# Mass spectrometry

> IB Chemistry SL · IB Chemistry SL
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u1-mass-spectrometry/

This module covers how mass spectrometry separates isotopes by mass, how to interpret mass spectra, and how to calculate the relative atomic mass of an element, a core SL exam topic.

**Prerequisites:** [Understanding of isotopes and relative atomic mass](https://www.owlsprep.com/study/ib-chemistry-sl-u1-isotopes-atomic-mass/)

## Learning objectives

- Describe the main components and functions of a mass spectrometer
- Interpret mass spectra to identify isotopes and their relative abundances
- Calculate relative atomic mass from mass spectrum data

## Components and Principles of Mass Spectrometry

A mass spectrometer processes samples to separate ions by their mass-to-charge ($m/z$) ratio. For IB SL, you only need to recall the function of each main processing stage, not detailed instrumental design.

**Mass-to-charge ratio** — For almost all elemental analysis ions, the charge is +1, so $m/z$ equals the mass of the ion in atomic mass units (u).

*Notation:* m/z

1. Vaporization: The solid/liquid sample is converted to gaseous atoms.
2. Ionization: Gaseous atoms are bombarded with electrons to form positive ions.
3. Acceleration: Ions are accelerated by an electric field to constant kinetic energy.
4. Deflection: Ions are deflected by a magnetic field: lighter ions are deflected more than heavier ions of the same charge.
5. Detection: Ions of different $m/z$ are detected, and their relative abundance is recorded.

**Worked example:** Match the mass spectrometer stage to its function: which stage separates ions by mass?

1. Recall the core function of each processing stage:
2. Ionization only creates ions, acceleration gives constant kinetic energy, and detection records signal.
3. Conclusion: Deflection by the magnetic field is the stage that separates ions by mass.

> **Exam tip**
>
> Multiple choice questions often ask to order the stages or match components to functions.

## Interpreting a Mass Spectrum

A mass spectrum plots relative abundance of ions on the y-axis against $m/z$ ratio on the x-axis. For a natural elemental sample, each distinct peak corresponds to a different stable isotope of the element.

**Worked example:** Chlorine has two isotopes: $^{35}\text{Cl}$ and $^{37}\text{Cl}$, with a natural abundance ratio of 3:1. Describe how this looks on a mass spectrum.

1. Each isotope gives one peak, at $m/z$ equal to its mass number for +1 ions.
2. The relative height of the peaks matches the abundance ratio.
3. $$\text{Peak 1: } m/z = 35, \text{ abundance } 75\% \\ \text{Peak 2: } m/z = 37, \text{ abundance } 25\%$$
4. The peak at 35 will be three times taller than the peak at 37.

**Check your understanding**

Test your basic interpretation

1. Which peak corresponds to the most abundant isotope?

   - The tallest peak
   - The peak at highest m/z
   - The peak at lowest m/z

   *Why:* Correct! Y-axis is relative abundance, so taller peak = higher abundance.

## Calculating Relative Atomic Mass

The relative atomic mass ($A_r$) of an element is the weighted average of the mass of its isotopes, weighted by their natural relative abundances. This is the value reported on the IB periodic table.

**Relative atomic mass formula** — For percentage abundances, the formula for $A_r$ is:

$$A_r = \dfrac{\sum (\text{isotopic mass} \times \text{percent abundance})}{100}$$

**Worked example:** Boron has two isotopes: $^{10}\text{B}$ (19.9% abundance) and $^{11}\text{B}$ (80.1% abundance). Calculate $A_r$ of boron to two decimal places.

1. Substitute the values into the formula:
2. $$A_r = \dfrac{(10 \times 19.9) + (11 \times 80.1)}{100}$$
3. Calculate the numerator:
4. $$(10 \times 19.9) = 199; (11 \times 80.1) = 881.1 \\ 199 + 881.1 = 1080.1$$
5. Divide by 100 to get the final result:
6. $$A_r = \dfrac{1080.1}{100} = 10.80$$

**Exam command terms**

Common IB SL command terms for this topic:

- **Calculate** — You must show all working steps to earn full marks, even for simple calculations. *(For $A_r$ calculations, always write the substituted formula before giving the final answer.)*

- **Deduce** — Use given data to logically arrive at the answer, working is recommended for partial credit.

## Common pitfalls

- **Wrong:** Forgetting to divide by 100 when using percentage abundances
  - Why it fails: If you use percentages, the sum of abundances equals 100, so skipping division gives an answer 100x too large.
  - Correct: Always check the sum of abundances: if they add to 100, divide by 100; if they add to 1 (fractional abundances), no division is needed.
- **Wrong:** Claiming heavier ions are deflected more than lighter ions
  - Why it fails: Most students mix up the relationship between mass and deflection, incorrectly associating more mass with more movement.
  - Correct: For the same charge, lighter ions have greater deflection in the magnetic field, heavier ions deflect less.
- **Wrong:** Using the element's relative atomic mass instead of individual isotopic masses in calculations
  - Why it fails: The relative atomic mass is the average you are trying to calculate, not the mass of a single isotope.
  - Correct: Always use the $m/z$ value (mass number) of each individual isotope for the calculation.
- **Wrong:** Including impurity or fragment peaks when calculating $A_r$
  - Why it fails: Extra peaks from contaminants or fragments can throw off your weighted average.
  - Correct: Check that your final calculated $A_r$ matches the value on the periodic table to confirm you used the correct peaks.

## Cheatsheet

| Concept | Key Fact/Formula |
| --- | --- |
| Processing order | Vaporization → Ionization → Acceleration → Deflection → Detection |
| Deflection rule | Same charge: lighter ions = more deflection |
| m/z for +1 ions | $m/z = \text{isotope mass (u)}$ |
| Relative atomic mass (percent abundance) | $A_r = \frac{\sum (m \times \%\text{abundance})}{100}$ |

## What's next

Mass spectrometry is a foundational analytical technique that you will build on if you continue to IB Chemistry HL, where you will use it to identify molecular fragments and full molecular masses of organic compounds. Understanding how mass spectrometry separates isotopes reinforces your core knowledge of atomic structure, which is required for all later topics including stoichiometry, nuclear chemistry, and organic analysis. Mastery of relative atomic mass calculations is a common source of easy marks in both Paper 1 multiple choice and Paper 2 short answer questions, so regular practice of this skill will pay off in your final exam.

- [Basic electron arrangement in atoms](https://www.owlsprep.com/study/ib-chemistry-sl-u1-basic-electron-arrangement-in-atoms/)
- [Structure 2: Electron Configuration](https://www.owlsprep.com/study/ib-chemistry-sl-u2-overview/)
- [Atomic orbitals](https://www.owlsprep.com/study/ib-chemistry-sl-u2-atomic-orbitals/)

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