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Chemistry

Writing equations — word equations and symbol equations

SB p.38

There is a sharp line between the two types of chemical equation you could be asked to write. A word-equation names the substances: for example, magnesium + oxygen → magnesium oxide. A chemical-equation (usually called a symbol equation) uses formulae instead: 2Mg + O₂ → 2MgO. Symbol equations are what chemists normally use — you should only write a word equation if you are specifically asked to.

One rule sits above everything else in this section, and it applies to both types: all chemical equations must be balanced. An unbalanced equation is not a slightly imperfect answer — it is a wrong answer, because it asserts that atoms can appear or disappear. The next four sections build the balancing skill step by step, and the exam-hint below flags the single most common way candidates throw marks away.

Exam hint

Read the instruction precisely: you should only write a word equation if you are specifically asked to. If a question says "write an equation", the examiner wants the symbol equation — a word equation where a formula equation was expected scores little or nothing, even if the chemistry is right.

Common misconception

"A word equation and a symbol equation are two different reactions." No — they are two descriptions of the same reaction. Mg + O₂ → 2MgO and magnesium + oxygen → magnesium oxide carry exactly the same chemical information; the symbol version additionally carries the proportions, which is why it can (and must) be balanced.

Quick check

Question-bank join: spec 1.25 items — choosing word vs symbol equations and the must-be-balanced rule.

Key terms in this section

word equation: — An equation naming the substances in words, e.g. magnesium + oxygen -> magnesium oxide. Write one only when specifically asked; all chemical equations must be balanced.
chemical equation: — A symbol equation using formulae (e.g. 2Mg + O2 -> 2MgO); usually what is meant by 'write an equation'. Shows the proportions of reactants and products and must be balanced.

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What all the numbers in a formula mean

SB p.39

Before you can balance anything, you must be able to count atoms through a formula — and that means understanding the two very different kinds of number that appear in equations. Take the balanced equation

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

to make the point. A big number written in front of a formula (the 2 in 2HCl) is sometimes called a coefficient; it multiplies everything in the formula. A small subscript number (the 3 in CaCO₃) applies only to the atom immediately before it.

A second example, 2Cl → Cl₂, shows why the distinction matters chemically, not just arithmetically: 2Cl means two separate chlorine atoms, while Cl₂ means two chlorine atoms joined in a molecule. Same atoms, completely different substances.

Work through the counting example: in 2H₂SO₄ the big 2 gives you 2 molecules of sulfuric acid; each molecule has 2 hydrogens, 1 sulfur and 4 oxygens. Count the atoms in the whole expression and you find 4 hydrogens, 2 sulfurs and 8 oxygens. Brackets extend the subscript's reach: in Ca(OH)₂ the 2 applies to everything inside the bracket — 1 calcium, 2 oxygens, 2 hydrogens. (Chapter 7 (u1:c07) shows how formulae like this are worked out.)

Key point

A small subscript number applies only to the atom immediately before it in the formula. A subscript after brackets applies to everything inside the brackets. A big number in front applies to the whole formula.

Exam hint

Counting atoms through formulae with brackets is exactly where marks go missing under time pressure. Practise until 2H₂SO₄ → 4 H, 2 S, 8 O and Ca(OH)₂ → 1 Ca, 2 O, 2 H are instant, because every balancing question in this chapter — and every reacting-mass calculation later — stands on this counting skill.

Common misconception

"The 2 in 2H₂SO₄ means 2 sulfur atoms." It means 2 whole molecules of H₂SO₄ — so 2 sulfurs, 4 hydrogens, 8 oxygens in total. The big number multiplies everything; the subscripts describe each single molecule. Confusing the two scales of "two" is the root of most miscounts.

Quick check

Question-bank join: spec 1.25 items — counting atoms through coefficients, subscripts and brackets.

Key terms in this section

coefficient: — The big number written in front of a formula in an equation (the 2 in 2HCl); it multiplies every atom in the formula and gives the number of moles reacting.

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Balancing equations

SB p.39

Chemical reactions rearrange atoms into new combinations — they never create or destroy them. It follows that a finished equation must end up with exactly the same number of each type of atom that it started with; that is what makes it a balanced-equation. Watch the idea build with methane burning in oxygen:

CH₄ + O₂ → CO₂ + H₂O

Count the atoms on each side and the equation can't be right: the right side has gained an oxygen atom and lost two hydrogens. The substances are all correct — methane really does burn to carbon dioxide and water — so the proportions must be wrong. Fixing proportions, without touching the substances, is the whole skill.

One sentence is the single most important sentence in this chapter, and it deserves to be quoted in full: you must never, never change a formula when balancing an equation. All you are allowed to do is write big numbers in front of the formulae. The next section applies the fix to the methane equation, and Worked example 1 shows the systematic method in action.

Exam hint

This is the really-important one, and the emphasis is the point: never change a formula when balancing. If the equation needs another oxygen, you write a bigger big number — you never turn O₂ into O₃ or H₂O into H₂O₂. Changing a formula changes the substances, which changes the chemistry; changing a big number changes only the proportions, which is what balancing means.

Common misconception

"CH₄ + O₂ → CO₂ + H₂O is nearly balanced, so it is nearly right." There is no such thing as nearly balanced — an equation either conserves every atom type or it is wrong. Here the left side has 4 H and 2 O while the right has 2 H and 3 O. The chemistry (which substances) is fine; the proportions (how much of each) are not, and only the proportions may be edited.

Quick check

Question-bank join: spec 1.25 items — the conservation rule and the never-change-a-formula law.

Key terms in this section

balanced equation: — An equation with the same number of atoms of each element on both sides, balanced by writing big numbers in front of formulae — never by changing a formula.

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Try again — balancing the methane equation

SB p.40

Now fix the methane equation. With the proportions adjusted — and only the proportions — the balanced equation for the combustion of methane is:

CH₄ + 2O₂ → CO₂ + 2H₂O

Count each symbol as one atom of that element and check both sides: 1 carbon each side; 4 hydrogens each side (4 on the left inside CH₄, 2 × 2 on the right inside 2H₂O); 4 oxygens each side (2 × 2 on the left, 2 + 2 on the right). Every atom type balances — the equation is balanced.

Balancing is a skill with a method. Learn the systematic approach and apply it exactly:

Work across the equation from left to right, checking one element after another. If an element seems impossible to fix that way, leave it until the end — you will often find it has sorted itself out.
Count unchanged groups as a whole. If a group of atoms such as the sulfate group (SO₄) appears unchanged on both sides, count the group once instead of individual sulfurs and oxygens. It saves time.
Check everything at the end, to be sure you have not disturbed something you already counted.

Exam hint

The three-step method is worth adopting verbatim: left-to-right, one element at a time; unchanged atom-groups counted whole; full recount at the end. The "leave the awkward element until last" tip is the one that rescues equations involving O₂ — you will see exactly why in Worked example 2 (the ethane combustion).

Common misconception

"I balanced it left to right, so it must still be balanced when I finish." Not necessarily — fixing a later element can unbalance an earlier one (for instance, changing 2H₂O on the right alters both the H and the O counts). That is precisely why the method ends with check everything at the end: the last pass is not optional tidying, it is where the marks are secured.

Quick check

Question-bank join: spec 1.25 items — the systematic balancing method on familiar equations.

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Worked example 1 — balancing zinc + hydrochloric acid

SB p.40

Balance the equation for the reaction between zinc and hydrochloric acid:

Zn + HCl → ZnCl₂ + H₂

Run the systematic method from the previous section, left to right. Zinc: 1 atom on each side — no problem. Hydrogen: 1 atom on the left, 2 on the right; if there are 2 at the end, there must have been 2 at the start, and the only way to achieve that is 2HCl. (You must not change the formula to H₂Cl, because that substance does not exist.) Chlorine: with 2HCl there are now 2 on each side — good. Finally check everything again, and the equation is finished:

Zn + 2HCl → ZnCl₂ + H₂

The atom-count table for the finished equation:

	Zn	H	Cl	Zn	H	Cl
numbers of atoms	1	2	2	1	2	2

Key point

Halves are acceptable in balanced equations, and you can always remove one later by doubling everything — but you will rarely meet them at International GCSE. Worked example 2 shows the halving trick in its natural habitat: an odd number of oxygens.

Exam hint

A second hint settles two state-symbol worries before they start: don't worry if you don't yet know what the state symbols should be — that is not important at this stage. But do register the correction it adds: water is a liquid (l), not an aqueous solution (aq) — an aqueous solution is formed when something is dissolved in water.

Common misconception

"If I need 2 hydrogens on the left, I can write H₂Cl." You cannot — H₂Cl is not a substance. Hydrochloric acid is HCl, period. Balancing adjusts how many formula units take part (the big number 2HCl); it never rewrites what the substance is. This is the never-change-a-formula law from the last section, caught red-handed in exam conditions.

Quick check

Question-bank join: spec 1.25 items — left-to-right balancing with the atom-count check.

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State symbols

SB p.41

State symbols are often — but not always — written after each formula in an equation to show the physical state everything is in. The four you need to know:

Symbol	State
(s)	solid
(l)	liquid
(g)	gas
(aq)	in aqueous solution (dissolved in water)

A worked illustration:

2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g)

Read aloud, this says: solid potassium reacts with liquid water to make a solution of potassium hydroxide in water, and hydrogen gas. Notice how much extra information the state symbols carry — the same balanced equation without them would not tell you that the potassium hydroxide ends up dissolved, or that the hydrogen leaves as a gas.

Exam hint

The two symbols candidates mix up are (l) and (aq). Water in a beaker is H₂O(l); salt dissolved in that water is NaCl(aq). The previous section said it plainly: water is a liquid, not an aqueous solution — (aq) is reserved for something dissolved in water. State symbols are part of the 1.25 objective ("including state symbols"), so a perfectly balanced equation can still drop marks without them.

