Skip to content

Formulae, Equations and Amount of Substance

7 questions, 67 marks, every one with its mark scheme.

Or open them one at a time, as you finish each question.

Question 1

5 marks
1(a)[1 mark]

A sample of well water contains 0.0045 % fluoride ions by mass.

Which value gives the fluoride content in ppm?

Select one answer from A to D and put a cross in the box.

  • A0.45 ppm
  • B4.5 ppm
  • C45 ppm
  • D450 ppm
Mark scheme
  • 45 ppm1
1(b)[1 mark]

Washing soda crystals have the formula Na2CO3·10H2O. Which value is the relative formula mass of the crystals?

Select one answer from A to D and put a cross in the box.

  • A106.0
  • B124.0
  • C286.0
  • D466.0
Mark scheme
  • 286.01
1(c)[1 mark]

How many oxygen atoms are present in 4.0 g of oxygen gas, O2?

[Avogadro constant L = 6.02 × 1023 mol−1]

Select one answer from A to D and put a cross in the box.

  • A7.5 × 1022
  • B1.5 × 1023
  • C3.0 × 1023
  • D2.4 × 1024
Mark scheme
  • 1.5 × 10231
1(d)[1 mark]

Propane burns completely in oxygen.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

What volume of oxygen is needed to burn 20 cm3 of propane completely? All volumes are measured at the same temperature and pressure.

Select one answer from A to D and put a cross in the box.

  • A20 cm3
  • B60 cm3
  • C100 cm3
  • D160 cm3
Mark scheme
  • 100 cm31
1(e)[1 mark]

Ammonium nitrate, NH4NO3, is used as a fertiliser. Which value is the percentage of its mass that is nitrogen?

Select one answer from A to D and put a cross in the box.

  • A17.5 %
  • B35.0 %
  • C46.7 %
  • D60.0 %
Mark scheme
  • 35.0 %1

Question 2

13 marks

Lithium hydroxide is used to remove carbon dioxide from the air inside a spacecraft.

2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)

2(a)[2 marks]

A canister on a spacecraft contains 1.20 kg of lithium hydroxide.

Calculate the amount, in moles, of lithium hydroxide in the canister.

Mark scheme
  • M(LiOH) = 6.9 + 16.0 + 1.0 = 23.9 g mol−11
  • 1200 ÷ 23.9 = 50.2 (mol)1

Do not accept: a mass in mg or g treated as an amount in moles

2(b)[2 marks]

Calculate the maximum volume of carbon dioxide, in dm3, that the canister can absorb at room temperature and pressure.

[Molar volume of a gas at r.t.p. = 24 dm3 mol−1]

Mark scheme
  • mol CO2 = 50.2 ÷ 2 = 25.11
  • 25.1 × 24 = 602 (dm3)1
2(c)[3 marks]

One crew member breathes out 0.90 kg of carbon dioxide in a day.

Calculate the number of oxygen atoms in 0.90 kg of carbon dioxide.

[Avogadro constant L = 6.02 × 1023 mol−1]

Mark scheme
  • mol CO2 = 900 ÷ 44.0 = 20.451
  • mol O atoms = 20.45 × 2 = 40.91
  • 40.9 × 6.02 × 1023 = 2.46 × 10251
2(d)[1 mark]

The safe limit for carbon dioxide in the cabin air is 5000 ppm by volume.

State what is meant by ppm.

Mark scheme
  • parts per million1
2(e)[2 marks]

The cabin contains a total of 3.00 × 103 mol of gas.

Calculate the maximum amount, in moles, of carbon dioxide allowed in the cabin, and express the limit of 5000 ppm as a percentage.

Mark scheme
  • 5000 ÷ 106 × 3.00 × 103 = 15.0 (mol)1
  • 5000 ÷ 104 = 0.5 %1
2(f)[3 marks]

The lithium carbonate formed is tested by dissolving 0.370 g in distilled water to give 250 cm3 of solution.

Calculate the concentration of this solution in mol dm−3 and in g dm−3.

Mark scheme
  • mol Li2CO3 = 0.370 ÷ 73.8 = 5.01 × 10−31
  • 5.01 × 10−3 ÷ 0.250 = 0.0201 mol dm−31
  • 0.370 ÷ 0.250 = 1.48 g dm−31

Do not accept: g or dm3 mol−1 as the unit of a concentration in moles

Question 3

9 marks
3(a)[5 marks]

A sample of 25.0 cm3 of hydrogen peroxide solution, of concentration 1.50 mol dm−3, decomposes completely.

