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2.1.2Elements, compounds and mixtures

Relative Masses — OCR Gateway GCSE Combined Science A J250 - Chemistry Revision Notes & Topical Questions

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📚 OCR Gateway GCSE Combined Science A J250 - Chemistry Hub / Topical Revision
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Section 1 of 7Reading chemical formulae

  1. PART 1 OF 6

    A chemical formula records which elements are present and how many atoms of each are represented. A subscript belongs to the symbol immediately before it. If there is no subscript, the number is one. A subscript after a bracket multiplies every atom inside that bracket.

    Counting atoms correctly
    FormulaAtom countReason
    H2SO4H: 2 · S: 1 · O: 4The unnumbered S represents one atom.
    Ca(NO3)2Ca: 1 · N: 2 · O: 6The outside 2 multiplies N and O inside the bracket.
    (NH4)2SO4N: 2 · H: 8 · S: 1 · O: 4Only NH4 is multiplied by the bracket subscript.

    A coefficient before a formula multiplies the whole formula. Thus 3H2O represents three water molecules containing six hydrogen atoms and three oxygen atoms in total. It does not change the composition of each water molecule.

Section 2 of 7Building ionic formulae

  1. PART 2 OF 6

    An ionic compound is electrically neutral. Its formula gives the smallest whole-number ratio of positive and negative ions that makes the total charge zero. The formula is a ratio in a lattice, rather than a separate molecule.

    Verify the formula by checking charge
    CompoundPositive totalNegative totalFormula
    Magnesium chloride1 × (+2) = +22 × (−1) = −2MgCl2
    Aluminium oxide2 × (+3) = +63 × (−2) = −6Al2O3
    Calcium hydroxide1 × (+2) = +22 × (−1) = −2Ca(OH)2

    Use brackets when more than one polyatomic ion is needed. Ca(OH)2 contains two complete hydroxide ions; CaOH2 would represent a different atom count and is not the correct formula. Reduce ratios when appropriate: magnesium oxide is MgO, not Mg2O2.

Section 3 of 7Molecular and empirical formulae

  1. PART 3 OF 6

    A molecular formula gives the actual numbers of atoms in one molecule. An empirical formula gives their simplest whole-number ratio. To obtain the empirical formula from a molecular formula, divide all subscripts by their highest common factor. A missing subscript counts as one.

    Actual numbers compared with simplest ratios
    SubstanceMolecular formulaEmpirical formulaCommon factor
    EthaneC2H6CH32
    Hydrogen peroxideH2O2HO2
    WaterH2OH2O1

    Do not simplify only one element

    • every subscript must be divided by the same factor.
    • Different substances can share an empirical formula, so that ratio alone does not identify the substance.
    • Ionic lattices such as sodium chloride are described by formula units, not molecular formulae.

Section 4 of 7Relative atomic mass and isotopes

  1. PART 4 OF 6

    Relative atomic mass, Ar, compares the average mass of atoms of an element with one twelfth of the mass of a carbon-12 atom. The average is weighted by the relative abundance of each isotope. Because it is a ratio, Ar has no unit.

    Worked example: a simplified chlorine sample
    IsotopePercentage abundanceWeighted contribution
    35Cl75.0%35 × 75.0 = 2625
    37Cl25.0%37 × 25.0 = 925
    Total100.0%2625 + 925 = 3550

    For this sample, Ar = 3550 ÷ 100 = 35.5. This is not a claim that an individual atom has a mass number of 35.5: each atom belongs to an isotope, while the value describes the weighted average. If abundances are given as fractions adding to one, multiply by those fractions and do not divide by 100 again.

Section 5 of 7Relative molecular and formula mass

  1. PART 5 OF 6

    Relative molecular mass, Mr, is the sum of Ar values for every atom in a molecule. For an ionic compound, the corresponding sum is called relative formula mass. Both are relative quantities without units. Molar mass is a different quantity and is expressed in g mol−1.

    Calculation using Ar: Mg = 24, N = 14, O = 16
    ElementAtom countArContribution
    Mg1241 × 24 = 24
    N2142 × 14 = 28
    O6166 × 16 = 96
    TotalMr = 148

    Write the atom count before doing the arithmetic; this separates a formula-reading error from a calculation error.

    • For ammonium sulfate, (NH4)2SO4, using N = 14, H = 1, S = 32 and O = 16 gives 28 + 8 + 32 + 64 = 132.
    • Neither 148 nor 132 should be followed by g when the question asks for a relative mass.

Section 6 of 7Balancing equations and removing spectator ions

  1. PART 6 OF 6

    A balanced equation has the same number of atoms of each element on both sides. Start with correct chemical formulae, then adjust coefficients. Changing a subscript changes the identity of a substance and is not a way to balance an equation.

    Atom check for 2Mg(s) + O2(g) → 2MgO(s)
    ElementReactantsProducts
    Mg22
    O22

    State symbols specify solid (s), liquid (l), gas (g), or dissolved in water (aq). In an ionic equation, unchanged spectator ions are omitted. Split aqueous ionic compounds into ions, retain solids, liquids and gases, cancel unchanged ions, then check both atoms and total charge.

    For a half-equation, include electrons to balance charge as well as atoms. Reduction of copper(II) ions is Cu2+ + 2e− → Cu. Oxidation of chloride ions is 2Cl− → Cl2 + 2e−.

Section 7 of 7Keep going

  1. Formulae & Relative Masses IGCSE Chemistry 0620: what to do next

    These Formulae & Relative Masses IGCSE Chemistry 0620 notes covered the key ideas, worked examples and exam technique for this topic. Check your Formulae & Relative Masses IGCSE Chemistry 0620 understanding with the practice questions below, then reread any Formulae & Relative Masses IGCSE Chemistry 0620 point you were unsure about before moving on.

    Formulae & Relative Masses IGCSE Chemistry 0620: revision checklist

    • Revise the main Formulae & Relative Masses IGCSE Chemistry 0620 definitions until you can write them from memory.
    • Practise Formulae & Relative Masses IGCSE Chemistry 0620 exam questions under timed conditions.

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