Relative masses and mass spectrometryEdexcel A-Level Chemistry: Revision notes

Atoms are far too light to weigh directly, so their masses are compared with a standard: 1/12 of the mass of an atom of carbon-12.

• Relative isotopic mass: the mass of an atom of an isotope relative to 1/12 of the mass of an atom of ¹²C.
• Relative atomic mass, Ar: the weighted mean mass of an atom of an element relative to 1/12 of the mass of an atom of ¹²C, taking account of the relative abundance of its isotopes.

Because Ar is a weighted mean, it is usually not a whole number (for example, chlorine is 35.5), even though each isotope has a whole-number mass number.

Relative molecular mass, Mr, is found by adding the relative atomic masses of all the atoms in the molecular formula. For compounds with giant structures (ionic or giant covalent) the term relative formula mass is used instead, for the formula unit.

Worked example: (NH₄)₂SO₄ (N = 14.0, H = 1.0, S = 32.1, O = 16.0). Mr = (2 × 14.0) + (8 × 1.0) + 32.1 + (4 × 16.0) = 28.0 + 8.0 + 32.1 + 64.0 = 132.1.

Relative masses have no units because they are ratios.

A mass spectrometer measures the m/z of each isotope and its relative abundance. Then:

Ar = Σ(isotopic mass × abundance) ÷ Σ(abundance)

Worked example (Mg): ²⁴Mg 78.6%, ²⁵Mg 10.1%, ²⁶Mg 11.3%. Ar = (24 × 78.6 + 25 × 10.1 + 26 × 11.3) ÷ 100 = 24.3.

Working backwards: boron has Ar = 10.8 and two isotopes, ¹⁰B and ¹¹B. Let x = % of ¹⁰B: 10x + 11(100 − x) = 1080, so 1100 − x = 1080 and x = 20. The sample is 20% ¹⁰B and 80% ¹¹B.

A diatomic molecule such as Cl₂ is ionised to give molecular ions, and each combination of isotopes gives a different m/z.

Chlorine is 75% ³⁵Cl and 25% ³⁷Cl (ratio 3 : 1). Probabilities: ³⁵Cl³⁵Cl = (3/4)² = 9/16 at m/z 70; ³⁵Cl³⁷Cl = 2 × 3/4 × 1/4 = 6/16 at m/z 72; ³⁷Cl³⁷Cl = (1/4)² = 1/16 at m/z 74. The peak heights are in the ratio 9 : 6 : 1.

For bromine (⁷⁹Br : ⁸¹Br ≈ 1 : 1), the peaks at m/z 158, 160 and 162 are in the ratio 1 : 2 : 1. The mixed combination always has the factor of 2, because it can arise in two ways.

In mass spectrometry the sample is ionised, and the ions are detected according to their mass-to-charge ratio (m/z). Most ions have a charge of 1+, so m/z equals the relative mass of the ion.

The molecular ion, M⁺, is formed when a whole molecule loses one electron. The mass of the lost electron is negligible, so the m/z of the M⁺ peak equals the relative molecular mass of the compound. For propanone, M⁺ appears at m/z 58, so Mr = 58.