole-number ratio, and those become your balancing numbers.

Reminding ourselves of the mole formula

To convert our masses into moles, we need the standard mole equation. You should be completely comfortable with rearranging this:

\text{Moles} = \frac{\text{Mass}}{\text{Relative formula mass}}

\frac{m}{M_r}

If you ever need to find the mass from the moles (which we won't do today, but is a vital skill), you rearrange it to make mass the subject: $\times M_r$ .

Common Mistake

Using the balancing number to calculate Mr

When you calculate the relative formula mass ( $M_r$ ) for this method, ignore the big numbers in front of the formula! You are trying to find the big numbers, so they don't exist yet. Just add up the atomic masses ( $A_r$ ) of the small subscript numbers. For example, the $M_r$ of $O2\text{O}_2$ is $16 \times 2 = 32$ .

The four-step method

We follow exactly four steps to get from raw mass data to a perfectly balanced symbol equation.

Masses: Write down the given mass of every reactant and product.
Mr / Ar: Calculate the relative formula mass (or relative atomic mass) for each substance.
Moles: Divide each mass by its $M_r$ to find the number of moles.
Ratio: Divide all your mole values by the smallest mole value you found. This forces the smallest value to become 1 and reveals the simplest whole-number ratio.

A clean, modern chemistry educational flowchart showing the steps to balance an equation from experimental masses.

Let's see this in action with a complete example.

Example

Deducing the balanced equation for a displacement reaction

In an experiment, 6.5 g of zinc ( $Zn\text{Zn}$ ) reacts exactly with 8.0 g of copper oxide ( $CuO\text{CuO}$ ) to produce 8.1 g of zinc oxide ( $ZnO\text{ZnO}$ ) and 6.4 g of copper ( $Cu\text{Cu}$ ).

Deduce the balanced symbol equation for this reaction.

(Relative atomic masses: $Zn=65\text{Zn} = 65$ , $Cu=64\text{Cu} = 64$ , $O=16\text{O} = 16$ )

Calculate the $M_{r}$ (or $A_{r}$ ) for each substance involved:
- $Zn=65\text{Zn} = 65$
- $CuO=64+16=80\text{CuO} = 64 + 16 = 80$
- $ZnO=65+16=81\text{ZnO} = 65 + 16 = 81$
- $Cu=64\text{Cu} = 64$
Divide the mass of each substance by its $M_{r}$ to find the moles ( $n = m / M_{r}$ ):
- Moles of $mol\text{Zn} = \frac{6.5}{65} = 0.1 \text{ mol}$
- Moles of $mol\text{CuO} = \frac{8.0}{80} = 0.1 \text{ mol}$
- Moles of $mol\text{ZnO} = \frac{8.1}{81} = 0.1 \text{ mol}$
- Moles of $mol\text{Cu} = \frac{6.4}{64} = 0.1 \text{ mol}$
Divide all the mole values by the smallest value (which is 0.1) to find the simplest ratio:
- $Zn:0.10.1=1\text{Zn}: \frac{0.1}{0.1} = 1$
- $CuO:0.10.1=1\text{CuO}: \frac{0.1}{0.1} = 1$
- $ZnO:0.10.1=1\text{ZnO}: \frac{0.1}{0.1} = 1$
- $Cu:0.10.1=1\text{Cu}: \frac{0.1}{0.1} = 1$
Write the final balanced equation using this 1:1:1:1 ratio. Because the numbers are all 1, we don't write any numbers in front of the formulas:

\text{Zn} + \text{CuO} \to \text{ZnO} + \text{Cu}

Dealing with awkward decimals

Sometimes, dividing by the smallest number of moles doesn't give you a perfect whole number. You might end up with a ratio like 1 : 1.5 : 1.

You cannot write $1.5$ as a balancing number in a standard GCSE chemical equation. Atoms react in whole numbers! If this happens, you must multiply all the numbers in your ratio by the same amount to clear the decimal.

Tip

Clearing decimals

If your ratio ends in .5, multiply everything by 2.
If your ratio ends in .33 or .67, multiply everything by 3.
If your ratio ends in .25 or .75, multiply everything by 4.

Example

Balancing an equation with an awkward ratio

22.4 g of iron ( $Fe\text{Fe}$ ) reacts with 42.6 g of chlorine gas ( $Cl2\text{Cl}_2$ ) to form 65.0 g of iron(III) chloride ( $FeCl3\text{FeCl}_3$ ).

Deduce the balanced symbol equation for this reaction.

