Topic 3: Stoichiometry / 3.1 Formulae

3.1 Formulae

Cambridge (CIE) IGCSE Chemistry Revision Notes
📚 Part of Topic 3 📝 Exam code: 0620 & 0971
📌 Topic 3: Stoichiometry
3.1 Formulae
1 State the formulae of the elements and compounds named in the subject content ▼

Diatomic Molecules (7 elements): H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂

Common Compounds:

表 CompoundFormulaCompoundFormula 表 WaterH₂OCarbon dioxideCO₂ AmmoniaNH₃MethaneCH₄ Sodium chlorideNaClSulfuric acidH₂SO₄ Hydrochloric acidHClSodium hydroxideNaOH Calcium carbonateCaCO₃Magnesium oxideMgO 表
2 Define the molecular formula of a compound as the number and type of different atoms in one molecule ▼
表 CompoundMolecular FormulaAtoms Present 表 WaterH₂O2 hydrogen, 1 oxygen MethaneCH₄1 carbon, 4 hydrogen AmmoniaNH₃1 nitrogen, 3 hydrogen EthanolC₂H₅OH2 carbon, 6 hydrogen, 1 oxygen GlucoseC₆H₁₂O₆6 carbon, 12 hydrogen, 6 oxygen 表
3 Deduce the formula of a simple compound from the relative numbers of atoms present in a model or a diagrammatic representation ▼
表 DiagramAtomsMolecular Formula 表 4 carbon + 10 hydrogenC₄H₁₀Butane 2 carbon + 4 hydrogenC₂H₄Ethene 1 carbon + 4 hydrogenCH₄Methane 2 hydrogen + 1 oxygenH₂OWater 1 nitrogen + 3 hydrogenNH₃Ammonia 表
💡 Count the number of each type of atom in the diagram to determine the molecular formula.
4 Construct word equations and symbol equations to show how reactants form products, including state symbols ▼

Word Equations: Reactants → Products (using full chemical names)

Example: magnesium + oxygen → magnesium oxide

Symbol Equations: Use chemical formulae instead of names

Example: 2Mg(s) + O₂(g) → 2MgO(s)

State Symbols:

  • (s) = solid
  • (l) = liquid
  • (g) = gas
  • (aq) = aqueous (dissolved in water)

Balancing Equations - Steps:

  1. Write unbalanced equation with correct formulae
  2. Count atoms of each element on both sides
  3. Add coefficients (numbers in front) to balance
  4. Check again

Worked Example: Magnesium + oxygen → magnesium oxide

  • Unbalanced: Mg + O₂ → MgO
  • Balance oxygen: Mg + O₂ → 2MgO
  • Balance magnesium: 2Mg + O₂ → 2MgO
5 Define the empirical formula of a compound as the simplest whole number ratio of the different atoms or ions in a compound Supplement ▼
表 CompoundMolecular FormulaEmpirical FormulaRatio 表 Hydrogen peroxideH₂O₂HO1:1 GlucoseC₆H₁₂O₆CH₂O1:2:1 EthaneC₂H₆CH₃1:3 EtheneC₂H₄CH₂1:2 BenzeneC₆H₆CH1:1 表
💡 Ionic compounds always have empirical formulas as their chemical formula (e.g., NaCl, MgO).
6 Deduce the formula of an ionic compound from the charges on the ions Supplement ▼

Swap and Drop Method:

表 IonsSwap and DropFormulaName 表 Na⁺ + Cl⁻Na¹ Cl¹NaClSodium chloride Mg²⁺ + Cl⁻Mg¹ Cl²MgCl₂Magnesium chloride Al³⁺ + O²⁻Al² O³Al₂O₃Aluminium oxide Fe³⁺ + SO₄²⁻Fe₂ (SO₄)₃Fe₂(SO₄)₃Iron(III) sulfate Ca²⁺ + OH⁻Ca¹ (OH)₂Ca(OH)₂Calcium hydroxide 表

Common Polyatomic Ions: CO₃²⁻ (carbonate), SO₄²⁻ (sulfate), OH⁻ (hydroxide), NO₃⁻ (nitrate), NH₄⁺ (ammonium)

7 Construct symbol equations with state symbols, including ionic equations Supplement ▼

Ionic Equations: Show only the ions that actually react. Remove spectator ions (ions that appear unchanged on both sides).

