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Stoichiometry — formulae & moles

Formulae, balanced equations and mole calculations, one step at a time — including limiting reactants, titration, empirical formulae and percentage yield.

FormulaeBalancing equationsMrThe moleReacting massesGas volumesConcentrationEmpirical formula
RSby Riyaz Sir
mentorriyaz.in
No tag = Core (everyone)EXTENDED Extended onlyhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A formula shows which atoms are in a substance. An equation shows what changes in a reaction.
  • Balanced = the same number of each atom on both sides. Change the numbers in front, never the small numbers.
  • Mr = add up the Ar values from the Periodic Table.
  • EXT Moles: n = mass ÷ Mr for solids, n = V ÷ 24 for gases (at r.t.p.), n = c × V for solutions.

By the end you can

  • Write formulae and balanced equations with state symbols.
  • Calculate relative formula mass and reacting masses.
  • Use moles to link mass, gas volume and concentration. EXT
  • Find the limiting reactant, empirical and molecular formulae, percentage yield and purity. EXT

1Formulae

What a formula tells you

  • A formula uses chemical symbols and small numbers (subscripts). The subscript shows how many atoms of the element before it: H₂O has 2 hydrogen atoms and 1 oxygen atom; CO₂ has 1 carbon and 2 oxygen.
  • If a group of atoms is repeated, it goes in brackets: Ca(OH)₂ has 1 Ca, 2 O and 2 H. (NH₄)₂SO₄ has 2 N, 8 H, 1 S and 4 O.
  • A molecular formula shows the actual number of each atom in a molecule: glucose C₆H₁₂O₆.
SubstanceFormulaSubstanceFormula
waterH₂Osodium chlorideNaCl
carbon dioxideCO₂magnesium oxideMgO
ammoniaNH₃calcium carbonateCaCO₃
methaneCH₄sodium hydroxideNaOH
hydrogen chlorideHClsulfuric acidH₂SO₄
oxygenO₂nitric acidHNO₃
  • Name endings: -ide = two elements only (sodium chloride, magnesium oxide). -ate = contains oxygen too (sulfate, nitrate, carbonate).

Ionic formulae from chargesEXTENDED

  • An ionic compound must have zero total charge. Balance the positive and negative charges.
CationsAnions
Na⁺ K⁺ Li⁺ Ag⁺ NH₄⁺Cl⁻ Br⁻ I⁻ OH⁻ NO₃⁻
Mg²⁺ Ca²⁺ Zn²⁺ Cu²⁺ Fe²⁺O²⁻ S²⁻ SO₄²⁻ CO₃²⁻
Al³⁺ Fe³⁺–
  • Cross-over trick: write the charges, swap the numbers (ignore the signs) and make them the subscripts.
Worked example

EXT Write the formula of aluminium sulfate.

  1. Ions: Al³⁺ and SO₄²⁻
  2. Swap the numbers: Al gets 2, SO₄ gets 3
  3. Put the sulfate in brackets because there are 3 of them: Al₂(SO₄)₃
  4. Check the charge: 2 × (+3) = +6 and 3 × (−2) = −6 → total 0
Worked example

EXT Write the formula of calcium hydroxide.

  1. Ca²⁺ and OH⁻
  2. One Ca²⁺ needs two OH⁻ to balance
  3. OH is a group, so use brackets: Ca(OH)₂
Common mistakePutting brackets round a single ion (CaCl₂ is correct, not Ca(Cl)₂) — and writing Ca(OH)₂ as CaOH₂.

State symbols

SymbolMeaning
(s)solid
(l)liquid
(g)gas
(aq)aqueous: dissolved in water
  • aqueous means dissolved in water. Pure water itself is (l).

