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How to use this booklet: Go back over some of the calculations topic if you need to: The videos I have provided

will not go through every type of mole calculation but will instead focus on useful

techniques for improving your confidence in calculations. Have a look at the list of

different types of calculation in Section A and recap anything you think you have

forgotten.

Choose which questions you are going to try:

• Section A: The Basics

These questions are clearly labelled and give examples of the most common types

of calculation encountered at A Level. There should be no sneaky surprises and

they are there for you to build your confidence or as a reminder for some equations

you may have forgotten. Written answers to these questions are at the back of the

booklet to allow you to self-check. Timing – allow about one minute per mark

• Section B: Applied Calculations

These questions are more challenging, unstructured questions where you are given

no extra clues. Pay attention to the marks awarded as they may guide you in the

expected number of steps. Also pay attention to any requirement for units or

significant figures. Timing – allow about one minute per mark

• Section C: Multiple Choice

Many students struggle with multiple choice calculation questions because they

don’t write it down and work it out fully! Never leap to conclusions as all the

alternative answers are deliberately designed to distract you. Timing – allow about

1.5 minutes per mark

Complete the questions under exam conditions: Once you’ve chosen your questions,

time yourself and don’t look at your notes! Even if you are struggling, try to stick to your

resolve and provide answers that are as complete as possible in the time limit provided.

Check your answers: Answers to Section A are provided at the end of the booklet with

some working out to help you see if you made a mistake. Answers to Sections B and C will

be given in video format to help talk through the calculations.

Materials required:

• Scientific calculator

• A Level periodic table, with relative atomic mass values given to 1 decimal place

Constants used:

Avogadro’s constant, NA: 6.02 x 1023 mol-1

Gas constant, R: 8.31 J mol-1 K-1

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Contents How to use this booklet: ........................................................................................................................ 1

Section A: Basic calculation practice – Total 50 marks ......................................................................... 3

A1. Moles, mass and molar mass ............................................................................................................ 3

A2. Gas volumes and the ideal gas equation .......................................................................................... 3

A3. Solution and concentration .............................................................................................................. 3

A4. Empirical formulae and calculations from experiments ................................................................... 3

A5. Percentage yield and atom economy ............................................................................................... 4

A6. Avogadro constant and non-standard units calculations ................................................................. 4

Section B: More complex calculations – Total 30 marks ...................................................................... 5

Section C: Multiple choice questions – Total 10 marks ........................................................................ 7

Answers to Section A: .......................................................................................................................... 10

A1. Moles, mass and molar mass .......................................................................................................... 10

A2. Gas volumes and the ideal gas equation ........................................................................................ 10

A3. Solution and concentration ............................................................................................................ 11

A4. Empirical formulae and calculations from experiments ................................................................. 11

A5. Percentage yield and atom economy ............................................................................................. 13

A6. Avogadro constant and non-standard units calculations ............................................................... 13



Section A: Basic calculation practice – Total 50 marks

A1. Moles, mass and molar mass a) Calculate the Mr of strontium nitrate, Sr(NO3)2 (1)

b) Calculate the mass in grams of 0.38 moles of ammonium sulfate, (NH4)2SO4 (1)

c) Calculate the number of moles in 2.65 grams of sodium carbonate, Na2CO3 (1)

d) 0.3 moles of a group 1 nitrate weighs 20.7 grams. Calculate its Mr and thus identify

the compound. (2)

A2. Gas volumes and the ideal gas equation This first section involves using a fixed value for molar gas volume of 24 dm3mol-1 at room

temperature and pressure (RTP). For some exam boards you will be given this value on

your data sheet; for others it will be given in the question. Some exam boards do not use

this value and only work from the ideal gas equation.

a) Calculate the number of moles in 1600 cm3 of ammonia gas at RTP. Give your

answer to three significant figures. (2)

b) Calculate the volume occupied, in dm3, by 6.05 grams of propane gas at RTP. (2)

This question requires use of the ideal gas equation. The gas constant used for these

examples is 8.31 J mol-1 K-1.

c) Find the mass of 200 cm3 of carbon monoxide, CO, at a temperature of 34 °C and a

pressure of 98.9 kPa. (3)

A3. Solution and concentration a) Calculate the concentration of a solution (in moldm-3) which contains 0.2 moles of

sodium chloride dissolved in 25cm3 (1)

b) A solution of ammonium iodide (NH4I) is made by dissolving 2.9 grams in 250 cm3 of

water. Calculate the concentration of the solution in mol dm-3 (2)

c) A solution of magnesium carbonate, MgCO3, has a concentration of 0.4 mol dm-3.

