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Chemistry calculations carry significant marks across all three papers. A clear, systematic approach to calculations will earn you method marks even when you make arithmetic errors — and prevent errors in the first place. This lesson covers the golden rules, every major calculation type with worked examples, and the common traps that cost students marks every year.
Chemistry mark schemes award method marks (M marks) and answer marks (A marks) separately. If you show your method clearly and make an arithmetic error, you lose only the final answer mark. If you show no working and get the answer wrong, you lose everything.
Always write:
Include units throughout your calculation, not just at the end. This helps you check that your answer makes sense and shows the examiner your reasoning. Units can also help you spot errors — if your intermediate result has impossible units, something has gone wrong.
Give your final answer to the same number of significant figures as the data in the question (or as specified). If the question says "give your answer to 3 significant figures," do exactly that. If no guidance is given, use 3 significant figures as a default.
Common error: Rounding intermediate steps. Keep full calculator values during the calculation and only round the final answer.
Another common error: Confusing significant figures with decimal places. For example, 0.00432 has 3 significant figures but 5 decimal places. The number 4320 also has 3 significant figures (unless a trailing zero is explicitly significant).
graph TD
A[Read the question] --> B[Identify what you need to find]
B --> C[Write the relevant formula]
C --> D[Extract values from the question]
D --> E{Do units need converting?}
E -->|Yes| F[Convert units - e.g. cm³ to dm³]
E -->|No| G[Substitute values into the formula]
F --> G
G --> H[Perform the calculation]
H --> I[Check: does the answer seem reasonable?]
I -->|Yes| J[Round to appropriate sig figs]
I -->|No| K[Re-check units and formula]
K --> G
J --> L[Write final answer with units]
The fundamental relationship: n = m / M
Where n = moles, m = mass in grams, M = molar mass in g mol⁻¹.
Worked Example 1: Calculate the number of moles in 5.85 g of NaCl (M = 58.5 g mol⁻¹). n = 5.85 / 58.5 = 0.100 mol
Worked Example 2: What mass of calcium carbonate contains 0.250 mol? (M = 100.1 g mol⁻¹) m = n × M = 0.250 × 100.1 = 25.0 g
Worked Example 3: Calculate the molar mass of a compound if 3.60 g contains 0.0400 mol. M = m / n = 3.60 / 0.0400 = 90.0 g mol⁻¹
Unit trap: If mass is given in kg, convert to g first. If the question gives milligrams, convert to grams.
At room temperature and pressure (RTP): V = n × 24.0 dm³ mol⁻¹
Worked Example: What volume of CO₂ is produced when 0.500 mol of CaCO₃ decomposes completely? CaCO₃ → CaO + CO₂ (1:1 ratio) V = 0.500 × 24.0 = 12.0 dm³
Using the ideal gas equation: pV = nRT
Worked Example: Calculate the volume of 0.100 mol of gas at 300 K and 100 kPa. V = nRT/p = (0.100 × 8.314 × 300) / (100 × 10³) V = 249.42 / 100000 = 2.49 × 10⁻³ m³ = 2.49 dm³
Unit trap: Pressure must be in Pa (not kPa) and volume in m³ when using R = 8.314.
c = n / V where c = concentration in mol dm⁻³, V = volume in dm³.
Remember: convert cm³ to dm³ by dividing by 1000.
Worked Example: 0.0200 mol of NaOH is dissolved in 250 cm³ of water. Calculate the concentration. V = 250 / 1000 = 0.250 dm³ c = 0.0200 / 0.250 = 0.0800 mol dm⁻³
Common error: Using volume in cm³ directly. This gives an answer 1000 times too large, which is an immediate red flag.
Worked Example: 25.00 cm³ of Ba(OH)₂ solution is titrated with 0.100 mol dm⁻³ HCl. The mean titre is 20.40 cm³. Calculate the concentration of Ba(OH)₂.
Equation: Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O Moles of HCl = 0.100 × (20.40/1000) = 2.040 × 10⁻³ mol Mole ratio Ba(OH)₂ : HCl = 1 : 2 Moles of Ba(OH)₂ = 2.040 × 10⁻³ / 2 = 1.020 × 10⁻³ mol Concentration = 1.020 × 10⁻³ / (25.00/1000) = 0.0408 mol dm⁻³
Common error: Forgetting to use the mole ratio, giving an answer that is double (or half) the correct value.
q = mcΔT where q = energy in joules, m = mass of solution in grams, c = 4.18 J g⁻¹ K⁻¹, ΔT = temperature change in K (or °C).
