Describe how a student could use the relative formula mass of a substance and a known number of moles to calculate the mass of that substance needed for an experiment. Use the compound calcium carbonate (CaCO₃) as your example throughout your answer.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Relative atomic masses: Ca = 40, C = 12, O = 16
Model answer (5 marks)
1. Add the relative atomic masses of the atoms in CaCO₃: Ca 40 + C 12 + 3×O 16 = 40 + 12 + 48 = 100.
2. The relative formula mass (Mr) of CaCO₃ is therefore 100.
3. The relationship between mass, moles and Mr is mass = moles × Mr (or moles = mass ÷ Mr).
4. The molar mass of CaCO₃ in g mol⁻¹ is numerically equal to its Mr, so 100 g mol⁻¹.
5. For example, to prepare 0.50 mol of CaCO₃, the required mass is 0.50 mol × 100 g mol⁻¹ = 50 g.
2. The relative formula mass (Mr) of CaCO₃ is therefore 100.
3. The relationship between mass, moles and Mr is mass = moles × Mr (or moles = mass ÷ Mr).
4. The molar mass of CaCO₃ in g mol⁻¹ is numerically equal to its Mr, so 100 g mol⁻¹.
5. For example, to prepare 0.50 mol of CaCO₃, the required mass is 0.50 mol × 100 g mol⁻¹ = 50 g.
Examiner tips
- Use the exact formula mass 100; show the addition. State the equation mass = moles × Mr. Mention that Mr = molar mass in g mol⁻¹. Give a numerical example with the calculation shown.
Common mistakes
- Using 48 for the O contribution instead of 3×16. Confusing Mr with molar mass units. Omitting the multiplication step in the example calculation.
Mark scheme (5 marks)
- Add up the relative atomic masses of all atoms in the formula to find the relative formula mass (Mr)
- Correct calculation of Mr of CaCO₃ = 100 (40 + 12 + 48)
- State the relationship: mass = moles × Mr (or moles = mass ÷ Mr)
- Explain that the molar mass in grams per mole is numerically equal to the Mr
- Apply the method to give a correct example mass for a stated number of moles of CaCO₃
Key terms in this question
Related
- All AQA A-Level Chemistry (7405) revision notes →
- How to answer a "Describe" question →
- Decode the mark scheme abbreviations →
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