Calculations You Must Know
Every calculation type in IGCSE Chemistry 0620: moles, concentration, gas volume, yield, purity, empirical formula, and energetics.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
This page lists every calculation type you might face in 0620, with the formula, a worked setup, and common pitfalls. Use alongside stoichiometry equations and mole concept explained.
1. Relative formula mass (Mr)
Formula: Mr = sum of all Ar values in the formula
Worked example: Mr of Ca(OH)2 = 40 + 2(16 + 1) = 40 + 34 = 74
Pitfall: forgetting to multiply by the subscript outside brackets.
2. Moles from mass
Formula: moles = mass (g) / Mr
Worked example: moles of 5.3 g of Na2CO3 (Mr = 106)
moles = 5.3 / 106 = 0.05 mol
Rearranged: mass = moles x Mr
3. Moles from concentration and volume
Formula: moles = concentration (mol/dm3) x volume (dm3)
Worked example: moles in 25.0 cm3 of 0.10 mol/dm3 HCl
Volume in dm3 = 25.0 / 1000 = 0.025 dm3
moles = 0.10 x 0.025 = 0.0025 mol
Pitfall: always convert cm3 to dm3 by dividing by 1000.
Rearranged: concentration = moles / volume (dm3)
4. Moles from gas volume (at r.t.p.)
Formula: moles = volume (dm3) / 24
At room temperature and pressure, 1 mol of any gas = 24 dm3 = 24000 cm3.
Worked example: moles in 600 cm3 of H2 at r.t.p.
Volume in dm3 = 600 / 1000 = 0.6 dm3
moles = 0.6 / 24 = 0.025 mol
Rearranged: volume (dm3) = moles x 24
5. Reacting mass calculations
Method:
- Write the balanced equation
- Calculate moles of the known substance
- Use the mole ratio
- Convert moles of the unknown to mass, volume, or concentration
Worked example: What mass of MgO is produced from 2.4 g of Mg?
2Mg + O2 -> 2MgO
- Moles of Mg = 2.4 / 24 = 0.1 mol
- Ratio Mg : MgO = 2 : 2 = 1 : 1
- Moles of MgO = 0.1 mol
- Mass of MgO = 0.1 x 40 = 4.0 g
6. Titration calculations
Worked example: 25.0 cm3 of NaOH is neutralised by 20.0 cm3 of 0.050 mol/dm3 H2SO4. Find the concentration of NaOH.
2NaOH + H2SO4 -> Na2SO4 + 2H2O
- Moles of H2SO4 = 0.050 x (20.0/1000) = 0.001 mol
- Ratio NaOH : H2SO4 = 2 : 1
- Moles of NaOH = 0.002 mol
- Concentration of NaOH = 0.002 / 0.025 = 0.08 mol/dm3
Pitfall: always check the mole ratio from the balanced equation — it is not always 1:1.
7. Empirical formula
Method:
- Write the mass (or percentage) of each element
- Divide each by its Ar
- Divide all by the smallest result
- Round to whole numbers
Worked example: a compound contains 40% Ca, 12% C, 48% O.
| Element | % | / Ar | Ratio | Simplified |
|---|---|---|---|---|
| Ca | 40 | 40/40 = 1.0 | 1.0 | 1 |
| C | 12 | 12/12 = 1.0 | 1.0 | 1 |
| O | 48 | 48/16 = 3.0 | 3.0 | 3 |
Empirical formula: CaCO3
See empirical formula.
8. Percentage composition
Formula: % of element = (total Ar of element in formula / Mr) x 100
Worked example: % of N in NH4NO3 (Mr = 80)
N contributes: 2 x 14 = 28
% N = (28/80) x 100 = 35%
9. Percentage yield
Formula: percentage yield = (actual yield / theoretical yield) x 100
Worked example: theoretical yield of CuO is 8.0 g; actual yield is 6.4 g.
% yield = (6.4 / 8.0) x 100 = 80%
Reasons yield < 100%: incomplete reaction, loss during transfer, side reactions, loss during purification. See percentage yield.
10. Percentage purity
Formula: percentage purity = (mass of pure substance / mass of impure sample) x 100
Worked example: 10.0 g of impure CaCO3 contains 9.2 g of CaCO3.
% purity = (9.2 / 10.0) x 100 = 92%
See percentage purity.
11. Water of crystallisation
Find x in CuSO4.xH2O given mass data.
Method: find moles of anhydrous CuSO4 and moles of H2O lost, then divide.
12. Energy change calculations
Formula: energy change = mass of solution (g) x specific heat capacity (4.2 J/g/C) x temperature change (C)
q = m x c x DeltaT
Worked example: 100 cm3 of solution rises by 6.5 C.
q = 100 x 4.2 x 6.5 = 2730 J = 2.73 kJ
Assume: density of solution = 1 g/cm3, so 100 cm3 = 100 g. See energy changes in reactions.
Common unit conversions
| From | To | Operation |
|---|---|---|
| cm3 | dm3 | Divide by 1000 |
| dm3 | cm3 | Multiply by 1000 |
| g | kg | Divide by 1000 |
| J | kJ | Divide by 1000 |
4.0 g of sodium hydroxide (Mr = 40) is dissolved in water to make 500 cm3 of solution. Calculate the concentration in mol/dm3. [3 marks]
- Moles of NaOH = 4.0 / 40 = 0.1 mol [1]
- Volume in dm3 = 500 / 1000 = 0.5 dm3 [1]
- Concentration = 0.1 / 0.5 = 0.2 mol/dm3 [1]
Calculate the volume of CO2 produced at r.t.p. when 5.0 g of CaCO3 reacts with excess HCl. (Ar: Ca = 40, C = 12, O = 16) [3 marks]
- CaCO3 + 2HCl -> CaCl2 + H2O + CO2
- Moles of CaCO3 = 5.0 / 100 = 0.05 mol; ratio 1:1, so moles of CO2 = 0.05 mol [1]
- Volume = 0.05 x 24 = 1.2 dm3 [1]
- (or 1200 cm3) [1]
Studying this yourself? Tutoring arrangements are normally made by a parent or guardian. Message us for the details to share with them, or send them this page.