2Equations

Word equations and symbol equations

  • A word equation shows the names. A symbol equation uses the formulae and shows how many of each particle react.
  • Example: magnesium + oxygen → magnesium oxide. In symbols: 2Mg(s) + O₂(g) → 2MgO(s)

Balancing: the method

+2+2CH₄2 O₂CO₂2 H₂Ocount each atom on both sidesC: 1 = 1 · H: 4 = 4 · O: 4 = 4 ✓ balancedchange only the big numbers in front, never the small numbers inside a formula
Burning methane: CH₄ + 2O₂ → CO₂ + 2H₂O. Atoms are never created or destroyed, so both sides must have the same number of each atom.
  • Step 1: write the correct formulae for everything. They can never be changed afterwards.
  • Step 2: count the atoms of each element on each side.
  • Step 3: add big numbers (coefficients) in front of formulae to balance. Do one element at a time. Leave the atoms that appear in several substances (usually O and H) until last.
  • Step 4: recount both sides. Add the state symbols.
Common mistakeChanging a small number to make it balance. H₂O → H₂O₂ is a different substance (hydrogen peroxide). Change only the large numbers in front.
Worked example

Balance: CH₄ + O₂ → CO₂ + H₂O

  1. C: 1 on each side — balanced
  2. H: 4 on the left, 2 on the right → put 2 in front of H₂O
  3. O: now 2 + 2 = 4 on the right, so put 2 in front of O₂
  4. CH₄ + 2O₂ → CO₂ + 2H₂O
Worked example

Balance: Fe₂O₃ + CO → Fe + CO₂

  1. Fe: 2 on the left → put 2 in front of Fe
  2. Fe₂O₃ gives away 3 oxygen atoms, and each CO can take only 1 → use 3 CO, which makes 3 CO₂
  3. check C: 3 = 3
  4. check O: 3 + 3 = 6 on the left, and 3 × 2 = 6 on the right
  5. Fe₂O₃ + 3CO → 2Fe + 3CO₂
Worked example

Balance: Al + HCl → AlCl₃ + H₂

  1. AlCl₃ has 3 Cl, so start with 3 HCl
  2. that gives 3 H, but H₂ comes in pairs — double everything so far: 2 AlCl₃, 6 HCl, 3 H₂
  3. 2 AlCl₃ contains 2 Al, so use 2 Al on the left
  4. check Al: 2 = 2, Cl: 6 = 6, H: 6 = 6
  5. 2Al + 6HCl → 2AlCl₃ + 3H₂
Worked example

Balance: C₃H₈ + O₂ → CO₂ + H₂O

  1. C: 3 on the left → 3 CO₂
  2. H: 8 on the left → 4 H₂O
  3. O on the right: 3 × 2 + 4 = 10, so 5 O₂ on the left
  4. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Ionic equationsEXTENDED

  • In an ionic equation you show only the ions that actually change. The ions that are the same on both sides are spectator ions and are left out.
Worked example

EXT Write the ionic equation for AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).

  1. Write everything that is dissolved as ions: Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
  2. Na⁺ and NO₃⁻ are the same on both sides → they are the spectator ions
  3. Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
  • The solid (precipitate) stays as a formula — it is not split into ions.
  • Neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l) is the ionic equation for any acid and alkali reacting.
Exam tipCheck charges balance as well as atoms in an ionic equation: Ag⁺ + Cl⁻ gives 0 on the left, and AgCl is neutral on the right.

3Relative masses

Ar and Mr

  • The relative atomic mass (Ar) is the mass of an atom compared with carbon-12. You can read it from the Periodic Table: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, Ca = 40, Fe = 56, Cu = 64.
  • EXT Formal definition: the average mass of the naturally occurring atoms of an element, on a scale where an atom of carbon-12 has a mass of exactly 12.

Relative formula mass (Mr) — the sum of the relative atomic masses of all the atoms in the formula. It has no units.

Worked example

Calculate the Mr of calcium carbonate, CaCO₃.

  1. Ca = 40, C = 12, O = 16 × 3 = 48
  2. 40 + 12 + 48 = 100
Worked example

Calculate the Mr of (NH₄)₂SO₄.

  1. N: 2 × 14 = 28
  2. H: 8 × 1 = 8
  3. S: 32
  4. O: 4 × 16 = 64
  5. 28 + 8 + 32 + 64 = 132

Reacting masses in simple proportions

  • A balanced equation shows the ratio of particles. Convert to masses using Ar or Mr, then use ratios.
Worked example

What mass of magnesium oxide forms when 6.0 g of magnesium burns completely in oxygen?

