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1.1.3Physical Chemistry

Relative Mass — Edexcel AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 3 of 42

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Part 1 of 6Four Relative Masses

1.1.3 Relative Mass

All these quantities are ratios, so they have no units.

Relative isotopic mass
  • The mass of an atom of one isotope compared with one twelfth of the mass of a carbon-12 atom.
Relative atomic mass, Ar
  • The weighted mean mass of the atoms of an element, taking account of isotope abundance, compared with one twelfth of carbon-12.
Relative molecular mass, Mr
  • The weighted mean mass of a molecule compared with one twelfth of the mass of a carbon-12 atom.
Relative formula mass
  • Used for ionic or giant compounds, found by adding the relative atomic masses in the formula unit.

Part 2 of 6Why Carbon-12 Is the Standard

Why Carbon-12 Is the Standard
Reference point
  • Carbon-12 is assigned a mass of exactly 12, and one twelfth of its mass is the unit against which all atoms are compared.
Ratio
  • Because it compares masses, a relative mass has no units, unlike molar mass which is given in g mol⁻¹.
Whole-number isotopes
  • Individual isotopes have relative isotopic masses very close to their mass numbers.
Averages
  • Elements with several isotopes give non-integer Ar values, such as 35.5 for chlorine.

Part 3 of 6Calculating Ar from Abundances

The relative atomic mass is a weighted average.

1
List each isotope's mass number and its percentage abundance.
2
Multiply each isotopic mass by the abundance of that isotope.
3
Add all of the products together to give a single total.
4
Divide the sum by 100 if abundances are percentages, or by the total of the ratio if they are given as a ratio.
5
Quote the answer to a sensible number of significant figures without units.

Part 4 of 6Calculating Mr and Formula Mass

Molecules
  • Add the Ar values of every atom in the molecular formula, so Mr of H₂O is 2(1.0) + 16.0 = 18.0.
Brackets
  • Multiply everything inside brackets by the subscript, so Ca(NO₃)₂ contains one Ca, two N and six O.
Hydrated salts
  • Add the water of crystallisation as a separate unit, multiplying the Mr of H₂O by the number of water molecules.

Part 5 of 6Worked Example

Question

  • Naturally occurring chlorine is 75.8% ³⁵Cl and 24.2% ³⁷Cl.
  • Calculate its relative atomic mass, and the relative formula mass of Ca(NO₃)₂ (Ar: Ca = 40.08, N = 14.01, O = 16.00).

  1. Sum of mass × abundance = (35 × 75.8) + (37 × 24.2) = 2653 + 895.4 = 3548.4.
  2. Divide by 100 to get Ar(Cl) = 35.484, which is 35.5 to 3 significant figures.
  3. Ca(NO₃)₂ contains 1 Ca, 2 N and 6 O atoms.
  4. Relative formula mass = 40.08 + (2 × 14.01) + (6 × 16.00) = 40.08 + 28.02 + 96.00 = 164.10.

Answer

  • Ar(Cl) = 35.5;
    • relative formula mass of Ca(NO₃)₂ = 164.1.

Part 6 of 6Exam Checklist & Tips

1.1.3 – Key Points for the Exam

  • Relative atomic mass Weighted mean mass of an element's atoms relative to one twelfth of a carbon-12 atom.
  • Relative molecular mass Weighted mean mass of a molecule relative to one twelfth of a carbon-12 atom.
  • Relative formula mass The sum of relative atomic masses in the formula unit of a compound.
  • Abundance The proportion of an element's atoms that are of a given isotope.

Common Mistakes

  • Giving relative masses units of grams; they are ratios and have no units.
  • Averaging the isotopic masses without weighting them by abundance.
  • Using Mr for an ionic compound, when relative formula mass is the correct term.
  • Forgetting to multiply atoms inside brackets by the subscript outside them.

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