Example: Neutralisation

  • Full equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
  • Ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)

Example: Displacement of chlorine with potassium iodide

  • Full equation: 2KI(aq) + Cl₂(aq) → 2KCl(aq) + I₂(aq)
  • Ionic equation: 2I⁻(aq) + Cl₂(aq) → 2Cl⁻(aq) + I₂(aq)
💡 Steps: 1. Write full balanced equation, 2. Replace ionic compounds with ions, 3. Remove spectator ions, 4. Write simplified ionic equation
8 Deduce the symbol equation with state symbols for a chemical reaction, given relevant information Supplement ▼

Worked Example 1: Aluminium burns in chlorine to form aluminium chloride.

  • Aluminium: Al(s)
  • Chlorine: Cl₂(g)
  • Aluminium chloride: AlCl₃(s)

Unbalanced: Al(s) + Cl₂(g) → AlCl₃(s)

Balance: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

Worked Example 2: Magnesium reacts with hydrochloric acid

  • Magnesium: Mg(s)
  • Hydrochloric acid: HCl(aq)
  • Magnesium chloride: MgCl₂(aq)
  • Hydrogen: H₂(g)

Unbalanced: Mg(s) + HCl(aq) → MgCl₂(aq) + H₂(g)

Balance: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

3.2 Relative masses of atoms and molecules
1 Describe relative atomic mass, Aᵣ, as the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of ¹²C ▼

Relative Atomic Mass (Aᵣ): The average mass of all isotopes of an element compared to 1/12th of a carbon-12 atom.

表 ElementAᵣMeaning 表 Hydrogen (H)11/12 the mass of carbon-12 Carbon (C)12Standard reference Oxygen (O)1616 times heavier than 1/12 of carbon-12 Magnesium (Mg)24Twice as heavy as carbon-12 Chlorine (Cl)35.5Average of isotopes Cl-35 and Cl-37 表
2 Define relative molecular mass, Mᵣ, as the sum of the relative atomic masses ▼
表 CompoundFormulaCalculationMᵣ 表 WaterH₂O(2 × 1) + 1618 Carbon dioxideCO₂12 + (2 × 16)44 Sodium chlorideNaCl23 + 35.558.5 Calcium carbonateCaCO₃40 + 12 + (3 × 16)100 Magnesium nitrateMg(NO₃)₂24 + (2 × 14) + (6 × 16)148 表
3 Calculate reacting masses in simple proportions ▼

Law of Conservation of Mass: Total mass of reactants = total mass of products

Worked Example: 2Ca + O₂ → 2CaO

  • 2 × 40 = 80g Ca reacts with 2 × 16 = 32g O₂
  • Forms 2 × (40 + 16) = 112g CaO
  • Ratio: 80 : 32 : 112 (simplified: 5 : 2 : 7)

Worked Example: Calculate mass of CO₂ from 32g CH₄

CH₄ + 2O₂ → CO₂ + 2H₂O

  • 16g CH₄ produces 44g CO₂
  • 32g CH₄ is double → 2 × 44 = 88g CO₂
3.3 The mole and the Avogadro constant Supplement
1 State that the mole is the unit of amount of substance and that one mole contains 6.02 × 10²³ particles (Avogadro constant) ▼

What is a Mole? The mole is the unit for amount of substance.

