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IB DemystifiedMYP Sciences

The mole and reacting masses

Chemists cannot count atoms one by one, so they weigh them. The mole links the mass you measure on a balance to the number of particles reacting, which lets you predict exactly how much product a reaction will give.

Recommended for MYP 4 · eAssessment priority · About 3 lessons · Criteria A, B, C and D

0.000.100.200.300.40mass of magnesium / g0.00.10.20.30.40.50.60.7mass of magnesium oxide / g
Figure 1. The mass of oxide is proportional to the mass of magnesium, because the formula MgO is fixed.
On this page
  1. Learning objectives
  2. Before you start
  3. Key vocabulary
  4. Understanding the ideas
  5. Calculations step by step
  6. Moles in the real world
  7. Worked examples
  8. In the eAssessment
  9. Check your understanding
  10. Practice questions
  11. Investigation
  12. Criterion-linked questions
  13. Challenge questions
  14. Topic check
  15. Review your mistakes
  16. Your progress

Learning objectives

By the end of this topic you should be able to:

  • calculate relative formula masses (Mr) from relative atomic masses (Ar)
  • convert between mass, moles and number of particles
  • balance equations and apply conservation of mass
  • calculate reacting masses and identify limiting reactants
  • calculate and explain percentage yield
  • evaluate experiments that measure reacting masses

Before you start

You will use these skills. If any feel shaky, review them first.

  • atoms, elements, compounds and chemical formulae
  • writing word and symbol equations
  • rearranging simple equations and using ratios

Key vocabulary

Relative atomic mass (Ar)
The average mass of an atom of an element compared with 1/12 of a carbon-12 atom.
Relative formula mass (Mr)
The sum of the Ar values of all atoms in a formula.
Mole
The amount of substance containing 6.02 × 10²³ particles; its mass in grams equals the Mr.
Conservation of mass
In a closed system, the total mass of products equals the total mass of reactants.
Limiting reactant
The reactant that is used up first and so limits the amount of product.
Percentage yield
Actual mass of product ÷ theoretical mass × 100.

Understanding the ideas

  1. What is it?

    A mole is a fixed number of particles (6.02 × 10²³), chosen so that one mole of any substance has a mass in grams equal to its Mr: 18 g of water, 44 g of carbon dioxide. So moles = mass ÷ Mr.

  2. Why does it happen?

    Balanced equations show the ratio in which particles react. Because atoms are only rearranged, not created or destroyed, mass is conserved, and the mole ratio in the equation lets you turn a mass of one substance into the mass of another.

  3. How do we know?

    Antoine Lavoisier showed in the 1770s, by careful weighing in sealed vessels, that mass is conserved in reactions. Chemists later found that elements always combine in fixed proportions by mass, which led to atomic theory and relative atomic masses.

  4. Why does it matter?

    Industry uses reacting masses to order the right amounts of raw materials, reduce waste and cost, and calculate yields. Doctors and pharmacists use the same ideas when making up medicines and doses.

  5. What does it connect to?

    The mole links to concentration and titration, energy changes and electrolysis in chemistry; to the particle model in physics; and to ratios and proportion in mathematics.

Calculations step by step

  1. Mr: add the Ar of every atom (remember brackets: Ca(OH)₂ = 40 + 2 × 17 = 74).
  2. Moles: moles = mass ÷ Mr; mass = moles × Mr.
  3. Reacting masses: mass of A → moles of A → use the equation ratio → moles of B → mass of B.
  4. Limiting reactant: compare the moles you have with the ratio the equation needs.
  5. Percentage yield: actual ÷ theoretical × 100.
  6. Gases: at room temperature and pressure, 1 mol of any gas occupies 24 dm³.

Moles in the real world

Pakistan's fertiliser factories calculate exactly how much natural gas and air they need to make each tonne of ammonia and urea. Steelworks calculate the iron ore and coke needed for each tonne of iron. Pharmacies prepare medicines by calculating masses precisely.

Worked examples

Example 1: reacting mass

What mass of carbon dioxide is produced when 12 g of carbon burns? C + O₂ → CO₂

  1. Moles of C = 12 ÷ 12 = 1.0 mol.
  2. Ratio C : CO₂ = 1 : 1, so 1.0 mol CO₂.
  3. Mass of CO₂ = 1.0 × 44 = 44 g.

Example 2: percentage yield

A reaction should make 8.0 g of product but gives 6.4 g. Calculate the percentage yield.

  1. Percentage yield = actual ÷ theoretical × 100.
  2. 6.4 ÷ 8.0 × 100 = 80%.

In the eAssessment

Mole questions reward clear, step-by-step working. Expect:

  • Calculate Mr, moles, reacting masses, gas volumes and percentage yields, showing each step.
  • Balance equations and use mole ratios.
  • Analyse experimental masses, spot anomalies and compare with theory.
  • Explain apparent mass changes in open and closed systems.

Common ways to lose marks: ignoring brackets in formulae; using Ar of H (1) for H₂; forgetting the mole ratio; reacting equal masses instead of mole ratios; and saying mass is destroyed when a gas escapes.

Check your understanding

Quick questions on the ideas above. Try each one before using a hint.

Practice questions

Show

Investigation: how much copper can iron displace?

Partially guided investigation · about 50 minutes · pairs

Research question
Does the mass of copper produced when iron filings react with excess copper sulfate solution match the mass predicted from the equation Fe + CuSO₄ → FeSO₄ + Cu?
Scientific background
Iron is more reactive than copper, so it displaces copper from copper sulfate solution. The equation shows that 1 mol of iron (56 g) should produce 1 mol of copper (63.5 g).
Hypothesis
Write your own prediction, with a scientific justification.
Variables
Identify your independent, dependent and control variables, and explain how you will control them.
Apparatus
Iron filings, 0.5 mol/dm³ copper sulfate solution, beakers, balance (0.01 g), stirring rod, filter funnel and paper, wash bottle of distilled water, drying oven or warm place.
Method
  1. Weigh about 1.0 g of iron filings accurately and add them to 50 cm³ of copper sulfate solution (an excess).
  2. Stir for 10 minutes until the reaction stops.
  3. Filter off the copper, wash it with distilled water, dry it and weigh it.
  4. Compare the mass of copper with your prediction and calculate the percentage yield.

Safety. Wear eye protection and gloves. Copper sulfate solution is harmful and irritating to eyes and skin; wash off splashes. The solution warms slightly as it reacts. Dispose of solutions as your teacher directs, not down the sink.

Then evaluate: why might your yield be above or below 100%, and how could you tell which errors mattered most?

Criterion-linked questions

Criterion B: inquiring and designing

Criterion C: processing and evaluating

Criterion D: reflecting on the impacts of science

Challenge questions

Harder problems in unfamiliar contexts. Plan before you calculate.

Topic check

Five questions picked at random from the whole topic. Take a new set whenever you like.

Review your mistakes

Questions you got wrong on this device appear here so you can try them again. Answer one correctly and it leaves the list.

Your progress

Tracked separately for each skill, on this device only.

SkillCorrectStatus

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