Stoichiometry & the mole concept prelim questions

Answer: C

This question is about the amount of atoms. You can compare the amount in terms of number of particles or moles. However, comparing in moles is more straightforward, as you can directly convert mass and gaseous volume to number of moles. Let’s consider the four options in turn.

• As helium is monoatomic, 1 mol of helium has exactly 1 mol of atoms.
• We can divide the given mass of water by its molar mass to get the number of moles: 18 ÷ 18 = 1 mol of water molecules. However, each molecule of water, H₂O comprises 3 atoms. Therefore, 1 mol of water molecules have 1 × 3 = 3 mol of atoms.
• We can divide the given gaseous volume of hydrogen by the molar volume (24 dm³ at rtp) to get the number of moles: 48 ÷ 24 = 2 mol of hydrogen gas. However, as hydrogen gas is diatomic, each molecule of hydrogen has 2 hydrogen atoms. Therefore, 2 mol of hydrogen molecules have 2 × 2 = 4 mol of atoms.
• We can divide the given number of oxygen atoms by the Avogadro constant (6×10²³) to get the number of moles: 1.2×10²⁴ ÷ 6×10²³ = 2 mol of atoms.

Answer: D

The Avogadro constant is 6×10²³, which is equal to 1 mol. Essentially, this question is asking you which option has exactly 1 mol of the stated particle.

• Nitrogen gas is diatomic. Thus, 24 dm³ of nitrogen gas has 1 mol of N₂ molecules but 2 mol of N atoms.
• The gaseous volume of 1 dm³ is way less than the molar volume of 24 dm³ at rtp, and hence there are fewer than 1 mol of molecules.
• 962 g of tranquillityite has 1 mol of Fe₃Zr₂Ti₃Si₃O₂₄, which has 2 mol of zirconium.
• 29.3 g has 0.5 mol of NaCl. As it is a binary compound, 0.5 mol of NaCl has 1 mol of ions.

If you know the percentage mass of copper in chalcopyrites, you can work backwards to find the mass of chalcopyrites containing 100 tonnes of copper.

• STEP 1: Find the relative molecular mass, M_r of CuFeS₂

  M_r = 64 + 56 + 32×2 = 184 g

• STEP 2: Compute the percentage mass of copper in CuFeS₂

  % mass of Cu = (64÷184) × 100 = 34.782%

• STEP 3: Find the mass of chalcopyrites needed

  Mass of chalcopyrites = (100÷34.782) × 100 = 288 tonnes (3 s.f.)

  There is no need to convert to grams here as percentage mass is a ratio. You can either have a ratio of mass in grams, or a ratio of mass in tonnes. Both are fine, so long as the units are consistent.

We can first find the percentage mass of phosphorous in P₂O₅. Then using the percentage of P₂O₅ in the fertiliser, we finally get the percentage mass of phosphorous in the fertiliser.

• STEP 1: Find the relative molecular mass, M_r of P₂O₅

  M_r = 31×2 + 16×5 = 142

• STEP 2: Find the relative mass of all the phosphorous atoms in P₂O₅

  Relative mass of phosphorous in P₂O₅ = 31×2 = 62

• STEP 3: Compute the percentage mass in P₂O₅

  % mass of P in P₂O₅ = (62÷142) × 100 = 43.662%

• STEP 4: Compute the percentage mass in the fertiliser

  % mass of P in the fertiliser = 43.662% × 30% = 13.1%

Therefore, the empirical formula is C₂H₅O. To find its molecular formula:

• STEP 1: Find the relative mass of atoms in the empirical formula of C₂H₅O

  Relative mass of atoms = 2×12 + 5×1 + 16 = 45

• STEP 2: Find the molar mass of compound X

  Molar mass = 75 cm³ × 1.20 g/cm³ = 90 g

• STEP 3: Given that the molecular formula is C_2nH_5nO_n, find the value of the multiplier, n

  n × 45 = 90
  n = 90 ÷ 45 = 2

• STEP 4: Write the molecular formula

  C₄H₁₀O₂

Answer: 47%

Na₂SO₄.xH₂O (s) ⟶ Na₂SO₄ (s) + xH₂O (g)

During the heating, water was lost as steam. This contributed to the decrease in mass from 15 g to 7.95 g. Therefore, we can find the mass of water of crystallisation by taking the difference between the initial mass and final mass.

