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FormulaeCambridge IGCSE Chemistry: Revision notes

Section 1

What are molecular and empirical formulae?

Molecular formula shows the actual number and type of atoms in one molecule of a compound. Empirical formula shows the simplest whole number ratio of different atoms or ions in a compound.

For example:

  • Hydrogen peroxide has a molecular formula of H₂O₂ but an empirical formula of HO
  • Glucose has a molecular formula of C₆H₁₂O₆ but an empirical formula of CH₂O

The molecular formula is always a multiple of the empirical formula. To find the molecular formula, you need to know the relative molecular mass (Mr) of the compound.

Key termsmolecular formulaempirical formularelative molecular mass
Common mistake

Students often confuse molecular and empirical formulae. Remember: empirical is always the simplest ratio, while molecular is the actual composition. The molecular formula is always a whole number multiple of the empirical formula.

Example

If a compound has an empirical formula of CH and a relative molecular mass of 78, find the molecular formula. First, find the Mr of CH: 12 + 1 = 13. Then divide: 78 ÷ 13 = 6. So the molecular formula is C₆H₆ (benzene).

Section 2

How do you deduce formulae from atoms or ions?

When given the relative numbers of atoms or ions in a model or diagram:

  1. Count the atoms or ions of each element present
  2. Write down the ratio of atoms/ions (e.g. 2:1:3)
  3. Simplify the ratio to the smallest whole numbers by dividing by the highest common factor
  4. Write the formula using the simplified ratio as subscripts

For ionic compounds, deduce the formula from the charges on the ions:

  • The positive charges must equal the negative charges
  • Work out how many ions of each type are needed to balance the charges
  • For example: magnesium oxide has Mg²⁺ and O²⁻ ions, so one of each is needed → MgO
  • Calcium chloride has Ca²⁺ and Cl⁻ ions, so two Cl⁻ ions are needed for one Ca²⁺ → CaCl₂
Key termsionic formulaion chargesvalency
Exam tip

When deducing ionic formulae, always check that the total positive charge equals the total negative charge. Examiners often check this working in mark schemes.

Example

A model shows 4 sodium atoms, 2 sulphur atoms, and 8 oxygen atoms. The ratio is 4:2:8. Divide by 2 to get 2:1:4. The formula is Na₂SO₄ (sodium sulphate).

Section 3

How do you write word equations and symbol equations?

Word equations show the names of reactants and products separated by an arrow:

Reactant + Reactant → Product + Product

Symbol equations show the chemical formulae of reactants and products:

Chemical formula + Chemical formula → Chemical formula + Chemical formula

Symbol equations must be balanced so that the number of atoms of each element is the same on both sides.

State symbols must be included in symbol equations:

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

Steps to write a balanced symbol equation:

  1. Write the unbalanced equation using correct formulae
  2. Count atoms of each element on both sides
  3. Add numbers (coefficients) in front of formulae to balance
  4. Check all elements are balanced
  5. Add state symbols

Example: Magnesium + oxygen → magnesium oxide Unbalanced: Mg + O₂ → MgO Balanced: 2Mg + O₂ → 2MgO

Key termsword equationsymbol equationstate symbolsbalanced equation
Common mistake

Do not change the formulae of compounds to balance an equation—only add numbers in front. For example, H₂O cannot become HO or H₃O; you must adjust coefficients instead.

Example

Copper reacts with oxygen to form copper oxide. Word equation: Copper + oxygen → copper oxide. Symbol equation (unbalanced): Cu + O₂ → CuO. Balanced: 2Cu + O₂ → 2CuO. With state symbols: 2Cu(s) + O₂(g) → 2CuO(s)

Section 4

What are ionic equations and when are they used?

Ionic equations show only the ions and atoms that actually participate in the chemical reaction, omitting spectator ions (ions that do not change during the reaction).

Steps to write an ionic equation:

  1. Write the full balanced symbol equation with state symbols
  2. Write the full ionic equation by showing all ionic compounds in their ionised form (only for compounds that are soluble and ionic)
  3. Identify spectator ions (ions that appear unchanged on both sides)
  4. Cancel spectator ions from both sides
  5. Write the net ionic equation showing only the particles that react

Example: Full equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Full ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l) Net ionic: H⁺(aq) + OH⁻(aq) → H₂O(l)

Na⁺ and Cl⁻ are spectator ions as they don't change.

Key termsionic equationspectator ionsnet ionic equationfull ionic equation
Exam tip

Examiners expect you to show your working: write the full equation first, then the full ionic equation, then cancel spectator ions clearly before writing the net ionic equation.

Section 5

How do you deduce symbol equations from given information?

You can work out a symbol equation if given:

  • The names of reactants and products
  • The state symbols
  • Information about the reaction (e.g. type of reaction, or partial equations)
  • Data such as masses, volumes, or molar amounts

Key steps:

  1. Identify the reactants (starting materials) and products (substances formed)
  2. Write the formulae of all substances (use your knowledge of common ions and compounds)
  3. Write the unbalanced equation
  4. Balance the equation by counting atoms of each element
  5. Add state symbols based on the reaction conditions

Common reaction types and products:

  • Combustion: fuel + oxygen → carbon dioxide + water (if contains C and H)
  • Oxidation: element + oxygen → oxide
  • Acid-base: acid + base → salt + water
  • Displacement: more reactive element + compound → new compound + less reactive element
  • Decomposition: compound → simpler substances

Example: Deduce the equation for the reaction between sodium carbonate solution and dilute hydrochloric acid: Reactants: Na₂CO₃(aq) and HCl(aq) Products: NaCl, CO₂, H₂O Unbalanced: Na₂CO₃(aq) + HCl(aq) → NaCl(aq) + CO₂(g) + H₂O(l) Balanced: Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)

Key termsreactantsproductscombustiondecompositiondisplacement
Example

Deduce the equation for burning magnesium in oxygen. Reactants: Mg and O₂. Product: MgO. Unbalanced: Mg + O₂ → MgO. Balanced: 2Mg + O₂ → 2MgO. With state symbols: 2Mg(s) + O₂(g) → 2MgO(s)

Must Know

  • Molecular formula shows the actual number of atoms in one molecule; empirical formula is the simplest whole number ratio
  • Deduce formulae from atom/ion numbers by counting, writing the ratio, simplifying to whole numbers, and writing the formula
  • For ionic compounds, balance positive and negative charges: Mg²⁺ + O²⁻ → MgO; Ca²⁺ + 2Cl⁻ → CaCl₂
  • Symbol equations must be balanced by adjusting coefficients (not formulae) and must include state symbols: (s), (l), (g), (aq)
  • Ionic equations show only reacting particles; identify and cancel spectator ions to write the net ionic equation
  • Deduce symbol equations by identifying reactants and products, writing correct formulae, balancing atoms, and adding state symbols
Key termsmolecular formulaempirical formulasymbol equationionic equationstate symbolsbalanced equation

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