Skip to content
InkFlaskCambridge IGCSE
Contains the answers

Answers · The mole

How many molecules are in a spoonful of water? — answer key

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

1. LQ1 · Practice

Model answer. One mole of any substance contains 6.02 × 10²³ particles, the Avogadro constant, so both samples hold 6.02 × 10²³ atoms. Their masses differ, because the atoms differ in mass: 1 mol of sulfur is 32 g and 1 mol of zinc is 65 g. Heavier atoms make a mole heavier, not bigger in number.

A full answer includes

  • I chose: both samples contain the same number of atoms.
  • I can say why: one mole of any substance holds 6.02 × 10²³ particles.
  • I did not choose the same mass: 1 mol of sulfur is 32 g and 1 mol of zinc is 65 g.

Also accept

  • Both samples contain the same number of atoms.

Watch for

  • “They have the same mass, because each is 1 mol.” — One mole always holds the same number of particles, but a zinc atom is about twice as heavy as a sulfur atom, so 1 mol of zinc (65 g) is heavier than 1 mol of sulfur (32 g).
  • “The zinc holds more atoms, because zinc atoms are heavier.” — A mole is a number of particles, whatever they weigh. Heavier atoms make the mole heavier, not more numerous.
  • “Each sample has a mass of 6.02 × 10²³ grams.” — 6.02 × 10²³ is a number of particles, not a number of grams. The mass of a mole is its molar mass.

2. LQ2 · Practice

Model answer. The molar mass has the same number as the Mr, with the unit g/mol. CO2: 12 + 2 × 16 = 44, so M = 44 g/mol. CaCO3: 40 + 12 + 3 × 16 = 100, so M = 100 g/mol.

Checked automatically

  • (a) Molar mass of CO2: 44 g/mol (exact value)
    • Common wrong answer 28 : Count both oxygen atoms: CO2 has two.
  • (b) Molar mass of CaCO3: 100 g/mol (exact value)
    • Common wrong answer 68 : CO3 has three oxygen atoms, not one.

A full answer includes

  • (a) I worked out 12 + 2 × 16.
  • (a) I got 44 g/mol.
  • (b) I worked out 40 + 12 + 3 × 16.
  • (b) I got 100 g/mol.
  • I did not leave out the unit g/mol.

Also accept

  • 44 g/mol; 100 g/mol

Watch for

  • “Molar mass has no unit, like Mr.” — Mr compares masses and has no unit, but molar mass is the mass of one mole, so it is in g/mol.

3. LQ3 · Practice

Model answer. Use n = m/M. (a) 14.9 g ÷ 74.5 g/mol = 0.200 mol. (b) Ammonia has M = 14 + 3 × 1 = 17 g/mol, so 4.25 g ÷ 17 g/mol = 0.250 mol. In both, the grams cancel and leave mol.

Checked automatically

  • (a) Amount of potassium chloride: 0.200 mol (3 significant figures)
    • Common wrong answer 5.00 : You divided the molar mass by the mass. Divide the mass by the molar mass.
    • Common wrong answer 1110 : You multiplied. To find an amount, divide the mass by the molar mass.
  • (b) Amount of ammonia: 0.250 mol (3 significant figures)
    • Common wrong answer 4.00 : You divided 17 by 4.25. Divide the mass by the molar mass.
    • Common wrong answer 72.3 : You multiplied. To find an amount, divide the mass by the molar mass.
    • Common wrong answer 0.283 : NH3 has three hydrogen atoms, so M = 14 + 3 × 1 = 17 g/mol.

A full answer includes

  • (a) I worked out 14.9 ÷ 74.5.
  • (a) I got 0.200 mol.
  • (b) I found M = 14 + 3 × 1 = 17 g/mol.
  • (b) I worked out 4.25 ÷ 17.
  • (b) I got 0.250 mol.
  • I did not divide the molar mass by the mass.

Also accept

  • 0.200 mol; 0.250 mol

Watch for

  • “Amount = molar mass ÷ mass.” — Divide the mass by the molar mass, n = m/M. The units show it: g ÷ g/mol leaves mol.
  • “NH3 has M = 15 g/mol.” — The 3 belongs to the hydrogen: three H atoms, so M = 14 + 3 × 1.

4. LQ4 · Practice

Model answer. (a) CuO: M = 64 + 16 = 80 g/mol, so n = m/M = 6.00 g ÷ 80 g/mol = 0.0750 mol. (b) MgO: M = 24 + 16 = 40 g/mol, so m = n × M = 0.350 mol × 40 g/mol = 14.0 g (the moles cancel).

Checked automatically

  • (a) Amount of copper(II) oxide: 0.0750 mol (3 significant figures)
    • Common wrong answer 13.3 : You divided 80 by 6.00. Divide the mass by the molar mass.
    • Common wrong answer 480 : You multiplied. To find an amount, divide the mass by the molar mass.
  • (b) Mass of magnesium oxide: 14.0 g (3 significant figures)
    • Common wrong answer 0.00875 : You divided the amount by the molar mass. For a mass, multiply: m = n × M.
    • Common wrong answer 114 : You divided the molar mass by the amount. For a mass, multiply: m = n × M.