Mass of water of crystallisation = 15 – 7.95 = 7.05 g
% mass of water in crystals = (7.05÷15) × 100 = 47%

In the second part, the unknown x can be found by finding the ratio between sodium sulfate and water in the crystals, much like how we find the ratio between elements in a compound.

Therefore, x = 7 and the chemical formula for the hydrated crystal is Na₂SO₄.7H₂O.

By asking for the “minimum mass“, the question wants you to use the mole ratio to find the exact amount of aluminium needed to react with 10 g of copper(II) oxide. This is like a titration, whereby both reactants are limiting and there is no excess.

Also, because the chemical equation is not given, you need to write it down first to make sense of the reactants, products, and their mole ratio.

• STEP 1: Write the chemical equation and highlight the substances you are interested in

  3CuO (s) + 2Al (s) ⟶ 3Cu (s) + Al₂O₃ (s)

• STEP 2: Convert any given data to number of moles

  Molar mass of CuO = 64 + 16 = 80 g/mol
  No. of moles of CuO = 10 ÷ 80 = 0.125 mol

• STEP 3: State the mole ratio of reactants

  Ratio of Al to CuO = 2:3

• STEP 4: Use the mole ratio to find the number of moles of Al needed

  No. of moles of Al = 2/3 × 0.125 = 0.083333 mol

• STEP 5: Convert number of moles to mass, as required

  Molar mass of Al = 27 g/mol
  Mass of Al = 0.083333 × 27 = 2.25 g

Air has an unlimited supply of oxygen. So we shall assume that oxygen is in excess, while nickel(II) sulfide is limiting. Let’s get cracking!

• STEP 1: Write the chemical equation and highlight the substances you are interested in

  2NiS + 3O₂ ⟶ 2NiO + 2SO₂

• STEP 2: Convert any given data to number of moles

  Molar mass of NiS = 59 + 32 = 91
  No. of moles of NiS = 200 ÷ 91 = 2.1978 mol

• STEP 3: State the mole ratio of reactants

  Ratio of SO₂ to NiS = 2:2 = 1:1

• STEP 4: Use the mole ratio to find the number of moles of SO₂ produced

  No. of moles of Al = 1/1 × 2.1978 = 2.1978 mol

• STEP 5: Convert number of moles to volume of gas, as required

  Molar volume at rtp = 24 dm³
  Volume of SO₂ gas = 24 × 2.1978 = 52.7 dm³ (or 52700 cm³)

Noticed that this question provides us with the mass of both reactants? That is a big hint to first identify the limiting reactant. It is the limiting reactant that will react fully to directly determine the amount of product formed.

• STEP 1: Convert any given data to number of moles

  Molar mass of Fe₃O₄ = 56×3 + 16×4 = 232 g/mol
  No. of moles of Fe₃O₄ = 60(1000) ÷ 232 = 259 mol

  Molar mass of Al = 27 g/mol
  No. of moles of Al = 15(1000) ÷ 27 = 556 mol

• STEP 2: State the mole ratio of reactants

  Ratio of Fe₃O₄ to Al = 3:8

• STEP 3: Determine the limiting reactant

  Consider the case that Fe₃O₄ were completely reacted:

  No. of moles of Al used = 8/3 × 259 = 691 mol

  Only 556 mol of Al was provided. Therefore, there was insufficient Al to completely react with Fe₃O₄. Therefore, Al was the limiting reactant while Fe₃O₄ was in excess.

• STEP 4: State the mole ratio of limiting reactant to product

  Ratio of Al to Fe = 8:9

• STEP 5: Use the limiting reactant to calculate the amount of product formed

  No. of moles of Fe formed = 9/8 × 556 = 625.5 mol

• STEP 6: Convert number of moles to mass, as required

  Mass of Fe formed = 625.5 × 56 = 35000 g (or 35 kg)