Common misconception

"(aq) means 'in water', so pure water is H₂O(aq)." No — (aq) means dissolved in water, which requires something doing the dissolving. Pure water is the solvent itself: H₂O(l). Only species dissolved in water — ions like KOH(aq) above — earn the (aq) label.

Quick check

Question-bank join: spec 1.25 items — the four state symbols and their correct assignment.

Key terms in this section

state symbol: — A symbol after a formula showing physical state: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water — e.g. 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g).
aqueous: — Dissolved in water, written (aq) after the formula. Formed when something dissolves in water; pure water itself is the liquid H2O, not aqueous.

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How much of each substance reacts? — the carbon-12 scale and Aᵣ

SB p.41

A practical question sets up the whole second half of the chapter: make iron(II) sulfide by heating iron with sulfur —

Fe + S → FeS

— and ask: in what proportions should you mix them? Certainly not equal masses, because iron and sulfur atoms do not weigh the same. An iron atom contains more protons and neutrons than a sulfur atom, so it is about one and three-quarter times heavier. Getting the proportions right therefore requires knowing the masses of the individual atoms taking part — which is exactly what the relative-atomic-mass delivers.

The carbon-12 scale. A single atom weighs about 1.67 × 10⁻²⁴ g — an absurd number to calculate with. Instead, chemists compare all atomic masses against a standard: the mass of an atom of the carbon-12 isotope (carbon-12-scale). On this scale one ¹²C atom weighs exactly 12 units, and the basic unit of the scale is one-twelfth of that — approximately the mass of the most common hydrogen atom. So fluorine-19 has a relative mass of 19 (19 times one-twelfth of a ¹²C atom), and magnesium's most common isotope has a relative mass of 24.

From isotopes to Aᵣ. Real samples contain a mixture of isotopes, so we need the average mass of an atom taking them all into account — the relative atomic mass. The definition, verbatim:

The relative atomic mass of an element is the weighted average mass of the isotopes of the element. It is measured on the carbon-12 scale, on which a carbon-12 (¹²C) atom has a mass of exactly 12.

Because these are relative masses, they have no units. Chlorine's Aᵣ is 35.45 (quoted 35.5), lithium's is 6.94 (quoted 7), sodium's is 22.99 (quoted 23) — averages over their isotope mixtures. The International GCSE convention: use the Periodic Table values, with decimals where given (Cl 35.5, Cu 63.5).

Key point

To work out a weighted average you have to know how much of each isotope is present in the sample — the weighted-average method from Chapter 3 (u1:c03) is the same mathematics again here.

Exam hint

One vocabulary line the examiner enforces: avoid 'relative molecular mass' for ionic substances — it can only properly describe actual molecules (covalent substances). You should not use it for magnesium oxide or sodium chloride. Relative formula mass covers everything, which is why the next section installs it as the default term.

Common misconception

"Aᵣ = 35.5 means a chlorine atom weighs 35.5 units." No single chlorine atom does — 35.5 is the weighted average across the ³⁵Cl/³⁷Cl mixture. Any one atom weighs 35 or 37; the sample average is 35.5. This is the Chapter 3 isotope story (weighted averages need how much of each isotope is present) reappearing as the foundation for Mᵣ.

Quick check

Question-bank join: spec 1.26 items — the carbon-12 scale and Aᵣ as weighted average.

Key terms in this section

relative atomic mass: — The weighted average mass of the isotopes of an element, measured on the carbon-12 scale on which a 12C atom is exactly 12. Relative masses have no units.
carbon-12 scale: — The comparison scale for atomic and formula masses: one atom of the carbon-12 isotope weighs exactly 12 units, so the scale's unit is one-twelfth of a 12C atom.

Watchblock 8

GCSE Chemistry - Relative Formula Mass | Mr & Percentage Mass Calculations (2027/28 exams)

Relative formula mass — adding Ar values the formula dictates, brackets included.

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Relative formula mass (Mᵣ)

SB p.42

Compounds get weighed on the same carbon-12 scale as atoms. A water molecule, H₂O, has a mass of 18 on this scale — 18 times the mass of one-twelfth of a ¹²C atom. For compounds the term is relative-formula-mass, given the symbol Mᵣ (it is sometimes called relative-molecular-mass — the previous section explains why formula mass is the safer phrase: it covers ionic substances too).

Calculating an Mᵣ is bookkeeping: add up the relative atomic masses of every atom in the formula, using the Periodic Table. The copper(II) sulfate crystals example, CuSO₄·5H₂O, shows the one twist you must be ready for. When some substances crystallise from solution, water becomes chemically bound to the salt — water-of-crystallisation — and the bound water is part of the formula: there are always 5 water molecules per CuSO₄ unit, and salts carrying it are hydrated. (Other examples: Na₂CO₃·10H₂O and MgCl₂·6H₂O.)

Key point

Relative atomic masses are always given in the exam — in the question or on the Periodic Table. You never calculate Aᵣ yourself; but if you use the Periodic Table, take the larger of the two numbers (the relative atomic mass, not the atomic number).

Exam hint

The classic miscount on hydrated formulae: work out the hydrogens and oxygens of the water as complete H₂O units, not separately. The common mistake is to get 10 hydrogens right (2 × 5) but then count only 1 oxygen instead of 5. Five waters = 10 H and 5 O. (The structure: five water molecules bonded to each copper(II) sulfate unit — and write copper(II) properly: capital I twice, not Il.)

Common misconception

"The ·5H₂O in CuSO₄·5H₂O is just wet salt — you can ignore it in calculations." You cannot. The water of crystallisation is chemically bound and part of the formula; the salt is hydrated, and its Mᵣ includes all five waters. Ignoring them turns every downstream percentage-composition and mole calculation on this salt into a wrong answer.

Quick check

Question-bank join: spec 1.26 items — Mᵣ from Aᵣ values, hydrated formulae included.

Key terms in this section

relative formula mass: — The sum of all the relative atomic masses in a formula unit, on the carbon-12 scale; symbol Mᵣ. Used for any compound, ionic or covalent — e.g. H2O = 18, CuSO4·5H2O = 249.5.
relative molecular mass: — The Mᵣ of a substance that exists as molecules (covalent substances). Properly applied only to molecules — use relative formula mass for ionic substances like NaCl or MgO.

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Using relative formula mass to find percentage composition

SB p.43

Once you have an Mᵣ, one division tells you what fraction of a compound's mass any part of it contributes — the percentage-composition. First worked case, magnesium carbonate:

To find the Mᵣ of MgCO₃ (Aᵣ values: C = 12, O = 16, Mg = 24):

Mᵣ = 24 + 12 + (3 × 16) = 24 + 12 + 48 = 84

You add the masses of 1 × Mg, 1 × C and 3 × O — the subscripts do the counting. The second case adds brackets to the arithmetic, calcium hydroxide:

To find the Mᵣ of Ca(OH)₂ (Aᵣ values: Ca = 40, O = 16, H = 1):

Mᵣ = 40 + 2 × (16 + 1) = 40 + 34 = 74

The bracketed (OH) group is counted as a unit (16 + 1 = 17) and then doubled — exactly the bracket rule from section 3, now earning its keep in a calculation. With the Mᵣ in hand, the percentage by mass of any element follows: that element's total contribution divided by the whole, × 100. The next two sections work that division through two complete examples.

Exam hint

Two habits are worth copying in the exam. First, write the Aᵣ values down before starting — list them ("Relative atomic masses: C = 12, O = 16, Mg = 24") even though they are given, because it makes the working checkable. Second, multiply first, add second: 3 × 16 = 48 as its own step stops the "C + O₃" addition errors.

Common misconception

"Mᵣ of Ca(OH)₂ is 40 + 16 + 1 + 2 = 59." The 2 after the bracket doubles everything inside it — both the O and the H. Correct working is 40 + 2 × (16 + 1) = 40 + 34 = 74. Forgetting to multiply the hydrogen is the bracket-rule failure wearing a calculation costume.

Quick check

Question-bank join: spec 1.26 items — Mᵣ with bracketed groups.

Key terms in this section

percentage composition: — The percentage by mass of an element in a compound: the element's total atom mass (atoms × Ar) divided by the compound's Mᵣ, × 100 — e.g. oxygen in Na2CO3 is 48/106 × 100.

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Worked example — Mᵣ of copper(II) sulfate crystals, CuSO₄·5H₂O

SB p.43

This formula looks far more complicated than the previous ones — the advice is to break it into its two natural pieces. First, why the water is there at all: when some substances crystallise from solution, water becomes chemically bound up with the salt. This is water-of-crystallisation, and salts containing it are hydrated (hydrated-salt). There are always 5 water molecules per 1 CuSO₄ unit — they are part of the formula, not impurities. (Other examples: Na₂CO₃·10H₂O and MgCl₂·6H₂O.)

To find the Mᵣ of CuSO₄·5H₂O (Aᵣ values: H = 1, O = 16, S = 32, Cu = 63.5).

The method: work out the relative formula mass of water first, then add the correct number of waters on to the rest of the formula:

Water: Mᵣ(H₂O) = (2 × 1) + 16 = 18
The salt part: Mᵣ(CuSO₄) = 63.5 + 32 + (4 × 16) = 63.5 + 32 + 64 = 159.5
Assemble: 159.5 + (5 × 18) = 159.5 + 90 = 249.5

The same section then points the tool at a single element — to find the percentage by mass of copper in copper(II) oxide, CuO (Aᵣ: O = 16, Cu = 63.5): Mᵣ(CuO) = 63.5 + 16 = 79.5, of which 63.5 is copper, so the copper percentage is (63.5 / 79.5) × 100 ≈ 79.9%. "Having found the relative formula mass of a compound, we can work out the percentage by mass of any part of it" — the bridge to the next section.

Key point

For a hydrated salt, compute the water's Mᵣ once (18 for H₂O) and multiply by the number of waters of crystallisation. The dot · in the formula means "plus this many whole waters per formula unit" — nothing more mysterious than that.