2H2O2(aq) → 2H2O(l) + O2(g)

Calculate the volume of oxygen, in cm3, formed at 35 °C and 98.0 kPa.

[Use R = 8.31 J mol−1 K−1]

Give your answer to 3 significant figures.

Mark scheme
  • mol H2O2 = 25.0 × 1.50 ÷ 1000 = 0.03751
  • mol O2 = 0.0375 ÷ 2 = 0.018751
  • p = 98 000 Pa and T = 308 K1
  • V = nRT ÷ p = 0.01875 × 8.31 × 308 ÷ 98 000 = 4.90 × 10−4 m31
  • 4.90 × 10−4 × 106 = 490 cm31
3(b)[4 marks]

A 1.00 dm3 flask is filled with a single unknown element, which is a gas. At 20 °C and 100 kPa the mass of the gas in the flask is 3.44 g.

Calculate the molar mass of the gas and identify the element.

[Use R = 8.31 J mol−1 K−1]

Mark scheme
  • V = 1.00 × 10−3 m3 and T = 293 K1
  • n = pV ÷ RT = 100 000 × 1.00 × 10−3 ÷ (8.31 × 293) = 0.0411 (mol)1
  • M = 3.44 ÷ 0.0411 = 83.8 g mol−11
  • krypton1

Question 4

9 marks

A student finds the molar volume of carbon dioxide using potassium hydrogencarbonate and excess dilute ethanoic acid. In each run a weighed portion of the solid is tipped into the acid in a boiling tube and the bung is quickly replaced. The gas is collected over water in an inverted 100 cm3 measuring cylinder. Every gas volume is read at room temperature and pressure.

KHCO3(s) + CH3COOH(aq) → CH3COOK(aq) + H2O(l) + CO2(g)

RunMass of KHCO3 / gVolume of CO2 / cm3
10.12027.6
20.18041.4
30.24047.5
40.30069.0
50.36082.7
60.480100.0

[Mr KHCO3 = 100.1. Take the molar volume as 24 000 cm3 mol−1.]

4(a)[2 marks]

Calculate the greatest mass of potassium hydrogencarbonate that can be used if the volume of carbon dioxide must not be more than 100 cm3.

Mark scheme
  • amount of CO2 = 100 ÷ 24 000 = 4.17 × 10−3 mol1
  • mass of KHCO3 = 4.17 × 10−3 × 100.1 = 0.417 g1

Do not accept: a mass in mg or g treated as an amount in moles

4(b)[2 marks]

Identify the two runs whose results should not be used to find the molar volume. Give a reason for each.

Mark scheme
  • run 6: 0.480 g is more than 0.417 g, so the volume of carbon dioxide produced would be greater than the measuring cylinder volume1
  • run 3: its volume is much lower than the trend shown by the other runs1
4(c)[1 mark]

Suggest why the volume of gas collected in run 3 is so low.

Mark scheme
  • some gas escapes before the bung is attached1

Do not accept: measurement errors as the reason for a low molar volume; suck-back as the reason for a low molar volume

4(d)[3 marks]

Use the result of run 4 to calculate the molar volume of carbon dioxide, in dm3 mol−1. Give your answer to 3 significant figures.

Mark scheme
  • amount of KHCO3 = amount of CO2 = 0.300 ÷ 100.1 = 2.997 × 10−3 mol1
  • molar volume = 69.0 ÷ 2.997 × 10−3 = 23 023 cm3 mol−11
  • 23.0 dm3 mol−11
4(e)[1 mark]

Runs 1, 2, 4 and 5 all give a molar volume below 24.0 dm3 mol−1, even though no gas escaped from the apparatus.

Suggest a reason for this.

Mark scheme
  • some carbon dioxide dissolved in the water1

Question 5

14 marks
5(a)[3 marks]

A white solid extracted from fossilised tree resin contains only carbon, hydrogen and oxygen. By mass it contains 40.7 % carbon and 5.1 % hydrogen.

Calculate the empirical formula of the solid.

Mark scheme
  • oxygen = 100 − 40.7 − 5.1 = 54.2 %1
  • C 40.7 ÷ 12.0 = 3.39, H 5.1 ÷ 1.0 = 5.1, O 54.2 ÷ 16.0 = 3.391
  • divide by the lowest: 1 : 1.5 : 1, so C2H3O21
5(b)[2 marks]

The relative molecular mass of the solid is 118.0.

Deduce its molecular formula.

Mark scheme
  • empirical formula mass = 59.0 so n = 118.0 ÷ 59.0 = 21
  • C4H6O41

Do not accept: a number written in front of the empirical formula

5(c)[2 marks]

A student finds the value of x in hydrated zinc sulfate, ZnSO4·xH2O, by heating it in a crucible.