  • 1 mole = 6.02 × 10²³ particles (Avogadro constant)
  • 1 mole of any element = its relative atomic mass in grams
  • 1 mole of any compound = its relative formula mass in grams

Examples:

  • 1 mole of carbon = 12 g (contains 6.02 × 10²³ atoms)
  • 1 mole of water = 18 g (contains 6.02 × 10²³ molecules)
  • 1 mole of sodium chloride = 58.5 g (contains 6.02 × 10²³ formula units)

Key Formula: Moles = Mass (g) / Molar Mass (g/mol)

2 Use the relationship: amount of substance (mol) = mass (g) / molar mass (g/mol) ▼

Worked Examples:

表 CalculateExampleAnswer 表 Moles from mass6g of carbon (Mᵣ=12)6/12 = 0.5 mol Mass from moles0.25 mol of water (Mᵣ=18)0.25 × 18 = 4.5 g Molar mass from moles0.1 mol has mass 5.6g5.6/0.1 = 56 g/mol Number of particles0.5 mol of CO₂0.5 × 6.02×10²³ = 3.01×10²³ molecules 表
3 Use the molar gas volume, taken as 24 dm³ at room temperature and pressure (r.t.p.), in calculations involving gases ▼

Molar Gas Volume: At r.t.p. (20°C, 1 atm), 1 mole of ANY gas occupies 24 dm³ (24,000 cm³).

Formulas: Volume (dm³) = Moles × 24 | Moles = Volume (dm³) / 24

表 GasMolesVolume (dm³)Volume (cm³) 表 Hydrogen24848,000 Carbon dioxide0.2566,000 Oxygen5.4129.6129,600 表
💡 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.
4 Calculate concentrations of solutions expressed in g/dm³ and mol/dm³ ▼

Concentration Formulas:

Concentration (g/dm³) = Mass of solute (g) / Volume of solution (dm³)

Concentration (mol/dm³) = Moles of solute / Volume of solution (dm³)

Conversions: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.

Worked Example: 80 g NaOH dissolved in 500 cm³ water

  • Molar mass NaOH = 40 g/mol
  • Moles = 80 / 40 = 2 mol
  • Volume = 500 / 1000 = 0.5 dm³
  • Concentration = 2 / 0.5 = 4 mol/dm³
5 Calculate empirical formulae and molecular formulae ▼

Steps to Calculate Empirical Formula:

  1. Write the elements
  2. Write given masses or percentages
  3. Divide by atomic mass to find moles
  4. Divide by the smallest number of moles
  5. Convert to whole numbers

Worked Example: 10 g H and 80 g O

  • H moles = 10 / 1 = 10
  • O moles = 80 / 16 = 5
  • Ratio = 10:5 = 2:1
  • Empirical formula = H₂O

Molecular Formula: (Empirical formula) × n, where n = Mᵣ(molecular) / Mᵣ(empirical)

Worked Example: Empirical formula CH₂, Mᵣ = 56

  • Mᵣ(CH₂) = 14
  • n = 56 / 14 = 4
  • Molecular formula = C₄H₈
6 Calculate percentage yield, percentage composition by mass and percentage purity ▼

Percentage Yield: % Yield = (Actual yield / Theoretical yield) × 100

Example: Actual yield = 1.6g, Theoretical = 2.0g → (1.6/2.0) × 100 = 80%

Percentage Composition by Mass: % of element = (Mass of element in compound / Mᵣ) × 100

Example: Fe in Fe₂O₃ → (112/160) × 100 = 70%

Percentage Purity: % Purity = (Mass of pure substance / Total mass) × 100

Example: Pure mass = 13.5g, Total = 15g → (13.5/15) × 100 = 90%

💡 Percentage yield is always ≤ 100%. Percentage composition of all elements in a compound adds up to 100%.
⚠️ Common Mistakes to Avoid:

• ❌ Forgetting to convert cm³ to dm³ → ✅ Always divide by 1000
• ❌ Using atomic mass instead of molecular mass for compounds → ✅ Use Mᵣ
• ❌ Confusing empirical and molecular formula → ✅ Empirical = simplest ratio; Molecular = actual atoms
• ❌ Forgetting to balance equations first → ✅ Always balance before reacting mass calculations
• ❌ Mixing up actual and theoretical yield → ✅ Actual = what you get; Theoretical = what you should get