Exam hint

Structure your Mᵣ working for hydrated salts in three parts: salt part, water part, assemble. Examiners award method marks for the stages — a bare "249.5" with no visible 159.5 + 90 can lose the working mark even with the right answer. And quote Mᵣ(CuSO₄·5H₂O) = 249.5 — this exact value returns in Worked example 7 (percentage yield of these very crystals).

Common misconception

"CuSO₄·5H₂O means CuSO₄ joined to 5 separate H₂O molecules floating around the crystal." The five waters are chemically bound into the crystal structure — bonded to each CuSO₄ unit. They are not loose water you could shake off; heating is needed to drive them out (that experiment is Worked example 6's neighbour, section 26).

Quick check

Question-bank join: spec 1.26 items — Mᵣ of hydrated salts.

Key terms in this section

water of crystallisation: — Water chemically bound into a salt's crystal structure as part of its formula (the ·5H2O in CuSO4·5H2O); always fixed whole numbers per formula unit. Salts containing it are hydrated.
hydrated salt: — A salt containing water of crystallisation as part of its formula, e.g. CuSO4·5H2O, Na2CO3·10H2O, MgCl2·6H2O. Its Mᵣ includes the bound water.

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The mole

SB p.44

In chemistry, the mole is a unit of the amount of substance. You can express an amount in grams or in moles — and the difference between the two expressions is that 1 mole of any substance has its own particular mass. Examples: 1 mole of water has a mass of 18 g, 1 mole of sulfur 32 g, 1 mole of magnesium oxide 40 g. The abbreviation for mole is mol.

So the same sample can be described both ways: 36 g of water is 2 mol (because 1 mol of water is 18 g); but 36 g of sulfur is only 1.125 mol (1 mol of sulfur is 32 g), and 36 g of magnesium is 1.5 mol (1 mol of magnesium is 24 g). Same 36 grams — different amounts of substance, because the particles weigh different amounts. On phrasing: "amount in moles" and "number of moles" mean the same thing, just as "amount in grams" and "number of grams" do.

Where the mass of 1 mole comes from: calculate the relative formula mass (Mᵣ) and attach the units, grams. Three worked out:

1 mole of oxygen gas, O₂ (Aᵣ: O = 16): Mᵣ = 32, so 1 mol has a mass of 32 g.
1 mole of calcium chloride, CaCl₂ (Aᵣ: Cl = 35.5, Ca = 40): Mᵣ = 40 + (2 × 35.5) = 111, so 1 mol has a mass of 111 g.
1 mole of iron(II) sulfate crystals, FeSO₄·7H₂O (Aᵣ: H = 1, O = 16, S = 32, Fe = 56): Mᵣ = 56 + 32 + (4 × 16) + 7 × [(2 × 1) + 16] = 56 + 32 + 64 + 126 = 278, so 1 mol has a mass of 278 g — the hydrated-crystal method from section 12, reused without ceremony.

Key point

Mass of 1 mole = Mᵣ expressed in grams. This is the molar-mass idea — the next section adds the terminology and its one fine-print distinction.

Exam hint

Notice the oxygen example specifies oxygen gas, O₂ — 32 g, not 16 g. Whenever a question says "1 mole of oxygen", check whether it means atoms (O, 16 g) or molecules (O₂, 32 g): the next section makes quoting the formula compulsory for exactly this reason.

Common misconception

"1 mole of every substance weighs the same." That is the property of 1 mole (it contains the same number of particles — section 17) — not of 1 mole's mass. 1 mol of water is 18 g, 1 mol of sulfur 32 g, 1 mol of MgO 40 g: equal particle counts, unequal masses, because their particles are differently heavy.

Quick check

Question-bank join: spec 1.27/1.28 items — mass of 1 mole from Mᵣ, hydrated crystals included.

Key terms in this section

mole: — The unit of amount of substance, abbreviated mol. 1 mole of a substance has a mass in grams equal to its Mᵣ and contains 6 × 10²³ particles.
molar mass: — The term commonly used for the mass of 1 mole of a substance. Strictly it has units of g/mol (the mass of 1 mole is in g), a distinction rarely needed at International GCSE.

Step throughblock 4

The mole ladder: magnesium, Ar = 24.3.

1•Moles in 48.6 g of Mg (n = m / Ar)
(mol)
2•Number of atoms in that sample (n x 6.02 x 10^23)
3•Mass of 0.25 mol of Mg (m = n x Ar)

Watchblock 9

GCSE Chemistry - Moles & Mass - Avogadro's Constant | Formula for Moles, Mass & Mr (2027/28 exams)

The mole — Avogadro's constant, moles from mass, and the conversion triangle the calculations paper runs on.

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Quoting the formula — and the molar mass vocabulary

SB p.44

Whenever you talk about a mole of something, you must quote its formula — otherwise there is a real risk of confusion. The classic example: "1 mole of oxygen" could mean

1 mole of oxygen atoms, O, with a mass of 16 g, or
1 mole of oxygen molecules, O₂, with a mass of 32 g.

Twice the mass, same words. The formula is what disambiguates, which is why exam answers never say "oxygen" without the formula when a mole calculation is in play.

This section assembles the vocabulary and the number sense you need:

'Molar mass' is often used instead of 'mass of 1 mole' — same practical meaning at this level.
Fine print (extension, for completeness): strictly, molar mass has units of g/mol while mass of 1 mole has units of g. You are unlikely to need the distinction at International GCSE.
Don't over-rely on learned formulae: if 1 mole weighs 100 g, it should be obvious that 2 moles weigh twice as much, 10 moles ten times as much, 0.2 moles 0.2 times as much.
Division as sense-check: if 1 mol weighs 18 g, the number of moles in 54 g is simply how many times 18 goes into 54. If you feel safer with a formula, use it — but never let the formula replace the sanity check.

Exam hint

The "how many times does it go in?" instinct is a genuine exam tool: in 54 g of water, 18 goes in 3 times → 3 mol. Do this mental division before reaching for moles = mass/Mᵣ, and you will catch flipped fractions (Mᵣ/mass) before they cost you the answer — 18/54 would plainly be nonsense here.

Common misconception

"'Moles of oxygen' is a complete answer." It is not — the examiner cannot mark what they cannot interpret. "1 mol oxygen" is ambiguous between 16 g (atoms) and 32 g (molecules); always write 1 mol O₂ or 1 mol O. The rule: quote the formula, every time.

Quick check

Question-bank join: spec 1.27/1.28 items — formula quoting and mole-mass sense checks.

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Simple calculations with moles

SB p.45

You need to be able to interconvert between a mass in grams and an amount in moles for a given substance. There is one formula to learn:

$\text{number of moles} = \frac{\text{mass (g)}}{\text{mass of 1 mole (g)}}$

You can rearrange it to find whatever you want. If rearranging causes problems, use a formula triangle: mass on top, number of moles and mass of 1 mole on the bottom — cover the quantity you want and the triangle shows the operation (top ÷ bottom, or bottom × bottom, or top ÷ bottom). Look at it carefully and make sure you can deduce all three equations from it.

Two worked cases, both with the Mᵣ found first:

Finding the mass of 0.2 mol of calcium carbonate, CaCO₃ (Aᵣ: C = 12, O = 16, Ca = 40):

Mᵣ(CaCO₃) = 40 + 12 + (3 × 16) = 100, so 1 mol has a mass of 100 g
0.2 mol has a mass of 0.2 × 100 = 20 g

Finding the number of moles in 54 g of water, H₂O (Aᵣ: H = 1, O = 16):

Mᵣ(H₂O) = (2 × 1) + 16 = 18, so 1 mol has a mass of 18 g
54 g of water is 54 / 18 = 3 mol

Key point

Number of moles = mass in grams ÷ mass of 1 mole in grams. Every remaining calculation in this chapter — reacting masses, yields, excess — is this line plus a chemical equation.

Exam hint

Always find (and show) the Mᵣ first. The two workings above each carry three visible stages — Mᵣ, then the mole-mass link, then the answer. That structure earns method marks even when an arithmetic slip lands late, and it makes your answer checkable by the "how many times does it go in?" instinct from the last section.

Common misconception

"moles = mass ÷ Mᵣ, so to get the mass I also divide by Mᵣ." Rearrange, don't repeat: mass = moles × mass of 1 mole. The two directions are the triangle's two different reads — 0.2 mol × 100 g/mol = 20 g versus 54 g ÷ 18 g/mol = 3 mol. Writing the division both ways is the single most common mole-arithmetic slip.

Quick check

Question-bank join: spec 1.28 items — mass ↔ moles interconversion.

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Moles and the Avogadro constant

SB p.46

Here is the number that makes the mole worth having. Imagine 1 mole of ¹²C: it has a mass of 12 g and contains about 6 × 10²³ carbon atoms — 6 followed by 23 noughts. That number of particles in 12 g of ¹²C is the avogadro-constant.

And the punchline: 1 mole of anything else contains this same number of particles. Examples:

1 mole of magnesium contains 6 × 10²³ magnesium atoms, Mg, and has a mass of 24 g
1 mole of water contains 6 × 10²³ water molecules, H₂O, and has a mass of 18 g

This is the reason we use moles: if we know the number of moles we also know how many particles are present — and if we know how many particles are present, we can work out how much of one substance reacts with a certain amount of another. Reactions happen particle-by-particle; the mole is the bridge from the grams you can weigh to the particles that actually react.

A scale check: approximately 1 × 10³² water molecules go over Niagara Falls every second in summer — 100 million million million million million molecules. Against that, 6 × 10²³ is a small, weighable, laboratory-sized crowd.

Key point

Equal moles = equal numbers of particles, regardless of substance. 1 mol of Mg and 1 mol of H₂O both carry 6 × 10²³ particles — the masses differ (24 g vs 18 g) because the particles differ; the counts do not.

