Mass spectrometryEdexcel International A Level Chemistry: Revision notes
Section 1
How a mass spectrometer works
A mass spectrometer separates ions by their mass-to-charge ratio, m/z. The stages are:
- Vaporisation: the sample is turned into a gas.
- Ionisation: high-energy electrons knock an electron from each atom or molecule, forming positive ions, e.g. X(g) + e⁻ → X⁺(g) + 2e⁻.
- Acceleration: an electric field accelerates the positive ions.
- Deflection: a magnetic field deflects the ions; lighter ions (lower m/z) are deflected more.
- Detection: the detector records the ions, and the signal size is proportional to the number of ions.
The whole instrument is kept under a vacuum so that ions do not collide with air molecules.
Learn the stages in order: vaporise, ionise, accelerate, deflect, detect. Mention the vacuum if the question asks why.
Section 2
Reading a mass spectrum
A mass spectrum plots relative abundance (peak height) against m/z. For an element, each peak is one isotope. Most ions have a charge of 1+, so m/z equals the mass number of the isotope.
A taller peak means a greater proportion of that isotope in the sample.
Section 3
Calculating relative atomic mass
Relative atomic mass is the weighted mean mass of the atoms of an element, compared with 1/12 of the mass of a carbon-12 atom.
Ar = Σ (isotope mass × abundance) ÷ Σ abundances
Worked example: a sample has peaks at 63 (69.2 %) and 65 (30.8 %). Ar = (63 × 69.2 + 65 × 30.8) ÷ 100 = 63.6.
Working backwards: boron, Ar = 10.8, with ¹⁰B and ¹¹B. Let the abundance of ¹⁰B be x %. 10x + 11(100 − x) = 1080, so 1100 − x = 1080 and x = 20 %. The sample is 20 % ¹⁰B and 80 % ¹¹B.
Do not use the simple mean of the isotope masses. Weight each mass by its abundance.
Section 4
Relative molecular mass and the molecular ion
When a molecule loses one electron it forms the molecular ion, M⁺. The peak with the highest m/z (ignoring small isotope peaks) gives the relative molecular mass, Mr.
For example, a molecular ion at m/z 72 shows Mr = 72, which fits pentane, C₅H₁₂, but not butane (58) or hexane (86). Comparing the m/z with the Mr of possible molecules identifies the compound.
Section 5
Ions with a 2+ charge
Sometimes two electrons are removed, forming an ion with a charge of 2+. Since m/z = mass ÷ charge, a ²⁴Mg²⁺ ion appears at m/z 24 ÷ 2 = 12. These peaks are much smaller than the 1+ peaks because forming a 2+ ion needs more energy, so fewer are formed.
Section 6
Predicting the spectrum of a diatomic molecule
A molecule such as Cl₂ contains two atoms, each of which may be either isotope. With ³⁵Cl (75 %) and ³⁷Cl (25 %):
- ³⁵Cl³⁵Cl⁺ (m/z 70): 0.75 × 0.75 = 0.5625
- ³⁵Cl³⁷Cl⁺ (m/z 72): 2 × 0.75 × 0.25 = 0.375
- ³⁷Cl³⁷Cl⁺ (m/z 74): 0.25 × 0.25 = 0.0625
The relative heights are 9 : 6 : 1. The middle peak includes a factor of 2 because the isotopes can be combined in two ways.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Mass spectrometry
- A forensic laboratory analyses a metal sample using a mass spectrometer. The sample gives two peaks: one at m/z 63 with a relative abundance of 69.2 % and one at m/z 65 with a relative abundance of 30.8 %. All the ions have a charge of 1+.Explain why the inside of a mass spectrometer is kept under a high vacuum.2 marks
- Boron has two isotopes, ¹⁰B and ¹¹B. A mass spectrum of a boron sample gives a relative atomic mass of 10.8.A second sample of boron contains a greater proportion of ¹⁰B than the first. State and explain how its relative atomic mass compares with 10.8.2 marks
- Chlorine consists of two isotopes, ³⁵Cl (75.0 %) and ³⁷Cl (25.0 %). A sample of chlorine gas, Cl₂, is analysed in a mass spectrometer, and molecular ions, Cl₂⁺, are detected.Identify the molecular ions responsible for the peaks at m/z 70, 72 and 74.3 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).