Edexcel International A Level Chemistry: Formulae, equations and amount of substance (YCH11)

This guide teaches Topic 1, Formulae, Equations and Amount of Substance, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers outcomes 1.1 to 1.12, including Core Practical 1 (outcome 1.11). The topic sits in Unit 1: Structure, Bonding and Introduction to Organic Chemistry, which is an IAS unit. The specification notes that ideas from this topic are applied in every other unit.

Pair this page with the revision notes and the practice questions to test yourself. The course hub lists every Edexcel IAL Chemistry page, the printable checklist lets you tick off outcomes, and a free 10-minute diagnostic will show where your gaps are.

Data used on this page (use these values): relative atomic masses H 1.0, C 12.0, N 14.0, O 16.0, Li 6.9, Mg 24.3, Na 23.0, S 32.1, Cl 35.5, Ti 47.9, Fe 55.8, Cu 63.5. Avogadro constant L = 6.02 × 10²³ mol⁻¹. Gas constant R = 8.31 J mol⁻¹ K⁻¹. Molar volume of a gas at room temperature and pressure = 24.0 dm³ mol⁻¹.

What this topic covers

Outcome	Skill or knowledge
1.1	Know the terms atom, element, ion, molecule, compound, empirical formula, molecular formula
1.2	Use the mole and the Avogadro constant L in calculations
1.3	Write balanced full and ionic equations with state symbols
1.4	Understand relative atomic mass (¹²C scale), relative molecular and formula mass, molar mass, ppm
1.5	Calculate concentration in mol dm⁻³ and g dm⁻³ (no titration calculations yet)
1.6	Calculate empirical and molecular formulae from experimental data
1.7	Calculate reacting masses from equations
1.8	Calculate gas volumes using amount of substance, molar volume and pV = nRT
1.9	Calculate percentage yield and percentage atom economy
1.10	Determine a formula or confirm an equation by experiment, and evaluate the data
1.11	Core Practical 1: measure the molar volume of a gas
1.12	Link full and ionic equations to observations in displacement, acid and precipitation reactions

1.1 The basic vocabulary

• Atom: the smallest particle of an element that keeps its chemical identity.
• Element: made of atoms with the same atomic (proton) number.
• Ion: a charged particle, made when an atom or a group of atoms gains or loses electrons.
• Molecule: a discrete group of atoms joined by covalent bonds.
• Compound: two or more elements chemically combined.
• Empirical formula: shows the lowest whole-number ratio in which the atoms of each element combine.
• Molecular formula: shows how many atoms of each element one molecule really contains.

Ethene, C₂H₄, has the empirical formula CH₂. Sodium chloride, a giant lattice, has only an empirical formula, NaCl.

1.2 The mole and the Avogadro constant

Chemists count particles in moles (mol), the unit of amount of substance. Each mole holds L = 6.02 × 10²³ particles; always say which particles.

Number of particles = amount in mol × L

Worked example. How many oxygen atoms are in 1.71 g of sucrose, C₁₂H₂₂O₁₁?

M(C12H22O11) = (12 × 12.0) + (22 × 1.0) + (11 × 16.0) = 342.0 g mol⁻¹
n = 1.71 / 342.0 = 0.00500 mol of sucrose molecules
O atoms = 0.00500 × 11 × 6.02 × 10²³ = 3.31 × 10²² atoms

1.3 Full and ionic equations

Balance atoms on both sides and add state symbols (s), (l), (g), (aq). An ionic equation leaves out spectator ions (unchanged on both sides), and charge must balance as well as atoms.

Worked example. Aluminium powder is added to copper(II) sulfate solution.

Full:   2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)
Ionic:  2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)

Charge: 6+ on each side. Sulfate is the spectator ion.

1.4 Relative masses, molar mass and ppm

Relative atomic mass, Ar, is the average mass of the element’s atoms, weighted for isotope abundance, on a scale where one carbon-12 atom has a mass of exactly 12. As a ratio, it has no units.

