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Mass spectrometry is one of the most powerful analytical techniques available to chemists. It allows you to determine the relative molecular mass of a compound and, with high-resolution instruments, to establish its molecular formula. Understanding how a mass spectrometer works and how to interpret the data it produces is essential for Edexcel A-Level Chemistry.
A mass spectrometer operates in four key stages: ionisation, acceleration, deflection, and detection. Each stage plays a distinct role in separating and identifying the ions produced from a sample.
flowchart LR
A["Sample\nVaporisation"] --> B["Ionisation\n(Electron Impact)"]
B --> C["Acceleration\n(Electric Field)"]
C --> D["Deflection\n(Magnetic Field)"]
D --> E["Detection\n(Ion Current)"]
E --> F["Mass Spectrum\n(m/z vs Abundance)"]
The sample is first vaporised (if it is not already a gas) and introduced into the ionisation chamber. Here, a beam of high-energy electrons is fired at the sample molecules. This is known as electron impact ionisation (EI).
When a high-energy electron strikes a sample molecule, it knocks out one of the molecule's own electrons, forming a positive ion with one unpaired electron — a radical cation:
M(g) + e⁻ → M⁺•(g) + 2e⁻
The ion M⁺• is called the molecular ion (or parent ion). It has the same mass as the original molecule (minus one electron, whose mass is negligible). In some instruments, electrospray ionisation (ESI) is used instead, which is gentler and particularly useful for large biomolecules.
The positive ions are accelerated by an electric field. Negatively charged plates attract the positive ions, accelerating them all to the same kinetic energy. Since they all have the same kinetic energy, ions with different masses will travel at different velocities — lighter ions move faster.
The accelerated ions pass through a magnetic field, which causes them to follow a curved path. The degree of deflection depends on the mass-to-charge ratio (m/z) of the ion. Lighter ions and ions with a higher charge are deflected more. Most ions produced by electron impact have a charge of +1, so m/z is effectively the mass of the ion.
Only ions with a specific m/z ratio will follow the correct curved path to reach the detector. The detector records the number of ions arriving (their relative abundance) at each m/z value. By varying the magnetic field strength, the spectrometer scans across a range of m/z values, building up a complete mass spectrum.
A mass spectrum is a bar chart with m/z on the horizontal axis and relative abundance (%) on the vertical axis. Two features are particularly important:
The peak at the highest m/z value (ignoring isotope peaks such as the M+1 peak) corresponds to the molecular ion — the intact molecule that has simply lost one electron. This gives you the relative molecular mass of the compound.
For example, if the molecular ion peak appears at m/z = 46, the compound has a relative molecular mass of 46. This is consistent with ethanol (C₂H₅OH, Mr = 46) or dimethyl ether (CH₃OCH₃, Mr = 46).
The tallest peak in the spectrum is called the base peak. It is assigned a relative abundance of 100%, and all other peaks are measured relative to it. The base peak represents the most stable (and therefore most abundant) ion formed during fragmentation. The base peak is not necessarily the molecular ion peak — in many spectra, a fragment ion is more abundant than the molecular ion.
The relative heights of the peaks reflect how many ions of each m/z value reach the detector. For elements with significant natural isotope distributions — such as chlorine (³⁵Cl and ³⁷Cl in a 3:1 ratio) and bromine (⁷⁹Br and ⁸¹Br in approximately 1:1 ratio) — characteristic patterns appear in the mass spectrum.
| Halogen | Isotopes | Ratio | M⁺ : M+2 pattern |
|---|---|---|---|
| Chlorine | ³⁵Cl, ³⁷Cl | 3 : 1 | M : M+2 = 3 : 1 |
| Bromine | ⁷⁹Br, ⁸¹Br | ~1 : 1 | M : M+2 ≈ 1 : 1 |
| Two Cl atoms | — | — | M : M+2 : M+4 = 9 : 6 : 1 |
| One Cl + one Br | — | — | M : M+2 : M+4 = 3 : 4 : 1 |
A compound containing one chlorine atom will show the molecular ion as two peaks: M⁺ and M+2, with relative intensities of approximately 3:1. A compound containing one bromine atom will show M⁺ and M+2 peaks of approximately equal intensity. Two chlorine atoms produce a distinctive M : M+2 : M+4 pattern of 9 : 6 : 1.
