Lesson 2: The Mole and Molar Mass

Chemists count particles the way grocers count eggs: in a standard batch. The batch for atoms is the mole: exactly 6.022 × 10²³ particles, a number called Avogadro's constant. One mole of any substance has a mass in grams numerically equal to its relative formula mass — its molar mass — and the mole is the bridge between the weighable world of grams and the invisible world of atoms. Almost every quantitative idea in this course stands on this lesson.

The mole and molar mass

The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all its atoms. For calcium carbonate, CaCO₃: 40.08 + 12.01 + (3 × 16.00) = 100.09, so its molar mass is 100.09 g/mol. The defining relationships are:

moles = mass ÷ molar mass          n = m ÷ M
particles = moles × 6.022 × 10²³    N = n × Nᴀ

Worked example 1: from grams to atoms

A tablet contains 25.0 g of CaCO₃. How many moles is that, and how many oxygen atoms does it contain? Step 1: n = m ÷ M = 25.0 ÷ 100.09 = 0.250 mol (3 s.f.). Step 2: each formula unit contains 3 oxygen atoms, so moles of O atoms = 3 × 0.250 = 0.749 mol. Step 3: N = 0.749 × 6.022 × 10²³ = 4.51 × 10²³ oxygen atoms. Notice how the mole ratio inside the formula (the subscript 3) does the counting automatically — subscripts are mole ratios, a fact stoichiometry exploits relentlessly.

Percent composition

Percent composition states what fraction of a compound's mass each element contributes: % element = (total mass of that element in the formula ÷ Mr) × 100. For CaCO₃: calcium 40.08 ÷ 100.09 = 40.0%; carbon 12.01 ÷ 100.09 = 12.0%; oxygen 48.00 ÷ 100.09 = 48.0%. The three values sum to 100.0% — always check this; a sum that misses 100 signals an arithmetic slip. Percent composition is how fertiliser bags report nitrogen content and how ore grades are quoted in mining.

Worked example 2: empirical and molecular formulas

A compound is 40.0% C, 6.7% H and 53.3% O, with Mr = 180. Find its empirical and molecular formulas. Step 1: assume 100 g, so masses are 40.0 g, 6.7 g, 53.3 g. Convert to moles: C 40.0 ÷ 12.01 = 3.33; H 6.7 ÷ 1.008 = 6.65; O 53.3 ÷ 16.00 = 3.33. Step 2: divide by the smallest (3.33): C 1.00, H 2.00, O 1.00 → empirical formula CH₂O, empirical mass 30.03. Step 3: 180 ÷ 30.03 = 6.0, so the molecular formula is 6 × CH₂O = C₆H₁₂O₆ — glucose. The empirical formula is the simplest ratio; only a molar-mass measurement can promote it to the molecular formula.

In the lab: water of crystallisation

Many salts crystallise with water locked into the lattice as water of crystallisation, shown in the formula with a dot: CuSO₄·5H₂O. The percentage of water can be found by weighing, heating to drive the water off, and reweighing. For copper(II) sulfate: Mr = 249.69, of which water is 5 × 18.02 = 90.10, so % water = 90.10 ÷ 249.69 × 100 = 36.1%. A student heats 2.50 g of the blue crystals and finds the white residue weighs 1.62 g — an apparent water loss of 35.2%. Two error sources explain the shortfall. First, incomplete heating: if heating stops before the mass is constant, some water remains and the measured loss is too small; the remedy is to heat, weigh, and repeat until two successive masses agree. Second, the residue is hygroscopic — it reabsorbs moisture from the air while cooling, again shrinking the apparent loss; cooling in a desiccator prevents it. A third, opposite error is overheating: scorching decomposes the sulfate itself, inflating the loss. Each balance reading carries ±0.01 g uncertainty, so the loss (a difference of two readings) carries ±0.02 g — on a 0.88 g loss, about ±2%. Reporting 35.2% ± 2% makes the discrepancy with 36.1% understandable rather than a "wrong answer".

Chemistry in the real world

Pharmacists dose by moles, even when they never say so: a 500 mg paracetamol tablet contains 0.500 ÷ 151.16 = 3.31 × 10⁻³ mol, and safe dosing limits are set from how many moles the liver's enzymes can process per hour. Anaesthetists likewise calculate gas doses in moles because equal moles of gas occupy equal volumes (Lesson 5).

Common mistakes

• Using the atomic mass of the wrong particle set. Mr of O₂ is 32.00, not 16.00; read subscripts before reaching for the calculator.
• Dividing by the largest mole value in empirical formulas instead of the smallest.
• Rounding mid-calculation. Carry at least one extra digit until the final step; premature rounding turns 35.45 into errors that compound through a whole problem.
• Forgetting the waters in hydrate molar masses. The "·5H₂O" is part of the mass you weighed.

Exam Experience

1. How many moles are in 44.0 g of carbon dioxide, CO₂ (Mr = 44.01)? A) 0.500 mol B) 1.00 mol C) 2.00 mol D) 0.250 mol Answer: B. n = 44.0 ÷ 44.01 = 1.00 mol. The other values come from halving, doubling or inverting the division.

