You can do every concentration calculation perfectly on paper and still fail an exam question — because the numbers are wrapped in a real-world scenario you don't recognise. This lesson teaches you to strip away the context, find the numbers, and apply the formulas you already know. That's the skill that separates 70% students from 90% students.
📚 Core Content
Real-world chemicals are rarely 100% pure. A bag of "sodium hydroxide" from a supply company might be 97% NaOH with 3% Na₂CO₃ (absorbed from air) and trace water. If you calculate moles assuming 100% purity, your answer will be wrong — and in an industrial or medical context, that error has consequences.
Percentage purity tells you what fraction of a sample is actually the substance you want:
When purity is involved, you must apply it before converting to moles. The workflow is:
HSC questions often use realistic scenarios to wrap the same calculation in unfamiliar language. These examples show you the range of contexts you might see:
Normal saline for intravenous infusion is 0.9% NaCl (by mass/volume — 9 g per 1000 mL). This is isotonic with blood plasma.
≈ 0.154 mol L⁻¹ NaClSafe pool chlorine levels are 1–3 ppm (mg/L) of free Cl₂. Above 5 ppm is irritating; below 0.5 ppm allows bacteria to grow.
≈ 1.4–4.2 × 10⁻⁵ mol L⁻¹Average ocean salinity is 35 g per kilogram of seawater (3.5% by mass). Composed mainly of NaCl with smaller amounts of MgCl₂, MgSO₄.
≈ 0.60 mol L⁻¹ NaClChildren's paracetamol suspension contains 250 mg of paracetamol per 5 mL. Dosing is weight-based to avoid overdose.
≈ 0.332 mol L⁻¹ (MM = 151.16 g mol⁻¹)🧮 Worked Examples
🧪 Activities
Show full working below:
Complete in your workbook.
Show full working below:
Complete in your workbook.
| Contaminant | Formula | MM (g mol⁻¹) | Limit (mg L⁻¹) | Limit (mol L⁻¹) — calculate |
|---|---|---|---|---|
| Nitrate | NO₃⁻ | 62.00 | 50.0 | ? |
| Fluoride | F⁻ | 19.00 | 1.50 | ? |
| Lead | Pb²⁺ | 207.20 | 0.010 | ? |
| Arsenic | As | 74.92 | 0.010 | ? |
Complete the table calculations and answer the questions below:
Complete in your workbook.
❓ Multiple Choice
1. A sample of K₂CO₃ is 94.0% pure. What mass of pure K₂CO₃ is present in 8.00 g of the sample?
2. A student dissolves 10.0 g of 90.0% pure NaOH in water to make 250 mL of solution. What is the concentration? (MM of NaOH = 39.997 g mol⁻¹)
3. A contaminant is present at 2.00 mg L⁻¹. Its molar mass is 50.0 g mol⁻¹. What is the molar concentration?
4. A chemist needs 0.300 mol of CaCO₃ from a 75.0% pure sample. What mass of impure sample must be weighed? (MM of CaCO₃ = 100.09 g mol⁻¹)
5. Normal saline is labelled "0.9% NaCl w/v", meaning 0.9 g per 100 mL. What is the molar concentration? (MM of NaCl = 58.44 g mol⁻¹)
✍️ Short Answer
6. A laboratory technician has a supply of sulfuric acid (H₂SO₄) that is labelled 98.0% pure (by mass). The solution has a density of 1.84 g mL⁻¹. (a) Calculate the mass of H₂SO₄ in 10.0 mL of this concentrated acid. (b) Calculate the number of moles of H₂SO₄ in 10.0 mL. (H = 1.008, S = 32.06, O = 15.999) 4 MARKS
Type your answer:
Answer in workbook.
7. A river water sample is tested and found to contain chloride ions (Cl⁻) at 250 mg L⁻¹. The river feeds into a reservoir with a capacity of 5.00 × 10⁸ L. (a) Express the chloride concentration in mol L⁻¹. (b) Calculate the total mass of Cl⁻ in the reservoir in kilograms. (Cl = 35.453) 4 MARKS
Type your answer:
Answer in workbook.
8. A student analyses a sample of impure magnesium oxide (MgO). They dissolve 2.50 g of the sample in excess acid and determine that the solution contains 0.0580 mol of Mg²⁺ ions. (a) Calculate the mass of pure MgO in the sample. (b) Calculate the percentage purity of the sample. (Mg = 24.305, O = 15.999) 4 MARKS
Type your answer:
Answer in workbook.
Scenario A:
n(NaCl) = 3.51 ÷ 58.443 = 0.06006 mol V = 0.500 L; c = 0.06006 ÷ 0.500 = 0.120 mol L⁻¹0.120 mol L⁻¹ is less concentrated than physiological saline (0.154 mol L⁻¹) — this is a hypotonic rehydration solution.
Scenario B:
n = c × V = 0.500 × 2.00 = 1.00 mol m(pure) = 1.00 × 100.09 = 100.1 g m(sample) = 100.1 ÷ 0.880 = 113.7 g ≈ 114 g(i) Highest molar limit: NO₃⁻. This doesn't mean it's least dangerous — molar concentration depends on MM. Nitrate has a higher limit because it is less acutely toxic than lead or arsenic at equivalent mass concentrations.
(ii) Pb²⁺: 0.030 mg/L = 3.0 × 10⁻⁵ g/L; c = 3.0 × 10⁻⁵ ÷ 207.20 = 1.45 × 10⁻⁷ mol L⁻¹This exceeds the ADWG limit of 4.83 × 10⁻⁸ mol L⁻¹ — the water is unsafe.
1. B — m(pure) = 8.00 × 0.940 = 7.52 g.
2. C — m(pure) = 10.0 × 0.900 = 9.00 g. n = 9.00 ÷ 39.997 = 0.225 mol. c = 0.225 ÷ 0.250 = 0.900 mol L⁻¹. (Note: options A and C are identical in this question — both correct at 0.900 mol L⁻¹; the intended distinction was meant to differentiate with/without purity step.)
3. D — 2.00 mg/L = 2.00 × 10⁻³ g/L. c = 2.00 × 10⁻³ ÷ 50.0 = 4.00 × 10⁻⁵ mol L⁻¹.
4. A — m(pure) = 0.300 × 100.09 = 30.03 g. m(sample) = 30.03 ÷ 0.750 = 40.0 g.
5. B — 0.9 g per 100 mL = 9.0 g L⁻¹. c = 9.0 ÷ 58.44 = 0.154 mol L⁻¹.
Q6 (4 marks):
(a) mass of 10.0 mL = density × V = 1.84 × 10.0 = 18.4 g m(pure H₂SO₄) = 18.4 × 0.980 = 18.03 g (b) MM(H₂SO₄) = 2(1.008) + 32.06 + 4(15.999) = 98.072 g mol⁻¹ n = 18.03 ÷ 98.072 = 0.184 molQ7 (4 marks):
(a) 250 mg/L = 0.250 g/L; c = 0.250 ÷ 35.453 = 7.05 × 10⁻³ mol L⁻¹ (b) Total mass = 0.250 g/L × 5.00 × 10⁸ L = 1.25 × 10⁸ g = 1.25 × 10⁵ kgQ8 (4 marks):
(a) MM(MgO) = 24.305 + 15.999 = 40.304 g mol⁻¹ m(MgO) = n × MM = 0.0580 × 40.304 = 2.338 g (b) % purity = (2.338 ÷ 2.50) × 100 = 93.5%Tick when you've finished all activities and checked your answers.