Pearson Edexcel International GCSE in Chemistry · 4CH1 · Free · no account needed

Empirical and Molecular Formulae

Working out a formula you were never told, from nothing but the masses on a balance.

On the specification this is empirical formula and molecular formula — 4CH1 statements 1.31 · 1.32 · 1.33 · 1.36.

Topic 1 · Principles of chemistry — one of 18 lessons in this topic, and one of 61 in Chemistry.

One of the 12 Chemistry required practicals.

This lesson ends with Justus von Liebig. Learn Empirical and Molecular Formulae, teach it, then defend it against Justus von Liebig.

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

  • 1.31 — Obtaining formulae of simple compounds experimentally, including water of crystallisation
  • 1.32 — The terms empirical formula and molecular formula
  • 1.33 — Calculating empirical and molecular formulae from experimental data
  • 1.36 — Practical: determine the formula of a metal oxide by combustion or reduction (required practical)

Specification section 1(e) Chemical formulae, equations and calculations — also taught in The Mole, Reacting Masses and Percentage Yield, Concentration and Gas Volumes, Ionic Bonding.

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

Every calculation so far has been given the formula. This page is the reverse: you are given masses from an experiment and asked what the compound IS. It is the closest school chemistry comes to genuine discovery, and it is the reason the mole was invented.

Two words first, because the whole page turns on the difference. The EMPIRICAL FORMULA is the simplest whole-number ratio of the atoms in a compound. The MOLECULAR FORMULA is the actual number of each kind of atom in one molecule. For water both are H₂O. For ethane the molecular formula is C₂H₆ but the empirical formula is CH₃, because the simplest ratio of carbon to hydrogen is one to three.

Think of it like a photograph described by its aspect ratio

Told that a photograph is 'three by two', you know its shape exactly and its size not at all — it could be a passport photo or a poster. That is an empirical formula: a true statement of the ratio, silent about the scale. The molecular formula is the actual measurement, 15 cm by 10 cm. This is why an experiment on masses can only ever give you the empirical formula: weighing tells you the proportions of the elements present, and proportions are shape without scale. To get from CH₃ to C₂H₆ you need one extra piece of information that no balance can supply — the relative molecular mass of the whole thing.

The four-line method for an empirical formula

  1. Write down the MASS of each element present, in grams. If you are given percentages instead, treat them as grams in a 100 g sample — the ratio is identical.
  2. DIVIDE each mass by that element's relative atomic mass, giving the moles of each.
  3. DIVIDE every answer by the SMALLEST of them. This turns the numbers into a ratio starting at 1.
  4. If any result is not a whole number, MULTIPLY the whole set until they all are — a 1 : 1.5 ratio becomes 2 : 3, never 1 : 2 by rounding.
  5. Write the ratio as the formula.

An empirical formula, then the molecular formula

(a) A compound contains 4.8 g of carbon and 1.0 g of hydrogen. Find its empirical formula. (C = 12, H = 1.) (b) Its relative molecular mass is 58. Find its molecular formula.

  1. (a) Moles of C = 4.8 ÷ 12 = 0.40. Moles of H = 1.0 ÷ 1 = 1.0.
  2. Divide both by the smallest, 0.40: C = 1, H = 2.5.
  3. Not whole numbers, so multiply both by 2: C = 2, H = 5. The empirical formula is C₂H₅.
  4. (b) The mass of one C₂H₅ unit is (2 × 12) + (5 × 1) = 29.
  5. 58 ÷ 29 = 2, so the molecule contains two of these units. The molecular formula is C₄H₁₀.
  6. Note that rounding 2.5 down to 2 in step (a) would have given CH₂, whose unit mass of 14 does not divide into 58 a whole number of times — the sign that the rounding was wrong.

Answer: (a) C₂H₅. (b) C₄H₁₀ — butane.

The rounding rule in that example is where most marks are lost, so it is worth stating on its own. A result of 1.5, 2.5 or 1.33 is not a measurement error to be tidied away — it is telling you the ratio needs doubling or trebling. Multiply the whole set. Rounding a 1.5 to 2 changes the compound.

The required practical, statement 1.36, is this calculation done with a balance. Heating a known mass of magnesium in a crucible until it all turns to oxide gives you two masses — the metal you started with, and the oxide you finished with. The difference between them is the mass of oxygen that combined, because oxygen is the only thing that has been added. Two masses, two elements, and the four-line method gives the formula.

The same reasoning handles WATER OF CRYSTALLISATION, which is water held within a crystal's structure and written after a dot, as in CuSO₄·5H₂O. Heating the crystals drives the water off and leaves the anhydrous salt behind. Weigh before and after: the loss in mass is the water, what remains is the salt, and dividing each by its relative formula mass gives the ratio of salt to water — the number in front of the H₂O.

