The Mole Higher
Atoms are far too small and too numerous to count one by one, so chemists count them in huge, fixed-size groups called moles — in the same way you might count eggs in dozens.
The mole (mol) — the unit for the amount of a substance. One mole of any substance contains the same number of particles (atoms, molecules or ions) as one mole of any other substance.
That number is the Avogadro constant: 6.02 × 10²³ per mole.
The link between moles and mass is the key to the whole topic:
The mass of one mole of a substance, in grams, is numerically equal to its relative formula mass (Mr). This is called the molar mass (units: g/mol).
So one mole of carbon (Ar = 12) has a mass of 12 g; one mole of water (Mr = 18) has a mass of 18 g.
number of moles = mass (g)Mr (g/mol)
Rearranging this one relationship lets you switch between mass and moles in either direction: mass = moles × Mr.
The measurement of amounts in moles applies to atoms, molecules, ions, electrons, formulae and equations. For example, one mole of carbon (C) contains the same number of atoms as the number of molecules in one mole of carbon dioxide (CO₂). Always show your working — you can pick up method marks even if the final number is wrong.
One mole of a compound contains 6.02 × 10²³ formula units of that substance, but each compound holds more than one atom or ion — so you must scale each of these by its subscript.
For example, one mole of calcium chloride, CaCl₂, contains one mole of calcium ions (Ca²⁺) but two moles of chloride ions (Cl⁻), because every CaCl₂ unit contains two chlorides. So one mole of CaCl₂ gives 6.02 × 10²³ calcium ions and twice as many chloride ions (1.204 × 1024).
(a) Find the mass of 0.250 mol of zinc. (Ar Zn = 65.)
mass = moles × Mr = 0.250 × 65 = 16.25 g
(b) Find the number of moles in 2.64 g of carbon dioxide, CO₂. (Ar: C = 12, O = 16, so Mr = 44.)
moles = massMr = 2.6444 = 0.06 mol
That 0.06 mol contains 0.06 × (6.02 × 10²³) = 3.61 × 10²² molecules of CO₂.
🧪 Exam-style questions
Calculate the number of moles in 5.4 g of water, H₂O.Use Ar: H = 1, O = 16.
Show answer
- Mr(H₂O) = (2 × 1) + 16 = 18 1 mark
- moles = massMr = 5.418 = 0.3 mol 1 mark
Calculate the mass of 0.25 mol of sodium carbonate, Na₂CO₃.Use Ar: C = 12, O = 16, Na = 23.
Show answer
- Mr(Na₂CO₃) = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 1 mark
- mass = moles × Mr = 0.25 × 106 = 26.5 g 1 mark
Balancing Equations Using Masses Higher
The reverse is also possible: if you know the masses of every reactant and product, you can work out the balancing numbers (the coefficients) of the equation.
- Convert each mass into moles (mass ÷ Mr).
- Write the moles as a ratio.
- Simplify to the smallest whole numbers — these are the balancing numbers. (If you get values like 1 : 2 : 2.5, multiply them all by the same number to clear the decimal.)
You also need to be able to change the subject of the mole equation confidently, since these questions often ask you to rearrange before you substitute.
When methanol burns, 64 g of CH₃OH reacts with 96 g of O₂ to make 88 g of CO₂ and 72 g of H₂O. Deduce the balanced equation. (Mr: CH₃OH = 32, O₂ = 32, CO₂ = 44, H₂O = 18.)
Step 1 — convert every mass to moles:
| Substance | Mass ÷ Mr | Moles |
|---|---|---|
| CH₃OH | 64 ÷ 32 | 2 |
| O₂ | 96 ÷ 32 | 3 |
| CO₂ | 88 ÷ 44 | 2 |
| H₂O | 72 ÷ 18 | 4 |
Step 2 — write the mole ratio: 2 : 3 : 2 : 4 (already whole numbers).
Step 3 — these are the balancing numbers:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
🧪 Exam-style questions
In the Haber process, 28 g of nitrogen (N₂) reacts with 6 g of hydrogen (H₂) to produce 34 g of ammonia (NH₃). Work out the moles of each substance, then complete the balanced equation.Use Mr: N₂ = 28, H₂ = 2, NH₃ = 17. Type the balancing numbers, then press Check.
Show answer
- moles: N₂ = 28 ÷ 28 = 1 · H₂ = 6 ÷ 2 = 3 · NH₃ = 34 ÷ 17 = 2 1 mark
- mole ratio N₂ : H₂ : NH₃ = 1 : 3 : 2 1 mark
- equation: N₂ + 3H₂ → 2NH₃ 1 mark (any equal multiple is accepted by the checker)
In a reaction, 0.5 mol of element A reacts with 1.5 mol of element B. What is the simplest whole-number mole ratio of A : B?Tick (✓) one box.
Reacting Masses Higher
A balanced equation can be read in moles. For example:
Mg + 2HCl → MgCl₂ + H₂
tells you that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to make 1 mole of magnesium chloride and 1 mole of hydrogen. These mole ratios let you calculate the mass of any product or reactant from the mass of another.
- Write the balanced equation (if it isn't already given).
- Convert the known mass into moles (mass ÷ Mr).
- Use the ratio in the equation to find the moles of the substance you want.
- Convert those moles back into a mass (moles × Mr).
Always work molar mass out from the formula, not from the big numbers in the equation. In 2MgO the molar mass of MgO is still 40 g/mol, not 80 — the 2 is a number of moles, not part of the formula.
