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InkFlaskCambridge IGCSE
Contains the answers

Answers · Concentration, gas volumes and yield

How much did you really make? — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. YQ17 · Practice

Model answer. The theoretical yield is 2.00 g and the actual yield 1.70 g, so the percentage yield is 1.70 ÷ 2.00 × 100 = 85.0 %.

Checked automatically

  • Percentage yield: 85.0 % (3 significant figures)
    • Common wrong answer 118 : You divided the theoretical yield by the actual yield. Put the actual yield on top.

A full answer includes

  • 1.70 ÷ 2.00 × 100 = 85.0 %.

Also accept

  • 85.0 %

Watch for

  • “Percentage yield = theoretical yield ÷ actual yield × 100.” — Put the mass you actually collected on top: actual ÷ theoretical × 100. A dry, pure product never gives more than 100 %.

2. YQ18 · Practice

Model answer. (a) Water has Mr = 2 × 1 + 16 = 18, and 16 of it is oxygen: 16 ÷ 18 × 100 = 88.9 %. (b) CaCO3 has Mr = 100, and 40 of it is calcium: 40.0 %.

Checked automatically

  • (a) Oxygen in water: 88.9 % (3 significant figures)
    • Common wrong answer 33.3 : You counted atoms: one in three. Use masses: 16 of the 18.
  • (b) Calcium in calcium carbonate: 40.0 % (3 significant figures)
    • Common wrong answer 20.0 : You counted atoms: one in five. Use masses: 40 of the 100.

A full answer includes

  • (a) Mr = 18; 16 ÷ 18 × 100 = 88.9 %.
  • (b) Mr = 100; 40 ÷ 100 × 100 = 40.0 %.

Also accept

  • 88.9 %; 40.0 %

Watch for

  • “Count the atoms, not their masses.” — One atom in three in water is oxygen, but the oxygen atom is much heavier. Percentage by mass uses Ar.

3. YQ19 · Practice

Model answer. An empirical formula gives the simplest whole-number ratio of the atoms of each element in a compound, such as CH3 for ethane. The molecular formula, C2H6, gives the actual numbers of atoms in one molecule.

A full answer includes

  • The simplest whole-number ratio of the atoms of each element in a compound.

Also accept

  • The simplest whole-number ratio of atoms

Watch for

  • “The empirical formula is the actual number of atoms in a molecule.” — That is the molecular formula. The empirical formula is the simplest ratio.
  • “It is the ratio of the masses.” — It is a ratio of atoms. Divide each mass by its Ar first.

4. YQ20 · Practice

Model answer. n(CuO) = 4.00 ÷ 80 = 0.0500 mol. CuO : Cu is 1 : 1, so the theoretical yield is 0.0500 × 64 = 3.20 g. Percentage yield = 2.88 ÷ 3.20 × 100 = 90.0 %. Copper is 64 of the 80 in one CuO: 64 ÷ 80 × 100 = 80.0 % by mass.

Checked automatically

  • (a) Theoretical yield of copper: 3.20 g (3 significant figures)
  • (b) Percentage yield: 90.0 % (3 significant figures)
    • Common wrong answer 72.0 : You divided by the mass of copper(II) oxide. Divide by the theoretical yield of copper, 3.20 g.
  • (c) Copper in copper(II) oxide: 80.0 % (3 significant figures)
    • Common wrong answer 20.0 : That is the oxygen. Put the copper on top: 64 ÷ 80 × 100.

A full answer includes

  • (a) n(CuO) = 4.00 ÷ 80 = 0.0500 mol, so 0.0500 mol of Cu: 0.0500 × 64 = 3.20 g.
  • (b) 2.88 ÷ 3.20 × 100 = 90.0 %.
  • (c) Cu is 64 of the 80 in CuO: 64 ÷ 80 × 100 = 80.0 %.

