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📘 GCSE Chemistry · Notes & worksheet

Quantitative Chemistry.

GCSE Higher (Grades 6-9)board-neutral (AQA / Edexcel / OCR)Worksheet: 28 marks

Notes

The mole

  • Moles = mass (g) ÷ Mr (or Ar for an element).
  • Relative formula mass (Mr) = sum of the Ar values. CO₂: 12 + 2 × 16 = 44.
  • One mole of any gas occupies 24 dm³ at room temperature and pressure.

Using equations

Balanced equations give the mole ratio. 2Mg + O₂ → 2MgO: 2 mol Mg makes 2 mol MgO.

Method:

  1. Write the balanced equation.
  2. Convert the given mass to moles.
  3. Use the ratio to find moles of the required substance.
  4. Convert to mass (or volume).

Limiting reactant

The reactant that is used up first. It sets the maximum amount of product. Work out moles of each reactant and compare against the equation ratio.

Concentration

Concentration (mol/dm³) = moles ÷ volume (dm³). Concentration (g/dm³) = mass ÷ volume (dm³). To convert cm³ to dm³, divide by 1000.

Titration calculations

  1. Moles of the known solution = concentration × volume (dm³).
  2. Use the equation ratio to find moles of the unknown.
  3. Concentration of unknown = moles ÷ its volume (dm³).

Yield and atom economy

  • Percentage yield = actual mass ÷ theoretical mass × 100.
  • Atom economy = Mr of desired product ÷ Mr of all reactants × 100 (using balanced quantities).

High atom economy means less waste and is more sustainable.

Empirical formula

Divide each mass by its Ar to get moles, divide by the smallest, then simplify to a whole-number ratio.

Common mistakes

  • Using cm³ instead of dm³ in concentration calculations.
  • Forgetting to use the balancing numbers when finding the ratio.
  • Using the Mr of the reactant when calculating moles of the product.
  • Forgetting to multiply by the balancing number in atom economy.

Exam tips

  • Lay out every calculation in steps: moles → ratio → answer.
  • Show units at every step and give the answer to the required significant figures.

Worksheet

Prints the questions only, without answers. Choose “Save as PDF” in the print window to download.

Total: 28 marks. Ar values: H = 1, C = 12, O = 16, Na = 23, Mg = 24, Fe = 56. Molar gas volume = 24 dm³.

  1. Calculate the number of moles in 11 g of carbon dioxide (CO₂). (2)
  2. 6.0 g of magnesium burns in oxygen: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide formed. (3)
  3. In an experiment, 7.5 g of a product was made. The theoretical yield was 10.0 g. Calculate the percentage yield. (1)
  4. Iron is extracted by Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the atom economy for making iron. (3)
  5. 4.0 g of sodium hydroxide (NaOH) is dissolved to make 250 cm³ of solution. Calculate the concentration in mol/dm³ and in g/dm³. (4)
  6. In a titration, 25.0 cm³ of NaOH solution was neutralised by 20.0 cm³ of 0.100 mol/dm³ HCl. Calculate the concentration of the NaOH in mol/dm³. HCl + NaOH → NaCl + H₂O. (4)
  7. 2.4 g of magnesium is added to 100 cm³ of 1.0 mol/dm³ hydrochloric acid. Mg + 2HCl → MgCl₂ + H₂.
    (a) Show that the acid is the limiting reactant. (3)
    (b) Calculate the volume of hydrogen produced at room temperature and pressure. (2)
  8. A compound contains 5.6 g of iron and 2.4 g of oxygen. Find its empirical formula. (3)
  9. Explain why a reaction with a high atom economy is more sustainable than one with a low atom economy. (3)
Show mark scheme and answers (tutor copy)

Try the worksheet first, then check your work here.

  1. Mr(CO₂) = 44 (M1). Moles = 11 ÷ 44 = 0.25 mol. (A1)
  2. Moles Mg = 6.0 ÷ 24 = 0.25 (M1). Ratio Mg : MgO = 1 : 1, so 0.25 mol MgO (M1). Mr(MgO) = 40, so mass = 0.25 × 40 = 10 g. (A1)
  3. 7.5 ÷ 10.0 × 100 = 75%. (B1)
  4. Mass of Fe in the equation = 2 × 56 = 112 (M1). Total Mr of reactants = 160 + (3 × 28) = 160 + 84 = 244 (M1). Atom economy = 112 ÷ 244 × 100 = 45.9% (about 46%). (A1)
  5. Moles = 4.0 ÷ 40 = 0.1 mol (M1). Volume = 0.25 dm³ (M1). Concentration = 0.1 ÷ 0.25 = 0.4 mol/dm³ (A1) = 0.4 × 40 = 16 g/dm³. (A1)
  6. Moles HCl = 0.100 × 0.0200 = 0.00200 mol (M1). Ratio 1 : 1, so moles NaOH = 0.00200 (M1). Concentration = 0.00200 ÷ 0.0250 (M1) = 0.0800 mol/dm³. (A1)
  7. (a) Moles Mg = 2.4 ÷ 24 = 0.10 (B1). Moles HCl = 1.0 × 0.100 = 0.10 (B1). 0.10 mol Mg needs 0.20 mol HCl but only 0.10 mol is present, so HCl is limiting. (B1)
    (b) Moles H₂ = 0.10 ÷ 2 = 0.05 mol (M1). Volume = 0.05 × 24 = 1.2 dm³. (A1)
  8. Moles Fe = 5.6 ÷ 56 = 0.1 and moles O = 2.4 ÷ 16 = 0.15 (M1). Ratio 0.1 : 0.15 = 2 : 3 (M1). Empirical formula Fe₂O₃. (A1)
  9. A high atom economy means a high proportion of reactant atoms end up in the useful product (B1). Less waste is produced, so less needs disposing of (B1). It conserves raw materials and often saves energy and money. (B1)
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