2. What is the empirical formula of a compound that is 52.2% C, 13.0% H and 34.8% O? A) C₂H₆O B) CH₃O C) C₂H₅O D) CHO Answer: A. Moles per 100 g: C 4.35, H 12.9, O 2.175; divide by 2.175 → C 2, H 6 (5.94 ≈ 6), O 1 → C₂H₆O (ethanol's molecular formula happens to be the same).

3. A student calculates the molar mass of Mg(OH)₂ as 42.3 g/mol. The error is: A) none — it is correct. B) magnesium's mass was taken as 12.3. C) the two OH groups were counted once — the correct value is 58.3 g/mol. D) oxygen's mass was doubled. Answer: C. Mg(OH)₂ = 24.31 + 2 × (16.00 + 1.008) = 58.3 g/mol. Counting OH once gives 24.31 + 17.01 = 41.3 ≈ the quoted wrong value's origin (bracket neglect).

Constructed response (5 marks). A sample of an unknown hydrate, BaCl₂·xH₂O, has a mass of 3.66 g. On heating to constant mass, 3.12 g of anhydrous BaCl₂ remain. (a) Calculate the mass of water lost. (b) Calculate the moles of BaCl₂ (Mr = 208.23) and of water. (c) Determine x. Scoring guide: (a) 1 mark: 3.66 − 3.12 = 0.54 g. (b) 1 mark: BaCl₂ = 3.12 ÷ 208.23 = 0.01498 mol; 1 mark: H₂O = 0.54 ÷ 18.02 = 0.02997 mol. (c) 1 mark: ratio 0.02997 ÷ 0.01498 = 2.00; 1 mark: x = 2 (BaCl₂·2H₂O). Withhold the final mark if the ratio is inverted (0.5) without correction.

Worked example 3: counting molecules in a drop of water

How many water molecules are in a single drop of water, mass 0.050 g? n = 0.050 ÷ 18.02 = 2.78 × 10⁻³ mol; N = 2.78 × 10⁻³ × 6.022 × 10²³ = 1.67 × 10²¹ molecules. Two septillion-ish molecules in one drop is why chemists never count particles directly — and why the mole was invented. Extend the same arithmetic: a 500 mL bottle holds 27.8 mol of water, and the hydrogen atoms number 2 × 27.8 × 6.022 × 10²³ = 3.35 × 10²⁵. If each atom were a grain of sand a millimetre wide, they would stretch to the nearest star and back many times over.

From moles to molarity — and to the tonne scale

The mole connects directly to solution work (Lesson 7): dissolve 0.250 mol of NaOH and make the volume up to 500.0 mL and the concentration is 0.250 ÷ 0.5000 = 0.500 mol/L. Industry scales the identical unit up rather than changing it: ammonia plants (Lesson 18) order nitrogen in kilomoles, where 1 kmol of N₂ is 28.02 kg, and a 1000-tonne-per-day plant processes about 1.24 × 10⁶ kmol of gas daily. Whether the quantity is 2.78 × 10⁻³ mol in a droplet or 10⁶ kmol through a reactor, the pipeline never changes: mass ↔ moles ↔ particles or volume, with Mr as the only conversion factor. Master this lesson's three-step chain and Lessons 4, 5, 7 and 10 become variations on a theme you already know.

The mole meets the gas laws early: molar volume

One mole of any gas at STP occupies 22.4 L (Lesson 5), so moles and gas volumes interconvert without any other data. A balloon holding 5.60 L of helium at STP contains 5.60 ÷ 22.4 = 0.250 mol — a mass of just 1.00 g of helium lifting a party balloon, because the displaced air (5.60 L ≈ 0.250 mol ≈ 7.25 g at Mr ≈ 29) outweighs it. The history under the number is worth knowing: Avogadro proposed in 1811 that equal volumes contain equal numbers of particles, but his hypothesis was ignored for half a century — partly because it required diatomic molecules (O₂, H₂) that Dalton's atomism resisted. Cannizzaro's advocacy at the Karlsruhe congress of 1860 finally standardised atomic masses on Avogadro's logic: the mole on your calculator descends from an argument chemists took fifty years to accept.

Primary standards: weighing something trustworthy

All the mole arithmetic in this course leans on knowing one amount exactly. A primary standard is a substance pure and stable enough to weigh into truth: anhydrous sodium carbonate, potassium hydrogen phthalate, and oxalic acid dihydrate are the classics — non-hygroscopic, high molar mass (so weighing errors shrink in percentage terms), and unchanging on the shelf. Sodium hydroxide fails every criterion: it absorbs water and CO₂ from air as you watch, so its solutions are always standardised *against* a primary standard by titration (Lesson 6) rather than trusted from the balance. The hierarchy — weigh the trustworthy, titrate the rest against it — is the quiet skeleton under every concentration quoted in this course, and Lesson 17's calibration curves are its instrumental descendant.