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1.2.1Physical Chemistry

Relative Atomic Mass and Relative Molecular Mass

5 min read 5 parts Lesson 7 of 63

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Part 1 of 5Comparing Masses

1.2.1 Relative Atomic Mass and Relative Molecular Mass

Relative masses are ratios, so they have no units.

Standard
  • Atoms are far too light to weigh singly, so their masses are compared with one-twelfth of a carbon-12 atom.
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 an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.
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
  • The same idea for ionic or giant compounds, using the formula unit; it is also given the symbol Mr.

Part 2 of 5Why Ar Is Not a Whole Number

Why Ar Is Not a Whole Number
Isotopes
  • Atoms of one element with the same number of protons but different numbers of neutrons, giving different masses.
Abundance
  • Each isotope occurs in a fixed proportion in natural samples, found using mass spectrometry.
Weighted mean
  • Ar is the sum of each isotopic mass multiplied by its percentage, divided by 100.
Chlorine
  • Chlorine has isotopes of mass 35 and 37 in roughly a 3 : 1 ratio, so Ar is about 35.5.

Part 3 of 5Calculating Mr

1
Write down the correct formula for the substance.
2
Count the atoms of each element, remembering that a number outside brackets multiplies everything inside them.
3
Multiply the Ar of each element by the number of atoms of that element present.
4
Add the contributions together to obtain Mr, with no units.
5
Note that molar mass is the same number but in g mol⁻¹, as it is a mass per mole.

Part 4 of 5Worked Example

Question

  • Magnesium consists of 78.99% ²⁴Mg, 10.00% ²⁵Mg and 11.01% ²⁶Mg.
  • Calculate its relative atomic mass, and find the Mr of (NH₄)₂SO₄ (Ar: N 14.0, H 1.0, S 32.1, O 16.0).

  1. Ar(Mg) = (24 × 78.99 + 25 × 10.00 + 26 × 11.01) / 100.
  2. = (1895.76 + 250.00 + 286.26) / 100 = 24.32.
  3. (NH₄)₂SO₄ contains 2 N, 8 H, 1 S and 4 O.
  4. Mr = 2(14.0) + 8(1.0) + 32.1 + 4(16.0) = 28.0 + 8.0 + 32.1 + 64.0 = 132.1.

Answer

  • Ar(Mg) = 24.3 (to 3 s.f.);
    • Mr of (NH₄)₂SO₄ = 132.1.

Part 5 of 5Exam Checklist & Tips

1.2.1 – Key Points for the Exam

  • Isotopes Atoms of the same element with the same proton number but different mass numbers.
  • Relative atomic mass The weighted mean mass of an atom of an element relative to one-twelfth of a carbon-12 atom.
  • Relative molecular mass The weighted mean mass of a molecule relative to one-twelfth of a carbon-12 atom.
  • Relative formula mass The sum of the Ar values of all atoms in the formula unit of a compound.

Common Mistakes

  • Giving Ar or Mr units; they are ratios and have none.
  • Forgetting that a subscript outside brackets, as in Ca(OH)₂, multiplies all atoms inside.
  • Using the mass number of one isotope instead of the weighted mean as Ar.
  • Using Mr for ionic compounds as if they existed as molecules; use relative formula mass.
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