⚖ Steps 1 and 2
Step 1: work out the Mr of every substance.
Step 2: change each mass into moles using moles = mass ÷ Mr.
Moles » Using Moles to Balance Equations
What you'll learn this session
Study time: 30 minutes
AQA spec: 4.3.2.3
- How masses from an experiment can give you the balancing numbers in an equation
- How to turn masses in grams into amounts in moles
- How to turn a mole ratio into a simple whole number ratio
- How to rearrange the moles equation to check your answer
Higher tier only
Everything in this lesson is Higher tier only. If you are sitting Foundation papers, you will not be asked to balance equations from masses.
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Unlock This CourseBalancing an equation from masses
Limestone chips fizzing in acid: weigh what reacts and what forms, convert to moles, and the balancing numbers pop out
You already know how to balance an equation by counting atoms. There is a second way. If you know the masses of the reactants and products, you can work out the balancing numbers with a calculation.
The spec says: the balancing numbers in a symbol equation can be calculated from the masses of reactants and products by converting the masses in grams to amounts in moles and converting the numbers of moles to simple whole number ratios.
Why does this work? Remember, the balancing numbers in an equation give the mole ratio of the substances. So if you find how many moles of each substance reacted or were made, you have found the ratio. Then you just tidy it up into small whole numbers.
This is useful because it is how chemists find out what really happens in a reaction. They measure masses in the lab, then work backwards to the equation.
Key terms:
- Simple whole number ratio: a ratio written using the smallest whole numbers possible, such as 2 : 3 instead of 0.2 : 0.3 or 1 : 1.5.
The four steps
Every question of this type uses the same method. Learn these four steps.
➗ Steps 3 and 4
Step 3: divide every number of moles by the smallest one.
Step 4: if any answer is not a whole number, multiply them all by the same number until they are. These are your balancing numbers.
Make sure every formula is correct before you start. You never change a formula to balance an equation. You only put numbers in front.
Worked examples
10.0 g of CaCO3 is 0.1 mol, and the gas escaping here is CO2: CaCO3 + 2HCl → CaCl2 + H2O + CO2
Worked example 1: limestone and acid
Some calcium carbonate reacts with hydrochloric acid. 10.0 g of CaCO3 reacts with 7.3 g of HCl to make 11.1 g of CaCl2, 1.8 g of H2O and 4.4 g of CO2. Balance the equation.
__CaCO3 + __HCl → __CaCl2 + __H2O + __CO2
(Ar: Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5)
- Step 1: Mr values: CaCO3 = 100, HCl = 36.5, CaCl2 = 111, H2O = 18, CO2 = 44
- Step 2: moles: CaCO3 = 10.0 ÷ 100 = 0.10, HCl = 7.3 ÷ 36.5 = 0.20, CaCl2 = 11.1 ÷ 111 = 0.10, H2O = 1.8 ÷ 18 = 0.10, CO2 = 4.4 ÷ 44 = 0.10
- Step 3: divide by the smallest (0.10): 1 : 2 : 1 : 1 : 1
- Step 4: these are already whole numbers.
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Sometimes step 3 gives you a number like 1.5. That is not a mistake. It just means you need step 4.
Worked example 2: when you need step 4
5.4 g of aluminium reacts with 21.3 g of chlorine to make 26.7 g of aluminium chloride. Balance the equation.
__Al + __Cl2 → __AlCl3
(Ar: Al = 27, Cl = 35.5)
- Step 1: Mr values: Al = 27, Cl2 = 71, AlCl3 = 27 + (3 × 35.5) = 133.5
- Step 2: moles: Al = 5.4 ÷ 27 = 0.20, Cl2 = 21.3 ÷ 71 = 0.30, AlCl3 = 26.7 ÷ 133.5 = 0.20
- Step 3: divide by 0.20: 1 : 1.5 : 1
- Step 4: 1.5 is not whole, so multiply everything by 2: 2 : 3 : 2
2Al + 3Cl2 → 2AlCl3
Count the atoms to check: 2 Al and 6 Cl on each side. It balances.
