Topic 1 of 5
Relative mass and the mole
Distinguish a relative mass, a molar mass and a number of particles.
A-Level 9476 (2026-2027)
Relative masses compare with one common reference
The reference is one twelfth of the mass of a carbon-12 atom.
| Quantity | Meaning |
|---|---|
| Relative isotopic mass | Mass of one atom of a specified isotope divided by one twelfth of the mass of a carbon-12 atom. |
| Relative atomic mass, Ar | Weighted mean mass of an atom of the element divided by the same reference mass, for the stated isotopic composition. |
| Relative molecular mass, Mr | Mass of one molecule divided by the same reference; equal to the sum of its constituent relative atomic masses. |
| Relative formula mass, Mr | Sum of relative atomic masses in one formula unit; use this term for ionic solids, which have no discrete molecules. |
The mass number of an isotope is an integer counting protons and neutrons. Its relative isotopic mass is a measured mass ratio and need not equal that integer exactly. For example, use the isotopic masses supplied in a question rather than silently replacing them with mass numbers.
A relative mass is dimensionless. A molar mass, M, is mass per mole and has a unit such as g mol-1. Their numerical values agree when M is expressed in g mol-1: CO2 has Mr = 44.0 and M = 44.0 g mol-1. For NaCl, say relative formula mass rather than relative molecular mass.
A mole counts specified entities
Always name whether the entities are atoms, molecules, ions or formula units.
One mole contains exactly 6.02214076 × 1023 specified elementary entities. This fixes the Avogadro constant NA = 6.02214076 × 1023 mol-1. Use N = nNA and n = m/M. The amount n is measured in mol, not in grams.
Worked example
One formula can count several kinds of entity
What does 0.200 mol of CaCl2 represent before and after complete dissolution?
- The solid contains 0.200 mol of CaCl2 formula units, rather than CaCl2 molecules.
- Dissolution gives 0.200 mol Ca2+ and 0.400 mol Cl-, assuming complete dissolution and ignoring any minor ion association.
- The total amount of ions is 0.600 mol; multiply the amount of the requested entity by the Avogadro constant.
There are 1.20 × 1023 formula units in the solid sample, and about 3.61 × 1023 dissolved ions under the stated model.
For isotopes, Ar = Σ(isotopic mass × fractional abundance). Percent abundances must first be divided by 100; relative peak intensities can instead be divided by their total. A weighted mean must lie between the masses used and closer to the more abundant isotope.
Worked example
Use relative abundance as the weighting
An element has isotopes of relative isotopic masses 79.0 and 81.0, with abundances 50.7% and 49.3%. Find its relative atomic mass.
- Convert the abundances to fractions 0.507 and 0.493; they sum to 1.
- Calculate 79.0(0.507) + 81.0(0.493) = 79.986.
- Round only the final answer to a precision consistent with the supplied masses.
Ar = 80.0 to three significant figures. The unrounded weighted value is slightly closer to 79.0 than to 81.0.