Mass spectrometry in organic analysisEdexcel A-Level Chemistry: Revision notes
Section 1
What a mass spectrum shows
In a mass spectrometer, organic molecules are ionised, usually by bombardment with high-energy electrons, which knock out one electron: M → [M]⁺ + e⁻. The positive ions are accelerated, separated according to their mass-to-charge ratio (m/z) and detected. Almost all ions carry a charge of 1+, so the m/z value equals the mass of the ion.
The spectrum plots the abundance of each ion against m/z. Only positive ions are detected.
Section 2
The molecular ion peak and relative molecular mass
The peak at the highest m/z is the molecular ion peak. Its m/z value is the relative molecular mass, Mr, of the compound.
Example: a molecular ion peak at m/z 58 means Mr = 58. Combined with elemental composition data, this gives the molecular formula (empirical formula mass × n = Mr).
The molecular ion peak is the peak of highest m/z, not the tallest peak. The tallest peak is the base peak and is usually a fragment.
Section 3
Fragmentation
The molecular ion has excess energy and often breaks into pieces: [M]⁺ → X⁺ + Y•. Only the positive fragment X⁺ is detected; the uncharged radical Y• is not.
The fragments formed depend on the structure, so the pattern of peaks acts as a fingerprint. Breaks tend to happen next to a functional group, such as either side of a C=O group or next to the C–OH carbon.
Section 4
Common fragments to recognise
- 15: [CH₃]⁺
- 29: [C₂H₅]⁺ or [CHO]⁺
- 31: [CH₂OH]⁺ (primary alcohols)
- 43: [C₃H₇]⁺ or [CH₃CO]⁺
- 45: [CH₃CHOH]⁺ or [CH₃CH₂O]⁺
- 57: [C₄H₉]⁺ or [CH₃CH₂CO]⁺
Useful differences: M − 15 means loss of CH₃; M − 17 loss of OH; M − 29 loss of C₂H₅ or CHO. Work out each fragment's formula by adding atomic masses (C = 12, H = 1, O = 16).
When a fragment and its partner add up to the molecular ion (e.g. 43 + 15 = 58), they came from the same break in the molecule.
Section 5
Suggesting a structure
- Read Mr from the molecular ion peak and, with any composition data, find the molecular formula.
- Identify the main fragment peaks and write the ion that fits each mass.
- Fit the pieces together to build a structure that would split that way.
- Check that other isomers would give different fragments.
Worked example: a compound with Mr 58 and fragments at 43 and 15. 43 is [CH₃CO]⁺ and 15 is [CH₃]⁺, so it is CH₃COCH₃, propanone. Propanal would give peaks at 29 and 57 instead.
Isomers: propan-1-ol gives a strong peak at 31 ([CH₂OH]⁺); propan-2-ol gives a strong peak at 45 ([CH₃CHOH]⁺). Both have Mr 60.
Isomers have the same molecular ion peak. Use the fragment peaks to tell them apart.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Mass spectrometry in organic analysis
- A chemist analyses a liquid that contains carbon, hydrogen and oxygen only. In its mass spectrum the peak with the highest m/z value is at 58, and there are strong fragment peaks at m/z 43 and m/z 15. There are no strong peaks at m/z 29 or 57.Deduce the identity of the compound, explaining how the peaks at m/z 58, 43 and 15 support your answer.2 marks
- Pentan-2-one and pentan-3-one are structural isomers with molecular formula C₅H₁₀O. A student uses mass spectrometry to tell them apart.Explain how the mass spectra would allow the student to distinguish between the two isomers.2 marks
- A student is given an unknown liquid X. Elemental analysis shows that it contains 64.9% carbon, 13.5% hydrogen and 21.6% oxygen by mass. In its mass spectrum the peak with the highest m/z value is at 74 and there is a strong fragment peak at 59. X is not oxidised by warm acidified potassium dichromate(VI). X reacts with sodium metal to give hydrogen gas.Determine the empirical formula and the molecular formula of X.3 marks
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).