These isotope patterns are extremely useful for identifying whether halogens are present in an unknown compound.
Low-resolution mass spectrometry gives m/z values as whole numbers, which means different molecular formulae can give the same nominal mass. For example, both CO (Mr = 28.0) and C₂H₄ (Mr = 28.0) appear at m/z = 28 at low resolution.
High-resolution mass spectrometry (HRMS) measures m/z to four or more decimal places. Since the exact atomic masses of isotopes are not exact whole numbers, different molecular formulae give slightly different exact masses.
| Isotope | Exact mass |
|---|---|
| ¹H | 1.00794 |
| ¹²C | 12.00000 |
| ¹⁴N | 14.0031 |
| ¹⁶O | 15.9949 |
| ³²S | 31.9721 |
| ³⁵Cl | 34.9689 |
| ⁷⁹Br | 78.9183 |
| Molecular formula | Exact mass |
|---|---|
| CO | 27.9949 |
| C₂H₄ | 28.0313 |
| N₂ | 28.0062 |
| CH₂O | 30.0106 |
High-resolution mass spectrometry can therefore distinguish between these and determine the molecular formula of an unknown compound, not just its molecular mass.
A compound gives a molecular ion at m/z = 44.0262 by HRMS. Possible formulae for nominal mass 44 include:
The measured value of 44.0262 matches C₂H₄O exactly. This is acetaldehyde (ethanal, CH₃CHO). The other formulae give different exact masses that do not match.
Mass spectrometry is often coupled with other techniques. GC-MS (gas chromatography–mass spectrometry) separates components of a mixture by gas chromatography and then identifies each component by its mass spectrum. This is invaluable in forensics, environmental chemistry, and pharmaceutical analysis.
The technique requires very small samples — often micrograms — and is extremely sensitive. However, it is a destructive technique: the sample is consumed during analysis.
Mistake 1: Assuming the base peak is always the molecular ion. The base peak is simply the most abundant ion — it can be a fragment ion. Always look for the highest m/z peak (excluding isotope peaks) for the molecular mass.
Mistake 2: Forgetting that the M+1 peak (from ¹³C) is NOT the same as an M+2 halogen peak. The M+1 peak due to ¹³C is typically very small (about 1.1% per carbon atom), while halogen M+2 peaks are much larger and have distinctive ratios.
Mistake 3: Confusing ionisation methods. Electron impact ionisation is the standard method discussed at A-Level. It produces fragment ions by transferring excess energy. Electrospray ionisation (ESI) is gentler and often produces [M+H]⁺ ions (molecular mass + 1) rather than M⁺• ions.
Mistake 4: Neglecting the nitrogen rule. A compound with an odd molecular mass must contain an odd number of nitrogen atoms. If M⁺ is odd and you have not considered nitrogen, revisit your molecular formula.
Mass spectrometry provides the relative molecular mass from the molecular ion peak, fragmentation information from the pattern of lower m/z peaks, and (with high-resolution instruments) the molecular formula. It is the first technique you should consider when trying to identify an unknown compound, because it immediately tells you how heavy the molecule is. Isotope patterns reveal halogen content, and the nitrogen rule helps identify nitrogen-containing compounds.
Edexcel 9CH0 specification, Topic 18 — Modern Analytical Techniques I, sub-strand 18.1 covers the principles of time-of-flight (TOF) mass spectrometry, including ionisation by electron impact and electrospray, acceleration of ions through a potential difference, drift through a field-free region, and detection of arrival times to determine mass-to-charge ratio (m/z); the use of the molecular ion peak [M]⁺• to obtain relative molecular mass; and the interpretation of isotope peaks (M, M+2 ratios) for elements with two stable isotopes such as chlorine (³⁵Cl/³⁷Cl ≈ 3:1) and bromine (⁷⁹Br/⁸¹Br ≈ 1:1) — refer to the official specification document for exact wording. This material is examined on Paper 2 (Core Organic and Physical Chemistry) and synoptically on Paper 3 (General and Practical Principles), often in combination with Topic 19 NMR data. The Edexcel data booklet provides isotope masses to four decimal places when needed for high-resolution interpretation.