Exam hint

Two forms of the same fact get tested: the Avogadro constant value (6 × 10²³ per mole) and the inference (moles → particles: 0.5 mol of Mg contains 0.5 × 6 × 10²³ = 3 × 10²³ atoms). Watch the precision habit in the examples — "about 6 × 10²³", "approximately" — the constant is quoted to one significant figure at this level.

Common misconception

"1 mole of magnesium has more particles than 1 mole of water, because magnesium atoms are heavier." Heavier particles mean a greater mass (24 g vs 18 g) — but the count is identical: 6 × 10²³ each. A mole is a count, not a weight; that is precisely why it can convert between the two.

Quick check

Question-bank join: spec 1.27/1.28 items — the Avogadro constant and moles-to-particles conversion.

Key terms in this section

Avogadro constant: — The number of particles in one mole of any substance: about 6 × 10²³ per mole. Equal moles of any substances contain equal numbers of particles.

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Formulae — from mole ratios to empirical and molecular formulae

SB p.46

Formulae come from mole ratios. Start with sulfur dioxide, SO₂: there are 2 O atoms for each S atom. Scale it up — if you had 1 mol of SO₂ you would still have twice as many O atoms as S atoms: 1 mol of S atoms and 2 mol of O atoms. And back down: if an experiment found a sample of sulfur dioxide containing 0.1 mol of S atoms, it must also contain 0.2 mol of O atoms (0.2 mol contains twice as many atoms as 0.1 mol — the same point holds with 0.02 vs 0.01 mol).

Run the logic forward and you can read formulae out of mole data:

A compound contains 0.2 mol Ca and 0.2 mol O → equal numbers of atoms → the formula is CaO.
A compound contains 0.4 mol Mn and 0.8 mol O → twice as many O atoms as Mn atoms → the formula is MnO₂.

Formulae found this way have names, and both are defined:

The empirical-formula shows the simplest whole number ratio of the atoms present in a compound.

The molecular-formula shows the actual number of atoms of each element present in a molecule (covalent compound) or formula unit (ionic compound) of a compound.

The two can coincide or differ: calcium oxide is CaO both ways; hydrogen peroxide has empirical formula HO but molecular formula H₂O₂. To get from one to the other you need one extra datum — the Mᵣ of the compound — which is exactly the machinery the next four sections build.

Key point

Mole ratios are atom ratios. The empirical formula is that ratio in simplest whole numbers; the molecular formula is the actual atom count, and it is always the empirical formula (or a whole-number multiple of it).

Exam hint

A significant-figures confession: more significant figures should have been used for the numbers of moles in these examples, but were omitted for simplicity. In the exam, carry more figures through the intermediate mole values (the practicals later give 0.0400 vs 0.040 for exactly this reason) and only round at the ratio.

Common misconception

"The empirical formula and molecular formula are always different." They are often the same — CaO is both. They differ only when the molecular formula is a multiple of the simplest ratio (HO vs H₂O₂). Treating "empirical" as automatically "not the real formula" is wrong; treating it as automatically "the real formula" is wrong too — you need the Mᵣ to decide (section 23).

Quick check

Question-bank join: spec 1.32/1.33 items — empirical vs molecular definitions and mole-ratio reading.

Key terms in this section

empirical formula: — The simplest whole-number ratio of the atoms of each element present in a compound, found from mole ratios — e.g. CH2 for the compound whose molecular formula is C4H8.
molecular formula: — The actual number of atoms of each element in one molecule (or formula unit). It is the empirical formula multiplied by a whole number — e.g. H2O2 is the molecular formula with empirical HO.

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Worked examples 3 and 4 — empirical formulae from masses

SB p.47

In order to find the empirical formula of a compound such as copper oxide, we need to know how many atoms of copper combine with how many atoms of oxygen — and we work atoms out from moles. If we know the ratio between the numbers of copper and oxygen atoms, we know the formula.

Example 3 — a sample of a compound contains 1.27 g of Cu and 0.16 g of O. Calculate the empirical formula (Aᵣ: Cu = 63.5, O = 16). It is easiest in columns, using a table:

	Cu	O
masses/g	1.27	0.16
moles of atoms = mass ÷ mass of 1 mole of atoms	1.27/63.5	0.16/16
number of moles of atoms	0.02	0.01
divide by the smaller number for the ratio	0.02/0.01	0.01/0.01
ratio of moles	2	1
empirical formula	Cu₂O

There are twice as many moles of Cu as O — so twice as many Cu atoms as O atoms in the compound. One warning that examiners love: we are counting atoms, so we divide by the mass of 1 mole of oxygen atoms (16), not by 32 (the mass of 1 mole of oxygen molecules).

Example 4 — a sample contains 0.78 g of K, 1.10 g of Mn and 1.28 g of O (Aᵣ: K = 39, Mn = 55, O = 16). Again, a table:

	K	Mn	O
masses/g	0.78	1.10	1.28
moles of atoms = mass ÷ mass of 1 mole of atoms	0.78/39	1.10/55	1.28/16
number of moles of atoms	0.02	0.02	0.08
divide by the smallest number for the ratio	0.02/0.02	0.02/0.02	0.08/0.02
ratio of moles	1	1	4
empirical formula	KMnO₄

Exam hint

On this page: the original masses were measured to 2 significant figures, so the numbers of moles should also be given to 2 significant figures (0.010). If you write 0.01 in the exam that is fine and still gives the correct answer — but match your figure count to your data, deliberately rather than accidentally.

Common misconception

"0.16 g of O is 0.16/32 = 0.005 mol, so the formula would be Cu₄O." The compound's formula counts atoms, and the oxygen in it arrived as individual atoms — divide by 16 (1 mol of O atoms), not 32 (1 mol of O₂ molecules). Worth stopping for: this is the single most common empirical-formula slip.

Quick check

Question-bank join: spec 1.33 items — empirical formulae from mass data in table form.

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Practicals — finding the formula of a metal oxide by combustion and by reduction

SB p.48

Spec 1.36 is a practical: know how to determine the formula of a metal oxide by combustion or by reduction. Both experiments are here, each ending in the same empirical-formula table.

By combustion — magnesium oxide

Burn magnesium in oxygen and watch how the mass changes. Procedure: weigh a crucible with its lid; add a ~10 cm piece of magnesium ribbon and weigh again; heat strongly (a roaring flame), lifting the lid every few seconds; cool and weigh when finished. The magnesium burns with a bright white flame to give magnesium oxide, a white powder. The lid stops the powder escaping — but it must be lifted to let oxygen in.

A set of results: crucible 32.46 g; + magnesium 32.70 g; + contents at the end 32.86 g.

mass of magnesium = 32.70 − 32.46 = 0.24 g
the mass increases because the magnesium combines with oxygen from the air: mass of oxygen gained = 32.86 − 32.70 = 0.16 g
the reaction type was combustion

The empirical-formula table (Aᵣ: Mg = 24, O = 16):

	Mg	O
masses/g	0.24	0.16
moles of atoms	0.24/24	0.16/16
number of moles	0.010	0.010
÷ smaller	1	1
empirical formula	MgO

Safety

Wear eye protection and take care not to get burned when raising the crucible lid with tongs. The crucible will stay very hot for some time.

By reduction — copper oxide

Alternatively, remove the oxygen and watch the mass fall. Pass hydrogen gas over copper oxide in a ceramic dish inside a tube; ignite the excess hydrogen at the small hole; heat strongly until pink-brown copper is seen. Results: dish 23.78 g; + copper oxide 26.96 g; + contents at end 26.32 g.

mass of copper oxide = 26.96 − 23.78 = 3.18 g
mass of copper at the end = 26.32 − 23.78 = 2.54 g
mass of oxygen removed = 3.18 − 2.54 = 0.64 g

The reaction, CuO + H₂ → Cu + H₂O, is a displacement-reaction (hydrogen is more reactive than copper and displaces it) — and can also be called a redox-reaction: the copper oxide has been reduced — oxygen removed, the reduction of the copper(II) oxide. Chapter 14 develops both.

The table (Aᵣ: Cu = 63.5, O = 16): 2.54/63.5 = 0.0400 mol Cu; 0.64/16 = 0.040 mol O; ratio 1 : 1 → empirical formula CuO.

Safety

The teacher demonstrating needs to wear a face shield and use safety screens; pupils require eye protection and should be no closer than 2 metres. If a drying agent is needed, use anhydrous calcium chloride — NOT concentrated sulfuric acid.

Key point

Before lighting the hydrogen stream, fill the whole tube with hydrogen first: a hydrogen–oxygen mixture is explosive. Let the gas flow through for a while to flush out all the oxygen, then light the stream (or the Bunsen burner).

Key point

How do you know the reaction has finished? Nothing more happens in the crucible — and you check by heating to constant mass: weigh, heat, weigh again, repeating until the mass stays constant. (Precision note: the original masses here are 2 significant figures, so quote the moles as 0.010 — the extra figure shows the precision honestly.)

Exam hint

In practice the ratio "is not very often" exactly 1:1 — the experimental culprits: some magnesium oxide powder may escape when the lid is lifted; not all the magnesium may react; and magnesium can also react with nitrogen in the air. Expect a question to hand you a slightly-off ratio and ask which of these explains it — that question is asking for this list.

Common misconception

"The mass went down in the reduction experiment, so matter was destroyed." The oxygen left the dish as water vapour — hydrogen carried it away. Every gram is conserved: 0.64 g of oxygen moved from copper oxide into the water stream. "Mass decrease" in this experiment means oxygen removed, i.e. reduction — not loss of matter.

Quick check

Question-bank join: spec 1.31/1.36 items — the two metal-oxide practicals and their mass-change logic.