(Relative atomic masses: $Fe=56\text{Fe} = 56$ , $Cl=35.5\text{Cl} = 35.5$ )

Calculate the $M_{r}$ (or $A_{r}$ ) for each substance:
- $Fe=56\text{Fe} = 56$
- $Cl2=35.5×2=71\text{Cl}_2 = 35.5 \times 2 = 71$
- $FeCl3=56+(35.5×3)=162.5\text{FeCl}_3 = 56 + (35.5 \times 3) = 162.5$
Calculate the moles of each substance by doing Mass $÷Mr\div M_r$ $\div M_{r}$ :
- Moles of $mol\text{Fe} = \frac{22.4}{56} = 0.4 \text{ mol}$
- Moles of $mol\text{Cl}_2 = \frac{42.6}{71} = 0.6 \text{ mol}$
- Moles of $mol\text{FeCl}_3 = \frac{65.0}{162.5} = 0.4 \text{ mol}$
Divide each by the smallest number of moles (0.4) to find the ratio:
- $Fe:0.40.4=1\text{Fe}: \frac{0.4}{0.4} = 1$
- $Cl2:0.60.4=1.5\text{Cl}_2: \frac{0.6}{0.4} = 1.5$
- $FeCl3:0.40.4=1\text{FeCl}_3: \frac{0.4}{0.4} = 1$
The ratio is $1 : 1.5 : 1$ . We have a .5 decimal, so multiply all numbers by 2 to get whole numbers. The true ratio is $2 : 3 : 2$ .
Write the equation with these big numbers in front of the corresponding formulas:

2\text{Fe} + 3\text{Cl}_2 \to 2\text{FeCl}_3

Exam technique

In the exam

Draw a little table or grid to keep your workings tidy. Make the columns your substances, and the rows: Mass, $M_r$ , Moles, Ratio.
Don't panic if your moles look like weird decimals (e.g., 0.125). That is totally normal! As long as you divide them all by the smallest one, a neat, whole-number ratio will pop out.
Once you've written your final equation, quickly check it like you would a normal balancing question. Do you have the same number of Iron atoms on both sides? Yes? Then you know for certain you've nailed the 4-6 marks.

Self review

Check yourself

Do we use the large balancing numbers when calculating the $M_r$ of a substance?
If you divide your moles and get a ratio of 1 : 2.5 : 2, what should your final balancing numbers be?
How do you rearrange $\frac{m}{M_r}$ to make mass ( $m$ ) the subject?

Using moles to balance equations (HT only)

Welcome! This is a Higher Tier (HT) only topic that flips what you already know about moles and chemical equations upside down.

What you'll learn:

How to convert the masses of all reactants and products into moles.
How to use those moles to find the simplest whole-number ratio.
How to turn that ratio into the large balancing numbers (stoichiometric coefficients) in a symbol equation.

The basic idea

Every chemical reaction follows a recipe. The large numbers written in front of formulas in a balanced chemical equation tell us the ratio of moles involved in the reaction.

Definition

Balancing number

Key Idea

The Golden Rule of Balancing with Masses

Reminding ourselves of the mole formula

To convert our masses into moles, we need the standard mole equation. You should be completely comfortable with rearranging this:

\text{Moles} = \frac{\text{Mass}}{\text{Relative formula mass}}

\frac{m}{M_r}

If you ever need to find the mass from the moles (which we won't do today, but is a vital skill), you rearrange it to make mass the subject: $\times M_r$ .

Common Mistake

Using the balancing number to calculate Mr

The four-step method

We follow exactly four steps to get from raw mass data to a perfectly balanced symbol equation.

Masses: Write down the given mass of every reactant and product.
Mr / Ar: Calculate the relative formula mass (or relative atomic mass) for each substance.
Moles: Divide each mass by its $M_r$ to find the number of moles.
Ratio: Divide all your mole values by the smallest mole value you found. This forces the smallest value to become 1 and reveals the simplest whole-number ratio.

A clean, modern chemistry educational flowchart showing the steps to balance an equation from experimental masses.

Let's see this in action with a complete example.

Example

Deducing the balanced equation for a displacement reaction

Deduce the balanced symbol equation for this reaction.

(Relative atomic masses: $Zn=65\text{Zn} = 65$ , $Cu=64\text{Cu} = 64$ , $O=16\text{O} = 16$ )

