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Relative mass and the mole

Relative masses compare atoms with carbon-12. The mole turns those relative masses into grams, so that weighing becomes a way of counting.

14 min read1.4(i)1.4(ii)1.21.4(iii)

Read first: Particles and formulae

Why chemists use a relative scale 1.4(i)

One atom of carbon-12 has a mass of about 2 × 10⁻²³ g. Numbers like that are useless on a balance and awkward on paper. Instead, every atom is compared with one standard: the carbon-12 atom, which is given a mass of exactly 12. One twelfth of a carbon-12 atom is the unit of the scale.

Each phrase in that definition earns its place:

Weighted mean: most elements are a mixture of isotopes. $A_\text{r}$ averages them according to how abundant each one is, which is why $A_\text{r}$ (Cl) = 35.5 and not a whole number.
Relative to one twelfth of carbon-12: it is a ratio of two masses, so $A_\text{r}$ has no units.

Relative molecular mass and relative formula mass 1.4(ii)

For substances with giant structures (ionic lattices such as NaCl, giant covalent structures such as SiO₂) there are no molecules, so the correct name is relative formula mass. It uses the same symbol, $M_\text{r}$ , and is calculated in exactly the same way: add up the $A_\text{r}$ of every atom in the formula.

Worked exampleRelative formula mass of calcium nitrate, Ca(NO₃)₂

List every atom. The bracket doubles everything inside it: 1 × Ca, 2 × N, 6 × O.
Add the relative atomic masses: 40.1 + 2(14.0) + 6(16.0) = 40.1 + 28.0 + 96.0.
$M_\text{r}$ = 164.1 (no units). It is ionic, so call it the relative formula mass.

Formula	Type	Working	$M_\text{r}$
H₂SO₄	molecular	2(1.0) + 32.1 + 4(16.0)	98.1
CO₂	molecular	12.0 + 2(16.0)	44.0
Mg(OH)₂	formula	24.3 + 2(16.0 + 1.0)	58.3
CaCO₃	formula	40.1 + 12.0 + 3(16.0)	100.1
CuSO₄·5H₂O	formula	63.5 + 32.1 + 4(16.0) + 5(18.0)	249.6

The mole and the Avogadro constant 1.2

That number is the Avogadro constant, symbol L (some books use $N_\text{A}$ ):

L = 6.02 \times 10^{23}\ \text{mol}^{-1} \qquad\qquad N = n \times L

where N is the number of particles and n is the amount in moles.

Here is why the number is so useful. One carbon-12 atom has $A_\text{r}$ = 12, and L of them weigh 12 g. One water molecule has $M_\text{r}$ = 18.0, so L water molecules weigh 18.0 g. So one mole of any substance has a mass in grams equal to its $A_\text{r}$ or $M_\text{r}$ . That is the exchange rate between the balance and the particles.

Molar mass 1.4(iii)1.2

n = \frac{m}{M} \qquad \begin{aligned} n &= \text{amount / mol}\\ m &= \text{mass / g}\\ M &= \text{molar mass / g mol}^{-1} \end{aligned}

Worked exampleAtoms in 9.00 g of water

Into moles: n = 9.00 ÷ 18.0 = 0.500 mol of H₂O molecules.
To particles: 0.500 × 6.02 × 10²³ = 3.01 × 10²³ molecules.
Each molecule has 3 atoms: 3 × 3.01 × 10²³ = 9.03 × 10²³ atoms.

Worked exampleIdentify an element from its molar mass

0.0500 mol of a metal has a mass of 3.18 g.

M = m ÷ n = 3.18 ÷ 0.0500 = 63.6 g mol⁻¹.
The closest $A_\text{r}$ is 63.5, so the metal is copper.

Masses are not always given in grams. Convert first: 1 kg = 1000 g, 1 tonne = 10⁶ g, 1 mg = 10⁻³ g.

Check yourself

Define relative atomic mass.
Calculate $M_\text{r}$ for (a) (NH₄)₂SO₄ (b) Al₂(SO₄)₃ (c) Na₂CO₃·10H₂O. Which of them should be called relative formula masses?
Calculate the amount in moles in 4.00 g of sodium hydroxide, NaOH.
What is the mass of 0.250 mol of calcium carbonate?
How many ions in total are in 10.0 g of calcium chloride, CaCl₂?
Calculate the mass in grams of one atom of iron.

Answers cover these until you have tried the questions

The weighted mean mass of an atom of an element, relative to one twelfth of the mass of an atom of carbon-12.
(a) 2(14.0 + 4 × 1.0) + 32.1 + 4(16.0) = 132.1. (b) 2(27.0) + 3(32.1 + 4 × 16.0) = 342.3. (c) 2(23.0) + 12.0 + 3(16.0) + 10(18.0) = 286.0. All three are ionic, so all three are relative formula masses.
M = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹. n = 4.00 ÷ 40.0 = 0.100 mol.
M = 100.1 g mol⁻¹. m = 0.250 × 100.1 = 25.0 g.
M = 40.1 + 2(35.5) = 111.1 g mol⁻¹. n = 10.0 ÷ 111.1 = 0.0900 mol. Each CaCl₂ gives 3 ions, so ions = 3 × 0.0900 × 6.02 × 10²³ = 1.63 × 10²³.
One mole of iron (6.02 × 10²³ atoms) has a mass of 55.8 g. One atom: 55.8 ÷ 6.02 × 10²³ = 9.27 × 10⁻²³ g.