Mass Spectrometry: Principles and Interpretation Revision Notes | A Level AQA Chemistry

1. Why Mass Spectrometry Matters

Picture this: you have a sample of an unknown element and you need to know exactly which isotopes it contains and in what proportions. A mass spectrometer answers both questions simultaneously, with extraordinary precision. That is why it sits at the heart of modern analytical chemistry.

At A Level, you need to understand how the instrument works, what the output spectrum tells you, and how to extract quantitative data from it. These skills connect directly to calculating relative atomic mass and identifying unknown substances — both of which appear regularly in AQA exam questions.

2. Principles of the Time of Flight (TOF) Mass Spectrometer

The TOF mass spectrometer separates ions by measuring how long they take to travel a fixed distance. The core logic is elegant: give every ion the same kinetic energy, then let them race. Lighter ions travel faster and arrive at the detector first; heavier ions arrive later. The arrival time tells you the mass.

There are four distinct stages you must be able to describe for AQA: ionisation, acceleration, ion drift, and ion detection. Each stage has a specific purpose, and examiners will ask you to describe or explain any one of them.

Once ions are formed, they are accelerated by an electric field. The accelerating voltage gives every ion the same kinetic energy, regardless of mass. This is the critical condition that makes the time-of-flight measurement meaningful.

After acceleration, the ions enter a field-free drift region — a long tube with no electric field. Here, kinetic energy is constant, so the velocity of each ion depends only on its mass. A lighter ion moves faster and crosses the drift region in less time than a heavier ion given the same kinetic energy.

At the end of the drift tube, ions strike a detector. Each impact generates a small electrical signal. The instrument records the time between ionisation and detection, then converts that time into a mass-to-charge ratio (m/z). Because most ions carry a 1+ charge, m/z is numerically equal to the relative mass of the ion.

• Stage 1 — Ionisation: sample particles are converted into positive gas-phase ions
• Stage 2 — Acceleration: ions are accelerated through an electric field; all ions gain the same kinetic energy
• Stage 3 — Ion drift: ions travel through a field-free region; lighter ions travel faster
• Stage 4 — Ion detection: ions hit the detector and generate an electrical signal; arrival time is recorded
• Stage 5 — Data analysis: the instrument converts time-of-flight data into a mass spectrum (m/z vs relative abundance)

The data analysis stage converts the raw time-of-flight measurements into the familiar mass spectrum: a plot of m/z on the x-axis against relative abundance (or percentage abundance) on the y-axis. Each peak corresponds to an ion of a particular mass reaching the detector.

The TOF design gives the instrument its key advantage: it can measure masses with very high accuracy, providing precise values for relative isotopic mass and reliable data on how abundant each isotope is. This is why mass spectrometry is the definitive technique for isotopic analysis.

3. Reading a Mass Spectrum of an Element

When a pure element is analysed, the mass spectrum shows one peak for each naturally occurring isotope. The position of each peak on the m/z axis gives the relative isotopic mass of that isotope. The height of each peak — more precisely, the area under it — gives the relative abundance of that isotope.

Consider chlorine as a concrete example. Chlorine has two stable isotopes: chlorine-35 and chlorine-37. A mass spectrum of chlorine atoms would show two peaks: one at m/z = 35 and one at m/z = 37. The peak at m/z = 35 is roughly three times taller than the peak at m/z = 37, reflecting the approximate 75%:25% natural abundance ratio.

From this spectrum you can read off two pieces of information directly: the relative isotopic masses (35 and 37) and the relative abundances (approximately 75% and 25%). These two pieces of data are all you need to calculate the relative atomic mass of chlorine.

Worked example: Calculating Relative Atomic Mass from a Mass Spectrum

A mass spectrum of boron shows two peaks: m/z = 10 with relative abundance 20% and m/z = 11 with relative abundance 80%. Calculate the relative atomic mass of boron.

1. Isotope 1: mass = 10, abundance = 20% Isotope 2: mass = 11, abundance = 80% — Read directly from the m/z axis (isotopic mass) and peak heights (relative abundance) of the spectrum.
2. 10 × 20 = 200 11 × 80 = 880 — This weights each isotope's contribution according to how common it is in nature.
3. 200 + 880 = 1080 — This is the total weighted contribution from all isotopes.
4. $A_r = \frac{1080}{100} = 10.8$ — Dividing by 100 converts the percentage-weighted sum into a mean mass. The answer has no units because relative atomic mass is a ratio.
5. 10.8 lies between 10 and 11, closer to 11 because boron-11 is more abundant (80%). — The weighted mean must always lie between the lightest and heaviest isotope, and must be closer to the more abundant one. If your answer falls outside this range, recheck your arithmetic.