Question (8 marks):
A pure sample of an unknown alkyl halide gives a low-resolution mass spectrum showing two prominent peaks at m/z = 108 and m/z = 110 in an intensity ratio of approximately 1:1, together with a peak at m/z = 29.
(a) State which halogen is present in the compound and justify your answer using the isotope-peak data. (3)
(b) Suggest the molecular formula of the compound and calculate its relative molecular mass to the nearest whole number using ¹H = 1, ¹²C = 12, ⁷⁹Br = 79. (3)
(c) Identify the fragment of m/z = 29 and explain how it is consistent with the proposed structure. (2)
Solution with mark scheme:
(a) Step 1. The mass difference between the two molecular-ion-region peaks is 110 − 108 = 2, characteristic of an isotope pair separated by 2 mass units. M1 — recognising the M / M+2 separation.
Step 2. A 1:1 intensity ratio is the signature of bromine, because natural bromine consists of approximately equal abundances of ⁷⁹Br and ⁸¹Br. Chlorine would give a 3:1 ratio. M1 — linking ratio to identity. A1 — concluding bromine is present (not chlorine, not iodine which is monoisotopic).
(b) Step 1. With ⁷⁹Br = 79, the rest of the molecule has mass 108 − 79 = 29. M1 — subtracting the halogen mass.
Step 2. A fragment of mass 29 with composition CₓHᵧ requires 12x + y = 29; with x = 2, y = 5, giving C₂H₅. M1 — identifying ethyl.
Step 3. The molecular formula is therefore C₂H₅Br (bromoethane), with Mr = 24 + 5 + 79 = 108. A1.
(c) The peak at m/z = 29 corresponds to loss of Br from [C₂H₅Br]⁺• to give the ethyl cation [C₂H₅]⁺. M1 — identifying the fragment. A1 — relating to homolytic / heterolytic cleavage of the weak C–Br bond.
Total: 8 marks (M5 A3).
Question (6 marks): A compound X gives a mass spectrum with the molecular ion peak at m/z = 78 and an M+2 peak at m/z = 80 in a 3:1 intensity ratio. The base peak appears at m/z = 43.
(a) Identify the halogen present and justify the choice. (2)
(b) Suggest a possible molecular formula for X. (2)
(c) Suggest the identity of the m/z = 43 fragment. (2)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Mass spectrometry questions on 9CH0 are AO2-heavy because each value must be interpreted, not merely recalled.
| AO | Typical share | Earned by |
|---|---|---|
| AO1 (knowledge / recall) | 20–30% | Stating the four TOF stages, recalling isotope ratios, naming the molecular ion |
| AO2 (application) | 50–60% | Calculating Mr, deducing halogens from ratios, suggesting formulas |
| AO3 (analysis / unfamiliar) | 15–25% | Combining MS with other techniques in extended problems |
Examiner-rewarded phrasing: "the molecular ion peak [M]⁺• occurs at m/z corresponding to the relative molecular mass"; "the M+2 peak arises from the heavier isotope ³⁷Cl / ⁸¹Br"; "ionisation by electron impact removes one outer electron". Phrases that lose marks include using "molecular weight" instead of "relative molecular mass", confusing m/z with mass alone, or describing the ion as having charge 2+ without justification. Avoid quoting any examiner report verbatim; describe the convention in your own words.
A specific Edexcel pattern: when a question gives an intensity ratio and asks for identification, state both the observed ratio and the diagnostic ratio for the assigned element ("3:1 matches the natural ³⁵Cl : ³⁷Cl abundance of approximately 75% : 25%"). Half-marks evaporate when the candidate writes only "Cl is present".
Question: Explain why the mass spectrum of methane shows a peak at m/z = 17 in addition to the molecular ion at m/z = 16.
