How Is Amount of Substance (Moles) Related to Number of Molecules and the Avogadro Constant?

The amount of substance $n$, measured in moles, equals the number of molecules $N$ divided by the Avogadro constant $N_A$:

$$n = \frac{N}{N_A}$$

This relationship is part of IB Physics B.3 Gas laws and applies at both SL and HL.

The Relationship

The Avogadro constant gives the number of specified particles in exactly one mole:

$$N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$$

Therefore, one mole contains $6.02 \times 10^{23}$ particles. Depending on the substance, these particles may be molecules, atoms, ions, or other specified entities.

This is a counting conversion, similar to using a dozen, but the counting unit is enormously larger. The unit $\text{mol}^{-1}$ means “particles per mole”. In $n=N/N_A$, the particle units cancel against particles per mole, leaving moles. In $N=$, moles cancel, leaving a pure count of particles.

Before calculating, check whether $N$ refers to molecules or another entity. For example, one mole of monatomic helium contains $N_A$ atoms, while one mole of oxygen gas contains $N_A$ molecules. The numerical particle count is the same, but the specified entities differ. Also keep $$ distinct from : varies with sample size, whereas is a constant. Since particle number is a count, has no unit; has unit mol.

The equation can be rearranged depending on the unknown quantity:

For example, suppose a gas sample contains $1.20 \times 10^{24}$ molecules. Its amount of substance is

$$n = \frac{1.20 \times 10^{24}}{6.02 \times 10^{23}} = 1.99\ \text{mol}$$

Conversely, a $0.500\ \text{mol}$ sample contains

$$N = (0.500)(6.02 \times 10^{23}) = 3.01 \times 10^{23}\ \text{molecules}.$$

A common misconception is that a mole is a unit of mass. It is not: the mole measures amount of substance, meaning the number of specified particles. Mass can be connected to moles using molar mass, but that is a separate relationship.

IB Exam Technique

In a Calculate question, show the equation, substitution, and numerical answer. Use $N$ for the number of particles and $n$ for the amount in moles, and take $N_A$ from the IB Physics data booklet. This relationship may also be combined with the ideal gas equations $PV = nRT$ and .