Mass spectrometry

IB Chemistry SL· Structure 1: Atomic Structure, Topic 3· 15 min read

1. Components and Principles of Mass Spectrometry★★☆☆☆⏱ 5 min

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.

📘 Definition

Mass-to-charge ratio

m/z

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

Vaporization: The solid/liquid sample is converted to gaseous atoms.
Ionization: Gaseous atoms are bombarded with electrons to form positive ions.
Acceleration: Ions are accelerated by an electric field to constant kinetic energy.
Deflection: Ions are deflected by a magnetic field: lighter ions are deflected more than heavier ions of the same charge.
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.

2. Interpreting a Mass Spectrum★★☆☆☆⏱ 4 min

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: $^{35} Cl$ and $^{37} Cl$ , 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.

✓ Quick check

Test your basic interpretation

Which peak corresponds to the most abundant isotope?
- The tallest peak
- The peak at highest m/z
- The peak at lowest m/z
Reveal answer
The tallest peak —
Correct! Y-axis is relative abundance, so taller peak = higher abundance.

3. Calculating Relative Atomic Mass★★★☆☆⏱ 6 min

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.

📘 Definition

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: $^{10} B$ (19.9% abundance) and $^{11} B$ (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
$A_r = \dfrac{1080.1}{100} = 10.80$

4. Common Pitfalls

Wrong move:

Forgetting to divide by 100 when using percentage abundances

Why:

If you use percentages, the sum of abundances equals 100, so skipping division gives an answer 100x too large.

Correct move:

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 move:

Claiming heavier ions are deflected more than lighter ions

Why:

Most students mix up the relationship between mass and deflection, incorrectly associating more mass with more movement.

Correct move:

For the same charge, lighter ions have greater deflection in the magnetic field, heavier ions deflect less.

Wrong move:

Using the element's relative atomic mass instead of individual isotopic masses in calculations

Why:

The relative atomic mass is the average you are trying to calculate, not the mass of a single isotope.

Correct move:

Always use the $m / z$ value (mass number) of each individual isotope for the calculation.

Wrong move:

Including impurity or fragment peaks when calculating $A_{r}$

Why:

Extra peaks from contaminants or fragments can throw off your weighted average.

Correct move:

Check that your final calculated $A_{r}$ matches the value on the periodic table to confirm you used the correct peaks.

5. Quick Reference 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	$m / z = isotope mass (u)$
Relative atomic mass (percent abundance)	$A_{r} = \frac{\sum ( m \times % abundance )}{100}$

6. Frequently Asked

Do I need to memorize all parts of a mass spectrometer for SL?

IB SL requires you to know the function of each main stage, not just names, but you will never need to draw the instrument.

Why are there multiple peaks in an elemental mass spectrum?

Each peak corresponds to a different stable isotope of the element, which has a unique mass, so separates by m/z ratio.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Calculate Ar from mass spectrum data
2023 · 2
Identify isotopes from spectrum
2021 · 1
Label mass spectrometer functions

Going deeper

Official SyllabusIB Chemistry SL GuideFull official syllabus reference

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.