Atomic particles, isotopes and mass spectra

What you’ll learn

• Determine particle numbers in atoms and ions.
• Calculate relative atomic mass from isotope abundance.
• Interpret charge and molecular-isotope patterns in mass spectra.

Understand the idea

The atomic number counts protons and defines the element; mass number counts protons plus neutrons. Isotopes have the same proton number but different neutron numbers. Forming an ion changes electron number, not the nucleus: a positive ion has lost electrons and a negative ion has gained them.

A mass spectrometer forms ions, separates them according to mass-to-charge ratio and detects their relative abundance. Relative atomic mass is an abundance-weighted mean on the carbon-12 scale. It can be non-integer even when the isotope mass numbers are integers. A doubly charged ion appears at half the mass-to-charge ratio of the corresponding singly charged ion.

Molecular isotope peaks combine the possible isotope choices at each atom. If two isotopes have fractions $p$ and $q$, the three mass combinations of a diatomic molecule occur in proportions $p^2:2pq:q^2$. The factor 2 counts the two ways to form the mixed-isotope molecule.

Key formulae

$$N_{\mathrm{neutrons}}=A-Z$$

$$N_{\mathrm{electrons}}=Z-q\quad(q\text{ is signed ionic charge})$$

$$A_r=\frac{\sum(\text{isotopic mass}\times\text{abundance})}{\sum\text{abundance}}$$

$$\text{diatomic isotope pattern}=p^2:2pq:q^2$$

Worked example

An element has isotopic masses 10.0 and 11.0 with abundances 20.0% and 80.0%. Find its relative atomic mass.

1. Convert the percentages to fractions $0.200$ and $0.800$.
2. Multiply each mass by its fraction: $10.0(0.200)=2.00$ and $11.0(0.800)=8.80$.
3. Add the contributions to obtain the weighted mean.

Answer: $A_r=10.8$.