# The nuclear atom and mass spectrometry

> IB Chemistry HL · S1: Models of the particulate nature of matter
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u1-the-nuclear-atom-and-mass/

This sub-topic covers Rutherford's nuclear model of the atom, properties of subatomic particles, isotopic composition, and the use of mass spectrometry to calculate relative atomic mass of elements.

**Prerequisites:** [Basic particulate nature of matter](https://www.owlsprep.com/study/ib-chemistry-hl-u1-basic-particulate-nature/)

## Learning objectives

- Describe the structure of the nuclear atom and properties of subatomic particles
- Interpret mass spectra to identify isotopes and calculate relative atomic mass
- Distinguish between isotopes based on their nuclear composition and properties
- Explain the key principles of mass spectrometry for elemental analysis

## 1. Structure of the Nuclear Atom

Ernest Rutherford's gold foil experiment disproved the earlier 'plum pudding model' of the atom, leading to the nuclear model that is accepted today.

**Nuclear atom** — A model where most of the atom's mass and all its positive charge is concentrated in a small dense nucleus, surrounded by negatively charged electrons that occupy most of the atom's volume.

*Example:* A neutral carbon-12 atom has a 6-proton, 6-neutron nucleus, with 6 electrons orbiting outside.

| Subatomic Particle | Relative Charge | Relative Mass | Location |
| --- | --- | --- | --- |
| Proton | $+1$ | $1$ | Nucleus |
| Neutron | $0$ | $1$ | Nucleus |
| Electron | $-1$ | $\approx 1/1836$ | Electron cloud outside nucleus |

**Worked example:** State the number of protons, neutrons and electrons in the ion $\ce{^{27}_{13}Al^3+}$

1. The lower value $Z$ (atomic number) equals the number of protons, so:
2. $$\text{Protons} = Z = 13$$
3. The upper value $A$ (mass number) is protons + neutrons, so calculate neutrons:
4. $$\text{Neutrons} = A - Z = 27 - 13 = 14$$
5. The $3+$ charge means 3 electrons are lost from the neutral atom, so:
6. $$\text{Electrons} = 13 - 3 = 10$$

> **Exam tip:** Proton number always defines the element, only electron count changes for ions.

## 2. Isotopes

**Isotopes** — Atoms of the same element that have the same number of protons (same atomic number) but different numbers of neutrons (different mass number).

*Example:* Chlorine has two naturally occurring stable isotopes: $\ce{^{35}Cl}$ and $\ce{^{37}Cl}$.

Isotopes have identical chemical properties because they have the same electron configuration, which governs chemical bonding and reactivity. They differ in physical properties such as mass, density, melting point and rate of diffusion, due to their different masses.

> **info**
>
> Unstable isotopes (radioisotopes) are widely used in medicine for diagnostic imaging and cancer treatment.

**Worked example:** Boron has atomic number 5. Compare the atomic structure and reactivity of boron-10 and boron-11.

1. Both isotopes are boron, so they have the same number of protons:
2. $$\text{Protons for both} = 5$$
3. Calculate neutrons for each isotope:
4. Boron-10: $10 - 5 = 5$ neutrons; Boron-11: $11 - 5 = 6$ neutrons
5. Neutral atoms of both have 5 electrons, so electron configuration is identical.
6. Conclusion: They have different nuclear composition but identical chemical reactivity.

## 3. Mass Spectrometry and Relative Atomic Mass Calculation

Mass spectrometry separates ions of different mass based on their mass-to-charge ratio ($m/z$). The resulting mass spectrum plots relative abundance of each isotope against $m/z$, which is equal to the isotopic mass for ions with charge +1 (the most common case).

**Relative atomic mass ($A_r$)** — The weighted average mass of a naturally occurring sample of an element, relative to $\frac{1}{12}$ the mass of a single carbon-12 atom.

**Worked example:** Chlorine has two isotopes with peaks at $m/z = 35$ (75.77% abundance) and $m/z = 37$ (24.23% abundance). Calculate the relative atomic mass of chlorine.

1. Multiply each isotopic mass by its decimal abundance:
2. $$(35 \times 0.7577) + (37 \times 0.2423)$$
3. Calculate the sum of the products:
4. $$26.5195 + 8.9651 = 35.4846$$
5. Round to 3 significant figures per IB conventions:
6. $$A_r(\text{Cl}) = 35.5$$

**Exam command terms**

Common IB command terms for this topic:

- **Calculate** — Obtain a numerical answer showing all working steps *(Calculate the relative atomic mass from mass spectrum data)*

- **Deduce** — Reach a conclusion from the given information *(Deduce the number of neutrons in a given isotope)*

**Check your understanding**

Test your calculation skill:

1. An element has two isotopes: 10X (20% abundance) and 11X (80% abundance). What is its relative atomic mass?

   - 10.0
   - 10.2
   - 10.8
   - 11.0

   *Why:* Correct! $(10 \times 0.2) + (11 \times 0.8) = 10.8$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Counting electrons in an ion as equal to the proton number regardless of charge
  - Why it fails: Only neutral atoms have equal numbers of protons and electrons; ions gain or lose electrons
  - Correct: Subtract positive charge from proton number, add negative charge to get the number of electrons
- **Wrong:** Claiming isotopes have different chemical properties
  - Why it fails: Chemical reactivity depends entirely on electron arrangement, which is identical for isotopes
  - Correct: State that isotopes have identical chemical properties and different physical properties
- **Wrong:** Forgetting to convert percentage abundances to decimals before calculation
  - Why it fails: This results in a final value 100 times larger than the correct answer
  - Correct: Divide all percentage abundances by 100, or divide the final sum by 100
- **Wrong:** Using whole number mass numbers when the question gives actual isotopic masses
  - Why it fails: Actual isotopic masses are slightly different from whole numbers, leading to small calculation errors
  - Correct: Always use the $m/z$ values given in the question's mass spectrum

## Cheatsheet

| Key Concept | Formula / Rule |
| --- | --- |
| Atomic number ($Z$) | Number of protons = defines the element |
| Mass number ($A$) | $A = Z + \text{number of neutrons}$ |
| Electrons in ion | Electrons = $Z - \text{charge}$ |
| Isotopes | Same $Z$, different $A$, same chemical properties |
| Relative atomic mass | $A_r = \frac{\sum (\text{isotopic mass} \times \text{abundance})}{\sum \text{abundance}}$ |
| Subatomic particle charges | Proton $+1$, Neutron $0$, Electron $-1$ |

## What's next

The nuclear atom and mass spectrometry is the foundation for all further chemistry study. The concept of relative atomic mass is core to stoichiometry, the basis of all quantitative chemistry calculations that you will explore next. Isotopic composition also connects to nuclear chemistry, where you will learn about radioactivity and nuclear reactions later in the IB HL syllabus. Mass spectrometry is also extended in organic chemistry, where it is used to identify molecular mass and structure of organic compounds.

- [Electron configurations](https://www.owlsprep.com/study/ib-chemistry-hl-u1-electron-configurations/)
- [The mole concept](https://www.owlsprep.com/study/ib-chemistry-hl-u1-the-mole-concept/)
- [Ideal gas behaviour](https://www.owlsprep.com/study/ib-chemistry-hl-u1-ideal-gas-behaviour/)

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