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Mass spectrometry in organic analysisAQA A-Level Chemistry: Revision notes

Section 1

The molecular ion peak

In a mass spectrometer, a sample is ionised and the positive ions are separated according to their mass-to-charge ratio, m/z. For a molecule M, the ion M⁺ formed by losing one electron is the molecular ion.

The molecular ion gives the peak with the highest m/z in the spectrum (ignoring very small peaks due to heavier isotopes such as carbon-13). Its m/z value equals the relative molecular mass, Mᵣ, of the compound. Peaks at lower m/z are fragment ions.

Key termsm/zmolecular ionrelative molecular mass
Common mistake

The molecular ion peak is the one with the highest m/z, not necessarily the tallest peak. The tallest peak is called the base peak.

Section 2

Low and high resolution

A low-resolution spectrometer gives m/z to the nearest whole number. Different compounds can have the same whole-number mass. For example, C₄H₁₀O and C₃H₆O₂ both have Mᵣ = 74.

A high-resolution (precise) spectrometer measures the mass to several decimal places. Because the precise masses of these formulae differ, the molecular formula can be found.

Key termslow resolutionhigh resolution

Section 3

Why precise masses differ

Only a carbon-12 atom has a mass that is exactly a whole number (12.0000). The precise masses of the other atoms are not whole numbers:

  • H = 1.0078
  • N = 14.0031
  • O = 15.9949

So different combinations of atoms with the same whole-number mass add up to slightly different precise masses. For example, CO = 27.9949, N₂ = 28.0062 and C₂H₄ = 28.0312.

Key termsprecise atomic mass

Section 4

Finding a molecular formula

  1. Use the whole-number mass to list possible formulae.
  2. Calculate the precise Mᵣ for each, using the precise atomic masses given.
  3. The formula whose precise mass matches the measured value is the molecular formula.

Worked example: a compound of C, H and O has precise Mᵣ 88.0524.

  • C₄H₈O₂ = 4 × 12.0000 + 8 × 1.0078 + 2 × 15.9949 = 88.0522
  • C₅H₁₂O = 5 × 12.0000 + 12 × 1.0078 + 15.9949 = 88.0885

The measured value matches C₄H₈O₂.

Key termsmolecular formula
Exam tip

Show every calculation in full with the numbers substituted, and give the answer to the same number of decimal places as the data.

Section 5

Limitations

A precise mass gives the molecular formula, not the structure. Isomers have the same molecular formula, so they have the same precise Mᵣ. For example, butanal and butanone are both C₄H₈O.

To identify the compound, combine mass spectrometry with other evidence, such as infrared spectra or test-tube reactions.

Key termsisomers

Must Know

  • The molecular ion peak has the highest m/z and gives Mᵣ
  • Low resolution gives whole-number masses; high resolution gives precise masses
  • Calculate precise Mᵣ for each candidate formula and match to the measured value
  • Isomers cannot be separated by precise mass

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Mass spectrometry in organic analysis

  1. A pure organic compound is analysed by mass spectrometry. In the mass spectrum, the peak with the highest mass-to-charge ratio (ignoring any very small peaks above it) is at m/z = 74.
    The compound is known to contain only carbon, hydrogen and oxygen. Explain why the m/z value of 74 alone cannot be used to find its molecular formula.2 marks
  2. High-resolution mass spectrometry measures relative molecular masses to four decimal places. Precise relative atomic masses: H = 1.0078, C = 12.0000, N = 14.0031, O = 15.9949.
    A compound has a precise relative molecular mass of 46.0054. Calculate the precise relative molecular masses of C₂H₆O and CH₂O₂ and decide which is the molecular formula of the compound.2 marks
  3. A pharmaceutical analyst finds that a compound containing only carbon, hydrogen and oxygen has a precise relative molecular mass of 88.0524 by high-resolution mass spectrometry. Precise relative atomic masses: H = 1.0078, C = 12.0000, O = 15.9949.
    The compound has a nominal mass of 88. Calculate the precise relative molecular masses of C₄H₈O₂ and C₅H₁₂O and deduce the molecular formula of the compound.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).