ChemistryICSE Class 9 · The Language of Chemistry

Balancing Equations, Relative Molecular Mass and Percentage Composition (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Balance an equation by hit and trial: write the correct formulae, put coefficients (never change subscripts) until each element has the same number of atoms on both sides, and reduce to the lowest whole numbers. Relative molecular mass is the sum of the relative atomic masses; percentage of an element = (mass of the element in the formula ÷ Mr) × 100.

Count each element on both sides: only coefficients are changed to balance (4 Al and 6 O each side), never the subscripts inside formulae.

Illustration drawn for QuestionBuddy.

Understand Balancing Chemical Equations, Relative Masses and Percentage Composition

Chemical equation. Reactants on the left, products on the right, with symbols and formulae; it must be balanced to obey the law of conservation of mass (equal numbers of each kind of atom on both sides).

Hit and trial method.
1. Write the correct formulae first; never change a subscript (that changes the substance).
2. Begin with the element in the largest or most complicated formula; leave H, O and free elements for last.
3. Place coefficients in front, recount every element, adjust.
4. Reduce the coefficients to the lowest whole numbers; add state or conditions if required.

Examples: $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ ; $F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}$ ; $N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$ .

Relative atomic mass (atomic weight) is how many times an atom is heavier than $\frac{1}{12}$ of the mass of a carbon-12 atom. It has no unit. Relative molecular mass $M_{r}$ is the sum of the relative atomic masses of all atoms in the formula; in a bracket multiply by its subscript. Hydrates include the water of crystallisation: $C u S O_{4} \cdot 5 H_{2} O$ : $M_{r} = 63.5 + 32 + 64 + 5 \times 18 = 249.5$ .

Percentage composition.

$% of an element = \frac{mass of the element in one formula}{M _{r}} \times 100,$

and the percentages of all elements add up to 100. For a hydrate, $% water = \frac{n \times 18}{M _{r}} \times 100$ .

Mass from percentage. Mass of an element in a given mass of compound $= mass of compound \times \frac{mass of the element in the formula}{M _{r}}$ .

Ratios in an equation. The coefficients and the relative masses give mass ratios: for $2 M g + O_{2} \to 2 M g O$ , 48 parts of magnesium combine with 32 parts of oxygen to give 80 parts of magnesium oxide.

Careful with. (i) Atomic masses must be taken as given in the question (for example Cl = 35.5). (ii) Count atoms inside brackets by multiplying by the subscript. (iii) In a hydrate the dot means "together with": the mass of the water is added to $M_{r}$ , not multiplied by it.

Balancing Chemical Equations, Relative Masses and Percentage Composition: formulas and key facts

Relative molecular mass: $M_{r} = \sum (atomic mass \times number of atoms)$; Brackets multiply.
Percentage composition: $% = \frac{mass of element in formula}{M _{r}} \times 100$; Totals 100%.
Water of crystallisation: $% H_{2} O = \frac{n \times 18}{M _{r} of hydrate} \times 100$; n = number of water molecules.
Mass from the formula: $m_{e l e m e n t} = m_{co m p o u n d} \times \frac{mass of element in formula}{M _{r}}$; Same ratio as the percentage.

How to solve Balancing Chemical Equations, Relative Masses and Percentage Composition questions

Count atoms, then balance; compute masses with the given atomic masses only.

1List atoms of each element on both sides.
2Balance the most complex formula first, H and O last.
3Recount every element after each change; reduce coefficients.
4For Mr, multiply each atomic mass by its number of atoms (include brackets).
5Divide the element’s total by Mr and multiply by 100 for the percentage.

Balancing Chemical Equations, Relative Masses and Percentage Composition in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Which ore has more iron?

Suppose a miner compares two iron ores, haematite (Fe₂O₃) and magnetite (Fe₃O₄), taking Fe = 56 and O = 16.

Which ore has the larger percentage of iron?