Also accept

  • 3.20 g; 90.0 %; 80.0 %

Watch for

  • “Percentage yield = theoretical yield ÷ actual yield × 100.” — Put the mass you actually collected on top: actual ÷ theoretical × 100. A dry, pure product never gives more than 100 %.

5. YQ21 · Practice

Model answer. n(CO2) = 0.960 ÷ 24 = 0.0400 mol, and the ratio is 1 : 1, so the sample held 0.0400 mol of CaCO3: 4.00 g. Percentage purity = 4.00 ÷ 5.00 × 100 = 80.0 %.

Checked automatically

  • (a) Mass of calcium carbonate: 4.00 g (3 significant figures)
    • Common wrong answer 23.0 : You multiplied the volume by 24. Divide: 0.960 ÷ 24 = 0.0400 mol.
    • Common wrong answer 2300 : You multiplied the volume by 24. Divide: 0.960 ÷ 24 = 0.0400 mol.
  • (b) Percentage purity: 80.0 % (3 significant figures)
    • Common wrong answer 125 : You divided the sample by the pure mass. Put the pure mass on top.

A full answer includes

  • (a) n(CO2) = 0.960 ÷ 24 = 0.0400 mol, so 0.0400 mol of CaCO3: 0.0400 × 100 = 4.00 g.
  • (b) 4.00 ÷ 5.00 × 100 = 80.0 %.

Also accept

  • 4.00 g; 80.0 %

Watch for

  • “Percentage purity = sample ÷ pure substance × 100.” — Put the pure substance on top: a sample cannot be more than 100 % pure.

6. YQ22 · Practice

Model answer. C: 2.40 ÷ 12 = 0.200 mol. H: 0.600 ÷ 1 = 0.600 mol. Dividing by the smaller, 0.200, gives C : H = 1 : 3, so the empirical formula is CH3.

Checked automatically

  • (a) Carbon atoms: 0.200 mol (3 significant figures)
  • (b) Hydrogen atoms: 0.600 mol (3 significant figures)

A full answer includes

  • (a) 2.40 ÷ 12 = 0.200 mol.
  • (b) 0.600 ÷ 1 = 0.600 mol.
  • C : H = 0.200 : 0.600 = 1 : 3, so CH3.

Also accept

  • 0.200 mol; 0.600 mol; CH3

Watch for

  • “Use the ratio of the masses, 2.40 : 0.600 = 4 : 1.” — A formula counts atoms. Divide each mass by its Ar first.

7. YQ23 · Practice

Model answer. Fe: 11.2 ÷ 56 = 0.200 mol; O: 4.80 ÷ 16 = 0.300 mol. Dividing by 0.200 gives 1 : 1.5. A ratio ending in .5 is doubled: 2 : 3, so the empirical formula is Fe2O3.

A full answer includes

  • Fe = 11.2 ÷ 56 = 0.200 mol; O = 4.80 ÷ 16 = 0.300 mol.
  • Fe : O = 1 : 1.5 = 2 : 3, so Fe2O3.

Also accept

  • Fe2O3

Watch for

  • “Round a ratio such as 1 : 1.5 to the nearest whole number.” — Only round a number very close to a whole one, such as 2.98. Multiply a ratio ending in .5 by 2, and one ending in .33 or .67 by 3.

8. YQ24 · Practice

Model answer. One CH2 unit has a mass of 14. The molecule has Mr = 42, so it holds 42 ÷ 14 = 3 units: C3H6.

A full answer includes

  • CH2 has a mass of 12 + 2 = 14.
  • 42 ÷ 14 = 3, so 3 × CH2 = C3H6.

Also accept

  • C3H6

Watch for

  • “The molecular formula is the same as the empirical formula.” — Only when Mr equals the mass of one empirical formula. Here 42 ÷ 14 = 3.

9. YQ25 · Practice

Model answer. Below 100 %: some product sticks to the apparatus or the filter paper when it is moved, some stays dissolved, or the reaction does not finish. Above 100 %: the extra mass is not product. It is water in a product that is still wet, or another substance mixed with it.

