Chemical Equations and Calculations

Summary: Balancing chemical equations using state symbols (s, l, g, aq). Writing ionic equations by removing spectator ions. Converting between mass, moles, concentration, and gas volume. Tags: igcse chemistry stoichiometry Created: 2026-07-14 Last Updated: 2026-07-16

Balancing Chemical Equations

A balanced chemical equation shows the formulae of reactants and products and the relative amounts of each substance involved in a reaction. Balancing ensures that the equation obeys the law of conservation of mass — atoms cannot be created or destroyed, so the number of each type of atom must be the same on both sides of the equation.

The Golden Rule

Only change the coefficients (the large numbers in front of each formula). Never change the chemical formulae themselves.

Changing H₂O to H₂O₂ to balance oxygen would be wrong — H₂O and H₂O₂ are different substances with different properties. The formula is fixed; only the number of molecules/formula units can be adjusted.

State Symbols

Every substance in a chemical equation should have a state symbol in parentheses after the formula:

The distinction between (l) and (aq) is important:

• H₂O(l) is pure liquid water
• NaCl(aq) is sodium chloride dissolved in water — an aqueous solution, not a pure liquid

Worked Balancing Examples

Example 1: Simple Combination

Balance: H₂ + O₂ → H₂O

Example 2: Metal + Non-metal

Balance: Fe + Cl₂ → FeCl₃

Example 3: Combustion of a Hydrocarbon

Balance: C₃H₈ + O₂ → CO₂ + H₂O

Example 4: Acid-Base Reaction

Balance: H₂SO₄ + NaOH → Na₂SO₄ + H₂O

Strategy for Balancing Equations

A systematic approach helps when balancing unfamiliar equations:

1. Balance metals first (e.g., Fe, Na, Ca, Mg)
2. Balance non-metals next (except H and O — e.g., Cl, S, N, P)
3. Balance hydrogen second-to-last
4. Balance oxygen last — oxygen often appears in multiple substances, so doing it last avoids repeatedly rebalancing

If an element appears in multiple places on one side, balance it last. If stuck, use the LCM (lowest common multiple) approach: find the LCM of the atom counts on each side and work from there.

For combustion equations specifically: balance C first, then H, then O last.

Ionic Equations

Many reactions in aqueous solution involve ions. A full chemical equation shows all reactants and products as complete formula units, but in solution, soluble ionic compounds exist as separate ions. An ionic equation shows only the species that actually take part in the reaction.

Spectator Ions

Spectator ions are ions that appear unchanged on both sides of the equation. They are present in the solution but do not participate in the chemical reaction. Spectator ions are cancelled (omitted) when writing a net ionic equation.

Method for Writing Ionic Equations

1. Write the full balanced chemical equation with state symbols
2. Split all aqueous ionic compounds into their constituent ions (leave solids, liquids, and gases as molecules/formula units)
3. This gives the full ionic equation — every aqueous ion is shown separately
4. Identify and cancel spectator ions — any ion that appears identically on both sides
5. Write the net ionic equation containing only the species that actually change

Worked Example 1: Precipitation

Full equation:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Step 2 — Split aqueous ionic compounds into ions:

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

AgCl(s) is not split — it is an insoluble solid, not free ions.

Step 3 — Identify spectator ions:

• Na⁺(aq) appears on both sides — spectator
• NO₃⁻(aq) appears on both sides — spectator

Step 4 — Cancel spectator ions and write net ionic equation:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

The net ionic equation shows that silver ions and chloride ions combine to form an insoluble precipitate of silver chloride. The sodium and nitrate ions were merely ‘watching’.

Worked Example 2: Neutralisation

Full equation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Full ionic equation:

H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)

H₂O(l) is not split — water is a covalent liquid, not free ions.

Cancel spectator ions (Na⁺ and Cl⁻):

H⁺(aq) + OH⁻(aq) → H₂O(l)

This is the universal net ionic equation for any strong acid–strong base neutralisation, regardless of which specific acid and base are used.