  1. 2Mg + O₂ → 2MgO
  2. Masses: 2 × 24 = 48 g of Mg makes 2 × 40 = 80 g of MgO
  3. 6.0 g is 6.0 ÷ 48 = 0.125 of the amount
  4. 0.125 × 80 = 10 g
Exam tipSet out the three lines: equation → mass ratio → scale to the amount in the question. Even if you get stuck, you will pick up method marks.

4The mole EXT

Counting by weighing

Mole — the amount of substance that contains 6.02 × 10²³ particles (atoms, molecules or ions). This number is the Avogadro constant.

Picture itA dozen is 12 of anything. A mole is 6.02 × 10²³ of anything. You cannot count atoms, but you can weigh them: one mole of any substance weighs its Mr in grams.
  • 1 mol of carbon (Ar = 12) weighs 12 g. 1 mol of water (Mr = 18) weighs 18 g. 1 mol of CO₂ (Mr = 44) weighs 44 g.
molesn (mol)mass (g)gas volume(dm³ at r.t.p.)concentration × volume(mol/dm³) × (dm³)number of particles(6.02 × 10²³ per mol)÷ Mr× Mr÷ 24× 24c = n ÷ Vn = c × V× 6.02×10²³÷ 6.02×10²³
The mole is the central quantity. From moles you can reach mass, gas volume, number of particles and solution concentration.

Four equations

n = mMr
moles (mol) = mass (g) ÷ molar mass (g/mol)
V = n × 24
gas volume (dm³) = moles × 24 dm³/mol (at room temperature and pressure, r.t.p.)
c = nV
concentration (mol/dm³) = moles ÷ volume (dm³)
N = n × 6.02 × 1023
number of particles = moles × Avogadro constant
  • Units: 1 dm³ = 1000 cm³. To change cm³ to dm³ divide by 1000.
  • Concentration in g/dm³ = mass of solute (g) ÷ volume (dm³). To convert between the two: g/dm³ = mol/dm³ × Mr.
Worked example

EXT How many moles and molecules are in 11 g of carbon dioxide? (Mr = 44)

  1. n = 11 ÷ 44 = 0.25 mol
  2. molecules = 0.25 × 6.02 × 10²³
  3. = 1.5 × 10²³ (1.505 × 10²³)
Worked example

EXT What volume of hydrogen at r.t.p. forms when 4.8 g of magnesium reacts with excess hydrochloric acid?

  1. Mg + 2HCl → MgCl₂ + H₂
  2. n(Mg) = 4.8 ÷ 24 = 0.20 mol
  3. mole ratio Mg : H₂ = 1 : 1 → n(H₂) = 0.20 mol
  4. V = 0.20 × 24 = 4.8 dm³
Worked example

EXT 4.0 g of sodium hydroxide is dissolved to make 500 cm³ of solution. Find the concentration in mol/dm³ and g/dm³.

  1. Mr(NaOH) = 23 + 16 + 1 = 40
  2. n = 4.0 ÷ 40 = 0.10 mol
  3. V = 500 ÷ 1000 = 0.500 dm³
  4. c = 0.10 ÷ 0.500 = 0.20 mol/dm³
  5. g/dm³ = 4.0 ÷ 0.500 = 8.0 g/dm³ (check: 0.20 × 40 = 8.0 ✓)

The standard route

  • Step 1: write the balanced equation.
  • Step 2: convert what you are given into moles (n = m ÷ Mr, or n = V ÷ 24, or n = c × V).
  • Step 3: use the mole ratio from the equation.
  • Step 4: convert moles into what is asked for (mass, volume, concentration).
Common mistakeUsing cm³ instead of dm³. Always divide by 1000 first.
Common mistakeForgetting the mole ratio. In 2H₂ + O₂ → 2H₂O, 1 mol of O₂ makes 2 mol of water, not 1.

Titration calculations

  • In a titration you find the unknown concentration of one solution by reacting a known volume of it with a solution of known concentration. (The practical method is in experimental techniques and acids, bases & salts.)
Worked example

EXT 25.0 cm³ of sodium hydroxide is neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. Find the concentration of the sodium hydroxide.

