At a glance
- Convert units before substituting into a relationship.
- Use a balanced equation to connect reacting amounts in moles.
- Check the reasoning and the final units, not only the calculator result.
Three relationships to organise your thinking
Amount in moles = mass ÷ molar mass. For a solution, amount in moles = concentration × volume, when concentration is in mol dm⁻³ and volume is in dm³. A balanced reaction then tells you the ratio of reacting amounts in moles.
These ideas sit within amount-of-substance work in AQA’s A-Level chemistry specification. Check your own awarding body’s specification for the content and assessment requirements that apply to you.
Example 1: mass to amount in moles
Question: What amount is present in 5.85 g of sodium chloride? Use a molar mass of 58.5 g mol⁻¹.
Amount = mass ÷ molar mass
Amount = 5.85 g ÷ 58.5 g mol⁻¹
Amount = 0.100 mol
The units help explain the operation: dividing grams by grams per mole leaves moles. As a check, one tenth of the molar mass should give one tenth of a mole.
Example 2: solution concentration and volume
Question: How many moles are in 25.0 cm³ of a 0.200 mol dm⁻³ solution?
Volume = 25.0 ÷ 1000 = 0.0250 dm³
Amount = 0.200 mol dm⁻³ × 0.0250 dm³
Amount = 0.00500 mol
Using 25.0 without converting the volume would make the answer 1000 times too large. Write the conversion as part of the working so it is easy to check.
Example 3: use the balanced equation
For the reaction Mg + 2HCl → MgCl₂ + H₂, one mole of magnesium reacts with two moles of hydrogen chloride.
If 0.0200 mol of magnesium reacts completely with sufficient acid, it requires 0.0400 mol of HCl and produces 0.0200 mol of hydrogen. The coefficients give mole ratios, not direct mass ratios.
If a problem supplies amounts for both reactants, check which limits the reaction before calculating the product. Do not assume the other reactant is in excess unless the question supports it.
Try a fresh question
Calculate the amount in 40.0 cm³ of a 0.150 mol dm⁻³ solution. Write the unit conversion before multiplying.
Show the working and answer
40.0 cm³ = 0.0400 dm³. Amount = 0.150 × 0.0400 = 0.00600 mol.
These are original worked examples, not questions taken from an exam paper. Numerical answers are given to three significant figures to match the example data.
Find the step that needs support
If the calculation is still difficult, identify whether the uncertainty is about the chemistry, selecting a relationship, rearranging it, converting units or using a ratio. That is a clearer starting point than trying more long questions without feedback.
Explore one-to-one A-Level chemistry tuition to discuss support with concepts and calculations. We confirm your course and a suitable tutor before agreeing lessons.
Sources and further reading
External resources explain the wider topic; their inclusion does not imply an endorsement of The Tutoring Hub.
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