Finding the water of crystallisation

5.00 g of hydrated magnesium sulfate, MgSO₄·xH₂O, is heated until it reaches constant mass, leaving 2.44 g of anhydrous MgSO₄. Find x. (MgSO₄ = 120, H₂O = 18.)

  1. Mass of water driven off = 5.00 − 2.44 = 2.56 g.
  2. Moles of MgSO₄ = 2.44 ÷ 120 = 0.0203 mol.
  3. Moles of H₂O = 2.56 ÷ 18 = 0.142 mol.
  4. Divide both by the smaller, 0.0203: MgSO₄ = 1, H₂O = 7.0.
  5. So x = 7, and the compound is MgSO₄·7H₂O.

Answer: x = 7 — magnesium sulfate heptahydrate.

'Heated to constant mass' in that problem is not a stylistic phrase. It means heating, cooling, weighing, and repeating until two successive weighings agree — which is the only way to be sure all the water has gone. Stop too early and some water remains, so the calculated x comes out too small.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

Statement 1.32 — the two formulae

Empirical formula
The simplest whole-number ratio of the atoms of each element in a compound. Ethane's is CH₃.
Molecular formula
The actual number of atoms of each element in one molecule. Ethane's is C₂H₆.
Water of crystallisation
Water held within the structure of a crystal, written after a dot: CuSO₄·5H₂O. Driven off by heating.
Anhydrous
Containing no water of crystallisation — what remains after the hydrated salt has been heated.

Calculating an empirical formula

  1. Write the MASS of each element in grams (percentages can be treated as grams in a 100 g sample).
  2. DIVIDE each mass by that element's relative atomic mass.
  3. DIVIDE every result by the SMALLEST of them.
  4. If any value is not a whole number, MULTIPLY the whole set until all are — 1 : 1.5 becomes 2 : 3.
  5. Write the ratio as the formula.

Going from empirical to molecular formula

  1. Work out the mass of ONE empirical formula unit.
  2. Divide the relative molecular mass of the compound by that figure.
  3. The answer is a whole number — how many empirical units are in one molecule.
  4. Multiply every subscript in the empirical formula by that number.
  5. Check: the new formula's relative molecular mass should equal the value given.

Required practical 1.36 — the formula of magnesium oxide

  1. Weigh a clean, dry crucible with its lid.
  2. Add a coil of cleaned magnesium ribbon and weigh again; the difference is the mass of magnesium.
  3. Heat the crucible strongly, lifting the lid briefly at intervals with tongs to admit air.
  4. Continue until the magnesium has fully reacted and no further change is seen.
  5. Allow the crucible to cool, then weigh it; the increase over the empty crucible is the mass of magnesium oxide.
  6. Reheat, cool and reweigh until two successive masses agree — heating to constant mass.
  7. Mass of oxygen = mass of oxide − mass of magnesium; divide each mass by its relative atomic mass and simplify the ratio.

Variables

Independent (changed) — The mass of magnesium ribbon used
Dependent (measured) — The mass of magnesium oxide formed

Control variableWhy it must be held constant
The same crucible and lid throughouta different crucible has a different mass, so the differences would be meaningless
Heating to constant massstopping early leaves unreacted magnesium and gives too little oxygen
Magnesium cleaned before usean existing oxide layer adds oxygen that was never measured as part of the reaction

Sources of error

TypeWhat goes wrongWhat to do
SystematicThe lid is lifted too often and white smoke of magnesium oxide escapes.Lift the lid only briefly and as seldom as possible.
SystematicHeating stops before the reaction is complete, so some magnesium is unreacted.Reheat, cool and reweigh until the mass is constant.
RandomWeighing the crucible while still warm gives a low reading from convection currents.Cool the crucible fully before each weighing.

Model answer [4 marks]

A compound contains 2.4 g of magnesium and 1.6 g of oxygen. Determine its empirical formula. (Mg = 24, O = 16.) [4]

Moles of Mg = 2.4 ÷ 24 = 0.10 mol. Moles of O = 1.6 ÷ 16 = 0.10 mol. Dividing both by the smaller value, 0.10, gives a ratio of 1 : 1. Both are already whole numbers, so the empirical formula is MgO.

Model answer [3 marks]

A compound has the empirical formula CH₂ and a relative molecular mass of 70. Determine its molecular formula. [3]

The mass of one CH₂ unit is 12 + (2 × 1) = 14. Dividing the relative molecular mass by this gives 70 ÷ 14 = 5, so there are five empirical units in one molecule. Multiplying each subscript by 5 gives the molecular formula C₅H₁₀.

