How to do mole calculations
The three-step route from mass to moles to mass, one worked example using a balanced equation, and the ratio mistake that loses the method marks.
What the examiner is testing
Whether you can move between grams and moles, and whether you use the balanced equation to get from one substance to another. The marks are spread across the steps, so showing the route matters more than the final number.
The one formula you need:
$$ \text{moles} = \frac{\text{mass in g}}{M_r} $$
The method
- Balance the equation. Nothing works without this.
- Work out \( M_r \) for the substance you were given data about.
- Convert its mass to moles.
- Use the balancing numbers to get moles of the substance you want.
- Convert those moles back to mass with that substance's \( M_r \).
Every question of this type is the same three moves: to moles, across the equation, back to mass.
Worked example
What mass of magnesium oxide is produced when 6.0 g of magnesium burns completely in oxygen?
(\( A_r \): Mg = 24, O = 16)
Step 1 — balance.
$$ 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} $$
Step 2 — \( M_r \) of what you were given. Magnesium is an element: \( M_r = 24 \).
Step 3 — mass to moles.
$$ n(\text{Mg}) = \frac{6.0}{24} = 0.25 \text{ mol} $$
Step 4 — across the equation. The ratio of Mg to MgO is \( 2 : 2 \), which is \( 1 : 1 \):
$$ n(\text{MgO}) = 0.25 \text{ mol} $$
Step 5 — moles back to mass. \( M_r(\text{MgO}) = 24 + 16 = 40 \):
$$ \text{mass} = n \times M_r = 0.25 \times 40 = 10 \text{ g} $$
Answer: 10 g of magnesium oxide.
Check by conservation of mass: 6.0 g of magnesium gained 4.0 g of oxygen. That is 0.25 mol of O atoms at 16 g/mol — consistent. Mass is conserved, so the product must weigh more than the metal you started with. If your answer is smaller than 6.0 g, something has gone wrong.
The three mistakes that lose marks
1. Skipping the balancing. With \( \text{Mg} + \text{O}_2 \rightarrow \text{MgO} \) the ratio still looks like 1:1 by luck, but try it with aluminium and it falls apart. Balance first, every time.
2. Using the wrong \( M_r \) in step 5. You need the \( M_r \) of the substance you are calculating, not the one you started with. Label each number with what it belongs to.
3. Forgetting brackets. \( M_r \) of \( \text{Ca(OH)}_2 \) is \( 40 + 2(16 + 1) = 74 \), not 57. The subscript applies to everything in the bracket.
30-second recap
Balance, find \( M_r \), divide mass by \( M_r \) to get moles, cross the equation using the balancing numbers, multiply back by the new \( M_r \). Then check the mass makes sense — products of a combination reaction always weigh more than the element you started with.