  1. HCl + NaOH → NaCl + H₂O (ratio 1 : 1)
  2. n(HCl) = 0.100 × 20.0 ÷ 1000 = 0.00200 mol
  3. n(NaOH) = 0.00200 mol
  4. c(NaOH) = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³
Worked example

EXT 20.0 cm³ of 0.50 mol/dm³ sodium hydroxide neutralises 25.0 cm³ of sulfuric acid. Find the concentration of the acid.

  1. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (ratio 1 : 2)
  2. n(NaOH) = 0.50 × 20.0 ÷ 1000 = 0.0100 mol
  3. n(H₂SO₄) = 0.0100 ÷ 2 = 0.00500 mol
  4. c(H₂SO₄) = 0.00500 ÷ 0.0250 = 0.20 mol/dm³

Limiting reactant

beforeafterreact4 H₂ + 3 O₂4 H₂O + 1 O₂ left overH₂ is the limiting reactant: it is all used up and decides the amount of productO₂ is in excess: some is left over
Hydrogen and oxygen. There are 4 H₂ but only 3 O₂. The H₂ runs out first, so 1 O₂ is left over.

Limiting reactant — the reactant that is used up first. It decides how much product forms. The other reactant is in excess.

Worked example

EXT 5.6 g of iron is heated with 6.4 g of sulfur: Fe + S → FeS. Which is limiting, and what mass of FeS forms?

  1. n(Fe) = 5.6 ÷ 56 = 0.10 mol
  2. n(S) = 6.4 ÷ 32 = 0.20 mol
  3. the ratio is 1 : 1, so iron (0.10 mol) runs out first → iron is limiting
  4. n(FeS) = 0.10 mol, Mr = 88 → mass = 0.10 × 88 = 8.8 g
  5. sulfur left over = 0.10 mol × 32 = 3.2 g
Exam tipCompare moles divided by the coefficient — the smaller one is the limiting reactant. Always use the limiting reactant for the calculation.
Picture itYou have 4 slices of bread and 3 slices of cheese and make cheese sandwiches (2 bread + 1 cheese each). Bread limits you to 2 sandwiches; cheese would allow 3. The bread runs out first and 1 slice of cheese is left over.

5Formulae from data and yields EXT

Empirical and molecular formula

Empirical formula — the simplest whole-number ratio of atoms of each element in a compound.

Molecular formula — the actual number of atoms of each element in one molecule. It is a whole-number multiple of the empirical formula.

  • Method: (1) write the mass or % of each element; (2) divide each by its Ar to get moles; (3) divide all by the smallest number; (4) turn the ratio into whole numbers.
Worked example

EXT A hydrocarbon is 85.7% carbon and 14.3% hydrogen by mass. Mr = 56. Find the empirical and molecular formulae.

  1. C: 85.7 ÷ 12 = 7.14
  2. H: 14.3 ÷ 1 = 14.3
  3. divide by 7.14: C = 1, H = 2 → empirical formula CH₂ (mass 14)
  4. 56 ÷ 14 = 4, so multiply by 4 → molecular formula C₄H₈
Worked example

EXT An iron oxide contains 69.9% iron and 30.1% oxygen. Find its formula.

  1. Fe: 69.9 ÷ 56 = 1.248
  2. O: 30.1 ÷ 16 = 1.881
  3. divide by 1.248: Fe = 1, O = 1.507
  4. that is 1 : 1.5, so double both: 2 : 3
  5. Fe₂O₃
Exam tipIf your ratio is 1 : 1.5 multiply by 2; 1 : 1.33 → 3; 1 : 1.25 → 4. Do not round 1.5 down to 1 or up to 2.

Percentage yield, composition and purity

% yield = actual yieldtheoretical yield × 100
the actual yield is what you really got; the theoretical yield is the maximum from the equation
% purity = mass of pure substancemass of sample × 100
how much of the sample is the wanted substance
% of an element = Ar × number of atomsMr × 100
percentage composition by mass
Worked example

EXT 10.0 g of calcium carbonate is heated: CaCO₃ → CaO + CO₂. 4.2 g of calcium oxide is obtained. Find the % yield.

