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2.1.3Elements, compounds and mixtures

Empirical Formula — 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 6Empirical and molecular formulae

Empirical formula
  • Empirical formula The simplest whole-number ratio of atoms of each element in a compound.
  • Ionic compounds are always written as their empirical formula, for example NaCl or MgCl2.
Molecular formula
  • Molecular formula The actual number of atoms of each element in one molecule.
    • Hexane is C6H14, and ethanoic acid is C2H4O2.
  • Sometimes the two formulae are the same, as in CO or H2O.
Going from molecular to empirical
  • Going from molecular to empirical Divide all the atom numbers by their highest common factor.
    • C6H14 divides by 2 to give C3H7.
    • C2H4O2 divides by 2 to give CH2O.
  • For a diagram, first count the atoms of each element, then simplify. Butane, with 4 carbons and 10 hydrogens, has the empirical formula C2H5.

Section 2 of 6Empirical formula from masses or percentages

1
Write down the mass (or percentage) of each element. If percentages do not add to 100, the missing element is found by subtraction.
2
Divide each mass by the relative atomic mass Ar of that element to get the number of moles.
3
Divide every answer by the smallest number of moles to find the simplest ratio.
4
If a ratio is not a whole number, such as 1.5, multiply all of the numbers by 2 (or 3) to get whole numbers.
5
Write the symbols with these whole numbers as subscripts.

Section 3 of 6Worked example

StepHydrogenOxygen
Mass in sample10 g80 g
Moles = mass ÷ Ar (Ar H = 1, O = 16)10 ÷ 1 = 1080 ÷ 16 = 5
Divide by the smallest (5)10 ÷ 5 = 25 ÷ 5 = 1
Empirical formulaH2O

Section 4 of 6Percentage composition

Equation
  • Equation % mass of an element = (Ar × number of atoms of that element) ÷ Mr of the compound × 100.
Worked example
  • Worked example Calcium carbonate, CaCO3: Mr = 40 + 12 + (3 × 16) = 100.
    • Calcium is 40 ÷ 100 × 100 = 40% of the mass.
  • Check by working out the carbon (12%) and oxygen (48%) too: the total should be 100%.
Ionic lattices
  • Ionic lattices For a lattice diagram, identify the ions, balance charges so the total is zero, and the result is the empirical formula.
    • Na+ and Cl− give NaCl; Mg2+ and Cl− give MgCl2.

Section 5 of 6Key terms

TermMeaning
Empirical formulaThe simplest whole-number ratio of atoms of each element in a compound.
Molecular formulaThe actual number of atoms of each element in one molecule.
Relative atomic mass (Ar)The mean mass of an atom of an element compared with one-twelfth of a carbon-12 atom.
MoleThe amount of substance containing 6.02 × 1023 particles.
Formula unitThe simplest ratio of ions in an ionic compound.

Section 6 of 6Exam technique

1
Dividing masses by each other instead of dividing by Ar first — the ratio must be in moles.
2
Rounding 1.5 down to 1 or up to 2; multiply the whole ratio to reach whole numbers instead.
3
Forgetting to multiply by 100 when calculating percentage by mass.
4
Treating the empirical formula of a molecular compound as its actual formula, such as writing ethanoic acid as CH2O.

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