IGCSE Chemistry 0620 · Topic 3

IGCSE Chemistry: Stoichiometry Practice Questions

Stoichiometry uses the mole to link mass, particle number and reaction equations. One mole of a substance has a mass in grams equal to its relative formula mass, and the balanced equation gives the ratio in which substances react.

Cambridge IGCSE Chemistry (0620) · Topic 3: Stoichiometry

Topic 3 of Cambridge IGCSE Chemistry 0620 is the highest scoring calculation topic on the Extended papers, and it is entirely method driven. Every question follows the same route: moles of what you know, ratio from the equation, moles of what you want, then convert back. The questions below use that route each time.

What you need to know for Stoichiometry

Relative formula massThe sum of the relative atomic masses of all the atoms in the formula. For H₂O it is 2 x 1 plus 16, which is 18.
The moleMoles = mass in grams divided by relative formula mass. Rearranged: mass = moles x Mr.
Using the equation ratioThe large numbers in front of the formulae in a balanced equation give the ratio of moles in which substances react and are produced.
Gas volumesAt room temperature and pressure, one mole of any gas occupies 24 dm³, which is 24000 cm³. Volume in dm³ = moles x 24.
ConcentrationConcentration in mol/dm³ = moles divided by volume in dm³. Remember that 1 dm³ equals 1000 cm³.
Empirical formulaThe simplest whole number ratio of atoms in a compound. Divide the mass or percentage of each element by its relative atomic mass, then divide all the answers by the smallest.

IGCSE Chemistry Stoichiometry questions and answers

4 exam-style questions written to the 0620 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

Calculate the relative formula mass of calcium carbonate, CaCO₃, and the number of moles in 25.0 g of it. Use Ca = 40, C = 12, O = 16.

Show the worked answer

Answer: Mr = 100 and there are 0.250 mol.

Add the relative atomic masses: calcium 40, carbon 12, and three oxygens at 16 each, giving 48.
Mr = 40 + 12 + 48 = 100.
Use moles = mass divided by Mr.
Moles = 25.0 divided by 100 = 0.250 mol.

How the marks are awarded. 1 mark for Mr = 100. 1 mark for using moles = mass divided by Mr. 1 mark for 0.250 mol.

Where students lose the mark. Forgetting to multiply the oxygen mass by three. Count every atom in the formula, including those inside brackets.

Question 2[4 marks]

Calcium carbonate decomposes on heating: CaCO₃ gives CaO + CO₂. Calculate the mass of calcium oxide produced when 25.0 g of calcium carbonate decomposes completely. Use Ca = 40, C = 12, O = 16.

Show the worked answer

Answer: 14.0 g

Moles of CaCO₃ = 25.0 divided by 100 = 0.250 mol.
The equation shows a 1 to 1 ratio between CaCO₃ and CaO, so 0.250 mol of CaO is produced.
Mr of CaO = 40 + 16 = 56.
Mass = moles x Mr = 0.250 x 56 = 14.0 g.

How the marks are awarded. 1 mark for moles of calcium carbonate. 1 mark for using the 1 to 1 mole ratio. 1 mark for Mr of CaO = 56. 1 mark for 14.0 g.

Where students lose the mark. Using the mass ratio directly instead of converting to moles first. Always go through moles, even when the ratio is 1 to 1.

Question 3[4 marks]

A compound contains 40.0 per cent carbon, 6.7 per cent hydrogen and 53.3 per cent oxygen by mass. Calculate its empirical formula. Use C = 12, H = 1, O = 16.

Show the worked answer

Answer: CH₂O

Assume 100 g of the compound, so the percentages become masses in grams.
Divide each mass by the relative atomic mass: carbon 40.0 divided by 12 = 3.33, hydrogen 6.7 divided by 1 = 6.7, oxygen 53.3 divided by 16 = 3.33.
Divide each result by the smallest, 3.33: carbon 1, hydrogen 2.01, oxygen 1.
Round to the nearest whole numbers, giving a ratio of 1 carbon to 2 hydrogen to 1 oxygen. The empirical formula is CH₂O.

How the marks are awarded. 1 mark for dividing each percentage by the relative atomic mass. 1 mark for the three values 3.33, 6.7 and 3.33. 1 mark for dividing by the smallest value. 1 mark for CH₂O.

Where students lose the mark. Dividing by the largest value instead of the smallest. Dividing by the smallest is what produces a ratio starting from 1.

Question 4[3 marks]

Calculate the volume of carbon dioxide, measured at room temperature and pressure, produced when 0.250 mol of calcium carbonate decomposes completely. One mole of gas occupies 24 dm³ at room temperature and pressure.

Show the worked answer

Answer: 6.00 dm³

The equation CaCO₃ gives CaO + CO₂ shows a 1 to 1 ratio, so 0.250 mol of carbon dioxide is produced.
Volume in dm³ = moles x 24.
Volume = 0.250 x 24 = 6.00 dm³.
In cubic centimetres this is 6000 cm³, since 1 dm³ equals 1000 cm³.

How the marks are awarded. 1 mark for 0.250 mol of carbon dioxide from the 1 to 1 ratio. 1 mark for using volume = moles x 24. 1 mark for 6.00 dm³ with the unit.

Where students lose the mark. Mixing up dm cubed and cm cubed. Check the unit the question asks for and convert by a factor of 1000.

Common mistakes in this topic

Working with masses in the equation ratio instead of converting to moles first.
Forgetting to count atoms inside brackets, for example in Ca(OH)₂.
Using an unbalanced equation. Balance it before taking any ratio.
Confusing dm cubed with cm cubed in concentration and gas volume questions.
Rounding intermediate values too early, which shifts the final answer.

Exam tips

Follow one route every time: moles of the known substance, ratio from the balanced equation, moles of the unknown, convert to mass or volume.
Write the equation and check it is balanced before any calculation. An unbalanced equation makes every subsequent mark unreachable.
Show the moles line explicitly. It is nearly always worth a method mark on its own.
Quote answers to three significant figures unless told otherwise, and always include the unit.
For percentage yield, use actual yield divided by theoretical yield multiplied by 100, both measured in the same units.

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Stoichiometry FAQs

How do I calculate the number of moles?

Divide the mass in grams by the relative formula mass. For example, 25.0 g of calcium carbonate with an Mr of 100 gives 25.0 divided by 100, which is 0.250 mol. Rearranged, mass equals moles multiplied by Mr.

How do I do a reacting mass calculation?

Convert the known mass to moles, use the large numbers in the balanced equation to find the mole ratio, apply that ratio to find the moles of the substance you want, then multiply by its relative formula mass to get the mass. Never apply the equation ratio directly to masses.

What volume does one mole of gas occupy?

At room temperature and pressure one mole of any gas occupies 24 dm3, which is 24000 cm3. This is the same for every gas regardless of its identity, so volume in dm3 equals the number of moles multiplied by 24.

How do I find an empirical formula from percentages?

Assume 100 g so the percentages become masses. Divide each mass by the relative atomic mass of that element. Divide all the resulting values by the smallest one. Round to whole numbers, and multiply through if you get a value close to a simple fraction such as 1.5.

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What you need to know for Stoichiometry

IGCSE Chemistry Stoichiometry questions and answers

Common mistakes in this topic

Exam tips

Practise 20 more questions like this, free

Stoichiometry FAQs

Continue through the IGCSE Chemistry syllabus

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