Empirical Formulae & Formulae of Ionic Compounds (Cambridge (CIE) IGCSE Chemistry)

Empirical formulae

Extended tier only

• The empirical formula is the simplest whole number ratio of the atoms of each element present in one molecule or formula unit of the compound 

• The empirical formula of an organic molecule is often different to its molecular / chemical formula

  • For example, ethanoic acid has the chemical formula CH₃COOH or C₂H₄O₂ but its empirical formula is CH₂O

• The molecular / chemical formula of an ionic compound is always its empirical formula 

  • For example, sodium chloride has the chemical formula NaCl, which is also its empirical formula

Deducing formulae of ionic compounds

Extended tier only

• Metals and non-metals react together to form ionic compounds

  • Ionic compounds are not simple molecules 

  • Remember: Simple molecules are formed when non-metals react together to form compounds

• Ionic compounds involve the metal losing electrons and the non-metal gaining electrons to form ions

• Some ions that you will be expected to be able to use, because they are stated in the exam specification, include:

  • Hydrogen ions, H⁺ - sometimes referred to as protons

  • Group 1 ions, e.g. Li⁺, Na⁺, K⁺ 

  • Group 7 ions, F^–, Cl^–, Br^– 

  • Copper(II) ions, Cu²⁺ 

  • Iron(II) ions, Fe²⁺ 

  • Iron(III) ions, Fe³⁺ 

• There are some polyatomic (containing more than one atom) ions stated in the exam specification: 

  • Carbonate ions, CO₃^2– 

  • Sulfate ions, SO₄^2– 

  • Hydroxide ions, OH^– 

  • Nitrate ions, NO₃^–

  • Ammonium ions, NH₄⁺ 

How to determine the formulae of ionic compounds

• Ionic compounds typically have no overall charge

  • This means that the size of any positively charged ion is cancelled by the size of any negatively charged ion

  • Careful: This should not be confused with an atom having no overall charge

Direct comparison

• The formula of an ionic compound can be determined by directly comparing the charges of the ions:

  • For example, iron(II) sulfate

    • The iron(II) ion is Fe²⁺, which means that it has a 2+ or +2 charge

    • The sulfate ion is SO₄^2–, which means that it has a 2– or –2 charge

    • The charges cancel each other out

      Mathematically, (+2) + (–2) = 0

    • This means that one SO₄^2– ion is needed to cancel the +2 charge on Fe²⁺ 

    • Therefore, the formula of iron(II) sulfate is FeSO₄ 

The swap-and-drop method

• When the ions in the ionic compound have different charges, it can be easier to use the swap-and-drop method

  • Careful: If you use this method with ions that have the same charge, then you must give the simplest whole number ratio to get the correct answer

• For example, copper(II) chloride:

  • The copper(II) ion is Cu²⁺, which means that it has a 2+ or +2 charge

  • The chloride ion is Cl^–, which means that it has a 1– or –1 charge

  • The size of the charge on the copper(II) ion indicates the number of chloride ions needed, and the size of the charge on the chloride ion indicates the number of copper(II) ions needed

Determining the formula of copper(II) chloride

The charges swap from element to element and drop down. The positive and negative signs are removed and there is no need for the number 1.

• This gives the overall formula of copper(II) chloride as CuCl₂