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Mass spectrometry is one of the most powerful analytical tools in chemistry. It allows us to determine the relative masses and abundances of isotopes, calculate relative atomic masses, and identify unknown compounds from their fragmentation patterns. For Edexcel A-Level, you need to understand how the instrument works and how to interpret the data it produces.
A mass spectrometer separates particles based on their mass-to-charge ratio (m/z). There are four key stages:
flowchart LR
A[Sample] --> B[1. Ionisation]
B --> C[2. Acceleration]
C --> D[3. Deflection]
D --> E[4. Detection]
B -- "Electron gun knocks off electrons" --> B
C -- "Electric field accelerates ions" --> C
D -- "Magnetic field deflects ions" --> D
E -- "Ion current measured" --> E
The sample is vaporised (if not already gaseous) and then ionised. The most common method is electron impact ionisation: a beam of high-energy electrons knocks out an electron from each atom or molecule, forming positive ions.
X(g) -> X+(g) + e-
The ions must be positive so they can be accelerated by an electric field.
Why remove only one electron? The instrument is designed so that most atoms lose only one electron (forming +1 ions). Occasionally, atoms may lose two electrons (forming +2 ions), which appear at m/z = mass/2 on the spectrum. These peaks are usually small but can cause confusion.
The positive ions are accelerated by a strong electric field (a potential difference between charged plates). All ions of the same charge gain the same kinetic energy, but lighter ions achieve higher velocities:
KE = (1/2)mv-squared = qV
Where q is the charge and V is the potential difference.
Rearranging for velocity: v = sqrt(2qV / m)
This shows that for the same KE and charge, lighter ions travel faster.
The accelerated ions pass through a magnetic field. The magnetic field exerts a force perpendicular to the ion's direction of travel, causing the ion to follow a curved path.
The radius of curvature depends on the mass-to-charge ratio (m/z):
By varying the magnetic field strength, ions of different m/z values can be brought to the detector one at a time.
Ions of a specific m/z reach the detector, where they generate an electrical signal. The size of the signal is proportional to the number of ions (abundance). The mass spectrum is then plotted as relative abundance against m/z.
The mass spectrometer operates under high vacuum (very low pressure). This is essential because:
A mass spectrum is a bar chart with:
Chlorine has two isotopes: 35-Cl (75.8%) and 37-Cl (24.2%).
The mass spectrum of atomic chlorine shows two peaks:
When Cl2 is analysed, the spectrum is more complex because different isotopic combinations are possible:
| Species | m/z | Isotopes | Expected Abundance |
|---|---|---|---|
| 35-Cl+ | 35 | Fragment | High |
| 37-Cl+ | 37 | Fragment | Medium |
| 35-Cl-35-Cl+ | 70 | Molecular ion | High |
| 35-Cl-37-Cl+ | 72 | Molecular ion | Medium |
| 37-Cl-37-Cl+ | 74 | Molecular ion | Low |
The relative abundances of the molecular ion peaks can be predicted using probability: if P(35-Cl) = 0.758 and P(37-Cl) = 0.242, then:
The relative atomic mass is the weighted mean of the isotopic masses:
Ar = Sum(abundance x mass) / Sum(abundance)
Copper has two isotopes: 63-Cu (69.2%) and 65-Cu (30.8%).
Ar = (69.2 x 63 + 30.8 x 65) / 100 Ar = (4359.6 + 2002.0) / 100 Ar = 6361.6 / 100 Ar = 63.6
A mass spectrum shows the following peaks for germanium:
| m/z | Relative Abundance |
|---|---|
| 70 | 20.5 |
| 72 | 27.4 |
| 73 | 7.7 |
| 74 | 36.5 |
| 76 | 7.9 |
Total abundance = 20.5 + 27.4 + 7.7 + 36.5 + 7.9 = 100.0
Ar = (20.5 x 70 + 27.4 x 72 + 7.7 x 73 + 36.5 x 74 + 7.9 x 76) / 100.0 Ar = (1435.0 + 1972.8 + 562.1 + 2701.0 + 600.4) / 100.0 Ar = 7271.3 / 100.0 Ar = 72.7
Sometimes you are given the Ar and asked to find the percentage abundance of each isotope. Bromine has two isotopes, 79-Br and 81-Br, with Ar = 79.9.
Let x = % abundance of 79-Br. Then (100 - x) = % abundance of 81-Br.
79.9 = (79x + 81(100 - x)) / 100 7990 = 79x + 8100 - 81x 7990 = 8100 - 2x 2x = 110 x = 55
Therefore: 79-Br = 55.0% and 81-Br = 45.0%.
Standard mass spectrometry measures m/z to the nearest whole number. High-resolution mass spectrometry can distinguish between species with the same nominal mass but different exact masses.
For example:
But their exact masses differ:
High-resolution mass spec can separate these, allowing identification of molecular formulae.
The exact masses of atoms are not exact integers because of mass defect -- the small amount of mass converted to binding energy that holds the nucleus together (E = mc-squared). Only carbon-12 has an exact integer mass by definition. All other isotopes have masses that deviate slightly from whole numbers.
| Isotope | Exact Mass (u) |
|---|---|
| H-1 | 1.00783 |
| C-12 | 12.00000 (by definition) |
| N-14 | 14.00307 |
| O-16 | 15.99491 |
| Cl-35 | 34.96885 |
When a molecule is ionised by electron impact, it forms a molecular ion (M+). The m/z of the molecular ion gives the relative molecular mass of the compound.