Iron is extracted from iron(III) oxide in the blast furnace. Calculate the mass of iron produced from 160 g of iron(III) oxide, Fe₂O₃. (Ar: O = 16, Fe = 56.)
Draw a table under the equation. Put Ratio, Moles, Mass and Mr down the left, and the two substances you care about across the top. Write in what you already know, then work out the rest one box at a time.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
| Fe₂O₃known | Fefind this | |
|---|---|---|
| Ratio | 11 | 12 |
| Moles | 41160 ÷ 160 | 521 × 2 |
| Mass | 2160 ggiven | 6112 g2 × 56 |
| Mr | 3160(2×56)+(3×16) | 356 |
Filling the table in order:
- Ratio — read the big numbers in the equation: Fe₂O₃ : Fe = 1 : 2.
- Mass you know — write 160 g under Fe₂O₃.
- Mr of each — work it out from the formula: Fe₂O₃ = (2×56)+(3×16) = 160; Fe = 56.
- Moles you know — mass ÷ Mr = 160 ÷ 160 = 1 mol.
- Moles you want — scale by the ratio (×2): 2 mol of Fe.
- Answer — moles × Mr = 2 × 56 = 112 g of iron.
Now work through a fresh example the same way. You can only fill one box at a time — each box unlocks once the one before it is correct. The prompt under the table tells you what to do next.
Calculate the mass of water produced when 4 g of methane, CH₄, burns completely in oxygen. (Ar: H = 1, C = 12, O = 16.)
CH₄ + 2O₂ → CO₂ + 2H₂O
| CH₄known | H₂Ofind this | |
|---|---|---|
| Ratio | ||
| Moles | ||
| Mass | 4 ggiven | g |
| Mr |
🧪 Exam-style questions
Calculate the mass of magnesium oxide formed when 6 g of magnesium burns completely in oxygen.
Show answer
- moles of Mg = 6 ÷ 24 = 0.25 mol 1 mark
- ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.25 mol 1 mark
- mass of MgO = 0.25 × 40 = 10 g 1 mark (Mr of MgO = 24 + 16 = 40)
Aluminium is extracted industrially by the electrolysis of aluminium oxide. Calculate the mass of aluminium produced from 51 tonnes of aluminium oxide, Al₂O₃.
Show answer
- Mr(Al₂O₃) = (2 × 27) + (3 × 16) = 54 + 48 = 102 1 mark
- moles of Al₂O₃ = 51 ÷ 102 = 0.5 1 mark (working in tonnes throughout)
- ratio Al₂O₃ : Al = 2 : 4 = 1 : 2, so moles of Al = 2 × 0.5 = 1.0 1 mark
- mass of Al = 1.0 × 27 = 27 tonnes 1 mark
The mole ratio works in any unit of mass, as long as you use the same unit on both sides.
Limiting Reactants Higher
When two reactants are mixed, they are rarely present in the exact ratio the equation needs. Usually one is added in excess to make sure the other is completely used up.
The limiting reactant is the reactant that is completely used up in a reaction. It is called “limiting” because once it runs out the reaction stops, so it controls (limits) the amount of product that can form. The reactant left over is said to be in excess.
The amount of product is always proportional to the amount of the limiting reactant: use more of it (and keep the other in excess) and you make more product. To find which reactant is limiting, convert both to moles and compare them against the ratio in the balanced equation.
0.5 g of hydrogen reacts with 8 g of oxygen to form water. Identify the limiting reactant and calculate the mass of water formed. (Mr: H₂ = 2, O₂ = 32, H₂O = 18.)
2H₂ + O₂ → 2H₂O
- Convert both reactants to moles: H₂ = 0.5 ÷ 2 = 0.25 mol; O₂ = 8 ÷ 32 = 0.25 mol.
- Compare with the equation ratio (H₂ : O₂ = 2 : 1): 0.25 mol of H₂ needs only 0.125 mol of O₂. We have 0.25 mol of O₂ — more than enough — so O₂ is in excess and H₂ is the limiting reactant.
- Base the product on the limiting reactant: H₂ : H₂O = 2 : 2 = 1 : 1, so moles of H₂O = 0.25 mol. mass = 0.25 × 18 = 4.5 g.
Try It: Which Reactant Runs Out First?
Choose how many hydrogen (H₂) and oxygen (O₂) molecules to start with, then press React. They combine in the ratio 2H₂ + O₂ → 2H₂O — whatever is left over at the end is the reactant that was in excess, and the one that ran out is the limiting reactant.
These molecules drift freely. Choose how many H₂ and O₂ to start with, then press React.
🧪 Exam-style questions
Magnesium reacts with hydrochloric acid. 0.10 mol of magnesium is added to a solution containing 0.16 mol of HCl. Calculate the maximum mass of hydrogen gas that can be produced.
Show answer
- The ratio Mg : HCl is 1 : 2, so 0.10 mol of Mg would need 0.20 mol of HCl. Only 0.16 mol of HCl is present, so HCl is the limiting reactant. 1 mark
- ratio HCl : H₂ = 2 : 1, so moles of H₂ = 0.16 ÷ 2 = 0.08 mol 1 mark
- mass of H₂ = 0.08 × 2 = 0.16 g 1 mark (you must base the answer on the limiting reactant, not the magnesium)
0.3 mol of carbon is heated with 0.2 mol of oxygen. Which is the limiting reactant?
Tick (✓) one box.