Grade C response (~140 words):
The m/z = 16 peak is the molecular ion of methane. The m/z = 17 peak is because of the carbon-13 isotope. About 1.1% of carbon atoms are ¹³C, so a small fraction of methane molecules contain one ¹³C atom and these have mass 17. This is why the small M+1 peak appears.
Examiner commentary: 3/3. Identifies the molecular ion correctly, attributes M+1 to ¹³C, and quantifies the abundance. The reasoning chain is complete.
Grade A response (~190 words):*
The peak at m/z = 16 is the molecular ion [CH₄]⁺• formed when an electron is knocked out of the methane molecule by electron-impact ionisation, producing a radical cation with mass equal to the relative molecular mass. The peak at m/z = 17 — known as the M+1 peak — arises because natural carbon contains approximately 1.1% ¹³C alongside 98.9% ¹²C. A methane molecule that incorporates one ¹³C atom in place of ¹²C therefore has mass 13 + 4 = 17 and contributes to the M+1 peak. The intensity ratio M : M+1 ≈ 100 : 1.1 reflects this isotopic abundance directly. (The contribution from ²H is negligible because deuterium is only 0.015% abundant, four orders of magnitude below ¹³C in natural carbon.)
Examiner commentary: 3/3 with confident extension. The candidate explicitly names the [M]⁺• radical cation, quantifies the isotopic abundance, and dismisses deuterium as a competing source — the kind of secondary-precision reasoning that signals A*.
Question: A compound has molecular ion peaks at m/z = 122 and 124 in a 1:1 intensity ratio. The compound also shows a peak at m/z = 105. Suggest a molecular formula and identify the m/z = 105 fragment.
Grade B response (~210 words):
The 1:1 ratio at M and M+2 means the compound contains one bromine atom. With ⁷⁹Br = 79, the rest of the molecule has mass 122 − 79 = 43. C₃H₇ has mass 43, so the formula is C₃H₇Br. The m/z = 105 peak is loss of 17 (which is OH), so this could be loss of OH.
Examiner commentary: 4/6. Bromine identification and molecular formula correct (4 marks secured). However, the candidate suggests OH loss without checking that the molecular formula contains an OH group — C₃H₇Br has no oxygen. Loss of 17 from m/z = 122 would actually give 105 only if OH were present, contradicting the proposed formula. Loses 2 marks for inconsistency.
Grade A response (~250 words):*
The molecular ion peaks at m/z = 122 and 124 in a 1:1 ratio are diagnostic of one bromine atom (⁷⁹Br : ⁸¹Br ≈ 1:1). With ⁷⁹Br contributing 79 mass units, the rest of the molecule has mass 122 − 79 = 43, consistent with a C₃H₇ residue (12 × 3 + 7 = 43). The molecular formula is therefore C₃H₇Br, which corresponds to either 1-bromopropane or 2-bromopropane.
The m/z = 105 fragment cannot result from loss of OH because the molecule contains no oxygen. The mass loss is 122 − 105 = 17, but this cannot correspond to OH; it must instead be re-examined — actually 17 cannot be CH₃ (15) or NH₃ (also no nitrogen). Reconsidering: the next-most-likely interpretation is that m/z = 105 contains the heavier ⁸¹Br isotope (so the 1:1 doublet is at 105/107), and the underlying neutral loss from m/z = 122 is to a different fragment. With ⁸¹Br accompanying, the m/z = 105 species cannot easily be interpreted without higher-resolution data; the candidate would flag this on the page.
Examiner commentary: 6/6. The candidate diagnoses the inconsistency and refuses to over-commit. Examiners reward this honest reasoning over confident-but-wrong identification.
Question: Describe how a time-of-flight mass spectrometer ionises, separates and detects a sample of an unknown organic compound, explaining how the resulting spectrum yields the relative molecular mass and information about the elements present.