Worked Example 3: Acid + Carbonate

Full equation:

2HCl(aq) + Na₂CO₃(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)

Full ionic equation:

2H⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → 2Na⁺(aq) + 2Cl⁻(aq) + CO₂(g) + H₂O(l)

Cancel spectator ions (Na⁺ and Cl⁻):

2H⁺(aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l)

Worked Example 4: Metal + Acid

Full equation:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Full ionic equation:

Mg(s) + 2H⁺(aq) + 2Cl⁻(aq) → Mg²⁺(aq) + 2Cl⁻(aq) + H₂(g)

Mg(s) and H₂(g) are not split — they are not aqueous ions.

Cancel spectator ions (Cl⁻):

Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)

This shows the actual electron transfer: Mg atoms donate electrons to H⁺ ions, forming Mg²⁺ ions and H₂ gas.

The Mole and Key Relationships

The mole (symbol: mol) is the SI unit for amount of substance. One mole contains 6.02 x 10²³ elementary entities (Avogadro’s constant).

The Three Key Mole Equations

1. Moles from mass:

n = m / Mr

where n = number of moles (mol), m = mass (g), Mr = relative formula mass (g/mol)

2. Moles from concentration and volume (solutions):

n = c × V

where n = number of moles (mol), c = concentration (mol/dm³), V = volume (dm³)

If the volume is given in cm³, convert to dm³ first: V(dm³) = V(cm³) / 1000

3. Moles from gas volume (at RTP):

n = V / 24

where n = number of moles (mol), V = volume (dm³), 24 = molar gas volume at RTP (dm³/mol)

At room temperature and pressure (RTP) — 20 °C, 1 atm — one mole of any gas occupies 24 dm³. This is the molar gas volume.

Unit Conversions

Worked Calculation Examples

Finding Moles from Mass

Question: How many moles are in 8.0 g of sulfur dioxide, SO₂? (Ar: S = 32, O = 16)

Solution:

Mr(SO₂) = 32 + (2 × 16) = 32 + 32 = 64 g/mol
n = m / Mr = 8.0 / 64 = 0.125 mol

Finding Mass from Moles

Question: What is the mass of 0.25 moles of calcium carbonate, CaCO₃? (Ar: Ca = 40, C = 12, O = 16)

Solution:

Mr(CaCO₃) = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 g/mol
m = n × Mr = 0.25 × 100 = 25.0 g

Finding Moles from Concentration and Volume

Question: How many moles of HCl are in 25.0 cm³ of 2.0 mol/dm³ hydrochloric acid?

Solution:

V = 25.0 cm³ = 25.0 / 1000 = 0.0250 dm³
n = c × V = 2.0 × 0.0250 = 0.050 mol

Finding Concentration from Moles and Volume

Question: 0.10 moles of NaOH is dissolved in water to make 250 cm³ of solution. What is the concentration in mol/dm³?

Solution:

V = 250 cm³ = 250 / 1000 = 0.250 dm³
c = n / V = 0.10 / 0.250 = 0.40 mol/dm³

Finding Gas Volume from Moles

Question: What volume does 0.50 moles of CO₂ gas occupy at RTP?

Solution:

V = n × 24 = 0.50 × 24 = 12 dm³

Finding Moles from Gas Volume

Question: How many moles are in 6.0 dm³ of oxygen gas at RTP?

Solution:

n = V / 24 = 6.0 / 24 = 0.25 mol

Reacting Mass Calculation

Question: What mass of magnesium oxide (MgO) is produced when 6.0 g of magnesium burns completely in oxygen? (Ar: Mg = 24, O = 16)

Solution:

1. Write the balanced equation: 2Mg(s) + O₂(g) → 2MgO(s)
2. Mr(Mg) = 24, Mr(MgO) = 24 + 16 = 40
3. Moles of Mg: n = m / Mr = 6.0 / 24 = 0.25 mol
4. Mole ratio from equation: Mg : MgO = 2 : 2 = 1 : 1
   So moles of MgO formed = 0.25 mol
5. Mass of MgO: m = n × Mr = 0.25 × 40 = 10.0 g

Titration Calculation

Question: 25.0 cm³ of NaOH solution of unknown concentration is neutralised by 30.0 cm³ of 0.10 mol/dm³ HCl. Find the concentration of the NaOH solution.