Then: ΔH = -q / n (negative sign for the convention: exothermic = negative ΔH).
Worked Example: 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.8°C. Calculate the enthalpy of neutralisation.
Total volume = 100.0 cm³, so mass ≈ 100.0 g (assuming density of water) q = 100.0 × 4.18 × 6.8 = 2842.4 J Moles of HCl = 1.00 × (50.0/1000) = 0.0500 mol ΔH = -2842.4 / 0.0500 = -56848 J mol⁻¹ = -56.8 kJ mol⁻¹
Common errors:
Worked Example 1 (Strong acid): Calculate the pH of 0.0500 mol dm⁻³ HCl. [H⁺] = 0.0500 mol dm⁻³ (strong acid, fully dissociates) pH = -log(0.0500) = 1.30
Worked Example 2 (Weak acid): Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³). [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³ pH = -log(1.32 × 10⁻³) = 2.88
Worked Example 3 (Buffer): A buffer contains 0.200 mol dm⁻³ ethanoic acid and 0.150 mol dm⁻³ sodium ethanoate. Ka = 1.74 × 10⁻⁵. Calculate the pH. pH = -log(1.74 × 10⁻⁵) + log(0.150/0.200) pH = 4.76 + log(0.750) pH = 4.76 + (-0.125) = 4.63
Common error: Using the strong acid formula for a weak acid. A weak acid with c = 0.100 does NOT have pH = 1.00.
Worked Example: Experiments show the rate equation is rate = k[A][B]². When [A] = 0.100 mol dm⁻³ and [B] = 0.200 mol dm⁻³, the rate is 3.20 × 10⁻⁴ mol dm⁻³ s⁻¹. Calculate k and give its units.
k = rate / ([A][B]²) = 3.20 × 10⁻⁴ / (0.100 × 0.200²) k = 3.20 × 10⁻⁴ / (0.100 × 0.0400) = 3.20 × 10⁻⁴ / 4.00 × 10⁻³ k = 0.0800
Units: (mol dm⁻³ s⁻¹) / (mol dm⁻³)(mol dm⁻³)² = mol⁻² dm⁶ s⁻¹
| Overall order | Units of k |
|---|---|
| 0 | mol dm⁻³ s⁻¹ |
| 1 | s⁻¹ |
| 2 | mol⁻¹ dm³ s⁻¹ |
| 3 | mol⁻² dm⁶ s⁻¹ |
Worked Example (ICE table): For the reaction A ⇌ 2B, the initial concentration of A is 1.00 mol dm⁻³ and B is 0. At equilibrium, [A] = 0.60 mol dm⁻³. Calculate Kc.
A 2B I 1.00 0 C -0.40 +0.80 E 0.60 0.80 Kc = [B]² / [A] = (0.80)² / 0.60 = 0.64 / 0.60 = 1.07 mol dm⁻³
If your calculated answer seems unreasonable (a pH of 25, a molar mass of 3 million, a negative number of moles), do not cross it out and leave a blank. Instead:
| Quantity | Reasonable range at A-Level |
|---|---|
| pH | 0 to 14 (usually 1 to 13) |
| Molar mass | 1 to ~500 g mol⁻¹ |
| Concentration | 0.001 to 2 mol dm⁻³ (usually) |
| ΔH (neutralisation) | -50 to -60 kJ mol⁻¹ |
| ΔH (combustion) | -200 to -5000 kJ mol⁻¹ |
| Bond energy | 150 to 1000 kJ mol⁻¹ |
| Activation energy | 30 to 300 kJ mol⁻¹ |
Calculation marks are the most reliably bankable marks on Edexcel 9CH0. They reward structure, not creativity. A candidate with a fluent calculation technique can secure 30–40 marks across the three papers almost mechanically, leaving cognitive bandwidth for the harder application and evaluation questions. This deeper strategy section turns calculation execution into a repeatable five-step ritual that protects every available method-mark.
Calculations on Edexcel 9CH0 fall into seven recurring families:
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