A recipe for water says: two hydrogen atoms for every oxygen atom. But nobody can count atoms; a single drop holds more than a thousand billion billion of them. Chemists solve this the way a bank counts coins: by weighing. If you know what one coin weighs, a bag's weight tells you how many coins are inside. The mole is the unit that makes this work for atoms, and it turns every formula into a recipe you can follow on a balance.
By the end you can
- Say what a mole is and why Avogadro's number has the value it has.
- Compute a molar mass from a formula and convert between grams, moles and particles.
- Compute the percent composition of a compound.
- Find an empirical and a molecular formula from lab data.
A bank knows that one 1-euro coin weighs 7.50 g. A bag of coins that weighs 750 g therefore holds 100 coins, and nobody had to count them. Atoms work the same way, with one complication: they are absurdly light. One carbon-12 atom weighs 12 u, which is 1.993 × 10⁻²³ g. Weighing out any useful amount, a gram or so, means handling around 10²³ atoms. So chemists use a counting unit big enough to match: the mole.
A mole (symbol mol) is a count, like a dozen, only much larger. Since 2019 the SI defines it as exactly 6.022 140 76 × 10²³ entities, the Avogadro number. "Entities" means whatever you are counting: atoms, molecules, ions, electrons. A mole of water is 6.022 × 10²³ water molecules, just as a dozen eggs is 12 eggs. The amount of substance n, measured in moles, is the SI's seventh base quantity.
Why such a strange number? It was chosen to make the atomic mass unit and the gram fit together. One mole of atoms that each weigh 1 u weighs (to ten significant figures) exactly 1 g, because N_A is the number of u in a gram. That is the whole trick of the mole: a mass in u, read in grams, is the mass of one mole.
The molar mass M of a substance is the mass of one mole of it, in grams per mole. For an element it is the atomic mass from the periodic table, read in g/mol: iron is 55.845 u per atom, so 55.845 g/mol. For a compound, add up the atomic masses of every atom in the formula, counting each as many times as it appears.
Pick a substance and slide the mass. The molar mass comes straight from the formula and the periodic table: the mass of one mole, in grams, equals the formula's mass in u. Dividing by it turns grams into moles; multiplying by Avogadro's number turns moles into particles. Try 18 g of water: exactly one mole, about a tablespoon. Even a milligram of anything holds more particles than anyone could ever count.
Almost every calculation in this track passes through moles. Grams and particles are the things you can weigh or think about; moles are the bridge between them. Each arrow is one multiplication or division.
| From | To | Do |
|---|---|---|
| mass m (g) | moles n | divide by the molar mass M |
| moles n | mass m (g) | multiply by M |
| moles n | particles N | multiply by N_A |
| particles N | moles n | divide by N_A |
| molecules | atoms of one element | multiply by that element's count in the formula |
Example: a 5.00 g sugar cube. n = 5.00 / 342.30 = 0.0146 mol of sucrose. N = 0.0146 × 6.022 × 10²³ = 8.80 × 10²¹ molecules. Each molecule has 12 + 22 + 11 = 45 atoms, so the cube holds 3.96 × 10²³ atoms, and 0.0146 × 12 = 0.175 mol of carbon atoms, 2.10 g of carbon.
The first chapter computed that water is 11.19 % hydrogen by mass. The same reasoning works for any compound, with molar masses: each element's share is its total mass in one mole, divided by the molar mass.
Now the reverse question, the one a chemist faces with a new substance: the lab measures the percentages, what is the formula? The answer comes in two stages. The empirical formula gives the simplest whole-number ratio of the atoms; for glucose it is CH₂O. The molecular formula gives the actual number of each atom in one molecule; for glucose it is C₆H₁₂O₆, six times the empirical formula. Percentages alone can only give the ratio, since CH₂O and C₆H₁₂O₆ have exactly the same percentages. To find the multiple, you also need the molar mass, measured separately (with a mass spectrometer, for example).