Key terms in this section

combustion: — Burning a substance in oxygen — the reaction type used to find the formula of magnesium oxide: the magnesium gains oxygen, so the crucible's contents increase in mass.
reduction: — The removal of oxygen from a substance — used to find the formula of copper oxide by passing hydrogen over it: the oxide loses oxygen, so the dish's contents decrease in mass.
displacement reaction: — A reaction in which a more reactive element takes the place of a less reactive one in a compound — hydrogen displaces copper from copper oxide in the reduction practical.
redox reaction: — A reaction in which reduction and oxidation happen together — e.g. CuO + H2 -> Cu + H2O: the copper oxide is reduced (oxygen removed) while the hydrogen is oxidised.

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Balancing chemical equations — the examiner's ground rules.

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Determining the formula of water

SB p.51

Modify the reduction apparatus slightly and the same experiment determines the formula of water: this time the water vapour produced by the reaction between copper(II) oxide and hydrogen is condensed in a U-tube and collected. The dish is weighed before and after as before, and the mass of water collected is measured.

The results table adds one row to the reduction experiment's:

	g
mass of empty dish	23.78
mass of dish + copper oxide	26.96
mass of dish + contents at end	26.32
mass of water collected	0.72

The reasoning, step by step: the mass of oxygen lost from the copper(II) oxide is 0.64 g (26.96 − 26.32). All that oxygen combines with hydrogen to form water — so the water contains 0.64 g of oxygen. Since 0.72 g of water was collected, the mass of hydrogen in it must be 0.72 − 0.64 = 0.08 g. Now the familiar table:

	H	O
masses/g	0.08	0.64
number of moles of atoms	0.08/1	0.64/16
number of moles	0.08	0.040
÷ smaller	2	1
empirical formula	H₂O

Two hydrogen atoms to every oxygen atom — the experiment lands on the formula you have written since Chapter 1, but now you have derived it from masses.

Exam hint

The whole experiment hinges on one conservation sentence worth quoting in answers: all the oxygen lost from the copper(II) oxide ends up in the water. That is why oxygen's share of the water is read off the dish (0.64 g) and only hydrogen's must be inferred (0.72 − 0.64). Questions love asking which mass comes from which measurement.

Common misconception

"0.72 g of water means 0.72/18 = 0.04 mol of water, so we're done." That shortcut uses the answer's formula — circular. The experiment's point is to derive H₂O without assuming it: weigh the oxygen that left the oxide (0.64 g), weigh the water formed (0.72 g), infer the hydrogen (0.08 g), and let the mole table say 2 : 1. Only after the table agrees with H₂O does the 18 g/mol check make sense as a confirmation.

Quick check

Question-bank join: spec 1.31 items — determining the formula of water experimentally.

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Working out formulae using percentage composition figures

SB p.51

The worked examples and practicals so far determined empirical formulae from masses. In the exam you are often given percentages by mass instead. The next worked example shows the one extra step — and it is a pleasant one.

Example 5. Find the empirical formula of a compound containing 82.7% C and 17.3% H by mass (Aᵣ: H = 1, C = 12).

The percentage figures apply to any amount you choose — so choose 100 g. Then the percentages simply become masses: 82.7% of 100 g is 82.7 g. The table does the rest:

	C	H
percentages/%	82.7	17.3
masses in 100 g/g	82.7	17.3
number of moles of atoms	82.7/12	17.3/1
number of moles	6.89	17.3
÷ smallest to get ratio	6.89/6.89	17.3/6.89
ratio of moles	1	2.5

The ratio comes out 1 : 2.5 — and here is the exam-thinking: The empirical formula is the whole-number ratio, so a fraction is not an error, it is a instruction to scale up: multiply both numbers by 2 to get C₂H₅, the empirical formula.

Key point

Percentages → assume 100 g → the percentages become masses → the familiar table runs unchanged. The only new decision is at the ratio: 1 : 2.5 is not "rounded to" 1 : 2 or 1 : 3 — it is multiplied up to whole numbers (2 : 5).

Exam hint

The exact line between legitimate rounding and corruption: calculating 17.3/6.89 gives 2.51 — rounding that to 2.5 (or 1.01 to 1) is fine; rounding 2.51 to 3 is not. And a ratio like 1 : 1.33 is not rounded down to 1 : 1 — you multiply by 3 to obtain the whole-number ratio. Halves ×2, thirds ×3, quarters ×4: recognise the fraction, clear it honestly.

Common misconception

"A ratio of 1 : 2.5 must be experimental error — round it to 1 : 2." No: with correct data, 2.5 is telling you the simplest whole-number ratio needs scaling. 1 : 2.5 → 2 : 5 → C₂H₅. Rounding a clean fraction away instead of multiplying it up produces a formula that is not just imprecise but chemically impossible to reconcile with the given percentages.

Quick check

Question-bank join: spec 1.33 items — empirical formulae from percentage composition.

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Converting empirical formulae into molecular formulae

SB p.52

The C₂H₅ from the last section cannot be a real compound's formula — as you will recognise once you meet organic chemistry in Unit 4. The molecular-formula (the actual number of each atom present in a molecule) has to be a multiple of the empirical formula, like C₄H₁₀.

The conversion needs exactly one extra piece of information: the relative formula mass of the compound — or the mass of 1 mole, which is just the Mᵣ in grams. Continuing the previous example: suppose you are told the relative formula mass is 58.

Mᵣ of C₂H₅ = (2 × 12) + (5 × 1) = 29 (Aᵣ: H = 1, C = 12)
How many times does 29 go into 58? 58 / 29 = 2
So the molecule contains 2 lots of C₂H₅: the molecular formula is C₄H₁₀

That is the whole method: divide the given Mᵣ by the empirical-formula Mᵣ, multiply the empirical formula by the answer. The ratio of elements is untouched (both C and H scale by the same factor), which is why the molecular formula always "cancels down" to the empirical formula.

Key point

Molecular formula = empirical formula × n, where n = Mᵣ(compound) ÷ Mᵣ(empirical formula). If n comes out as a whole number, the data is consistent; if it does not, something upstream (the percentage table or the Mᵣ) needs re-checking.

Exam hint

Show the division as its own line (58 / 29 = 2) and then the scaled formula — method marks live in the n step, not just the final C₄H₁₀. And note: knowing "a bit more organic chemistry" (Unit 4) is how you'd spot that C₂H₅ is impossible — but you do not need that knowledge in the exam; the Mᵣ does the deciding for you.

Common misconception

"The molecular formula is C₂H₅ doubled in mass, so 2 × 29 = 58 — therefore the molecular formula is 58." The Mᵣ is a number; the formula is atoms. The factor n multiplies the formula (C₂H₅ → C₄H₁₀), not the mass figure. Writing "the molecular formula is 2" or "is 58" confuses the scaling factor with the formula itself.

Quick check

Question-bank join: spec 1.32/1.33 items — empirical-to-molecular conversion via Mᵣ.

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Worked example 6 — and into the crucible: heating barium chloride crystals

SB p.52

Example 6. A compound has the empirical formula CH₂. If the relative formula mass is 56, work out the molecular formula.

Mᵣ of CH₂ = 12 + (2 × 1) = 14
56 / 14 = 4
The molecule contains 4 lots of CH₂: the molecular formula is C₄H₈

The closing check: the ratio of elements in the molecular formula must be the same as in the empirical formula — and indeed C₄H₈ cancels down to CH₂.

The section then turns to the crucible work that the next sections calculate. Two reminders set the vocabulary: when some substances crystallise from solution, water becomes chemically bound to the salt — water of crystallisation — and the salt is said to be hydrated; anhydrous means without water. Picture barium chloride crystals being heated in a crucible: the experiment that will hand you the mass data for finding n in BaCl₂·nH₂O.

Safety

Barium chloride is toxic, but a Bunsen flame is not hot enough to release it into the atmosphere.

Exam hint

The practical question to be ready for: how can you be sure all the water has been driven off? Heat to constant mass — heat the crucible, weigh it, and repeat until the mass remains constant. If the mass is still falling, water remains; when it is constant, all the water has gone. The same constant-mass discipline appeared in the metal-oxide practicals (section 20); it is the standard answer wherever a crucible experiment is examined.

Common misconception

"CH₂ with Mᵣ 56 must be C₂H₄ — because 2 × 14 = 28 and 28 is close to 56, so it 'rounds up'." n must be an exact division, not an approximation: 56 / 14 = 4 exactly, giving C₄H₈. If your division does not land on a whole number, the error is upstream (wrong Mᵣ or a percentage-table slip) — never "round n to the nearest whole number and hope".

Quick check

Question-bank join: spec 1.32/1.33 items — molecular formula from empirical formula + Mᵣ.

Key terms in this section

anhydrous: — Without water: the salt left when all the water of crystallisation has been driven off by heating, e.g. anhydrous barium chloride, BaCl2 (no ·nH2O).

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Interpreting symbols in equations in terms of moles

SB p.53

Take the methane equation you balanced in section 5:

CH₄ + 2O₂ → CO₂ + 2H₂O

Read at molecular level, it says 1 molecule of CH₄ combines with 2 molecules of O₂ to form 1 molecule of CO₂ and 2 molecules of H₂O. But molecules are inconveniently small for calculation — and the mole fixes that, because the same ratios hold for moles:

1 mol CH₄ reacts with 2 mol O₂ to form 1 mol CO₂ and 2 mol H₂O

Nothing about the equation's meaning has changed — the ratios are still the same. The equation's big numbers do double duty: they balance the atoms and they announce the proportions in moles.

The arithmetic, pinned: the big numbers in front of the formulae tell us the number of moles of each substance that react. And although there does not appear to be a big number in front of the CH₄ and CO₂, there is — a '1' that we simply do not write.

Key point

A chemical equation can always be read as a mole recipe: the balancing numbers are the moles. An invisible '1' counts as a big number (CaCO₃ → CaO + CO₂ means 1 mol : 1 mol : 1 mol).

