Questions on the mole, the Avogadro constant and molar mass

Question 1

Describe how a student could determine the value of x in a hydrated metal sulfate, MSO₄·xH₂O, whose anhydrous molar mass is known. Include how x is calculated and how the method shows that the result can be relied on.

Accepted Answer

Indicative content

weigh a clean, dry crucible and lid, add two to three grams of the hydrated salt and weigh again, so the mass of hydrate is known by difference
heat the crucible on a pipeclay triangle, gently at first so that the solid does not spit out, then more strongly for several minutes
let the crucible cool with its lid on, or in a desiccator, so that the anhydrous salt cannot absorb water from the air before it is weighed
reweigh, then repeat heating, cooling and weighing until two successive masses agree within 0.01 g, which shows that all the water has been driven off
the mass of water is the mass of hydrate minus the final mass of anhydrous salt, and each mass is divided by its molar mass to give an amount in moles
x is the moles of water divided by the moles of anhydrous MSO₄, rounded to a whole number; a larger sample reduces the percentage uncertainty in the masses

Levels

Level 3 (5–6 marks) weighing by difference, gentle then stronger heating, cooling with the lid or in a desiccator, heating to constant mass as the check that all water is removed, and the calculation of x from the two amounts in moles are all present; the answer explains why each step makes the result reliable
Level 2 (3–4 marks) the crucible is weighed, heated and reweighed and x comes from the mass of water lost; heating to constant mass, cooling away from moist air, or putting both masses into moles is missed or unexplained
Level 1 (1–2 marks) the hydrate is heated once and the mass loss found; x is often taken as the mass ratio of water to salt rather than the mole ratio

Question 2

Calculate the value of x in hydrated magnesium sulfate, MgSO₄·xH₂O, from a student's results, given in the table. The crucible was heated, cooled and weighed four times. Use Mg 24.3, S 32.1, O 16.0, H 1.0.

Accepted Answer

B1 heatings 3 and 4 agree, so constant mass is reached and 25.68 g is the mass of crucible and anhydrous salt M1 mass of anhydrous MgSO₄ = 25.68 − 24.36 = 1.32 g and mass of water = 27.05 − 25.68 = 1.37 g M1 n(MgSO₄) = 1.32 ÷ 120.4 = 0.01096 mol and n(H₂O) = 1.37 ÷ 18.0 = 0.07611 mol M1 ratio n(H₂O) ÷ n(MgSO₄) = 0.07611 ÷ 0.01096 = 6.94 A1 x = 7, a whole number with no unit; reject 6.94 given as the final answer

Question 3

Calculate the total number of ions in solution when a student dissolves 8.55 g of aluminium sulfate, Al₂(SO₄)₃, in water and it dissociates completely into its ions. Use Al 27.0, S 32.1, O 16.0 and L = 6.022 × 10²³ mol⁻¹.

Accepted Answer

M1 M(Al₂(SO₄)₃) = 2 × 27.0 + 3 × (32.1 + 4 × 16.0) = 342.3 g mol⁻¹ M1 n(Al₂(SO₄)₃) = 8.55 ÷ 342.3 = 0.02498 mol M1 each formula unit gives 2 Al³⁺ and 3 SO₄²⁻, five ions, so n(ions) = 5 × 0.02498 = 0.1249 mol A1 N = 0.1249 × 6.022 × 10²³ = 7.52 × 10²² ions, to three significant figures; allow error carried forward from a wrong Mr

Question 4

Calculate the value of x in hydrated sodium carbonate, Na₂CO₃·xH₂O, given that 2.86 g of the crystals leaves 1.06 g of anhydrous Na₂CO₃ after heating to constant mass. Use Na 23.0, C 12.0, O 16.0, H 1.0.

Accepted Answer

M1 mass of water = 2.86 − 1.06 = 1.80 g M1 n(Na₂CO₃) = 1.06 ÷ 106.0 = 0.0100 mol M1 n(H₂O) = 1.80 ÷ 18.0 = 0.100 mol, and the ratio H₂O : Na₂CO₃ = 0.100 ÷ 0.0100 A1 x = 10, a whole number with no unit, so the formula is Na₂CO₃·10H₂O

Question 5

Evaluate this conclusion. A student heats 0.50 g of CuSO₄·xH₂O once for two minutes, weighs it once on a two-decimal-place balance, calculates x = 4.2, and concludes that the formula is CuSO₄·4H₂O.

Accepted Answer

B1 a single heating and weighing does not show constant mass, so water may remain and x is likely to be low B1 the mass of water from 0.50 g is under 0.2 g, and a mass by difference on this balance is two readings of ±0.01 g, ±0.02 g, an uncertainty of over 10% B1 incomplete heating can only make x too low, so the true value is above 4.2 and rounding down to 4 is not justified B1 the conclusion is not supported; heating to constant mass with a larger sample would give x close to 5

Question 6

Calculate the mass of washing soda, Na₂CO₃·10H₂O, that a student must weigh out to obtain 0.0150 mol of sodium carbonate, and the mass of water that this washing soda contains. Use Na 23.0, C 12.0, O 16.0, H 1.0.

Accepted Answer

M1 M(Na₂CO₃·10H₂O) = 106.0 + 10 × 18.0 = 286.0 g mol⁻¹; allow 286 g mol⁻¹ A1 mass of washing soda = 0.0150 × 286.0 = 4.29 g, to three significant figures; error carried forward from M1 A1 mass of water = 0.0150 × 10 × 18.0 = 2.70 g, to three significant figures

Question 7

Calculate the total number of atoms in 4.40 g of carbon dioxide. L = 6.022 × 10²³ mol⁻¹.

