Define relative atomic and molecular mass, the mole and the Avogadro constant, and interconvert mass, amount in moles and number of particles

What this dot point is asking

SEAB wants you to define relative atomic mass and relative molecular (or formula) mass, define the mole and the Avogadro constant, and move confidently between three quantities: the mass of a substance, the amount in moles, and the number of particles. These conversions are the engine of every quantitative chemistry calculation, so fluency here pays off across the whole subject.

The answer

Relative atomic and molecular mass

Atoms are far too small to weigh individually, so chemists compare their masses to a standard. The relative atomic mass ($A_r$) of an element is the average mass of its atoms compared with $\tfrac{1}{12}$ the mass of a carbon-12 atom. It has no units. Values such as $A_r(\text{H}) = 1$, $A_r(\text{C}) = 12$, $A_r(\text{O}) = 16$ are taken from the Periodic Table.

The relative molecular mass ($M_r$), or relative formula mass for an ionic compound, is the sum of the relative atomic masses of all the atoms in the formula. For example, $M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$.

The mole and the Avogadro constant

A mole is the amount of substance that contains the same number of particles as there are atoms in $12$ g of carbon-12. That number is the Avogadro constant, $6.0 \times 10^{23}$ per mole. So one mole of any substance contains $6.0 \times 10^{23}$ particles (atoms, molecules or formula units).

The useful link is that the mass of one mole of a substance, in grams, equals its relative molecular (or atomic) mass. So one mole of water has a mass of $18$ g, and one mole of carbon atoms has a mass of $12$ g.

The two key conversions

Everything in this dot point reduces to two equations:

where $n$ is the amount in moles, $m$ the mass in grams, $M_r$ the relative molecular mass, $N$ the number of particles and $L = 6.0 \times 10^{23}$ per mol the Avogadro constant. Rearrange them as needed: $m = n \times M_r$ to find a mass, or $n = N/L$ to find moles from a particle count.

Choosing the right route

A typical question gives you one quantity and asks for another. Work through moles as the central hub: convert what you are given into moles first, then out to whatever is asked. Mass to particles, for instance, goes mass to moles ($n = m/M_r$) then moles to particles ($N = nL$).

Examples in context

Example 1. Counting atoms in a diamond. A small diamond weighing $0.60\ \text{g}$ is pure carbon ($A_r = 12$), so it contains $0.60/12 = 0.050$ mol of carbon atoms, which is $0.050 \times 6.0 \times 10^{23} = 3.0 \times 10^{22}$ atoms. The mole lets chemists count an astronomical number of atoms simply by weighing.

Example 2. Scaling up a recipe. To make a batch of a compound, a chemist needs the moles of each reactant, then converts to a mass to weigh out. The conversion between mass and moles is the everyday tool that turns a balanced equation into an actual quantity to measure on a balance.

Try this

Q1. Calculate the relative molecular mass of carbon dioxide, $\text{CO}_2$. (Ar: C = 12, O = 16.) [1 mark]

• Cue. $M_r = 12 + (2 \times 16) = 44$.

Q2. Calculate the amount, in moles, in $20.0\ \text{g}$ of sodium hydroxide, $\text{NaOH}$. (Ar: Na = 23, O = 16, H = 1.) [2 marks]

• Cue. $M_r = 23 + 16 + 1 = 40$; $n = 20.0/40 = 0.500\ \text{mol}$.

Q3. Calculate the number of molecules in $0.20$ mol of methane. (Avogadro constant $= 6.0 \times 10^{23}$ per mol.) [1 mark]

• Cue. $N = 0.20 \times 6.0 \times 10^{23} = 1.2 \times 10^{23}$ molecules.