Relative molecular mass, Mr, is the same comparison for a molecule. For compounds with giant structures (ionic lattices such as MgO, or giant covalent substances such as SiO₂), the specification says to use relative formula mass, because no separate molecule exists. You find either by adding the relative atomic masses in the formula.

Molar mass, M, is the mass per mole, in g mol⁻¹: numerically equal to Mr, but with units.

Parts per million (ppm) expresses tiny proportions. For atmospheric gases it is usually by volume (1 cm³ of gas per 10⁶ cm³ of air); for solutes in water, usually by mass (1 mg per kg).

Worked examples. An air sample contains 0.0420% carbon dioxide by volume. As ppm: 0.0420 / 100 × 10⁶ = 420 ppm. A 0.150 kg water sample contains 0.36 mg of nitrate ions: 0.36 mg / 0.150 kg = 2.4 ppm by mass.

1.5 Concentration of solutions

concentration (mol dm⁻³) = amount (mol) / volume (dm³)
concentration (g dm⁻³)   = mass (g) / volume (dm³)  =  mol dm⁻³ × M

Worked example. 3.955 g of sodium thiosulfate, Na₂S₂O₃, is dissolved and made up to 200 cm³.

M = (2 × 23.0) + (2 × 32.1) + (3 × 16.0) = 158.2 g mol⁻¹
n = 3.955 / 158.2 = 0.0250 mol
c = 0.0250 / 0.200 = 0.125 mol dm⁻³
mass concentration = 3.955 / 0.200 = 19.8 g dm⁻³

The specification states that titration calculations are not required at this stage.

1.6 Empirical and molecular formulae from data

Method: mass (or %) of each element ÷ Ar → divide by the smallest → whole-number ratio. Then measured Mr ÷ empirical formula mass gives the multiplier.

Worked example. Heating 1.116 g of iron in chlorine gives 3.246 g of a chloride. The vapour of the chloride has Mr ≈ 325.

mass Cl = 3.246 − 1.116 = 2.130 g
Fe: 1.116 / 55.8 = 0.0200 mol      Cl: 2.130 / 35.5 = 0.0600 mol
ratio Fe : Cl = 0.0200 : 0.0600 = 1 : 3   → empirical formula FeCl3 (162.3)
325 / 162.3 = 2.00                         → molecular formula of vapour Fe2Cl6

1.7 Reacting masses

Steps: balanced equation → moles of the known → mole ratio → mass of the unknown.

Worked example. Titanium is made industrially by reducing titanium(IV) chloride with magnesium: TiCl₄ + 2Mg → Ti + 2MgCl₂. What mass of magnesium is needed per 1.00 kg of titanium?

n(Ti) = 1000 / 47.9 = 20.88 mol
n(Mg) = 2 × 20.88 = 41.75 mol
mass Mg = 41.75 × 24.3 = 1015 g ≈ 1.01 kg

1.8 Gas volumes

Molar volume. Under the same conditions, a mole of every gas takes up an equal volume: here volume = n × 24.0 dm³. So gas volumes react in the same ratio as the equation’s coefficients.

Worked example. Volume of hydrogen at room temperature and pressure from 0.350 g of lithium: 2Li(s) + 2H₂O(l) → 2LiOH(aq) + H₂(g).

n(Li) = 0.350 / 6.9 = 0.0507 mol     n(H2) = 0.0507 / 2 = 0.02536 mol
V = 0.02536 × 24.0 = 0.609 dm³

Ideal gas equation. pV = nRT. Put pressure in Pa, volume in m³ and temperature in K, with R = 8.31 J mol⁻¹ K⁻¹. It also works for the vapour of a volatile liquid, giving a molar mass: M = mRT / pV.

Worked example. 0.172 g of a volatile liquid evaporates completely to give 62.8 cm³ of vapour at 100 °C and 101 kPa.

T = 373 K   p = 101 000 Pa   V = 62.8 × 10⁻⁶ m³
M = (0.172 × 8.31 × 373) / (101 000 × 62.8 × 10⁻⁶) = 84.1 g mol⁻¹

This fits cyclohexane, C₆H₁₂ (84.0).