Try it first, then show the solutionSolution

Fe₂O₃: $M_{r} = 112 + 48 = 160$ ; Fe $= \frac{112}{160} \times 100 = 70%$ .
Fe₃O₄: $M_{r} = 168 + 64 = 232$ ; Fe $= \frac{168}{232} \times 100 \approx 72.4%$ .

Answer: Magnetite (72.4% against 70%).

Memory hook: Compute both, compare.

Tricks to remember Balancing Chemical Equations, Relative Masses and Percentage Composition

Short memory aids. Say each one aloud once, then use it on a question the same day.

Subscripts are sacred
Only coefficients may change while balancing.
Use it when: Balancing.
H and O last
Balance metals and others first; H and O at the end.
Use it when: Hit and trial.
Part over whole
Percentage = part ÷ whole × 100, the whole being Mr.
Use it when: Percentage composition.

Balancing Chemical Equations, Relative Masses and Percentage Composition: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Coefficients · 1 mark · MCQBoard level

The balanced equation for the burning of magnesium in oxygen has the coefficients (in order Mg, O₂, MgO):
(a) 1, 1, 1
(b) 2, 1, 2
(c) 1, 2, 1
(d) 2, 2, 2

Try it first, then show the solutionSolution

$2 M g + O_{2} \to 2 M g O$ : 2 Mg and 2 O on each side.

Answer: (b) 2, 1, 2

Example 2: Relative molecular mass · 2 marks · Short answerBoard level

Calculate the relative molecular mass of (a) calcium hydroxide, Ca(OH)₂, and (b) ammonium sulphate, (NH₄)₂SO₄. (Ca = 40, O = 16, H = 1, N = 14, S = 32)

Try it first, then show the solutionSolution

(a) $40 + 2 (16 + 1) = 74$ . (1)
(b) $2 (14 + 4) + 32 + 64 = 132$ . (1)

Answer: (a) 74 (b) 132

Example 3: Balance three equations · 3 marks · Short answerHarder board level

Balance: (a) Fe₂O₃ + CO → Fe + CO₂, (b) Al + HCl → AlCl₃ + H₂, (c) Na₂CO₃ + HCl → NaCl + H₂O + CO₂.

Try it first, then show the solutionSolution

(a) $F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}$ . (1)
(b) $2 A l + 6 H C l \to 2 A l C l_{3} + 3 H_{2}$ . (1)
(c) $N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$ . (1)

Answer: See working

Example 4: Which ore is richer? · 3 marks · Short answerHarder board level

Calculate the percentage of iron in (a) Fe₂O₃ and (b) Fe₃O₄ (Fe = 56, O = 16) and say which has more iron.

Try it first, then show the solutionSolution

(a) $M_{r} = 160$ ; Fe $= \frac{112}{160} \times 100 = 70%$ . (1)
(b) $M_{r} = 232$ ; Fe $= \frac{168}{232} \times 100 \approx 72.4%$ . (1)
Fe₃O₄ has the greater percentage. (1)

Answer: 70%; 72.4%; Fe₃O₄

Example 5: Washing soda · 4 marks · Case-basedHOTS / competency

Washing soda is Na₂CO₃·10H₂O (Na = 23, C = 12, O = 16, H = 1).
(a) Find its relative molecular mass.
(b) Find the percentage of water of crystallisation.
(c) Find the mass of anhydrous sodium carbonate in 28.6 g of the crystals.
(d) Write the equation for the decomposition of the crystals on strong heating and the mass of water driven off from 28.6 g.

Try it first, then show the solutionSolution

(a) Na₂CO₃ $= 46 + 12 + 48 = 106$ ; $10 H_{2} O = 180$ ; $M_{r} = 286$ . (1)
(b) $\frac{180}{286} \times 100 \approx 62.9%$ . (1)
(c) $28.6 \times \frac{106}{286} = 10.6$ g. (1)
(d) $N a_{2} C O_{3} \cdot 10 H_{2} O \to N a_{2} C O_{3} + 10 H_{2} O$ ; water $= 28.6 - 10.6 = 18.0$ g. (1)

Answer: (a) 286 (b) 62.9% (c) 10.6 g (d) 18.0 g

Where students lose marks in Balancing Chemical Equations, Relative Masses and Percentage Composition

The slips examiners see most often, and what to write instead.