Calculate its concentration in g dm-3 (2)

d) A solution was made by bubbling 72 cm3 of ammonia gas (at room temperature and

pressure) into 50 cm3 of water. Calculate the concentration of the resulting

solution assuming all the ammonia dissolved. (2)

A4. Empirical formulae and calculations from experiments a) Find the empirical formula of a compound containing 5.85 g potassium, 2.10 g

nitrogen and 4.80 g oxygen. (2)

b) Find the empirical formula of a hydrocarbon which contains 83.7% carbon by mass.

If the Mr of the compound is 86 g mol-1, deduce the molecular formula of the

compound. (4)

c) 25.0 cm3 of a 0.200 mol dm-3 solution of sodium carbonate was neutralised by 20.0

cm3 of dilute hydrochloric acid. Find the concentration, in g dm-3, of the acid using

the equation for the reaction below.

Na2CO3 (aq) + 2HCl (aq) → 2NaCl (aq) + CO2 (g) + H2O (l) (3)



d) 15.00 g of the hydrated salt Na2SO4.xH2O was found to contain 7.05 g of water.

Calculate the value of x in the formula. (2)

e) Iron will react with sulphuric acid according to the following equation:

2Fe(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3H2(g)

i) Calculate the mass of iron required to fully react with 100 cm3 of a sulphuric acid

solution of concentration 0.500 moldm-3. (3)

ii) Calculate the volume of hydrogen gas that would be produced in this reaction at

RTP. (2)

iii) The resulting iron (III) sulfate salt is hydrated. If 0.025 moles of the salt weighs

12.25 grams, calculate the molar mass of the salt and thus deduce its formula. (4)

A5. Percentage yield and atom economy a) A student produced some magnesium sulfate crystals, MgSO4.7H2O by reacting 1.20

g magnesium with a slight excess of sulfuric acid. After crystallisation the mass of

crystals collected was 9.84 g. Calculate the percentage yield of the reaction. (3)

b) Titanium is manufactured by heating titanium (IV) chloride with magnesium (at

1200 °C) according to the following equation:

TiCl4 (g) + 2Mg (l) → Ti (s) + 2MgCl2 (l)

Calculate the atom economy of this process, and suggest one way in which the

atom economy can be improved. (3)

A6. Avogadro constant and non-standard units calculations Avogadro constant used in these calculations = 6.02 x 1023 mol-1. You are not required to

learn the Avogadro constant; therefore make sure that you use the value given to you in

the question or on your data sheet rather than memorising the value.

a) Calculate the number of hydrogen atoms in 1 g of water. (2)

b) How many nitrate ions are there in one drop of a 0.05 mol dm-3 solution of calcium

nitrate, Ca(NO3)2? (Assume one drop has a volume of 0.050 cm3.) (3)



Section B: More complex calculations – Total 30 marks

These questions are longer and there are no clues about which calculations to perform.

You may still need to refer to the Avogadro constant, Gas constant and of course the

periodic table!

Question 1

Compound Y contains carbon, hydrogen and oxygen. The following experiments were

carried out to determine its molecular formula:

When 0.540 g of Y is heated to 100 °C it vaporises to produce 200 cm3 of gas at a pressure

of 95 kPa.

When 3.96 g of Y is fully combusted in excess oxygen, and the products collected and

weighed, the reaction produced 7.92 g of carbon dioxide and 3.24 g of water.

From this information, calculate the empirical formula and molecular mass of Y and hence

determine its molecular formula.

(10 marks)

Question 2

A sample of sodium hydrogen carbonate crystals, Na2CO3.10H2O, had been heated too

strongly and lost some of its waters of crystallisation. 2.696 g of the solid were dissolved

in water and made up to 250 cm3 in a volumetric flask. In a series of titrations, 20.0 cm3

portions of the solution were titrated with 0.10 mol dm-3 hydrochloric acid, giving the

results shown below:

Titration number

1 (rough) 2 3

Final burette reading / cm3

22.00 23.00 22.15

Initial burette reading / cm3

1.00 2.35 1.60

Determine the percentage of loss of mass from the crystals using these titration results.

(10 marks)



Question 3

An experiment was carried out to determine the percentage by mass of calcium carbonate

in a sample of limestone.

A 1.510 g sample of limestone was crushed and added to 50.00 cm3 1.0 mol dm-3 solution

of hydrochloric acid. When the reaction with the calcium carbonate was complete, the

resulting solution was decanted into a volumetric flask and made up to a volume of 250

cm3.