The method has five steps: assume 100 g of the compound, so the percentages become grams; divide each by its atomic mass to get moles of atoms; divide by the smallest number of moles; if a ratio ends in .33, .5 or .67, multiply everything by 3, 2 or 3; finally scale up by k. Step through it below, and try rounding too early.
Step through the method with "next step". Taking exactly 100 g turns every percentage into grams; dividing by the atomic mass turns grams into moles of atoms, and only moles can be compared, because a formula counts atoms, not grams. Dividing by the smallest gives the ratio. When a ratio ends in .33, .5 or .67, multiply everything by 3, 2 or 3 to clear it. Now break it: round the ratio immediately. For glucose nothing goes wrong, but vitamin C becomes CHO and rust FeO₂, and the check row shows those formulas would not have the measured percentages.
Combustion analysis
How does a lab measure the percentages of a carbon compound? It burns a weighed sample in pure oxygen. All the carbon ends up in carbon dioxide and all the hydrogen in water, which are trapped and weighed. Their masses give back the carbon and hydrogen; oxygen is whatever mass is left.
Solids and liquids pack their particles tightly, so a mole of each takes a different volume. Gases are different: their molecules are so far apart that their own size hardly matters, and one mole of any gas takes almost the same volume at the same temperature and pressure: 24.5 L at 25 °C and normal atmospheric pressure (22.4 L at 0 °C). Avogadro guessed this in 1811, long before anyone could count molecules. The gas-laws chapter derives it; for now, it is why a mole of oxygen in the figure above is a box almost 30 cm on each side.
1. How many moles are in 10.0 g of table salt, NaCl? M = 22.990 + 35.45 = 58.44 g/mol, so n = 10.0 / 58.44 = 0.171 mol. That is 0.171 mol of Na⁺ ions and 0.171 mol of Cl⁻ ions.
2. What is the mass of 0.250 mol of calcium carbonate, CaCO₃? M = 40.078 + 12.011 + 3 × 15.999 = 100.09 g/mol, so m = 0.250 × 100.09 = 25.0 g.
3. What does one water molecule weigh? M / N_A = 18.015 g/mol ÷ 6.022 × 10²³ mol⁻¹ = 2.99 × 10⁻²³ g.
A formula from percentages
4. A compound is 85.6 % carbon and 14.4 % hydrogen, and its molar mass is 56.1 g/mol. Moles in 100 g: C 85.6 / 12.011 = 7.13, H 14.4 / 1.008 = 14.3. Ratio 1 : 2.00, so the empirical formula is CH₂ (14.03 g/mol). k = 56.1 / 14.03 = 4, so the molecule is C₄H₈, butene.
| Wrong | Right | Why |
|---|---|---|
| a mole is a mass | a mole is a count | a mole of hydrogen and a mole of gold weigh 2 g and 197 g, but hold the same number of molecules or atoms |
| n = M / m | n = m / M | check the units: g ÷ g/mol = mol |
| the molar mass of O₂ is 16.00 g/mol | it is 32.00 g/mol | oxygen gas is made of O₂ molecules; count both atoms |
| rounding 1.5 to 2 in a ratio | multiply everything by 2 | formulas need whole numbers, reached by multiplying, not rounding |
| CH₂O must be glucose | it is only the empirical formula | formaldehyde, acetic acid and glucose all share it |
Key ideas
- A mole is exactly 6.022 140 76 × 10²³ particles; N = n N_A.
- Avogadro's number makes the mass of a particle in u equal to the mass of a mole in grams.
- The molar mass M is the sum of the atomic masses in the formula, in g/mol; n = m / M.
- Grams ↔ moles ↔ particles: divide or multiply by M, then by N_A.
- Percentages give the empirical formula (simplest ratio); the molar mass gives the molecular formula.
- One mole of any gas takes about 24.5 L at 25 °C and 1 atm.