IB Chemistry Structure 1 1.2 The Nuclear Atom 1.2.2
1.2.2

Isotopes

Same element, different mass, and why it matters for physical properties.

📘 IB Definition

Isotopes are atoms of the same element that possess an identical number of protons (atomic number) but a different number of neutrons (mass number).

Identical Chemistry, Different Physics

Because isotopes have identical nuclear charges and therefore identical electron configurations, they exhibit completely indistinguishable chemical reactivity. They form the exact same bonds and undergo the same reactions.

However, their varying mass numbers result in distinctly different physical properties:

🧊 Classic Example. Heavy Water

In heavy water (D₂O), the standard protium isotope (¹H) is replaced by deuterium (²H), which has an extra neutron. Heavy water molecules are substantially more massive, so:

  • Solid heavy water ice has a higher density than normal liquid water, so it sinks rather than floats
  • Heavy water ice has a slightly higher melting point than the 0°C threshold

This is a favourite examiner example because it challenges the assumption that "all ice floats".

Common Isotope Examples

Element Isotope Protons Neutrons Use
Hydrogen \(^1_1\text{H}\) (Protium) 1 0 Normal hydrogen
Hydrogen \(^2_1\text{H}\) (Deuterium) 1 1 Heavy water, NMR
Carbon \(^{12}_6\text{C}\) 6 6 Standard for Ar scale
Carbon \(^{14}_6\text{C}\) 6 8 Radiometric dating
Chlorine \(^{35}_{17}\text{Cl}\) and \(^{37}_{17}\text{Cl}\) 17 18 / 20 Mass spectrometry problems

Relative Atomic Mass (\(A_r\))

The relative atomic mass is the weighted mean of all naturally occurring isotopes of an element, measured relative to ¹⁄₁₂ of the mass of carbon-12.

⚠️ Examiner Tip: \(A_r\) Has No Units

\(A_r\) is a dimensionless ratio. It is a comparison to carbon-12, not a measurement with units. Never write "g" or "amu" after \(A_r\). Use the data booklet values and report your answer to 2 decimal places unless told otherwise.

Reverse Calculation: Finding % Abundance

If given the \(A_r\) and asked to find the percentage abundance of two isotopes, use algebra:

Example: Chlorine has \(A_r\) = 35.45, with isotopes ³⁵Cl and ³⁷Cl.

Let the fraction of ³⁵Cl = \(x\), so the fraction of ³⁷Cl = \(1 - x\).

\(35x + 37(1-x) = 35.45\)

\(35x + 37 - 37x = 35.45\)

\(-2x = -1.55\)

\(x = 0.775\), so ³⁵Cl = 77.5% and ³⁷Cl = 22.5%

← 1.2.1 Subatomic Particles 1.2.3 Mass Spectrometry →