A full answer includes

  • (a) I found M = 64 + 16 = 80 g/mol.
  • (a) I worked out 6.00 ÷ 80.
  • (a) I got 0.0750 mol.
  • (b) I found M = 24 + 16 = 40 g/mol.
  • (b) I worked out 0.350 × 40.
  • (b) I got 14.0 g.
  • (b) I did not divide the amount by the molar mass.

Also accept

  • 0.0750 mol; 14.0 g

Watch for

  • “Amount = molar mass ÷ mass.” — Divide the mass by the molar mass, n = m/M. The units show it: g ÷ g/mol leaves mol.
  • “Mass = amount ÷ molar mass.” — Rearrange n = m/M by multiplying both sides by M: m = n × M.

5. LQ5 · Practice

Model answer. M = m/n = 4.80 g ÷ 0.120 mol = 40.0 g/mol. The Ar is the same number without the unit, 40, and the element with Ar 40 in the Periodic Table is calcium.

Checked automatically

  • (a) Molar mass: 40.0 g/mol (3 significant figures)
    • Common wrong answer 0.0250 : You divided the amount by the mass: that is moles in each gram. Divide the mass by the amount.
    • Common wrong answer 0.576 : You multiplied. Molar mass is the mass divided by the amount.
  • (b) Relative atomic mass: 40 (exact value)

A full answer includes

  • (a) I worked out 4.80 ÷ 0.120.
  • (a) I got 40.0 g/mol.
  • (b) I wrote Ar = 40, with no unit.
  • (c) I named calcium.
  • (a) I did not divide 0.120 by 4.80.

Also accept

  • 40.0 g/mol; Ar 40; calcium

Watch for

  • “Molar mass has no unit, like Mr.” — Mr compares masses and has no unit, but molar mass is the mass of one mole, so it is in g/mol.
  • “Molar mass = amount ÷ mass.” — That gives mol/g, the moles in each gram. Molar mass is grams in each mole: mass ÷ amount.

6. LQ6 · Practice

Options as numbered on the pupil sheet: 1 — 1 mol; 2 — 2 mol; 3 — 3 mol; 4 — 4 mol.

Model answer. One mole of formulae holds one mole of each particle the formula has once, two moles of a particle it has twice, and so on. NH3 has 1 + 3 = 4 atoms; MgCl2 gives one Mg2+ and two Cl− ions, 3 in all; argon is single atoms; Al2O3 has two Al3+ ions; SO2 has two oxygen atoms.

A full answer includes

  • I matched the atoms in ammonia, NH3, to 4 mol.
  • I matched the ions in magnesium chloride, MgCl2, to 3 mol.
  • I matched the atoms in argon to 1 mol.
  • I matched the Al3+ ions in aluminium oxide, Al2O3, to 2 mol.
  • I matched the oxygen atoms in sulfur dioxide, SO2, to 2 mol.
  • I did not count one mole of molecules as one mole of atoms.

Also accept

  • NH3 4, MgCl2 3, Ar 1, Al2O3 2, SO2 2

Watch for

  • “A mole of molecules holds 6.02 × 10²³ atoms.” — It holds 6.02 × 10²³ molecules. Multiply by the number of atoms in each molecule to count the atoms.
  • “MgCl2 gives two ions: Mg2+ and Cl2−.” — Chloride ions are single, Cl−. Each MgCl2 gives one Mg2+ and two separate Cl− ions.
  • “Argon is a gas, so it is made of Ar2 molecules.” — The noble gases are single atoms. Their formula has no 2.
  • “Al2O3 holds one aluminium ion in each formula.” — The small 2 after Al means two Al3+ ions in each formula unit.

7. LQ7 · Practice

Model answer. Molecules: 0.350 × 6.02 × 10²³ = 2.107 × 10²³, which is 2.11 × 10²³ to three significant figures. Each NO2 molecule has one nitrogen and two oxygen atoms, three atoms, so there are 3 × 2.107 × 10²³ = 6.32 × 10²³ atoms. Use the unrounded 2.107 in the second step.

Checked automatically

  • (a) Molecules: 2.11×10^23 (3 significant figures)
    • Common wrong answer 1.72e24 : You divided 6.02 × 10²³ by 0.350. Multiply the amount by the Avogadro constant.
  • (b) Atoms: 6.32×10^23 (3 significant figures)
    • Common wrong answer 2.11e23 : That is the number of molecules. Each NO2 molecule has three atoms.
    • Common wrong answer 4.21e23 : NO2 has one nitrogen atom as well as two oxygen atoms: three atoms in each molecule.

A full answer includes

  • (a) I worked out 0.350 × 6.02 × 10²³.
  • (a) I got 2.11 × 10²³ molecules.
  • (b) I counted three atoms in each NO2 molecule.
  • (b) I worked out 3 × 2.107 × 10²³.
  • (b) I got 6.32 × 10²³ atoms.
  • (b) I did not give the number of molecules as the number of atoms.

