AQA GCSE Combined Science (8464), Higher tier · Chemistry › Quantitative chemistry › Use of amount of substance in relation to masses of pure
Practise Using moles to balance equations (HT only). 11 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
If you know the masses of the reactants and products in a reaction, you can work out the balancing numbers in its equation. Convert each mass to moles, then find the simplest whole-number ratio. This subtopic is Higher tier only.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
6
Convert each reacting mass into molesDivide each mass by its Mr, using O2, H2, Cl2 or N2 for gaseous elements.
7
Find the simplest whole-number mole ratioDivide every number of moles by the smallest one.
7
Write the balanced equation from the ratioThe whole numbers are the balancing numbers in front of each formula.
8
Turn a non-whole ratio into whole numbersIf you get 1 : 1.5, multiply every number by 2 to get 2 : 3.
9
Find a missing mass, then balanceUse conservation of mass to find the unknown mass before converting all masses to moles.
Notes
The method
Step 1: write the formula of each reactant and product and work out each Mr.
Step 2: moles of each substance = mass ÷ Mr.
Step 3: divide every number of moles by the smallest of them.
Step 4: if any answer is not a whole number, multiply them all by the same number (e.g. a 1.5 means multiply everything by 2).
Step 5: use the whole numbers as the balancing numbers, then check that the atoms balance.
Worked example
4.8 g of magnesium reacts with 3.2 g of oxygen, O2, to make 8.0 g of magnesium oxide, MgO.
Equation: 2Mg + O2 → 2MgO. Check: 2 Mg atoms and 2 O atoms on each side.
Useful points
Use the formula of the gas as it exists: oxygen gas is O2 (Mr 32), not O (16).
The masses obey conservation of mass (here 4.8 + 3.2 = 8.0 g), so you can use this to find a missing mass.
You may need to rearrange moles = mass ÷ Mr to find a mass or an Mr.
A ratio very close to a whole number (e.g. 2.01 : 1) comes from rounding: treat it as 2 : 1.
Cheatsheet
moles = mass ÷ Mr
Divide all the mole values by the smallest one
×2 if you get a .5; ×3 if you get .33 or .67
Whole-number mole ratio = balancing numbers
Gaseous elements: O2, H2, N2, Cl2
Missing mass: total mass of reactants = total mass of products
How to answer each type of question
Balance an equation from reacting masses
3 to 4 marks7
Work out the moles of every substance: mass ÷ Mr.
Divide by the smallest number of moles.
Multiply up if you need whole numbers.
Write the balanced equation.
Example. A student reacted 5.6 g of iron with 10.65 g of chlorine, Cl2. 16.25 g of iron chloride, FeCl3, formed. Use these masses to balance the equation. ....Fe + ....Cl2 → ....FeCl3 Ar: Cl = 35.5; Fe = 56 [4 marks]
Show the model answer
moles of Fe = 5.6 ÷ 56 = 0.10 (1) moles of Cl2 = 10.65 ÷ 71 = 0.15 and moles of FeCl3 = 16.25 ÷ 162.5 = 0.10 (1) ratio 0.10 : 0.15 : 0.10 = 1 : 1.5 : 1 = 2 : 3 : 2 (1) 2Fe + 3Cl2 → 2FeCl3 (1)
Find a missing mass, then balance
4 marks8
Use conservation of mass to find the missing mass.
Convert every mass to moles.
Divide by the smallest and write the balanced equation.
Example. 2.8 g of nitrogen, N2, reacted with hydrogen, H2, to make 3.4 g of ammonia, NH3. Use these masses to balance the equation. ....N2 + ....H2 ⇌ ....NH3 Ar: H = 1; N = 14 [4 marks]
Find the moles of the fuel, oxygen and each product.
Divide by the smallest number of moles.
Double everything if you get a .5, then write the equation.
Example. When 1.16 g of butane, C4H10, burns completely, it reacts with 4.16 g of oxygen, O2, and produces 3.52 g of carbon dioxide and 1.80 g of water. Use these masses to write a balanced equation for the reaction. Ar: H = 1; C = 12; O = 16 [4 marks]
Use a table: substance | mass | Mr | moles | ÷ smallest | whole number. It keeps every step tidy.
Conservation check: the reactant masses add up to the product masses. If one mass is missing, find it this way first.
Decimals to spot: .5 → ×2; .33 or .67 → ×3; .25 or .75 → ×4.
Always finish by counting atoms: if they balance, your ratio is right.
Where marks are lost
Using the Ar of O (16) for oxygen gas instead of the Mr of O2 (32).
Dividing by the largest number of moles instead of the smallest.
Rounding 1.5 up to 2 instead of doubling every value.
Using the masses themselves as the ratio instead of converting them to moles.
Multiplying only some of the values when scaling to whole numbers.
Exam technique
Show the moles of every substance: each is often a separate mark.
Write the final balanced equation in full, not just the ratio.
Keep 2 or 3 significant figures in the moles so the ratio is clear.
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The balancing numbers in an equation can be found from the numbers of moles of the substances that react and are produced. number of moles = mass in g ÷ Mr
(a) In a reaction, 0.2 mol of hydrogen, H2, reacted with 0.1 mol of oxygen, O2, to produce 0.2 mol of water, H2O. Give the simplest whole number ratio of moles of H2 : O2 : H2O.[1]
(b) Use this ratio to write the balanced equation for the reaction.[1]
(c) Calculate the number of moles in 4.4 g of carbon dioxide. Mr of CO2 = 44[1]
(d) In another reaction, 0.3 mol of sodium reacted with 0.15 mol of chlorine, Cl2, to produce 0.3 mol of sodium chloride, NaCl. Write the balanced equation for this reaction.[1]
Show the answer and mark scheme
(a)Answer: 2 : 1 : 2
2 : 1 : 2
(b)Answer: 2H2 + O2 → 2H2O
2H2 + O2 → 2H2O
(c)Answer: 0.1 mol
0.1 (mol)
(d)Answer: 2Na + Cl2 → 2NaCl
2Na + Cl2 → 2NaCl
Question 2Medium5 marks
A student burned 2.4 g of magnesium in oxygen. 1.6 g of oxygen reacted and 4.0 g of magnesium oxide was produced. Relative atomic masses (Ar): O = 16, Mg = 24
(a) Calculate the number of moles of magnesium (Mg), oxygen (O2) and magnesium oxide (MgO) involved in the reaction.[3]
(b) Use your answers to write the balanced equation for the reaction.[2]
Show the answer and mark scheme
(a)Answer: Mg 0.1 mol, O2 0.05 mol, MgO 0.1 mol
Mg: 2.4 ÷ 24 = 0.1 (mol)
O2: 1.6 ÷ 32 = 0.05 (mol)
MgO: 4.0 ÷ 40 = 0.1 (mol)
(b)Answer: 2Mg + O2 → 2MgO
ratio 0.1 : 0.05 : 0.1 = 2 : 1 : 2
2Mg + O2 → 2MgO
Question 3Hard6 marks
Iron reacts with chlorine to form an iron chloride. A student found that 5.60 g of iron reacted completely with 10.65 g of chlorine, Cl2. Relative atomic masses (Ar): Cl = 35.5, Fe = 56
(a) Calculate the mass of iron chloride produced.[1]
(b) The relative formula mass of the iron chloride is 162.5. Show that the formula of the iron chloride is FeCl3.[1]
(c) Use the masses to find the balancing numbers in the equation for the reaction. Write the balanced equation.[4]