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Learn7405Paper 1 · Paper 1: Physical and inorganic chemistryAmount of substanceThe mole and equations
Amount of substance · Subtopic 2 of 3

The Mole, Formulae and Equations

Relative masses are measured against one-twelfth of a carbon-12 atom. One mole contains 6.02×10236.02\times10^{23} particles (the Avogadro constant, LL), and has a mass in grams equal to MrM_r: n=mMn=\frac{m}{M}. Empirical formulae come from composition by mass (or combustion data); the molecular formula is a whole-number multiple matching MrM_r. Equations balance atoms and charge; ionic equations omit spectator ions. Water of crystallisation is found from mass lost on heating. Atom economy =Mr of desired product∑Mr of all products×100=\frac{M_r\text{ of desired product}}{\sum M_r\text{ of all products}}\times100.

Cambridge 9701 Paper 1Cambridge 9701 Paper 2Edexcel YCH11 WCH11AQA 7405 Paper 1≈ 45 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Define relative masses using carbon-12.
  • Use the mole and the Avogadro constant.
  • Find empirical and molecular formulae, including from combustion data and hydrated salts.
  • Write balanced and ionic equations; calculate atom economy.
Pause and think

Burning 1.00 g of an unknown hydrocarbon gives 3.08 g of CO₂. How can that tell you its formula?

Section 2 of 14

Core idea

Start hereFrom a burnt sample to a formula

Chemists identify a new compound by burning a few milligrams and weighing the CO₂ and water produced. Moles turn those masses into a formula.

The big idea

Moles connect mass, particle number and formula. Combustion data, composition and heating hydrates all give ratios of moles, and ratios give formulae.

1. Combustion analysis

Worked exampleIdentifying a compound

0.880 g of a compound (C, H, O) burns to give 1.76 g CO₂ and 0.720 g H₂O. Mr=88M_r=88.

Goal
Molecular formula.
Strategy
Moles of C and H, O by difference, then scale.
  1. C:1.7644=0.0400 mol=0.480 g\mathrm{C}: \tfrac{1.76}{44}=0.0400\ \text{mol}=0.480\ \text{g}
    All C ends up in CO₂.
  2. H:2×0.72018=0.0800 mol=0.0800 g\mathrm{H}: 2\times\tfrac{0.720}{18}=0.0800\ \text{mol}=0.0800\ \text{g}
    Two H per water.
  3. O:0.880−0.560=0.320 g=0.0200 mol\mathrm{O}: 0.880-0.560=0.320\ \text{g}=0.0200\ \text{mol}
    By difference.
  4. C2H4O (Mr=44)⇒C4H8O2\mathrm{C_2H_4O}\ (M_r=44)\Rightarrow\mathrm{C_4H_8O_2}
    Ratio 2 : 4 : 1; double to reach 88.
AnswerC4H8O2\mathrm{C_4H_8O_2}
✓ Check4(12)+8(1)+2(16)=884(12)+8(1)+2(16)=88 ✓.

2. Atom economy

Yield vs atom economy

Two different measures of efficiency.

Common misconceptionHydrogen in water mol H = 2 × mol H₂O. Forgetting the 2 halves the hydrogen.
Pause and think

Can a reaction have 100% yield but low atom economy?

Summary

  • n=m/Mn=m/M, N=nLN=nL.
  • Combustion and hydrate calculations.
  • Atom economy from the equation.
Key words
Empirical formula
Simplest whole-number ratio of atoms.
Water of crystallisation
Water bound in a crystal lattice.
Atom economy
Percentage of reactant mass that ends up in the desired product.
HistoryDefining the mole. Since 2019 the mole has been defined by fixing the Avogadro constant at exactly 6.022 140 76×10236.022\,140\,76\times10^{23}, no longer by the mass of carbon-12.
Fun factA mole of seconds. A mole of seconds is about 19 million billion years, over a million times the age of the universe.
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