A quick guide for step 4: if you get a number ending in .5, multiply by 2. If you get a number ending in .33 or .67, multiply by 3.
Sometimes a question leaves out one mass. Remember, mass is conserved in a reaction, so the total mass of the products equals the total mass of the reactants. You can use that to find the missing mass first.
Worked example 3: a missing mass
2.8 g of ethene burns in 9.6 g of oxygen. It makes 8.8 g of carbon dioxide and some water. Balance the equation.
__C2H4 + __O2 → __CO2 + __H2O
(Ar: C = 12, H = 1, O = 16)
- Missing mass: total of reactants = 2.8 + 9.6 = 12.4 g, so water = 12.4 − 8.8 = 3.6 g
- Step 1: Mr values: C2H4 = 28, O2 = 32, CO2 = 44, H2O = 18
- Step 2: moles: C2H4 = 2.8 ÷ 28 = 0.10, O2 = 9.6 ÷ 32 = 0.30, CO2 = 8.8 ÷ 44 = 0.20, H2O = 3.6 ÷ 18 = 0.20
- Step 3: divide by 0.10: 1 : 3 : 2 : 2
C2H4 + 3O2 → 2CO2 + 2H2O
Changing the subject of the equation
The spec also says you should be able to change the subject of a mathematical equation. In this topic, that means using the moles equation in any of its three forms:
➤ Find moles
moles = mass ÷ Mr
➤ Find mass
mass = moles × Mr
➤ Find Mr
Mr = mass ÷ moles
To rearrange, do the same thing to both sides. Start with moles = mass ÷ Mr. Multiply both sides by Mr and you get mass = moles × Mr. Then divide both sides by moles and you get Mr = mass ÷ moles.
The second form is a great way to check your answer. In worked example 2, 2 mol of Al has a mass of 2 × 27 = 54 g. 3 mol of Cl2 is 3 × 71 = 213 g. 2 mol of AlCl3 is 2 × 133.5 = 267 g. And 54 + 213 = 267. These masses are ten times the masses in the question, in the same ratio, so the balancing numbers are right.
Common mistakes
- Using the masses as the ratio. 5.4 g and 21.3 g do not tell you the ratio. You must change masses into moles first.
- Using the wrong Mr for a gas. Oxygen is O2, so its Mr is 32. Chlorine is Cl2, so its Mr is 71.
- Rounding 1.5 to 2. Never round a ratio like this. Multiply everything by 2 instead.
- Forgetting to divide by the smallest. 0.20 : 0.30 : 0.20 is a ratio, but it is not the balancing numbers.
- Changing a formula. AlCl3 stays AlCl3. Only the numbers in front change.
Exam-style question
Iron is made by reacting iron(III) oxide with carbon monoxide. In a test, 16.0 g of Fe2O3 reacted with 8.4 g of CO to make 11.2 g of Fe and 13.2 g of CO2.
__Fe2O3 + __CO → __Fe + __CO2
Use the masses to balance the equation. You must show your working. (Ar: Fe = 56, O = 16, C = 12) [4 marks]
Model answer
- Mr values: Fe2O3 = 160, CO = 28, Fe = 56, CO2 = 44 (1)
- Moles: Fe2O3 = 16.0 ÷ 160 = 0.10, CO = 8.4 ÷ 28 = 0.30, Fe = 11.2 ÷ 56 = 0.20, CO2 = 13.2 ÷ 44 = 0.30 (1)
- Divide by the smallest (0.10): 1 : 3 : 2 : 3 (1)
- Fe2O3 + 3CO → 2Fe + 3CO2 (1)
Exam tip
Lay out your working in a table with one column for each substance and rows for mass, Mr, moles and ratio. It keeps your numbers tidy and earns method marks even if you slip up on one value.