However, the molecular ion often breaks apart (fragmentation) into smaller charged and neutral fragments. The fragmentation pattern provides structural information:
In organic mass spectrometry, a small peak at M+1 often appears due to the natural abundance of carbon-13 (1.1% of all carbon). For a molecule with n carbon atoms, the M+1 peak is approximately n x 1.1% of the M+ peak height. This can be used to estimate the number of carbon atoms.
Misconception: "The mass spectrometer separates isotopes by mass."
Correction: The mass spectrometer separates ions by mass-to-charge ratio (m/z), not by mass alone. An ion with mass 40 and charge +2 would appear at the same m/z position as an ion with mass 20 and charge +1 (both at m/z = 20). For singly charged ions, m/z equals the mass, but this assumption breaks down when multiply charged ions are present.
| Quantity | Formula | When to Use |
|---|---|---|
| Relative atomic mass | Ar = Sum(mass x abundance) / Sum(abundance) | Calculating Ar from isotopic data |
| Kinetic energy of ions | KE = (1/2)mv-squared = qV | Understanding acceleration stage |
| Velocity of ions | v = sqrt(2qV / m) | Why lighter ions travel faster |
| Isotope abundance (2 isotopes) | x = (Ar - m2) / (m1 - m2) x 100 | Reverse calculation from Ar |
Always verify:
Edexcel 9CH0 specification Topic 1, sub-topic 1.4 covers the principles and use of time-of-flight mass spectrometry: ionisation (electron impact and electrospray), acceleration through an electric field to constant kinetic energy, drift through a field-free region, and detection; the calculation of the relative atomic mass of an element from isotopic abundance data; and the interpretation of mass spectra, including the use of m/z values and isotope ratio patterns (e.g. for Cl, Br) (refer to the official Pearson Edexcel specification document for exact wording). Mass spectrometry is examined in Paper 1 as a stand-alone analytical technique and in Paper 2 / Paper 3 as a tool for organic structure determination (Topics 18–19), where the M⁺, M+1, M+2 peaks of organic molecules require the same isotope-arithmetic introduced here.
Question (8 marks):
A sample of pure chlorine gas, Cl₂, is analysed by time-of-flight mass spectrometry. Two ionisation peaks are observed for Cl⁺ at m/z = 35 and m/z = 37 with relative intensities 3:1. Two peaks are also observed for Cl₂⁺ molecular ions.
(a) Predict the m/z values and relative intensities of the three Cl₂⁺ peaks expected. (4) (b) State why electrospray ionisation (ESI) might be preferred over electron impact (EI) for analysis of a fragile biomolecule, but EI for analysis of a small inorganic species like Cl₂. (2) (c) Calculate the time of flight for a Cl⁺ ion of mass 35 u accelerated through a potential difference of 4000 V across a drift region of length 1.50 m. (Use m(¹u) = 1.66 × 10⁻²⁷ kg, e = 1.60 × 10⁻¹⁹ C.) (2)
Solution with mark scheme:
(a) M1 — the three Cl₂⁺ peaks come from the three combinations: ³⁵Cl–³⁵Cl (m/z = 70), ³⁵Cl–³⁷Cl (m/z = 72), ³⁷Cl–³⁷Cl (m/z = 74). M1 — relative probabilities, given a 3:1 abundance of ³⁵Cl:³⁷Cl, are: P(35–35) = 3/4 × 3/4 = 9/16; P(35–37) + P(37–35) = 2 × (3/4 × 1/4) = 6/16; P(37–37) = 1/4 × 1/4 = 1/16. M1 — relative intensities are therefore 9:6:1. A1 — peaks at m/z = 70 (intensity 9), 72 (intensity 6), 74 (intensity 1). Common error: candidates write 9:3:1 (forgetting the two ways of getting 35–37) or 9:1:1.
(b) B1 — ESI is a soft ionisation method that does not fragment fragile molecules, allowing the molecular ion of a biomolecule to be observed. B1 — EI is a harder ionisation method that produces fragments; for a robust small species like Cl₂ this is acceptable and provides additional structural information.
(c) Step 1 — find the kinetic energy. Eₖ = eV = 1.60 × 10⁻¹⁹ × 4000 = 6.40 × 10⁻¹⁶ J.
Step 2 — find the velocity. Eₖ = ½mv², so v = √(2Eₖ/m) where m = 35 × 1.66 × 10⁻²⁷ = 5.81 × 10⁻²⁶ kg.
v = √(2 × 6.40 × 10⁻¹⁶ / 5.81 × 10⁻²⁶) = √(2.20 × 10¹⁰) = 1.48 × 10⁵ m s⁻¹.
Step 3 — find time of flight. t = d/v = 1.50 / 1.48 × 10⁵ = 1.01 × 10⁻⁵ s (≈ 10.1 μs).
M1 — correct rearrangement and substitution. A1 — answer to appropriate s.f.
Total: 8 marks (M3 A1 + B2 + M1 A1).
Question (6 marks): A mass spectrum of an unknown halogen X₂ shows three molecular-ion peaks at m/z = 158, 160, 162 with relative intensities 1:2:1.
(a) Identify the halogen and justify your answer. (3) (b) Sketch the appearance of the corresponding atomic-ion (X⁺) peaks and predict the m/z values and relative intensities. (2) (c) State one practical implication of the symmetric isotope pattern for organic chemistry analysis. (1)
Mark scheme decomposition by AO:
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