Relative and Reacting Mass Calculation

The relative formula mass (Mr) of a compound is simply the sum of the relative atomic masses (Ar) of all the atoms in its formula. You will be given the Ar values (or read them from the Periodic Table provided in the exam).

To find Mr:

1. Count how many atoms of each element the formula contains.
2. Multiply each Ar by that number of atoms.
3. Add all the values together.

Watch the subscripts. A subscript multiplies the atom directly in front of it:

• Mr(H₂O) = 2(1) + 16 = 18
• Mr(CO₂) = 12 + 2(16) = 44
• Mr(NaOH) = 23 + 16 + 1 = 40
• Mr(CaCO₃) = 40 + 12 + 3(16) = 100

Watch the brackets — the most common slip. A subscript outside a bracket multiplies everything inside the bracket:

• Mr(Ca(OH)₂) = 40 + 2 × (16 + 1) = 40 + 34 = 74
• Mr((NH₄)₂SO₄) = 2 × (14 + 4) + 32 + 4(16) = 36 + 32 + 64 = 132

Tip: Mr has no units — it is a relative mass (a comparison), so never write "g" after an Mr value.

Atoms are far too tiny to count one by one, so chemists count them in a huge "package" called the mole.

One mole of any substance contains 6.02 × 10²³ particles. This number is the Avogadro constant (symbol L). The "particles" could be atoms, molecules or ions, depending on the substance.

The clever part: one mole of a substance has a mass equal to its Mr, in grams. That is what links the invisible world of particles to the masses you can actually weigh out:

• 1 mole of water (Mr = 18) weighs 18 g and contains 6.02 × 10²³ water molecules.
• 1 mole of CO₂ (Mr = 44) weighs 44 g.
• 1 mole of CaCO₃ (Mr = 100) weighs 100 g.

Number of particles. To find how many particles are in a sample:

$\text{particles} = \text{moles} \times 6.02 \times 10^{23}$

For example, 0.50 mol of water contains 0.50 × 6.02 × 10²³ = 3.01 × 10²³ molecules.

The link between moles (n), mass (m, in grams) and relative formula mass (Mr) is:

$n = \dfrac{m}{M_r}$

This rearranges two ways, so you can find whichever quantity is missing:

• Mass from moles: m = n × Mr
• Mr from mass and moles: Mr = m ÷ n

The triangle below helps you rearrange: cover the quantity you want and read off the rest.

Worked example (moles from mass). How many moles in 24.0 g of NaOH?

• Mr(NaOH) = 23 + 16 + 1 = 40
• n = 24.0 ÷ 40 = 0.60 mol

Worked example (mass from moles). What is the mass of 0.50 mol of CaCO₃?

• Mr(CaCO₃) = 40 + 12 + 3(16) = 100
• m = 0.50 × 100 = 50 g

Units matter. Mass must be in grams before you substitute. If a question gives a mass in kilogrammes or tonnes, convert to grams first (× 1000 or × 1 000 000).

A balanced equation tells you the ratio in which substances react — the big numbers in front (the coefficients) give the mole ratio. This lets you predict the mass of product from a mass of reactant (or vice versa).

The reliable five-step route:

1. Balance the equation (the ratio comes from the coefficients).
2. Moles of the substance you know: n = m ÷ Mr.
3. Mole ratio: read it off the balanced equation.
4. Moles of the substance you want: apply the ratio.
5. Convert back to mass: m = n × Mr.

Worked example (simple 1 : 1 ratio). What mass of magnesium oxide forms when 6.0 g of magnesium burns in excess oxygen?

$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$

• n(Mg) = 6.0 ÷ 24 = 0.25 mol
• Ratio Mg : MgO = 2 : 2 = 1 : 1, so n(MgO) = 0.25 mol
• m(MgO) = 0.25 × 40 = 10 g

Worked example (non-1 : 1 ratio). What mass of chlorine reacts with 5.4 g of aluminium?

$2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$

• n(Al) = 5.4 ÷ 27 = 0.20 mol
• Ratio Al : Cl₂ = 2 : 3, so n(Cl₂) = 0.20 × (3 ÷ 2) = 0.30 mol
• m(Cl₂) = 0.30 × 71 = 21.3 g (Mr of Cl₂ = 2 × 35.5)

Limiting reactant. If you are given the masses of both reactants, one will run out first — the limiting reactant — and it controls how much product forms. Find moles of each, divide by its coefficient, and the smaller answer is the limiting reactant. Base the rest of the calculation on it.