Exam hint

This reading is the hinge between the first half of this chapter (equations) and the second half (reacting masses): every mass calculation from here on starts by interpreting the equation in terms of moles. When a question says "using the equation", that instruction means: write the mole row underneath the equation before touching any masses.

Common misconception

"The equation means 1 gram of CH₄ reacts with 2 grams of O₂." No — the balancing numbers are mole ratios, never gram ratios. In fact 1 mol CH₄ (16 g) reacts with 2 mol O₂ (64 g): the mass ratio is 16 : 64, not 1 : 2. Grams and moles are connected through Mᵣ, which is exactly why the next sections interpose a moles step between the equation and the masses.

Quick check

Question-bank join: spec 1.28 items — equations read as mole recipes.

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Empirical formula calculations involving water of crystallisation — finding n

SB p.53

Heat a salt that contains water of crystallisation and the water is driven off, leaving the anhydrous salt behind. Barium chloride — a hydrated-salt — is the commonly used example, because the barium chloride itself does not decompose even on quite strong heating — only the water leaves.

The experiment (heating the crystals in a crucible) hands you three masses:

mass of crucible = 30.00 g
mass of crucible + barium chloride crystals, BaCl₂·nH₂O = 32.44 g
mass of crucible + anhydrous barium chloride, BaCl₂ = 32.08 g

Finding n means finding the ratio of the number of moles of BaCl₂ to the number of moles of water — "it's just another empirical formula calculation" (Aᵣ: H = 1, O = 16, Cl = 35.5, Ba = 137; Mᵣ of BaCl₂ = 137 + 2 × 35.5 = 208, Mᵣ of H₂O = 18):

mass of BaCl₂ = 32.08 − 30.00 = 2.08 g
mass of water = 32.44 − 32.08 = 0.36 g

The mole table:

	BaCl₂	H₂O
masses/g	2.08	0.36
÷ Mᵣ for moles	2.08/208	0.36/18
number of moles	0.0100	0.020
ratio of moles	1	2
empirical formula	BaCl₂·2H₂O

The unknown n was 2 — found by the identical divide-by-Mᵣ, divide-by-smallest machinery as every other empirical formula in this chapter. The only novelty is that one of the "elements" is a whole water molecule.

Exam hint

Two Mᵣ values must be assembled before the table can start: BaCl₂ = 208 (137 + 2 × 35.5) and H₂O = 18. Show both constructions — the marks sit in the assembly, and the table cannot misfire if 208 and 18 are written explicitly. Note also which masses produce which subtraction: crystals-minus-crucible gives the hydrate, anhydrous-minus-crucible gives the salt, and their difference is the water.

Common misconception

"n is found by dividing the mass of crystals by the mass of water." That ignores the salt entirely: n is the ratio of moles — 0.020 mol water to 0.010 mol BaCl₂ — not of masses (0.36/2.08 = 0.17 would be nonsense). Convert each mass to moles with its own Mᵣ first; the ratio comes only after both conversions.

Quick check

Question-bank join: spec 1.31/1.33 items — finding n in hydrated salts.

Key terms in this section

anhydrous: — Without water: the salt left when all the water of crystallisation has been driven off by heating, e.g. anhydrous barium chloride, BaCl2 (no ·nH2O).
hydrated salt: — A salt containing water of crystallisation as part of its formula, e.g. CuSO4·5H2O, Na2CO3·10H2O, MgCl2·6H2O. Its Mᵣ includes the bound water.

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Calculations using moles, chemical equations and masses of substances

SB p.53

Section 25 taught you to read an equation in moles. This section cashes that in for masses. Take the methane equation again:

CH₄ + 2O₂ → CO₂ + 2H₂O

Interpreted in moles: 1 mol CH₄ reacts with 2 mol O₂ to form 1 mol CO₂ and 2 mol H₂O. The molecular bookkeeping beneath it: 2 mol of O₂ contains twice as many molecules as 1 mol of CH₄ — the proportions you weigh out are the proportions that the molecules react in.

Now attach masses, using the Mᵣ values (CH₄ = 16, O₂ = 32, CO₂ = 44, H₂O = 18):

1 mol CH₄ = 16 g and 2 mol O₂ = 64 g react exactly...
...to form 1 mol CO₂ = 44 g and 2 mol H₂O = 36 g

Check the bookkeeping with the conservation instinct from section 4: 16 + 64 = 80 g on the left; 44 + 36 = 80 g on the right. Mass conserved, moles proportioned — the equation is now a quantitative recipe. The first full problem, previewed here, arrives in the next section: limestone, impure calcium carbonate, decomposes on heating to calcium oxide (quicklime) and carbon dioxide — a thermal decomposition with Mᵣ(CaCO₃) = 100 and Mᵣ(CaO) = 56, the two numbers you will need.

Key point

The mass scale of any reaction comes from its mole recipe: multiply each mole count by the substance's mass of 1 mole. Conservation is the built-in check — total mass of products must equal total mass of reactants.

Exam hint

The four masses (16 g, 64 g, 44 g, 36 g) are worth deriving rather than memorising: each is moles × Mᵣ, straight from the equation's big numbers. Examiners ask this pattern constantly — "what mass of oxygen reacts with 16 g of methane?" — and the answer is always the same three moves: equation → moles → mass.

Common misconception

"If the equation says CH₄ + 2O₂, I need equal masses of methane and oxygen." The equation proportions are moles (1 : 2), not grams. Converting to masses gives 16 g : 64 g — oxygen's extra factor of 4 comes from its Mᵣ being double methane's while you need double the moles. Confusing the mole row with a mass row is the error that makes conservation look broken.

Quick check

Question-bank join: spec 1.28/1.29 items — mole-recipe to mass conversions.

Watchblock 10

GCSE Chemistry - Limiting Reactants & Balanced Equations (2027/28 exams)

Limiting reactants — which reagent runs out first and how it caps the product yield.

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Calculations involving only masses

SB p.54

The reacting-mass problems are now in view, and the exam's format is precise. Typical calculations give you a mass of starting material and ask how much product you are likely to obtain; you will also find them done in reverse — told the mass of product, asked how much starting material you would need. And in almost all the cases you will meet at International GCSE, you will be given the equation for the reaction.

That last sentence is worth reading twice. The exam does not test whether you can recall an equation mid-problem; it tests whether you can use the one supplied — read it in moles (section 25), convert the given mass to moles (section 16), walk the mole ratio, and convert back to mass. The next sections run that pipeline through three complete problems: heating calcium carbonate, extracting iron, and extracting lead — each with the method spelled out and the alternative route shown.

The three-step skeleton those problems share:

Calculate the number of moles using the mass you have been given.
Use the chemical equation to work out the number of moles of the substance you are interested in.
Convert the number of moles to a mass.

Exam hint

"When you have finished a chemistry calculation, the impression should nearly always be that there are a lot of words with a few numbers between them, not vice versa" — that is the description of a full-mark answer. Show the equation-reading and the unit changes as words and lines, not as a bare arithmetic stream: the words are where the method marks are parked.

Common misconception

"Reacting-mass questions require memorising which equation belongs to which reaction." At International GCSE, the equation is given in almost all cases — the skill being tested is the moles pipeline, not recall. Memorise the pipeline (mass → moles → ratio → moles → mass) rather than reaction equations, and read the supplied equation carefully for its balancing numbers.

Quick check

Question-bank join: spec 1.29 items — the three-step reacting-mass skeleton.

Keep goingCalculate Reacting Masses 1-2-elements-compounds-and-mixtures 1-3-atomic-structure 1-5-chemical-formulae-equations-calculations

A problem involving heating calcium carbonate — the three-step method

SB p.54

When calcium carbonate, CaCO₃, is heated, calcium oxide is formed. The task: calculate the mass of calcium oxide produced by heating 25 g of calcium carbonate (Aᵣ: C = 12, O = 16, Ca = 40):

CaCO₃ → CaO + CO₂

Step 1 — moles from the given mass. Mᵣ(CaCO₃) = 40 + 12 + (3 × 16) = 100, so:

number of moles = mass ÷ mass of 1 mole = 25 / 100 = 0.25 mol

Step 2 — the equation gives the target moles. From the equation, 1 mol of CaCO₃ decomposes to produce 1 mol of CaO — so 0.25 mol CaCO₃ produces 0.25 mol CaO.

Step 3 — moles back to mass. Mᵣ(CaO) = 40 + 16 = 56:

mass = number of moles × mass of 1 mole = 0.25 × 56 = 14 g

The reaction will produce 14 g of calcium oxide.

The alternative route (ratios). The same answer without the explicit moles: 1 mol CaCO₃ produces 1 mol CaO, so 100 g produces 56 g; therefore 1 g produces 56/100 = 0.56 g; therefore 25 g produces 25 × 0.56 = 14 g. "You do not have to learn both methods, just learn whichever method you are happier with."

Two more hints complete the section. On precision: the impression of a full-mark answer is "a lot of words with a few numbers between them" — and even if your mental arithmetic is fast enough to skip steps, you must still show all your calculations. On a half-factor trap (which returns in the iron problem): to find the mass of Fe you multiply moles of Fe by 56 — a common mistake is to multiply by the mass of 2Fe (112 g), but the '2' has already been used when the moles of Fe were worked out from the equation.

Key point

The Aᵣ of Fe is 56; the Mᵣ of Fe₂O₃ = (2 × 56) + (3 × 16) = 160. These are the two numbers the next problem runs on — assembled exactly as here, from the Periodic Table values.

Exam hint

The ratio method shines when the equation's coefficients are awkward or the arithmetic is round-numbered (100 → 56 here). The three-step method scales to every problem in the rest of the chapter, including the two-equation lead problem where the ratio method gets messier. Learn one primary method, keep the other as a checking tool — and remember to show the working either way.

Common misconception

"CaCO₃ → CaO + CO₂ shows 3 substances, so 1 mol CaCO₃ gives 3 mol of products." The big (or unwritten '1') numbers give 1 mol CaO and 1 mol CO₂ separately — the mole row is read per substance, not summed into a single "3". Step 2 needs the specific pair (CaCO₃ → CaO), which the balancing numbers state directly.