1.9 Percentage yield and atom economy

percentage yield = actual mass of product / theoretical mass × 100%
atom economy = M(desired product) / total M of all products in the equation × 100%

Yield measures one experiment. Atom economy is fixed by the equation.

Worked example (yield). A suggested practical in the specification is the preparation of the double salt ammonium iron(II) sulfate, (NH₄)₂Fe(SO₄)₂·6H₂O (M = 392.0). Starting from 1.40 g of iron, with one mole of salt formed per mole of iron, a student collects 7.10 g of dry crystals.

n(Fe) = 1.40 / 55.8 = 0.02509 mol → theoretical mass = 0.02509 × 392.0 = 9.84 g
percentage yield = 7.10 / 9.84 × 100 = 72.2%

Worked example (atom economy). Steam reforming of methane: CH₄ + H₂O → CO + 3H₂, with hydrogen as the desired product. Atom economy = (3 × 2.0) / (28.0 + 6.0) × 100 = 17.6%.

1.10 Formulae and equations from experiment

Worked example. A sample of black copper oxide (1.590 g) is heated in a stream of hydrogen until its mass stops changing, leaving 1.270 g of copper.

O lost = 1.590 − 1.270 = 0.320 g
Cu: 1.270 / 63.5 = 0.0200 mol    O: 0.320 / 16.0 = 0.0200 mol   → CuO

To evaluate, ask which way each error pushes the answer. Incomplete reduction, or hot copper re-oxidising in air, makes the “copper” mass too high and the Cu : O ratio above 1. Heating to constant mass and cooling in hydrogen deal with both.

1.11 Core Practical 1: molar volume of a gas

A typical approach: react a weighed mass of a metal or carbonate with excess acid in a flask joined to a gas syringe, and record the final gas volume, room temperature and pressure.

Worked example. 0.0720 g of magnesium ribbon reacts with excess hydrochloric acid and 70.2 cm³ of hydrogen is collected: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).

n(H2) = n(Mg) = 0.0720 / 24.3 = 0.002963 mol
molar volume = 70.2 cm³ / 0.002963 mol = 23 700 cm³ mol⁻¹ = 23.7 dm³ mol⁻¹

This is 1.3% below 24.0 dm³ mol⁻¹. Likely causes: an oxide coating on the ribbon, gas escaping before the bung is fitted, or leaks. Compare percentage uncertainties: with a balance reading of ±0.0005 g and mass found by difference, 0.0720 g carries about 1.4%; a ±0.5 cm³ syringe reading on 70.2 cm³ is about 0.7%. The mass is the weaker measurement, so a larger mass of magnesium (and a larger syringe) improves the result most.

1.12 Equations from test-tube observations

• Displacement: zinc in copper(II) sulfate solution. You see a pink-brown solid form and the blue colour fade. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
• Typical reactions of acids: a metal fizzes, giving hydrogen; a carbonate fizzes, giving carbon dioxide (limewater turns cloudy); an alkali shows no visible change but warms slightly. CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l) and H⁺(aq) + OH⁻(aq) → H₂O(l).
• Precipitation: mixing two solutions to give an insoluble solid. Barium chloride with a sulfate gives a white precipitate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Silver nitrate with chloride, bromide and iodide gives white, cream and yellow precipitates of AgCl, AgBr and AgI.

Common errors

• Writing Mr for an ionic compound such as CaCl₂ instead of relative formula mass.
• Leaving cm³ in a concentration calculation, which gives answers 1000 times too small.
• Using °C or kPa in pV = nRT. Convert to K and Pa, and cm³ to m³ (× 10⁻⁶).
• Forgetting to multiply by the number of atoms or ions per formula unit when counting particles.
• Rounding mid-calculation, so a ratio such as 1 : 2.96 is misread.
• Ionic equations that balance atoms but not charge, or that keep spectator ions.

Next steps

Read the revision notes, then work through the practice questions. Next in Unit 1: atomic structure revision notes. The exam preparation guide covers the whole course.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited: Topic 1, outcomes 1.1 to 1.12.