20 cm3 samples of this solution were titrated against a solution of 0.100 mol dm-3 sodium

hydroxide to determine the amount of hydrochloric acid remaining. The mean titre was

22.6 cm3.

Assuming that no other substances in the limestone reacted with the acid, use the

information above along with the balanced equations below to calculate the percentage

by mass of calcium carbonate in the limestone.

CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

(10 marks)



Section C: Multiple choice questions – Total 10 marks These questions are mixed up in terms of the calculations you need to perform.

Remember to use units and other clues to help you and allow 1.5 minutes per mark – so

this entire section should take 15 minutes (or more if you have extra time). Don’t forget

to write down your working!

1. There are 392 moles of pure gold in a bar measuring 10 cm by 10 cm by 40 cm.

What is the density of gold in g cm−3 ?

A. 193

B. 19.3

C. 1.93

D. 0.193

2. In a molecule of a hydrocarbon, the fraction by mass of carbon is 9/11.

What is the empirical formula of the hydrocarbon?

A. CH

B. CH3

C. C3H8

D. C5H12

3. Three flasks each contain 240 cm3 of a different gas. Each flask contains...

A the same mass of gas

B 0.1 mole of gas

C the same number of gas molecules

D the same number of atoms

4. If 5.85g of NaCl is dissolved in water to make exactly 250 cm3 of solution, what is

the concentration of the solution?

A 0.0004 moldm-3

B 0.4 moldm-3

C 0.025 moldm-3

D 0.0025 moldm-3

5. An average person has a total of 5 dm3 of blood, which contains sodium ions in a

concentration of 140 mmol dm-3. What total mass of sodium does the person’s

blood contain?

A 16.1 g

B 0.161 g

C 0.0161 g

D 1.6 g



6. Fermentation is used to make ethanol. The equation for the reaction is as follows. The Mr values of each compound are also given. C6H12O6(aq) = 2C2H6O(aq) + 2CO2(g) Mr values: g mol-1 180 46 44

Which expression represents the atom economy for the reaction? A (2 x 46 + 2 x 44) x 100

180 B (2 x 46) x 100 180 C 46 x 100 180 D (2 x 44) x 100 180

7. 5 dm3 of carbon monoxide, CO(g), and 2 dm3 of oxygen, O2(g), at the same

temperature and pressure are mixed together. Assuming complete reaction

according to the equation given, what is the maximum volume of carbon dioxide in

dm3 that can be formed?

2CO (g) + O2 (g) → 2CO2

A. 3

B. 4

C. 5

D. 7

8. Which of the following solutions would react exactly with a solution containing

0.0500 mol sulfuric acid?

A. 50.0 cm3 of 1.00 mol dm−3 KOH

B. 100.0 cm3 of 2.00 mol dm−3 KOH

C. 100.0 cm3 of 2.00 mol dm−3 Ba(OH)2

D 50.0 cm3 of 1.00 mol dm−3 Ba(OH)2

9. A student is provided with a 5.00 cm3 sample of 1.00 × 10–2 mol dm–3 hydrochloric

acid. The student is asked to devise a method to prepare a hydrochloric acid

solution with a concentration of 5.00 × 10–4 mol dm–3 by diluting the sample with

water. Which of these is the correct volume of water that should be added?

A. 45.0 cm3

B. 95.0 cm3

C. 100 cm3

D. 995 cm3



10. Which of the following contains the most chloride ions?

A. 10 cm3 of 3.30 × 10–2 mol dm–3 aluminium chloride solution

B. 20 cm3 of 5.00 × 10–2 mol dm–3 calcium chloride solution

C. 30 cm3 of 3.30 × 10–2 mol dm–3 hydrochloric acid

D. 40 cm3 of 2.50 × 10–2 mol dm–3 sodium chloride solution



Answers to Section A:

A1. Moles, mass and molar mass a) Calculate the Mr of strontium nitrate, Sr(NO3)2 (1)

87.6 + (2 x 14) + (6 x 16) = 211.6

b) Calculate the mass in grams of 0.38 moles of ammonium sulfate, (NH4)2SO4 (1)

Mr = (2 x 14) + (8 x 1) + 32.1 + (16 x 4) = 132.1

Mass = 132.1 x 0.38 = 50.198 / 50.20 g

c) Calculate the number of moles in 2.65 grams of sodium carbonate, Na2CO3 (1)