Calculate the $M_{r}$ (or $A_{r}$ ) for each substance involved:
- $Zn=65\text{Zn} = 65$
- $CuO=64+16=80\text{CuO} = 64 + 16 = 80$
- $ZnO=65+16=81\text{ZnO} = 65 + 16 = 81$
- $Cu=64\text{Cu} = 64$
Divide the mass of each substance by its $M_{r}$ to find the moles ( $n = m / M_{r}$ ):
- Moles of $mol\text{Zn} = \frac{6.5}{65} = 0.1 \text{ mol}$
- Moles of $mol\text{CuO} = \frac{8.0}{80} = 0.1 \text{ mol}$
- Moles of $mol\text{ZnO} = \frac{8.1}{81} = 0.1 \text{ mol}$
- Moles of $mol\text{Cu} = \frac{6.4}{64} = 0.1 \text{ mol}$
Divide all the mole values by the smallest value (which is 0.1) to find the simplest ratio:
- $Zn:0.10.1=1\text{Zn}: \frac{0.1}{0.1} = 1$
- $CuO:0.10.1=1\text{CuO}: \frac{0.1}{0.1} = 1$
- $ZnO:0.10.1=1\text{ZnO}: \frac{0.1}{0.1} = 1$
- $Cu:0.10.1=1\text{Cu}: \frac{0.1}{0.1} = 1$
Write the final balanced equation using this 1:1:1:1 ratio. Because the numbers are all 1, we don't write any numbers in front of the formulas:

\text{Zn} + \text{CuO} \to \text{ZnO} + \text{Cu}

Dealing with awkward decimals

Sometimes, dividing by the smallest number of moles doesn't give you a perfect whole number. You might end up with a ratio like 1 : 1.5 : 1.

Tip

Clearing decimals

If your ratio ends in .5, multiply everything by 2.
If your ratio ends in .33 or .67, multiply everything by 3.
If your ratio ends in .25 or .75, multiply everything by 4.

Example

Balancing an equation with an awkward ratio

22.4 g of iron ( $Fe\text{Fe}$ ) reacts with 42.6 g of chlorine gas ( $Cl2\text{Cl}_2$ ) to form 65.0 g of iron(III) chloride ( $FeCl3\text{FeCl}_3$ ).

Deduce the balanced symbol equation for this reaction.

(Relative atomic masses: $Fe=56\text{Fe} = 56$ , $Cl=35.5\text{Cl} = 35.5$ )

Calculate the $M_{r}$ (or $A_{r}$ ) for each substance:
- $Fe=56\text{Fe} = 56$
- $Cl2=35.5×2=71\text{Cl}_2 = 35.5 \times 2 = 71$
- $FeCl3=56+(35.5×3)=162.5\text{FeCl}_3 = 56 + (35.5 \times 3) = 162.5$
Calculate the moles of each substance by doing Mass $÷Mr\div M_r$ $\div M_{r}$ :
- Moles of $mol\text{Fe} = \frac{22.4}{56} = 0.4 \text{ mol}$
- Moles of $mol\text{Cl}_2 = \frac{42.6}{71} = 0.6 \text{ mol}$
- Moles of $mol\text{FeCl}_3 = \frac{65.0}{162.5} = 0.4 \text{ mol}$
Divide each by the smallest number of moles (0.4) to find the ratio:
- $Fe:0.40.4=1\text{Fe}: \frac{0.4}{0.4} = 1$
- $Cl2:0.60.4=1.5\text{Cl}_2: \frac{0.6}{0.4} = 1.5$
- $FeCl3:0.40.4=1\text{FeCl}_3: \frac{0.4}{0.4} = 1$
The ratio is $1 : 1.5 : 1$ . We have a .5 decimal, so multiply all numbers by 2 to get whole numbers. The true ratio is $2 : 3 : 2$ .
Write the equation with these big numbers in front of the corresponding formulas:

2\text{Fe} + 3\text{Cl}_2 \to 2\text{FeCl}_3

Exam technique

In the exam

Draw a little table or grid to keep your workings tidy. Make the columns your substances, and the rows: Mass, $M_r$ , Moles, Ratio.
Don't panic if your moles look like weird decimals (e.g., 0.125). That is totally normal! As long as you divide them all by the smallest one, a neat, whole-number ratio will pop out.
Once you've written your final equation, quickly check it like you would a normal balancing question. Do you have the same number of Iron atoms on both sides? Yes? Then you know for certain you've nailed the 4-6 marks.

Self review

Check yourself

Do we use the large balancing numbers when calculating the $M_r$ of a substance?
If you divide your moles and get a ratio of 1 : 2.5 : 2, what should your final balancing numbers be?
How do you rearrange $\frac{m}{M_r}$ to make mass ( $m$ ) the subject?

Using moles to balance equations (HT only)

The basic idea

Reminding ourselves of the mole formula

The four-step method

Dealing with awkward decimals

You've reached the end

Using moles to balance equations (HT only)

The basic idea

Reminding ourselves of the mole formula

The four-step method

Dealing with awkward decimals

You've reached the end

Using moles to balance equations (HT only)

The basic idea

Reminding ourselves of the mole formula

The four-step method

Dealing with awkward decimals

You've reached the end

Using moles to balance equations (HT only) Revision Guide

Using moles to balance equations (HT only)

The basic idea

Reminding ourselves of the mole formula

The four-step method

Dealing with awkward decimals

You've reached the end

Using moles to balance equations (HT only) Revision Guide