Quick check

Question-bank join: spec 1.29 items — the three-step method on thermal decomposition.

Keep goingCalculate Reacting Masses 1-2-elements-compounds-and-mixtures 1-3-atomic-structure 1-5-chemical-formulae-equations-calculations

A problem involving the extraction of lead — two equations and a tonne

SB p.56

Lead is extracted from galena, PbS. The ore is roasted in air to produce lead(II) oxide, PbO:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

The lead(II) oxide is then converted to lead by heating it with carbon in a blast furnace:

PbO(s) + C(s) → Pb(l) + CO(g)

The molten lead is tapped from the bottom of the furnace. Calculate the mass of lead that would be produced from 1 tonne of galena.

Step 1 — mind the units. Moles need grams, so convert: 1 tonne = 1 000 000 g (multiply by 1 000 000 — the number has to increase). Then, with Mᵣ(PbS) = 207 + 32 = 239:

number of moles of PbS = 1 000 000 / 239 = 4184 mol

Step 2 — trace the lead through BOTH equations. From the first: 2 mol PbS produce 2 mol PbO, so 4184 mol PbS → 4184 mol PbO. From the second: 1 mol PbO produces 1 mol Pb, so 4184 mol PbO → 4184 mol Pb.

Step 3 — back to mass, then to tonnes. Mass of Pb = 4184 × 207 = 866 000 g. Divide by 1 000 000: 0.866 tonne of lead.

The doubling trick explains a second route: double the second equation (2PbO + 2C → 2Pb + 2CO) so you can trace what happens to all the 2PbO from the first — and then notes the simplification: this could be reduced to 1 mol PbS produces 1 mol Pb, saving arithmetic, though it makes no difference to the answer. Its HINT weighs the two methods: the three-step route uses the same steps in every calculation but needs the tonne conversions; the ratio route has fewer steps but will not extend to the solution calculations of the next chapter.

Cross-curricular links

This flow sheet — roast the ore, reduce the oxide, tap the metal from the furnace's base — is the extractive-metallurgy pattern that powered the industrial revolution, and the blast furnace is still the reason iron and lead (Chapter 14 revisits both) have been cheap enough to build a metal-based civilisation on. The chemistry is old; the mole arithmetic you just did is what makes the yields predictable.

Exam hint

Two-equation problems are bookkeeping, not new chemistry: follow the substance (here PbS → PbO → Pb) through each equation's own mole ratio, in order. Write the mole row for each equation separately before combining — and keep the tonne conversion at the very edges of the problem (once in, once out), never in the middle of the mole arithmetic.

Common misconception

"The two equations both start with 2, so 2 mol PbS gives 1 mol Pb." Trace it: 2 mol PbS → 2 mol PbO (first equation), then 2 mol PbO → 2 mol Pb (second). The lead count never halves — 2 mol PbS yields 2 mol Pb. The doubling of the second equation exists precisely to make this chain visible; simplifying to 1 : 1 is a convenience, not a change.

Quick check

Question-bank join: spec 1.29 items — multi-step extraction problems with unit conversions.

Keep goingCalculate Reacting Masses 1-2-elements-compounds-and-mixtures 1-3-atomic-structure 1-5-chemical-formulae-equations-calculations

What is a percentage yield?

SB p.57

If you calculate how much product a reaction might produce, in real life you rarely obtain as much as you expected. Expect 100 g, collect 80 g — your percentage yield is 80%. The rest is lost somehow, and the standard loss mechanisms are: spillages; losses when transferring a liquid from one container to another; and — a big one — side reactions, which divert some of your starting materials into unwanted products — that "happens a lot during reactions in organic chemistry", where experiments making and purifying organic compounds rarely give high yields.

Percentage yield is also where the previous problem's arithmetic gets its second life. Revisiting the lead calculation with the ratios route: trace the lead through both equations — 2 mol PbS produces 2 mol PbO, then (doubling the second equation) 2 mol PbO produces 2 mol Pb — so every 2 mol of PbS produces 2 mol of Pb. Substitute masses: 2 × 239 g PbS produces 2 × 207 g Pb, that is 478 g PbS produces 414 g Pb. Tonnes asked? "It's much easier to think a bit, and realise that the ratio is always going to be the same, whatever the units": 478 tonnes PbS produces 414 tonnes Pb. Proportion: 1 tonne gives 414/478 = 0.866 tonne — the same answer as section 31, with no gram-conversion at all.

Key point

A proportion's ratio is unit-independent: 478 g : 414 g and 478 tonnes : 414 tonnes are the same 414/478. This is why the ratio method can work directly in tonnes — the units cancel inside the proportion.

Exam hint

The "ratio is the same whatever the units" insight is a genuine exam shortcut for any yield or reacting-mass question set in tonnes (or kg). Use it deliberately: state the proportion in the question's own units, then divide. It replaces the convert-in-convert-out arithmetic of the three-step method — at the cost of the generality that method's HINT in section 31 warned about.

Common misconception

"A low percentage yield means I did the calculation wrong." No — the calculated amount is the theoretical yield (what the equation promises); the low percentage yield measures the experiment's losses against it. Spills, transfer losses and side reactions are real chemistry, not arithmetic errors. The next section defines the two yields formally and does the division.

Quick check

Question-bank join: spec 1.30 items — why yields fall short of theory.

Watchblock 11

GCSE Chemistry - Percentage Yield - Theoretical & Actual Yield | Calculations (2027/28 exams)

Percentage yield — actual over theoretical, times one hundred, with the classic traps flagged.

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Keep goingCalculate Percentage Yield 1-2-elements-compounds-and-mixtures 1-5-chemical-formulae-equations-calculations 2-7-acids-bases-and-salt-preparations

Calculating percentage yields

SB p.57

Suppose a calculation says you should get 12.5 g of copper sulfate crystals, but the experiment delivers only 11.2 g. Name both numbers:

12.5 g is the theoretical-yield — obtained by carrying out a moles calculation from the equation; - 11.2 g is the actual-yield — what the experiment actually produced, and it will be given to you in the question.

The formula is then one division:

$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Here: (11.2 / 12.5) × 100 = 89.6%.

The sanity bound every marker looks for: the percentage yield should always come out as less than 100%. If it comes out above 100%, either you have made a mistake in the calculation, or the question is deliberately continuing by asking what went wrong experimentally. Its own example: if you are making a salt (Chapter 17), a percentage yield above 100% can mean the salt was not completely dry — extra water inflating the "product" mass.

Key point

Percentage yield = actual ÷ theoretical × 100. The actual yield is always the given number; the theoretical yield is always the one you calculate from the equation. Never divide the other way.

Exam hint

Identify which number is which before dividing — that is the entire skill in this calculation type. Clues: the theoretical yield's context is "calculate/could/might produce"; the actual yield's context is "she made/obtained/collected". Worked example 7 (next section) shows the theoretical yield being derived through a full two-equation mole chain before the division — that derivation is where most of the marks live.

Common misconception

"A percentage yield of 89.6% means 10.4% of the atoms vanished." Conservation of mass never breaks: the "missing" 10.4% of product is unreacted starting material, product lost in transfers or purification, or mass diverted into side-reaction products you did not count. The yield measures how much of your product ended up in the bottle, not whether atoms were destroyed.

Quick check

Question-bank join: spec 1.30 items — percentage yield calculations.

Key terms in this section

percentage yield: — actual yield ÷ theoretical yield × 100 — how much product was obtained compared with what the equation's moles promise. Should come out below 100%; above it suggests wet product or a mistake.
theoretical yield: — The mass of product a reaction could produce, calculated from the equation and the moles of starting material — e.g. 7.53 g of CuSO4·5H2O from 2.40 g of CuO.
actual yield: — The mass of product actually obtained in the experiment, given in the question — e.g. 5.21 g of crystals. Always divided by (not into) the theoretical yield for the percentage.

Keep goingCalculate Percentage Yield 1-2-elements-compounds-and-mixtures 1-5-chemical-formulae-equations-calculations 2-7-acids-bases-and-salt-preparations

Worked example 7 — percentage yield of copper(II) sulfate crystals

SB p.58

Example 7. A student reacted 2.40 g of copper(II) oxide (CuO) with hot sulfuric acid. She made 5.21 g of copper(II) sulfate crystals (CuSO₄·5H₂O). The two reactions run in sequence: the copper(II) oxide reacts with the sulfuric acid to form copper(II) sulfate in solution, and the copper(II) sulfate then crystallises from that solution with five waters of crystallisation attached — CuSO₄ → CuSO₄·5H₂O, one formula unit giving one formula unit of crystals.

Deriving the theoretical yield. First, the moles of copper(II) oxide — we have enough information to calculate them. Mᵣ(CuO) = 63.5 + 16 = 79.5:

number of moles of CuO = 2.40 / 79.5 = 0.0302 mol

From the first reaction, 0.0302 mol CuO produces 0.0302 mol CuSO₄ (one formula unit → one formula unit). From the crystallisation step, 0.0302 mol CuSO₄ produces 0.0302 mol CuSO₄·5H₂O.

Now convert to mass — the Mᵣ of the crystals is the one you built in section 12: Mᵣ(CuSO₄·5H₂O) = 249.5.

mass = 0.0302 × 249.5 = 7.53 g

This is the theoretical yield of copper(II) sulfate crystals.

The percentage yield. The actual yield is what the experiment obtained: 5.21 g.

percentage yield = (5.21 / 7.53) × 100 = 69.2%

Key point

A 1 : 1 : 1 mole chain (CuO → CuSO₄ → CuSO₄·5H₂O) means the moles never change through the two steps — only the mass does, when each substance's own Mᵣ is applied. The chain's job is to get you from the starting mass to the theoretical yield of the product as collected (crystals included).