Where the marks are lost

  • ROUNDING a 1.5 or a 2.5 instead of multiplying the whole ratio up — this changes the compound
  • Dividing by the WRONG element's relative atomic mass
  • Forgetting that the oxygen's mass is the DIFFERENCE between the oxide and the metal
  • Not heating to CONSTANT MASS, so the water or oxygen figure is too small
  • Giving the empirical formula when the question asked for the molecular formula

Not this: An experiment on masses can give you the molecular formula of a compound.

This: It can only give the EMPIRICAL formula, because masses reveal proportions and proportions have no scale. The molecular formula needs one further piece of information — the relative molecular mass — which no balance can supply.

Mark-losing trap. A ratio of 1 : 1.5 becomes 2 : 3 by MULTIPLYING. Rounding it to 1 : 2 gives a different compound.

Mark-losing trap. The mass of oxygen is the DIFFERENCE between the oxide and the metal — it is never weighed directly.

Mark-losing trap. 'Heat to constant mass' means reheat, cool and reweigh until two readings agree.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

  1. Grade 6 · Define [1 mark] — What is meant by the empirical formula of a compound?
  2. Grade 7 · Calculate [3 marks] — A compound contains 1.2 g of carbon and 0.4 g of hydrogen. Calculate its empirical formula. (C = 12, H = 1.)
  3. Grade 8 · Calculate [3 marks] — A compound has the empirical formula CH₂O and a relative molecular mass of 180. Calculate the number of empirical formula units in one molecule. (C = 12, H = 1, O = 16.)
  4. Grade 9 · Describe [6 marks] — Select every statement that belongs in a full-mark description of how to determine the formula of magnesium oxide experimentally by heating magnesium in a crucible.
  5. 9+ · Analyse [6 marks] — A student determines the formula of copper oxide by reduction, obtaining masses that give a ratio of copper to oxygen of 1 : 1.33. They round this to 1 : 1 and report the formula as CuO. Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one who supplied the missing step from ratio to real formula, and the one who showed a mass increase is a mass gained from somewhere. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

Amedeo Avogadro — the one who supplied the missing step from ratio to real formula

This page's central limitation — that masses give only the empirical formula — was the central confusion of chemistry for fifty years, and Avogadro's 1811 hypothesis is what resolved it. Equal volumes of gases, he proposed, contain equal numbers of particles, which lets relative molecular masses be measured and so lets a ratio be scaled up into a real formula. It required accepting that hydrogen and oxygen exist as PAIRS rather than single atoms, which Dalton and Berzelius both refused, and the idea sat ignored until Cannizzaro revived it at Karlsruhe in 1860, four years after Avogadro died.

  • “Why can masses alone not give the molecular formula?”
  • “What does equal volumes of gases containing equal numbers actually let you do?”
  • “Why did the elementary gases have to exist as pairs?”
  • “Why did nobody accept your hypothesis for fifty years?”
  • “What was chemistry unable to settle without it?”

Antoine Lavoisier — the one who showed a mass increase is a mass gained from somewhere

The required practical on this page rests entirely on one inference: that the increase in mass when magnesium is heated IS the mass of oxygen that combined with it. That reasoning is Lavoisier's, and it was not obvious. The prevailing theory held that burning released a substance called phlogiston — which made a metal gaining mass while burning so awkward that phlogiston was seriously proposed to have negative weight. Lavoisier weighed sealed vessels and showed the gain came from the air, and from a specific component of it that he named oxygen.

  • “Why does a metal get heavier when it burns?”
  • “How did weighing a sealed vessel settle the question?”
  • “What was phlogiston supposed to be?”
  • “Why would anyone propose a substance with negative weight?”
  • “How do you know the gain in mass is the oxygen?”

Think you understand it?

Reading is the start. On Incandio a topic counts as mastered only once you have taught it and defended it — the two things a revision guide cannot ask of you.

Incandio this topic → Prove it. Teach it. Defend it. What does “Incandio it” mean?

  1. Prove it — five questions on Empirical and Molecular Formulae, marked on the spot, that tell you what was missing rather than just what was wrong.
  2. Teach it — Explain Empirical and Molecular Formulae in your own words to Ember, an AI pupil who does not have it yet and asks the awkward question. Understanding is credited; wording is not required.
  3. Defend it — Argue Empirical and Molecular Formulae against Justus von Liebig — three rounds, marked against descriptors you can read before you begin.
  4. Then the exam — read all 11 authored exam-style questions on this unit, each with its command word and its tariff, then have them marked on Incandio with the mistake you actually made named.

Completely free. No subscriptions, no paywalls, no account needed to start.

  • The Mole — defines mole, which this lesson uses

Carry on through the course