  1. n(CaCO₃) = 10.0 ÷ 100 = 0.100 mol
  2. n(CaO) = 0.100 mol, Mr(CaO) = 56
  3. theoretical yield = 0.100 × 56 = 5.6 g
  4. % yield = 4.2 ÷ 5.6 × 100 = 75%
Worked example

EXT 5.00 g of impure calcium carbonate reacts with excess acid and gives 960 cm³ of carbon dioxide at r.t.p. Find the % purity.

  1. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
  2. n(CO₂) = 0.960 ÷ 24 = 0.0400 mol
  3. n(CaCO₃) = 0.0400 mol → mass = 0.0400 × 100 = 4.00 g
  4. % purity = 4.00 ÷ 5.00 × 100 = 80%
Worked example

EXT Find the percentage of nitrogen in ammonium nitrate, NH₄NO₃.

  1. Mr = 14 + 4 + 14 + 48 = 80
  2. two N atoms: 2 × 14 = 28
  3. % N = 28 ÷ 80 × 100 = 35%
Common mistakeA % yield above 100%. That means the product is wet or impure, or the calculation is wrong.

6Test yourself

Test yourself — cover the answers first

Quick check

  1. How many atoms are in one formula unit of Ca(OH)₂?
  2. Balance: Na + H₂O → NaOH + H₂
  3. Find the Mr of H₂SO₄.
  4. What are the four state symbols?
  5. EXT How many moles are in 36 g of water?
  6. EXT What volume do 0.50 mol of any gas occupy at r.t.p.?
  7. EXT Write the formula of magnesium nitrate.

Exam-style

  1. Calculate the mass of carbon dioxide formed when 5.0 g of calcium carbonate reacts with excess acid. [3]
  2. EXT 2.4 g of magnesium reacts with 50.0 cm³ of 2.0 mol/dm³ hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Which is limiting? What volume of H₂ forms at r.t.p.? [4]
  3. EXT A compound contains 40.0% C, 6.7% H and 53.3% O by mass, and has Mr = 180. Find the molecular formula. [4]
  4. EXT 25.0 cm³ of 0.200 mol/dm³ NaOH is titrated with HCl. 20.0 cm³ of acid is needed. Find the HCl concentration. [3]
Show answers
  1. 1 + 2 + 2 = 5 atoms (1 Ca, 2 O, 2 H)
  2. 2Na + 2H₂O → 2NaOH + H₂
  3. 2 + 32 + 64 = 98
  4. (s) solid, (l) liquid, (g) gas, (aq) aqueous
  5. 36 ÷ 18 = 2.0 mol
  6. 0.50 × 24 = 12 dm³
  7. Mg²⁺ and NO₃⁻ → Mg(NO₃)₂
  8. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
    n(CaCO₃) = 5.0 ÷ 100 = 0.050 mol → n(CO₂) = 0.050 mol
    mass = 0.050 × 44 = 2.2 g
  9. n(Mg) = 2.4 ÷ 24 = 0.10 mol
    n(HCl) = 2.0 × 0.0500 = 0.100 mol
    HCl needs 2 per Mg: 0.100 mol HCl reacts with only 0.050 mol Mg → HCl is limiting
    n(H₂) = 0.100 ÷ 2 = 0.050 mol → V = 0.050 × 24 = 1.2 dm³
  10. C: 40.0 ÷ 12 = 3.33; H: 6.7 ÷ 1 = 6.7; O: 53.3 ÷ 16 = 3.33
    ratio 1 : 2 : 1 → CH₂O (mass 30)
    180 ÷ 30 = 6
    C₆H₁₂O₆
  11. n(NaOH) = 0.200 × 0.0250 = 0.00500 mol
    1 : 1 ratio, so n(HCl) = 0.00500 mol
    c = 0.00500 ÷ 0.0200 = 0.250 mol/dm³
Practise this chapter
Formulae & equationsRelative masses & the moleConcentration & titration
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