Mr = (23 x 2) + 12 + (16 x 3) = 106

Moles = 2.65 / 106 = 0.025

d) 0.3 moles of a group 1 nitrate weighs 20.7 grams. Calculate its Mr and thus identify

the compound. (2)

Mr = 20.7 / 0.3 = 69

Formula of a group 1 nitrate = XNO3

Mr (NO3) = 14 + (3 x 16) = 62

Ar (X) = 69 – 62 = 7

Therefore X = Li; the compound is LiNO3

A2. Gas volumes and the ideal gas equation a) Calculate the number of moles in 1600 cm3 of ammonia gas at RTP. Give your

answer to three significant figures. (2)

1600 cm3 / 1000 = 1.6 dm3

1.6 / 24 = 0.06666…

= 0.0667 mol to 3 s.f.

b) Calculate the volume occupied, in dm3, by 6.05 grams of propane gas at RTP. (2)

Mr propane (C3H8) = 44

Moles = 6.05 / 44 = 0.1375 moles

Volume = 0.1375 x 24 = 3.3 dm3

This question requires use of the ideal gas equation. The gas constant used for these

examples is 8.31 J mol-1 K-1.

c) Find the mass of 200 cm3 of carbon monoxide, CO, at a temperature of 34 °C and a

pressure of 98.9 kPa. (3)

PV = nRT

n = PV / RT P = 98900 Pa; V = 2 x 10-4 m3; T = 407 K

n = (98900 x 2 x 10-4 ) / (8.31 x 407) = 5.85 x 10-3 moles

mass = 5.85 x 10-3 x 28 = 0.164 g



A3. Solution and concentration a) Calculate the concentration of a solution (in moldm-3) which contains 0.2 moles of

sodium chloride dissolved in 25cm3 (1)

25 / 1000 = 0.025 dm3

Concentration = 0.2 / 0.025 = 8 mol dm-3

b) A solution of ammonium iodide (NH4I) is made by dissolving 2.9 grams in 250 cm3 of

water. Calculate the concentration of the solution in mol dm-3 (2)

Mr NH4I = 144.9

Moles = 2.9 / 144.9 = 0.0200 vol = 250 / 1000 = 0.250 dm3

Concentration = 0.0200 / 0.25 = 0.0800 mol dm-3

c) A solution of magnesium carbonate, MgCO3, has a concentration of 0.4 mol dm-3.

Calculate its concentration in g dm-3 (2)

Mr MgCO3 = 84.3

Mass = 84.3 x 0.4 = 33.72

Concentration = 33.72 g dm-3

d) A solution was made by bubbling 72 cm3 of ammonia gas (at room temperature and

pressure) into 50 cm3 of water. Calculate the concentration of the resulting

solution assuming all the ammonia dissolved. (2)

Volume ammonia = 72 / 1000 = 0.072 dm3

Moles ammonia = 0.072 / 24 = 0.003 mol

Volume solution = 50 / 1000 = 0.05 dm3

Concentration = 0.003 / 0.05 = 0.06 mol dm-3

A4. Empirical formulae and calculations from experiments a) Find the empirical formula of a compound containing 5.85 g potassium, 2.10 g

nitrogen and 4.80 g oxygen. (2)

Element K N 0

Mass 5.87 2.10 4.80

Moles (mass / Ar) 0.15 0.15 0.3

Ratio 1 1 2

Empirical formula = KNO2

(marks are for calculating moles, then calculating ratio and formula)

b) Find the empirical formula of a hydrocarbon which contains 83.7% carbon by mass.

If the Mr of the compound is 86 g mol-1, deduce the molecular formula of the

compound. (4)

Element C H

Mass 83.7 16.3

Moles (mass / Ar) 6.975 16.3

Divide by smallest 1 2.33

Simplest whole number ratio

3 6.99

Empirical formula = C3H7

Calculating the empirical formula from here – when you divide down to get a whole

number ratio, it may not work out that one of them is a whole number, like the

2.33 example here. There are several scenarios – you need to look at the number



after the decimal place to see whether to round or multiply by another. The table

below gives the main three scenarios:

1.05, 2.96, 1.08 These numbers are close to whole numbers, so you would round them.

1.55, 2.45, 3.5 The decimal place digits are closest to ½, so you should multiply all numbers by 2 to get whole numbers

1.67, 2.33 The decimal place digits are closest to 1/3 or 1/6, so you should multiply all numbers by 3 to get whole numbers.

c) 25.0 cm3 of a 0.200 mol dm-3 solution of sodium carbonate was neutralised by 20.0

cm3 of dilute hydrochloric acid. Find the concentration, in g dm-3, of the acid using

the equation for the reaction below.