State what the student must do to make sure that all the water of crystallisation has been removed.

Mark scheme
  • heat, cool and reweigh1
  • repeat until constant mass1
5(d)[3 marks]

The student's results are shown.

MeasurementMass / g
empty crucible18.62
crucible and hydrated zinc sulfate21.49
crucible and solid at constant mass20.23

Calculate the value of x.

Mark scheme
  • mass of ZnSO4 = 1.61 g and mass of water = 1.26 g1
  • mol ZnSO4 = 1.61 ÷ 161.5 = 9.97 × 10−3 and mol H2O = 1.26 ÷ 18.0 = 0.07001
  • x = 0.0700 ÷ 9.97 × 10−3 = 71
5(e)[1 mark]

Another student obtains a value of x = 6.4 using the same method.

Suggest a reason for this low value.

Mark scheme
  • not enough water has been removed: the solid has not been heated long enough1

Do not accept: measurement errors as the reason x is wrong

5(f)[3 marks]

A sample of an oxide of tin is heated in a stream of hydrogen until only tin remains.

Mass of tin oxide = 2.67 g. Mass of tin = 2.10 g.

Calculate the empirical formula of the oxide.

Mark scheme
  • mass of oxygen = 2.67 − 2.10 = 0.57 g1
  • mol Sn = 2.10 ÷ 118.7 = 0.0177 and mol O = 0.57 ÷ 16.0 = 0.03561
  • SnO21

Question 6

8 marks
6(a)[2 marks]

Marble chips, CaCO3, fizz in dilute nitric acid. Give the balanced equation for this reaction. Include state symbols.

Mark scheme
  • CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O1
  • (s) (aq) → (aq) (g) (l)1
6(b)[1 mark]

A yellow precipitate forms when silver nitrate solution is mixed with potassium iodide solution. Give the ionic equation, with state symbols, for the formation of this precipitate.

Mark scheme
  • Ag+(aq) + I−(aq) → AgI(s)1

Do not accept: an ionic equation with uncancelled spectator ions

6(c)[1 mark]

A piece of zinc is placed in blue copper(II) sulfate solution.

Give the ionic equation, with state symbols, for the displacement.

Mark scheme
  • Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)1

Do not accept: an ionic equation with uncancelled spectator ions

6(d)[2 marks]

Give two observations when zinc reacts with copper(II) sulfate solution.

Mark scheme
  • blue to colourless solution1
  • pink-brown solid1

Do not accept: just the solution changes colour

6(e)[2 marks]

Give two observations when a granule of zinc is added to excess dilute sulfuric acid.

Mark scheme
  • effervescence1
  • the zinc dissolves1

Do not accept: a white precipitate when a metal reacts with an acid

Question 7

9 marks

Chromium is extracted from chromium(III) oxide by heating it with aluminium.

Cr2O3(s) + 2Al(s) → 2Cr(s) + Al2O3(s)

7(a)[3 marks]

A mixture of 50.0 g of chromium(III) oxide and 20.0 g of aluminium is ignited.

Show by calculation that chromium(III) oxide is the limiting reagent.

Mark scheme
  • mol Cr2O3 = 50.0 ÷ 152.0 = 0.3291
  • mol Al = 20.0 ÷ 27.0 = 0.7411
  • 0.741 ÷ 2 = 0.370, which is greater than 0.329, so aluminium is in excess1
7(b)[2 marks]

Calculate the theoretical mass of chromium formed.

Mark scheme
  • mol Cr = 0.329 × 2 = 0.6581
  • 0.658 × 52.0 = 34.2 g1
7(c)[1 mark]

The mass of chromium obtained is 29.6 g.

Calculate the percentage yield.

Mark scheme
  • 29.6 ÷ 34.2 × 100 = 86.5 %1

Do not accept: the mass of product divided by the mass of the starting material

7(d)[2 marks]

Calculate the percentage atom economy for the production of chromium by this reaction.

Mark scheme
  • 2 × 52.0 = 104.0 and 104.0 + 102.0 = 206.01
  • 104.0 ÷ 206.0 × 100 = 50.5 %1

Do not accept: the atom economy left as a decimal fraction

7(e)[1 mark]

The chromium obtained is less than the theoretical mass. Suggest a cause.

Mark scheme

Any 1 from:

  • reaction incomplete1
  • transfer losses1
  • side reactions1

Do not accept: spillages as the reason for a low yield

All Edexcel International A-level Chemistry question setsThe notes for this topicFlashcards for this topic