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1.1.4Physical Chemistry

Mass Spectrometry — Edexcel AS Chemistry Revision Notes & Topical Questions

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Section 1 of 5Stages of the Instrument

1.1.4 Mass Spectrometry

The instrument measures the mass-to-charge ratio and relative abundance of the ions in a sample.

1
Vaporisation: the sample is turned into a gas inside a vacuum so that the ions can travel without colliding with air molecules.
2
Ionisation: in electron impact ionisation, high-energy electrons knock an electron off each atom or molecule, forming positive ions, for example X(g) → X⁺(g) + e⁻.
3
Acceleration: the positive ions are accelerated by an electric field so they travel with a controlled energy.
4
Separation: ions are deflected by a magnetic field or separated by their flight time, with the path depending on the mass-to-charge ratio, m/z.
5
Detection: ions reaching the detector produce a current that is recorded, with larger signals for more abundant ions.

Section 2 of 5Reading a Mass Spectrum

Reading a Mass Spectrum
Horizontal axis
  • The x-axis shows m/z, which equals the mass number for ions carrying a single positive charge.
Vertical axis
  • The y-axis shows relative abundance or intensity, so taller peaks mean more ions of that mass.
Elements
  • Each peak represents one isotope, so a spectrum of an element gives its isotopes and their proportions.
Molecules
  • The peak with the highest m/z is usually the molecular ion M⁺, whose m/z equals the relative molecular mass.

Section 3 of 5Using Spectra for Calculations

Ar from an element
  • Multiply each m/z by the relative abundance of its peak, add them, then divide by the total abundance.
Molecular mass
  • The m/z value of the molecular ion peak gives Mr of the compound directly.
Fragments
  • Molecules can break into smaller positive ions, giving extra peaks at lower m/z that help to identify structure.
Diatomic elements
  • Chlorine spectra show several molecular ion peaks, because Cl₂⁺ ions can contain different combinations of ³⁵Cl and ³⁷Cl.

Section 4 of 5Worked Example

Question

  • A sample of magnesium gives peaks at m/z 24 (78.6%), 25 (10.1%) and 26 (11.3%).
  • Calculate the relative atomic mass.

  1. The abundances total 78.6 + 10.1 + 11.3 = 100.0.
  2. Multiply each m/z by its abundance: 24 × 78.6 = 1886.4; 25 × 10.1 = 252.5; 26 × 11.3 = 293.8.
  3. Add1886.4 + 252.5 + 293.8 = 2432.7.
  4. Divide by 100 to get 24.327, which is 24.3 to 3 significant figures.

Answer Ar(Mg) = 24.3.

Section 5 of 5Exam Checklist & Tips

1.1.4 – Key Points for the Exam

  • m/z The mass-to-charge ratio of an ion, equal to its mass number for singly charged ions.
  • Molecular ion The ion formed when a molecule loses one electron, whose m/z equals Mr.
  • Electron impact Ionisation by bombarding gaseous molecules with high-energy electrons.
  • Relative abundance The proportion of ions at each m/z value shown by peak height.

Common Mistakes

  • Writing that the sample is negatively charged; positive ions are the species detected.
  • Forgetting that the instrument is kept under vacuum to stop collisions with air.
  • Assuming the tallest peak must be the molecular ion, when the molecular ion is the one at highest m/z.
  • Calculating Ar without dividing by the total abundance of all the peaks.

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