Exam hint

Two details make this example exam-hard. First, the product's Mᵣ must include the water of crystallisation (249.5, not 159.5 for CuSO₄) — use the wrong one and the whole calculation dies at the last step (despite correct moles). Second, number discipline: 0.0302 mol carries three significant figures because the mass data does. The 69.2% then lands below 100% as it should (section 33's bound).

Common misconception

"The theoretical yield is 2.40 g — the copper(II) oxide all becomes product, so the masses match." The moles carry over 1 : 1, not the masses: 0.0302 mol of CuO weighs 2.40 g, but the same 0.0302 mol of CuSO₄·5H₂O weighs 0.0302 × 249.5 = 7.53 g — the sulfate and the five waters add enormous mass. Each step multiplies by its own substance's mass of 1 mole; only the mole count is conserved.

Quick check

Question-bank join: spec 1.30 items — theoretical yield derivation plus percentage yield.

Keep goingCalculate Percentage Yield 1-2-elements-compounds-and-mixtures 1-5-chemical-formulae-equations-calculations 2-7-acids-bases-and-salt-preparations

Which substance is in excess?

SB p.58

When you make something in the lab, you rarely mix things in exactly the right proportions. Usually something is in excess — there is more than enough of it to react with the other substance — and you remove the excess afterwards (by filtering, for example). A reagent is just something that reacts; the substance left over is the excess-reagent.

The trap: until now, every problem gave you enough information to calculate moles of one thing. What if you are given moles of both?

Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) 0.2 mol Mg is reacted with 0.2 mol HCl. Calculate the mass of hydrogen gas produced.

The tempting wrong answer: 0.2 mol Mg produces 0.2 mol H₂, so the mass of H₂ = 0.2 × 2 = 0.4 g. "Unfortunately this is not the correct answer!" Check against the equation: 1 mol Mg reacts with 2 mol HCl, so 0.2 mol Mg would need 0.4 mol HCl — but only 0.2 mol was used. The magnesium cannot all react, so its moles cannot be used to work out the hydrogen.

Start from the HCl instead: 2 mol HCl reacts with 1 mol Mg, so 0.2 mol HCl reacts with 0.1 mol Mg — and there is enough Mg (0.2 mol) to react with all of it. The Mg is in excess, which means all the HCl reacts, and HCl controls the product:

2 mol HCl produce 1 mol H₂ → 0.2 mol HCl produce 0.1 mol H₂

mass of H₂ = 0.1 × 2 = 0.2 g — the correct answer.

Key point

You always have to use the chemical equation to determine which substance is in excess — "more of it" is not the test. Variant: 0.2 mol Mg with 0.3 mol HCl looks like HCl is in excess (there is more of it), but 0.3 mol HCl only reacts with 0.15 mol Mg — still not enough HCl to use up all the Mg, so the Mg is still in excess.

Exam hint

The method closes in one line: to deduce which substance is in excess, try both numbers of moles in the chemical equation and compare. Run each reagent's moles through the ratio and see which one runs out first — the one that runs out (HCl here) is the limiting reagent, and it — never the excess one — feeds the product calculation.

Common misconception

"Whichever reagent I have more moles of must be the one in excess." Excess is a ratio property, not a size property: 0.3 mol of HCl against 0.2 mol of Mg still leaves HCl short, because the equation demands 2 mol HCl per 1 mol Mg. Always convert both amounts through the equation's ratio before naming the excess.

Quick check

Question-bank join: spec 1.29 items — limiting and excess reagent identification.

Key terms in this section

excess reagent: — The reagent present in more than enough to react with all of the other one — determined from the equation's ratios, not from which amount is bigger. Removed after the reaction (e.g. by filtering).
reagent: — Simply something that reacts. In excess calculations, try both reagents' moles in the equation: the one that runs out first (limiting) feeds the product calculation.

Keep goingCalculate Reacting Masses 1-2-elements-compounds-and-mixtures 1-3-atomic-structure 1-5-chemical-formulae-equations-calculations

Worked example 8 — excess with nitric acid, and the chapter close

SB p.59

Example 8. Copper reacts with concentrated nitric acid according to the equation:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2H₂O(l) + 2NO₂(g)

3.2 g of copper is reacted with 0.40 mol of concentrated nitric acid. Work out which reagent is in excess.

Convert the copper mass to moles first (Aᵣ: Cu = 63.5):

number of moles of Cu = 3.2 / 63.5 ≈ 0.050 mol

Run each reagent through the equation. The equation says 1 mol Cu reacts with 4 mol HNO₃, so:

0.050 mol Cu reacts with 4 × 0.050 = 0.20 mol HNO₃

We have more than 0.20 mol HNO₃ (we have 0.40 mol) — therefore HNO₃ is in excess (and the copper is the limiting reagent).

The follow-through: if the question went further and asked for the amount of NO₂ produced using these quantities, you would have to use the copper's moles — not all the HNO₃ is used up. From the equation, 2 mol Cu produce 2 mol NO₂, so 0.050 mol Cu gives 2 × 0.050 = 0.10 mol NO₂. This is exactly the section 35 law: the limiting reagent feeds the product calculation.

The closing skills to self-check: balancing equations, mole and molar-mass conversions, empirical formulae from percentages, reacting masses through CaCO₃ and the extraction reactions, and yield problems — this lesson's practice is wired to the question bank through the quick-checks above.

Summary

Equations must balance — same atoms out as in. Balance with big numbers only; never change a formula. State symbols: (s), (l), (g), (aq).
Big numbers = coefficients = moles; subscripts belong to the atom (or bracket) before them (2H₂SO₄ = 4 H, 2 S, 8 O).
Mᵣ = sum of the Aᵣ values (hydrated salts include their waters: CuSO₄·5H₂O = 159.5 + 90 = 249.5); mass of 1 mole = Mᵣ in grams.
1 mol = 6 × 10²³ particles (the Avogadro constant); moles = mass ÷ mass of 1 mole.
Empirical formula = simplest whole-number ratio (from masses or percentages → assume 100 g); molecular formula = empirical × n, n = Mᵣ(compound) ÷ Mᵣ(empirical).
Formulae from experiments: metal oxides by combustion or reduction (spec 1.36); water from collected masses; n in BaCl₂·nH₂O.
Reacting masses: mass → moles → equation ratio → moles → mass. The limiting reagent (never the excess one) feeds the product.
Percentage yield = actual ÷ theoretical × 100 — always < 100% unless something went wrong (or the product is wet).

Can you…?

balance any given equation left-to-right, with state symbols, and explain the never-change-a-formula rule?
count atoms through coefficients, subscripts and brackets (2H₂SO₄; Ca(OH)₂)?
calculate Mᵣ for simple and hydrated formulae (CuSO₄·5H₂O) and state the mass of 1 mole?
interconvert masses and moles, including the × 6 × 10²³ particle step?
find an empirical formula from masses and from percentages, then convert to the molecular formula given Mᵣ?
outline both metal-oxide practicals (combustion and reduction) and say how the mass change reveals the formula?
carry a reacting-mass problem through the three steps, including 2 : 1 ratios and tonne conversions?
identify which reagent is in excess and justify it from the equation?
calculate a percentage yield from an actual and a theoretical yield, and explain why it should fall below 100%?

Exam hint

One chapter, one spine: the mole converts between what you can weigh (grams) and what actually reacts (particles). Every exam question in this chapter is that conversion wearing a different coat — equation reading, Mᵣ assembly, empirical formulae, practicals, yields, excess. When a question looks unfamiliar, find which link of the chain it is testing and run that link; the quick-checks above map the coats one-to-one.

Common misconception

"This chapter is eight separate skills to memorise." It is one skill — mole bookkeeping — applied eight ways. The HINT from section 28 says the deep version: a full-mark answer is "a lot of words with a few numbers between them". If you can say which link each calculation runs through, the numbers follow; if you cannot say it, no formula-triangle will save the answer.

Quick check

Question-bank join: spec 1.29 items — excess-reagent problems with mixed mass/mole data (Example 8 pattern).

Keep goingCalculate Reacting Masses 1-2-elements-compounds-and-mixtures 1-3-atomic-structure 1-5-chemical-formulae-equations-calculations

Formulae and calculations (part 1)

Formulae, equations and calculations — the chapter map

Writing equations — word equations and symbol equations

What all the numbers in a formula mean

Balancing equations

Try again — balancing the methane equation

Worked example 1 — balancing zinc + hydrochloric acid

Worked example 2 — balancing the combustion of ethane

State symbols

How much of each substance reacts? — the carbon-12 scale and Aᵣ

Relative formula mass (Mᵣ)

Using relative formula mass to find percentage composition

Worked example — Mᵣ of copper(II) sulfate crystals, CuSO₄·5H₂O

Worked example — percentage by mass of oxygen in sodium carbonate

The mole

Quoting the formula — and the molar mass vocabulary

Simple calculations with moles

Moles and the Avogadro constant

Formulae — from mole ratios to empirical and molecular formulae

Worked examples 3 and 4 — empirical formulae from masses

Practicals — finding the formula of a metal oxide by combustion and by reduction

By combustion — magnesium oxide

By reduction — copper oxide

Determining the formula of water

Working out formulae using percentage composition figures

Converting empirical formulae into molecular formulae

Worked example 6 — and into the crucible: heating barium chloride crystals

Interpreting symbols in equations in terms of moles

Empirical formula calculations involving water of crystallisation — finding n

Calculations using moles, chemical equations and masses of substances

Calculations involving only masses

A problem involving heating calcium carbonate — the three-step method

A problem about extracting iron — a 2 : 1 mole ratio

A problem involving the extraction of lead — two equations and a tonne

What is a percentage yield?

Calculating percentage yields

Worked example 7 — percentage yield of copper(II) sulfate crystals

Which substance is in excess?

Worked example 8 — excess with nitric acid, and the chapter close