Na2CO3 (aq) + 2HCl (aq) → 2NaCl (aq) + CO2 (g) + H2O (l) (3)

Moles sodium carbonate = 0.2 x (25/1000) = 0.005

Moles HCl = 0.005 x 2 = 0.01

Concentration in mol dm-3 = 0.01 / (20/1000) = 0.5 mol dm-3

Mr HCl = 36.5

Concentration in g dm-3 = 0.5 x 36.5 = 18.25 g dm-3

d) 15.00 g of the hydrated salt Na2SO4.xH2O was found to contain 7.05 g of water.

Calculate the value of x in the formula. (2)

Formula Na2SO4 H2O

Mass 7.95 7.05

Mr 142.1 18

Moles 0.0559 0.392

Ratio 1 7

Therefore x = 7

e) Iron will react with sulphuric acid according to the following equation:

2Fe(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3H2(g)

iv) Calculate the mass of iron required to fully react with 100 cm3 of a sulfuric acid

solution of concentration 0.500 moldm-3. (3)

Moles sulfuric acid = 0.5 x 100/1000 = 0.05 mol dm-3

Ratio 2:3 iron : acid therefore moles iron = 0.05 x 2/3 = 0.0333 mol

Mass = 0.0333 x 55.8 = 1.86 g

v) Calculate the volume of hydrogen gas that would be produced in this reaction at

RTP. (2)

0.05 moles of acid would produce 0.05 moles of H2

Volume = 0.05 x 24 = 1.2 dm3 (you must include units here to differentiate from

volume in cm3)

vi) The resulting iron (III) sulfate salt is hydrated. If 0.025 moles of the salt weighs

12.25 grams, calculate the molar mass of the salt and thus deduce its formula. (4)

Hydrated salt Mr = mass / mol = 12.25 / 0.025 = 490

Mr Fe2(SO4)3 = 400 (to 3 s.f., actual answer 399.9)

Total Mr of water molecules = 490 – 400 = 90



90/18 = 5 therefore formula = Fe2(SO4)3.5H2O

A5. Percentage yield and atom economy a) A student produced some magnesium sulfate crystals, MgSO4.7H2O by reacting 1.20

g magnesium with a slight excess of sulfuric acid. After crystallisation the mass of

crystals collected was 9.84 g. Calculate the percentage yield of the reaction. (3)

Moles Mg = 1.2 / 24.3 = 0.0494

1 moles of Mg makes 1 mole of hydrated salt

Mr hydrated salt = 246.4

Maximum mass of hydrated salt = 0.0494 x 246.4 = 12.17 g

Percentage yield = 9.84 / 12.17 x 100 = 81 %

b) Titanium is manufactured by heating titanium (IV) chloride with magnesium (at

1200 °C) according to the following equation:

TiCl4 (g) + 2Mg (l) → Ti (s) + 2MgCl2 (l)

Calculate the atom economy of this process, and suggest one way in which the

atom economy can be improved. (3)

Ar Ti = 47.9 Total Mr products = 47.9 + (95.3 x 2) = 238.5 Atom economy = 47.9 / 238.5 x 100 = 20 % Ways to improve atom economy: Use a different, lighter metal such as lithium in place of magnesium, or find a use for the magnesium chloride so that both products are useful.

A6. Avogadro constant and non-standard units calculations Avogadro constant used in these calculations = 6.02 x 1023 mol-1. You are not required to

learn the Avogadro constant; therefore make sure that you use the value given to you in

the question or on your data sheet rather than memorising the value.

a) Calculate the number of hydrogen atoms in 1 g of water. (2)

moles water molecules = 1 / 18 = 0.0556

Number of water molecules = 0.0556 x 6.02 x 1023 = 3.34 x 1022

Number of hydrogen atoms = 3.34 x 1022 x 2 = 6.68 x 1022

b) How many nitrate ions are there in one drop of a 0.05 mol dm-3 solution of calcium

nitrate, Ca(NO3)2? (Assume one drop has a volume of 0.050 cm3.) (3)

moles calcium nitrate = 0.05 x (0.050/1000) = 2.5 x 10-6

Each mole of Ca(NO3)2 produces two moles of nitrate ions when dissolved so moles of

nitrate ions = 5.0 x 10-6

Number of nitrate ions = 5.0 x 10-6 x 6.02 x 1023 = 3.01 x 1013


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