Accepted Answer

M1 n(CO₂) = 4.40 ÷ 44.0 = 0.100 mol M1 each CO₂ molecule has three atoms, so n(atoms) = 3 × 0.100 = 0.300 mol, and N = 0.300 × 6.022 × 10²³ A1 1.81 × 10²³ atoms, to three significant figures

Question 8

Calculate the mass of magnesium oxide formed when 1.20 g of magnesium burns completely: 2Mg(s) + O₂(g) → 2MgO(s). Use Mg 24.3, O 16.0.

Accepted Answer

M1 n(Mg) = 1.20 ÷ 24.3 = 0.04938 mol M1 the ratio Mg : MgO is 1 : 1, so n(MgO) = 0.04938 mol, and M(MgO) = 40.3 g mol⁻¹ A1 mass of MgO = 0.04938 × 40.3 = 1.99 g, to three significant figures

Question 9

Compare the relative molecular mass of water, H₂O, with its molar mass, and say why the relative mass of sodium chloride is called a relative formula mass instead.

Accepted Answer

B1 Mr of water and its molar mass have the same numerical value, 18.0 B1 Mr is a ratio to one twelfth of the mass of a carbon-12 atom and has no unit, whereas molar mass is the mass of one mole, 18.0 g mol⁻¹; ignore a unit given with Mr B1 sodium chloride is a lattice of Na⁺ and Cl⁻ ions with no molecules, so its relative mass is that of one formula unit, NaCl

Question 10

Explain why a hydrated salt is heated to constant mass, rather than heated once for a fixed time, when finding its water of crystallisation.

Accepted Answer

B1 the aim is to remove all of the water of crystallisation B1 after a single heating there is no evidence that all the water has gone; some may remain in the solid B1 when two successive masses after heating and cooling agree, no more water is being lost, so the dehydration is complete

Question 11

Explain why one mole of water and one mole of carbon dioxide contain the same number of molecules but have different masses.

Accepted Answer

B1 one mole of any substance contains the same number of particles, 6.022 × 10²³ B1 a carbon dioxide molecule has a greater mass than a water molecule, Mr 44.0 against 18.0 B1 so the molar masses differ, 44.0 g mol⁻¹ against 18.0 g mol⁻¹, and the same number of molecules has a different mass

Question 12

Suggest why some of the white solid turns black when a student heats hydrated copper(II) sulfate very strongly, and the effect this has on the value of x the student calculates.

Accepted Answer

B1 the anhydrous copper(II) sulfate has started to decompose, to black copper(II) oxide and sulfur trioxide B1 the extra mass lost as sulfur trioxide is counted as water B1 the calculated moles of water are too large, so x comes out too high

Question 13

Calculate the number of chloride ions in 2.22 g of calcium chloride, CaCl₂. Use Ca 40.1, Cl 35.5 and L = 6.022 × 10²³ mol⁻¹.

Accepted Answer

M1 n(CaCl₂) = 2.22 ÷ 111.1 = 0.01998 mol M1 n(Cl⁻) = 2 × 0.01998 = 0.03996 mol, and N = 0.03996 × 6.022 × 10²³ A1 2.41 × 10²² chloride ions, to three significant figures

Question 14

Calculate the value of x in hydrated zinc sulfate, ZnSO₄·xH₂O, given that the crystals are 43.8% water by mass. M(ZnSO₄) = 161.5 g mol⁻¹; H₂O = 18.0 g mol⁻¹.

Accepted Answer

M1 in 100 g of crystals, n(H₂O) = 43.8 ÷ 18.0 = 2.433 mol M1 n(ZnSO₄) = (100 − 43.8) ÷ 161.5 = 56.2 ÷ 161.5 = 0.3480 mol, and the ratio is 2.433 ÷ 0.3480 = 6.99 A1 x = 7, a whole number, so the formula is ZnSO₄·7H₂O

Question 15

Explain why a car that burns 1.00 kg of octane, C₈H₁₈, gives off about three kilograms of carbon dioxide: 2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(l).

Accepted Answer

B1 each mole of octane gives eight moles of carbon dioxide, from the 2 : 16 ratio in the equation B1 one mole of octane, 114.0 g, gives 8 × 44.0 = 352 g of carbon dioxide, about 3.09 times the mass; accept a full calculation giving 3.09 kg B1 the extra mass is oxygen taken from the air, which ends up bonded to every carbon atom in the CO₂

Question 16

State what is meant by one mole of a substance, and state the value of the Avogadro constant with its unit.

Accepted Answer

B1 one mole is the amount of substance containing 6.022 × 10²³ particles, the same number as there are atoms in 12 g of carbon-12; allow either half alone B1 the Avogadro constant is 6.022 × 10²³ mol⁻¹; the unit is needed for the mark

Question 17

Suggest why hydrated iron(II) sulfate cannot be analysed by heating it in an open crucible to constant mass in the same way as hydrated copper(II) sulfate.

Accepted Answer

B1 iron(II) is oxidised to iron(III) by oxygen in the air when it is heated B1 the mass then changes for a reason other than the loss of water, so the mass loss does not give the mass of water

Question 18

Suggest why washing soda crystals, Na₂CO₃·10H₂O, left in an open dish for several weeks give a value of x below 10 when analysed by heating to constant mass.

Accepted Answer

B1 the crystals lose some of their water of crystallisation to the air while they stand in the open dish B1 less water is then driven off for each mole of sodium carbonate left, so the ratio of water to Na₂CO₃, x, comes out below 10

Weighing	Mass / g
Empty crucible	24.36
Crucible and hydrated salt	27.05
After heating 1	26.02
After heating 2	25.72
After heating 3	25.68
After heating 4	25.68