Grade A response (~340 words):*
A sample is introduced into the high-vacuum chamber and ionised, most commonly by electron impact (high-energy electrons knock outer electrons from the molecules, producing radical cations [M]⁺•) or by electrospray ionisation (the sample is sprayed through a charged needle, producing protonated molecular ions [M+H]⁺). The vacuum prevents collisions with air molecules that would deflect or neutralise the ions.
The ions are then accelerated through a fixed potential difference, gaining kinetic energy ½mv² = zeV. Because all ions of charge z gain the same kinetic energy, lighter ions emerge with greater velocity than heavier ions: v = √(2zeV/m).
The accelerated ions enter a field-free drift region of fixed length L. The time of flight t = L/v varies as √m for ions of equal charge: lighter ions arrive first, heavier ions later. The detector — typically an electron multiplier — records arrival times and converts each to a mass-to-charge ratio.
The output is a mass spectrum: relative abundance plotted against m/z. The molecular ion peak [M]⁺• is the peak with the highest m/z (ignoring isotope satellites), and its m/z value equals the relative molecular mass of the compound — the central piece of structural information.
Information about elements present comes from isotope patterns: a 3:1 doublet separated by 2 mass units in the molecular-ion region indicates chlorine (³⁵Cl/³⁷Cl ≈ 75/25); a 1:1 doublet indicates bromine; an M+1 peak about 1.1% of M for each carbon present allows estimation of carbon count. Fragment ions (lower m/z) reveal structural fragments: m/z = 29 suggests CHO or C₂H₅, m/z = 17 suggests OH, and so on.
Together, the molecular-ion mass, the isotope ratios and the fragment-ion pattern provide a near-complete elemental and structural picture without ever weighing the original sample, only identifying the masses of charged products of its ionisation and (occasionally) decomposition.
Examiner commentary: 9/9. All four TOF stages described, [M]⁺• correctly named, isotope and fragment information explained. Examiners reward the explicit "v = √(2zeV/m)" derivation.
Oxbridge interview prompt: "How would you tell the difference between butan-1-ol and butan-2-ol using only mass spectrometry?"
This lesson supports Core Practical 16 (CP16): Analysis of organic and inorganic unknowns by qualitative tests and spectroscopic methods. In CP16, candidates use IR spectra, where available mass spectra, and chemical tests (Tollens', Fehling's, Brady's, iodoform, sodium carbonate for carboxylic acids) to identify unknown compounds. Mass spectrometry contributes the molecular-ion mass and isotope patterns that pin down the molecular formula before chemical tests narrow the functional groups. In a typical CP16 procedure students record an IR spectrum, perform two or three targeted chemical tests, and (in some centres) interpret a provided mass spectrum to identify the compound. Risk assessment for CP16 includes Brady's reagent (toxic, irritant — wear goggles and gloves), Tollens' reagent (formed in situ; never store, as it can form explosive silver fulminate on standing), and any halogenated solvents used as IR or NMR media. The MS data in CP16-style problems is provided rather than measured (school instruments rarely include MS), but interpreting it is a stated assessment skill.
This content is aligned with the Pearson Edexcel GCE A Level Chemistry (9CH0) specification, Topic 18 (Modern Analytical Techniques I), sub-strand 18.1 — mass spectrometry — examined on Paper 2 (Core Organic and Physical) and Paper 3 (General and Practical Principles). For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.
graph TD
A["Sample introduced<br/>into TOF spectrometer"] --> B["Ionisation:<br/>EI or ESI"]
B --> C["Acceleration<br/>through potential V"]
C --> D["Drift region:<br/>v = √(2zeV/m)"]
D --> E["Detector:<br/>m/z vs abundance"]
E --> F{"Spectrum<br/>features"}
F -->|"Highest m/z"| G["Molecular ion<br/>[M]⁺• → Mr"]
F -->|"M+2 doublet"| H{"Ratio?"}
F -->|"Lower m/z"| I["Fragment ions →<br/>structural clues"]
H -->|"3:1"| J["Cl present"]
H -->|"1:1"| K["Br present"]
G --> L["Molecular formula<br/>+ DoU"]
I --> L
J --> L
K --> L
style G fill:#27ae60,color:#fff
style L fill:#3498db,color:#fff