Moles: Counting Particles by Weighing

What is a mole?

Atoms are far too small to count one by one. So chemists count them in huge bundles called moles (symbol mol). One mole of anything contains 6.02 × 10²³ particles. This number is called the Avogadro constant.

The trick: one mole of a substance has a mass in grams equal to its relative formula mass (Mr). Mr is found by adding the relative atomic masses (Ar) of all atoms in the formula.

• C: Ar = 12, so 1 mol C = 12 g.
• H₂O: Mr = 1 + 1 + 16 = 18, so 1 mol H₂O = 18 g.
• CO₂: Mr = 12 + 16 + 16 = 44, so 1 mol CO₂ = 44 g.

moles (n) = mass (m) ÷ Mr, and so mass = n × Mr.

Number of particles = moles × 6.02 × 10²³.

Amounts of substances in equations

The big numbers in a balanced equation tell you the mole ratio. In 2Mg + O₂ → 2MgO, 2 mol of magnesium react with 1 mol of oxygen to make 2 mol of magnesium oxide.

To find a reacting mass, use three steps:

1. Change the mass you know into moles (n = m ÷ Mr).
2. Use the ratio from the equation to find moles of the other substance.
3. Change those moles back into mass (m = n × Mr).

Example: how much MgO from 48 g Mg? n(Mg) = 48 ÷ 24 = 2 mol. Ratio 2 : 2, so 2 mol MgO. Mass = 2 × 40 = 80 g.

Mass is conserved: the total mass of reactants equals the total mass of products.

Using moles to balance equations

If you know the masses that reacted, you can work out the balanced equation:

1. Change each mass into moles.
2. Divide every answer by the smallest one.
3. The whole-number ratio gives the big numbers in the equation.

Example: 2.4 g of carbon burns to make 8.8 g of CO₂ using 6.4 g O₂. Moles: C = 2.4 ÷ 12 = 0.2, O₂ = 6.4 ÷ 32 = 0.2, CO₂ = 8.8 ÷ 44 = 0.2. Ratio 1 : 1 : 1, so C + O₂ → CO₂.

Limiting reactants

Often one reactant is added in excess (more than needed) so that the other one reacts completely. The reactant that is used up first is the limiting reactant. It decides how much product forms.

To find it: change both masses into moles, divide each by its number in the equation, and the smaller answer is limiting.

Example: 4 g of H₂ (2 mol) with 16 g of O₂ (0.5 mol) in 2H₂ + O₂ → 2H₂O. H₂: 2 ÷ 2 = 1; O₂: 0.5 ÷ 1 = 0.5. O₂ is limiting. Water formed = 2 × 0.5 = 1 mol = 18 g.

Concentration of solutions

A solution's concentration tells you how much solute is dissolved in a volume of solution. In grams per cubic decimetre:

concentration (g/dm³) = mass of solute (g) ÷ volume (dm³)

1 dm³ = 1 litre = 1000 cm³, so divide cm³ by 1000 to get dm³. Example: 10 g of salt in 250 cm³ of solution → 10 ÷ 0.25 = 40 g/dm³. More solute or less water makes it more concentrated.

Try it: count by weighing

Weigh 20 dry rajma beans (or any beans) on a kitchen scale and find the mass of one bean. Now weigh a big handful and work out how many beans are in it without counting. Count them to check! This is exactly what chemists do with moles: they know the mass of one 'bundle' (Mr in grams), so weighing tells them how many particles they have.

Key formulas and definitions

• 1 mol = 6.02 × 10²³ particles (Avogadro constant)
• Mr = sum of Ar of all atoms in the formula
• n = m ÷ Mr; m = n × Mr
• Number of particles = n × 6.02 × 10²³
• Equation numbers = mole ratio
• Concentration (g/dm³) = mass (g) ÷ volume (dm³); 1 dm³ = 1000 cm³

Worked examples

Common mistakes

• Using Ar instead of Mr: for O₂ use 32, not 16.
• Forgetting the ratio from the equation and assuming 1 : 1 every time.
• Choosing the limiting reactant by smaller mass instead of smaller moles ÷ equation number.
• Dividing by cm³ instead of dm³ for concentration: divide cm³ by 1000 first.