Answer: $A_r$ of boron = 10.8

4. Using Mass Spectrometry to Identify Elements

Because every element has a unique set of isotopes with characteristic masses and natural abundances, the pattern of peaks in a mass spectrum acts as a fingerprint. By comparing the m/z values and relative abundances of peaks against known data, an unknown element can be identified with confidence.

For example, if a spectrum shows two peaks at m/z = 63 and m/z = 65 in approximately a 2:1 ratio, this pattern is characteristic of copper, which has two naturally occurring isotopes: copper-63 (approximately 69%) and copper-65 (approximately 31%). No other element produces this exact combination of masses and abundances.

The mass spectrometer is particularly powerful here because it measures relative isotopic mass with high accuracy — far more precisely than chemical methods. This accuracy is what makes identification reliable, even for elements present in very small quantities in a sample.

5. Determining Relative Molecular Mass

Mass spectrometry is not limited to elements. When a molecular compound is analysed, the instrument can determine its relative molecular mass directly. The key is the molecular ion peak, also called the M+ peak (or [M+H]+ peak in electrospray ionisation).

When a molecule passes through the instrument and loses one electron (in electron ionisation), it forms a molecular ion with the same mass as the original molecule. This ion produces a peak at the highest m/z value in the spectrum — the molecular ion peak. Reading the m/z value of this peak directly gives the relative molecular mass of the compound.

In practice, molecules can also fragment inside the instrument, producing smaller ions at lower m/z values. These fragment peaks are useful for structural identification at A Level, but for the purpose of determining relative molecular mass, you focus on the molecular ion peak at the highest m/z value.

This comparison highlights a key distinction: when interpreting a spectrum, you first need to decide whether you are looking at an element or a compound. For an element, every peak corresponds to an isotope. For a compound, the highest m/z peak gives Mr, and lower peaks are fragments.

6. Worked Interpretation: Putting It All Together

Now that the four stages of the TOF instrument and the two main uses of mass spectra are established, it is worth working through a complete interpretation question — the kind that appears in AQA papers combining identification, abundance data, and relative atomic mass calculation in a single question.

Worked example: Interpreting a Mass Spectrum of Magnesium

A mass spectrum of magnesium shows three peaks: m/z = 24 (79.0%), m/z = 25 (10.0%), m/z = 26 (11.0%). (a) State the number of naturally occurring isotopes of magnesium shown in this spectrum. (b) Calculate the relative atomic mass of magnesium to one decimal place.

1. Three peaks are present in the spectrum. — Each peak in an atomic mass spectrum corresponds to one isotope. Three peaks means three naturally occurring isotopes: magnesium-24, magnesium-25, and magnesium-26.
2. $A_r = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}$ — Multiply each isotopic mass by its percentage abundance, sum the results, then divide by 100.
3. 24 × 79.0 = 1896.0 25 × 10.0 = 250.0 26 × 11.0 = 286.0 — Work through each multiplication separately to avoid arithmetic errors.
4. 1896.0 + 250.0 + 286.0 = 2432.0 — Add the three weighted contributions together.
5. $A_r = \frac{2432.0}{100} = 24.3$ — This gives the weighted mean mass. Rounded to one decimal place: 24.3.
6. 24.3 lies between 24 and 26, much closer to 24 because magnesium-24 has by far the highest abundance (79.0%). — The answer is consistent with the spectrum: the dominant peak at m/z = 24 pulls the weighted mean close to 24.

Answer: (a) 3 isotopes. (b) Ar of magnesium = 24.3

7. Exam Technique and Common Mistakes

Mass spectrometry questions on AQA papers tend to fall into three categories: describe the principles of the TOF instrument, interpret a given spectrum, or calculate relative atomic mass. Each type has its own mark-winning habits and its own traps.

When a question asks you to describe the TOF mass spectrometer, work through the stages in order: ionisation, acceleration, ion drift, detection, data analysis. Examiners award marks for each stage described correctly. A common error is to describe the stages out of order or to omit the data analysis stage entirely — that stage is explicitly on the AQA specification and can carry a mark.

For calculation questions, always show the weighted mean formula before substituting numbers. If you write the formula first, you can earn a method mark even if your arithmetic goes wrong. Never use the simple (unweighted) average of isotopic masses — this is the single most common error in this topic and it will cost you the accuracy mark.

Taken together, the TOF mass spectrometer provides two types of information from a single experiment: the precise relative isotopic masses of each isotope (from the m/z axis) and the relative abundances of those isotopes (from the peak heights). These two outputs underpin every calculation and identification task in this topic, and they are why mass spectrometry is described in the AQA specification as giving 'accurate information about relative isotopic mass and also about the relative abundance of isotopes'.