Also accept

  • 2.11 × 10²³ molecules; 6.32 × 10²³ atoms

Watch for

  • “A mole of molecules holds 6.02 × 10²³ atoms.” — It holds 6.02 × 10²³ molecules. Multiply by the number of atoms in each molecule to count the atoms.

8. LQ8 · Practice

Model answer. Divide the number of molecules by the Avogadro constant: 2.408 × 10²³ ÷ 6.02 × 10²³ = 0.400 mol. Methane has M = 12 + 4 × 1 = 16 g/mol, so m = n × M = 0.400 × 16 = 6.40 g.

Checked automatically

  • (a) Amount of methane: 0.400 mol (3 significant figures)
    • Common wrong answer 1.45e47 : You multiplied by the Avogadro constant. To go from particles to moles, divide.
  • (b) Mass of methane: 6.40 g (3 significant figures)
    • Common wrong answer 0.0250 : You divided the amount by 16. For a mass, multiply: m = n × M.
    • Common wrong answer 40.0 : You divided 16 by the amount. For a mass, multiply: m = n × M.
    • Common wrong answer 5.20 : CH4 has four hydrogen atoms, so M = 12 + 4 × 1 = 16 g/mol.

A full answer includes

  • (a) I worked out 2.408 × 10²³ ÷ 6.02 × 10²³.
  • (a) I got 0.400 mol.
  • (b) I found M = 12 + 4 × 1 = 16 g/mol.
  • (b) I worked out 0.400 × 16.
  • (b) I got 6.40 g.
  • (a) I did not multiply by the Avogadro constant.

Also accept

  • 0.400 mol; 6.40 g

Watch for

  • “Particles → moles: multiply by the Avogadro constant.” — Each mole holds 6.02 × 10²³ particles, so the number of moles is the particles ÷ 6.02 × 10²³.
  • “Amount = molar mass ÷ mass.” — Divide the mass by the molar mass, n = m/M. The units show it: g ÷ g/mol leaves mol.

9. LT1 · Apply

Model answer. Glucose: M = 6 × 12 + 12 × 1 + 6 × 16 = 180 g/mol, so n = 5.40 ÷ 180 = 0.0300 mol and 0.0300 × 6.02 × 10²³ = 1.81 × 10²² molecules. Urea: M = 12 + 16 + 2 × (14 + 2 × 1) = 60 g/mol, so n = 5.40 ÷ 60 = 0.0900 mol and 0.0900 × 6.02 × 10²³ = 5.42 × 10²² molecules. 5.418 × 10²² ÷ 1.806 × 10²² = 3: the urea pot holds three times as many molecules. The reason: a glucose molecule is 180 ÷ 60 = 3 times as heavy, so the same mass holds only a third as many of them.

Checked automatically

  • (a) Molecules of glucose: 1.81×10^22 (3 significant figures)
    • Common wrong answer 3.25e24 : You multiplied the mass by the Avogadro constant. Find the amount first: n = m/M.
    • Common wrong answer 5.42e22 : That is the urea. Glucose has M = 180 g/mol, so 5.40 g of it is fewer moles.
  • (b) Molecules of urea: 5.42×10^22 (3 significant figures)
    • Common wrong answer 3.25e24 : You multiplied the mass by the Avogadro constant. Find the amount first: n = m/M.
    • Common wrong answer 7.39e22 : The 2 after the bracket doubles the NH2: two NH2 groups, so M = 12 + 16 + 2 × (14 + 2 × 1) = 60 g/mol.
    • Common wrong answer 1.81e22 : That is the glucose. Urea has the smaller molar mass, so the same mass is more moles.
  • (c) Times as many: 3 (within ±0.05)
    • Common wrong answer 0.333 : You divided the glucose count by the urea count. The urea pot holds more molecules, so the answer is bigger than 1.
    • Common wrong answer 1 : The pots have the same mass, not the same number of molecules. Compare the counts you found in (a) and (b).

A full answer includes

  • (a) I found M = 180 g/mol for glucose.
  • (a) I worked out 5.40 ÷ 180 = 0.0300 mol.
  • (a) I got 1.81 × 10²² molecules.
  • (b) I found M = 60 g/mol for urea.
  • (b) I worked out 5.40 ÷ 60 = 0.0900 mol.
  • (b) I got 5.42 × 10²² molecules.
  • (c) I worked out 5.418 × 10²² ÷ 1.806 × 10²² (or 180 ÷ 60).
  • (c) I got 3.
  • (c) I did not get 1: the same mass does not hold the same number of molecules.

Also accept

  • 1.81 × 10²²; 5.42 × 10²²; 3 times

Watch for

  • “The same mass holds the same number of molecules.” — Heavier molecules mean fewer of them in the same mass. Compare amounts in moles, not masses.
  • “CO(NH2)2 has one NH2 group.” — The 2 after the bracket doubles everything inside it: two N and four H atoms.