Mistake: Subscript altered to balance
It changes the compound.
Wrong working
H₂ + O₂ → H₂O₂ to balance oxygen.
Right working
2H₂ + O₂ → 2H₂O.
Do this: Only change coefficients.
Usually costs: The equation.
Mistake: Bracket multiplier forgotten in Mr
Every atom inside the bracket is multiplied.
Wrong working
Ca(OH)₂ = 40 + 16 + 1 + 2 = 59
Right working
40 + 2(16 + 1) = 74
Do this: Expand the bracket first.
Usually costs: The answer.
Mistake: Percentage taken over the wrong total
The denominator is Mr of the whole formula.
Wrong working
% Fe in Fe₂O₃ = 112/48
Right working
112/160 × 100 = 70%
Do this: Part ÷ whole.
Usually costs: The answer.

Exam tips

Show atom counts on both sides when asked to verify balance.
Write Mr working line by line.
Use the atomic masses given in the question; quote them first.
Check percentages of all elements add up to 100.

Practice questions on Balancing Chemical Equations, Relative Masses and Percentage Composition

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
While balancing an equation by hit and trial, you may change:
(a) the subscripts of formulae
(b) the coefficients in front of formulae
(c) the formulae of products
(d) the symbols of elements
Hint
What you change must not change the substance.
Answer
(b)
Q2 · 1 mark · Very short answerBoard level
Does relative atomic mass have a unit?
Hint
It is a ratio.
Answer
No.
Q3 · 2 marks · Short answerHarder board level
Find the relative molecular mass of aluminium sulphate, Al₂(SO₄)₃ (Al = 27, S = 32, O = 16).
Hint
Brackets multiply.
Answer
$54 + 3 (32 + 64) = 54 + 288 = 342$ .
Q4 · 3 marks · Short answerHarder board level
Balance and state the number of reactants and products: (a) Zn + HCl → ZnCl₂ + H₂ (b) NH₃ + O₂ → NO + H₂O (c) Al(OH)₃ → Al₂O₃ + H₂O.
Hint
Balance H and O last.
Answer
(a) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$ (2 reactants, 2 products). (b) $4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$ (2, 2). (c) $2 A l (O H)_{3} \to A l_{2} O_{3} + 3 H_{2} O$ (1, 2).
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): While balancing, we change coefficients and not subscripts.
Reason (R): Changing a subscript changes the substance itself.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
H₂O versus H₂O₂.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
Nitrous oxide N₂O, nitric oxide NO and nitrogen dioxide NO₂ are three oxides of nitrogen (N = 14, O = 16). Arrange them in order of percentage of nitrogen (highest first) and find the mass of nitrogen in 11 g of N₂O.
Hint
Compute each percentage.
Answer
N₂O: $\frac{28}{44} = 63.6%$ ; NO: $\frac{14}{30} = 46.7%$ ; NO₂: $\frac{14}{46} = 30.4%$ . Order: N₂O > NO > NO₂. Nitrogen in 11 g of N₂O: $11 \times \frac{28}{44} = 7$ g.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Why must an equation be balanced?Show answer
To obey the law of conservation of mass.
2.Relative molecular mass?Show answer
Sum of the relative atomic masses of all atoms in the formula.
3.Percentage composition formula?Show answer
(mass of element in formula ÷ Mr) × 100.
4.Mr of CuSO₄·5H₂O?Show answer
249.5.
5.What may be changed while balancing?Show answer
Only the coefficients.

Balancing Chemical Equations, Relative Masses and Percentage Composition: frequently asked questions

How do you balance a chemical equation by hit and trial?

Write correct formulae, then add coefficients in front of formulae, recounting atoms of each element until both sides match, leaving H and O for last.

How do you calculate the percentage composition?

Divide the total mass of the element in the formula by the relative molecular mass and multiply by 100.

How do you find the percentage of water of crystallisation?

n × 18 divided by the Mr of the hydrate, multiplied by 100.