A full answer includes

  • (a) Any two: product lost when moved (on the filter paper, beaker or rod); some stays dissolved in the solution; the reaction did not finish or another reaction used up some reactant.
  • (b) The product is still wet, or it contains other substances (impurities).

Also accept

  • (a) two losses (b) wet or impure

Watch for

  • “A yield over 100 % means extra product was made.” — The equation sets the most product there can be. Extra mass is water or an impurity.

10. YT3 · Apply

Model answer. In 100 g: C 82.8 ÷ 12 = 6.90 mol, H 17.2 ÷ 1 = 17.2 mol. Dividing by 6.90 gives 1 : 2.49, which is 1 : 2.5; double it to 2 : 5, so the empirical formula is C2H5 with mass 2 × 12 + 5 = 29. 58 ÷ 29 = 2, so the molecular formula is C4H10.

Checked automatically

  • (a) Carbon atoms in 100 g: 6.90 mol (3 significant figures)
  • (b) Hydrogen atoms in 100 g: 17.2 mol (3 significant figures)
  • (c) Empirical-formula units in one molecule: 2 (exact value)
    • Common wrong answer 4 : 1 : 2.49 is really 1 : 2.5, not 1 : 2 or 1 : 3. Double it to C2H5, mass 29: 58 ÷ 29 = 2.

A full answer includes

  • (a) 82.8 ÷ 12 = 6.90 mol.
  • (b) 17.2 ÷ 1 = 17.2 mol.
  • C : H = 1 : 2.49 ≈ 1 : 2.5 = 2 : 5, so the empirical formula is C2H5 (mass 29).
  • (c) 58 ÷ 29 = 2, so the molecular formula is C4H10.

Also accept

  • 6.90 mol; 17.2 mol; 2 units: C2H5, C4H10

Watch for

  • “Round a ratio such as 1 : 1.5 to the nearest whole number.” — Only round a number very close to a whole one, such as 2.98. Multiply a ratio ending in .5 by 2, and one ending in .33 or .67 by 3.

11. YT4 · Apply

Model answer. Mr(C7H6O3) = 7 × 12 + 6 + 3 × 16 = 138, so n = 13.8 ÷ 138 = 0.100 mol. The ratio is 1 : 1, and Mr(C9H8O4) = 180, so the theoretical yield is 18.0 g and the percentage yield 15.3 ÷ 18.0 × 100 = 85.0 %. The tablet is 0.300 g of aspirin in 0.325 g: 92.3 % pure. The rest is other substances that hold the tablet together.

Checked automatically

  • (a) Theoretical yield of aspirin: 18.0 g (3 significant figures)
    • Common wrong answer 13.8 : The mass does not stay the same from reactant to product. Go through moles.
  • (b) Percentage yield: 85.0 % (3 significant figures)
    • Common wrong answer 111 : You divided by the mass of salicylic acid. Divide by the theoretical yield of aspirin, 18.0 g.
    • Common wrong answer 118 : You divided the theoretical yield by the actual yield. Put the actual yield on top.
  • (c) Percentage purity of the tablet: 92.3 % (3 significant figures)
    • Common wrong answer 108 : You divided the tablet by the aspirin. Put the pure aspirin on top.

A full answer includes

  • (a) Mr(C7H6O3) = 138; n = 13.8 ÷ 138 = 0.100 mol, so 0.100 mol of aspirin: 0.100 × 180 = 18.0 g.
  • (b) 15.3 ÷ 18.0 × 100 = 85.0 %.
  • (c) 0.300 ÷ 0.325 × 100 = 92.3 %.

Also accept

  • 18.0 g; 85.0 %; 92.3 %

Watch for

  • “Percentage yield = theoretical yield ÷ actual yield × 100.” — Put the mass you actually collected on top: actual ÷ theoretical × 100. A dry, pure product never gives more than 100 %.
  • “The product has the same mass as the reactant.” — Only the total mass is conserved. Go through moles: change to moles, use the equation, change back.