Solution:

1. Equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)   (1:1 ratio)
2. Moles of HCl: n = c × V = 0.10 × (30.0 / 1000) = 0.0030 mol
3. From the equation, NaOH : HCl = 1 : 1
   So moles of NaOH = 0.0030 mol
4. Concentration of NaOH: c = n / V = 0.0030 / (25.0 / 1000) = 0.0030 / 0.0250 = 0.12 mol/dm³

Gas Volume from Reacting Masses

Question: What volume of CO₂ gas (at RTP) is produced when 10.0 g of calcium carbonate is heated until it decomposes completely? CaCO₃(s) → CaO(s) + CO₂(g). (Ar: Ca = 40, C = 12, O = 16)

Solution:

1. Mr(CaCO₃) = 40 + 12 + 48 = 100 g/mol
2. Moles of CaCO₃: n = m / Mr = 10.0 / 100 = 0.10 mol
3. From the equation, CaCO₃ : CO₂ = 1 : 1
   So moles of CO₂ = 0.10 mol
4. Volume of CO₂ at RTP: V = n × 24 = 0.10 × 24 = 2.4 dm³

Key Concepts from Past Papers

• One mark each for any 3 of:
• (limonene) particles go from liquid to vapour
• random movement of particles / particles move anywhere / particles move in all directions
• spreading out of particles / intermingling of particles / mixing of particles / particles collide / particles bounce off each other / particles go all over
• (bulk) movement of particles from higher to lower concentration / movement of particles down concentration gradient
• mark each for any 3 of:
• (HCl) molecules escape from solution
• molecules in (constant) movement / molecules collide / molecules travel

Keywords from Past Papers

molecules, particles, concentration, movement, rate, reaction, lower, spread, koh, move, collide, mix, volume, amount, increases

Related Notes

• Mole Concept
• Empirical and Molecular Formulae
• Percentage Yield and Purity
• Stoichiometry
• Titrations

Sources

• OpenStax Chemistry 2e, Chapter 4 (Stoichiometry of Chemical Reactions) — Sections 4.1–4.3 on balancing equations, reaction stoichiometry, and limiting reactants — openstax.org/books/chemistry-2e/
• BBC Bitesize GCSE Chemistry, Chemical Equations and Calculations — bbc.co.uk/bitesize/topics/
• Cambridge IGCSE Chemistry 0620 Syllabus, Topic 4: Stoichiometry
• CK-12 Chemistry, Balancing Equations and Mole Calculations — ck12.org/chemistry/

Past Paper Sources

• 0620/31 May/June 2015: Q22(b)(iv) (0m), Q44(a)(iii) (0m), Q55(c)(i) (0m) (+3 more)
• 0620/32 Feb/March 2017: Q33(b)(v) (2m)
• 0620/32 May/June 2018: Q11(b)(i) (1m)
• 0620/32 May/June 2020: Q66(b)(i) (1m)
• 0620/33 May/June 2016: Q66(d)(i) (1m)
• 0620/33 May/June 2017: Q66(b)(iv) (1m)
• 0620/33 May/June 2021: Q22(d)(ii) (3m)
• 0620/33 May/June 2022: Q33(b)(ii) (1m)
• 0620/33 May/June 2023: Q88(c)(v) (0m)
• 0620/33 May/June 2024: Q66(c)(ii) (1m)
• 0620/33 October/November 2015: Q77(b)(i) (2m), Q77(b)(ii) (0m)
• 0620/33 October/November 2018: Q22(a)(i) (2m)

Common Misconceptions