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

These revision notes cover 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: outcomes 1.1 to 1.12, including Core Practical 1. The topic belongs to Unit 1: Structure, Bonding and Introduction to Organic Chemistry (IAS). The full teaching, with longer examples, is in the study guide; to test yourself, use the practice questions.

Also useful: the course hub, the printable checklist and the free 10-minute diagnostics.

Use these values throughout: H 1.0, C 12.0, N 14.0, O 16.0, Na 23.0, S 32.1, Ca 40.1; L = 6.02 × 10²³ mol⁻¹; R = 8.31 J mol⁻¹ K⁻¹; molar volume at room temperature and pressure 24.0 dm³ mol⁻¹.

Definitions (1.1, 1.4)

Term	Meaning
Atom	Smallest particle of an element that keeps its chemical identity
Element	Substance whose atoms all have the same atomic number
Ion	Particle carrying a charge because electrons were gained or lost
Molecule	Discrete particle of atoms joined by covalent bonds
Compound	Two or more elements chemically combined
Empirical formula	Lowest whole-number ratio of the atoms present
Molecular formula	Real atom count per molecule, element by element
Relative atomic mass, Ar	Abundance-weighted average atomic mass, on the scale where carbon-12 = 12 exactly
Relative molecular mass, Mr	Sum of Ar values for a molecule (simple molecular substances)
Relative formula mass	The same sum for a compound with a giant structure, e.g. NaCl, SiO₂
Molar mass, M	Mass per mole of a substance, in g mol⁻¹
ppm	Parts per million: by volume for atmospheric gases, by mass for solutes

Ar, Mr and relative formula mass are ratios, so they have no units. Molar mass has units.

Formula sheet

Quantity	Expression	Units to use
Amount from mass (1.2, 1.7)	n = m / M	g, g mol⁻¹
Particles (1.2)	N = n × L	mol, mol⁻¹
Concentration (1.5)	c = n / V	mol, dm³
Mass concentration (1.5)	g dm⁻³ = c × M	mol dm⁻³, g mol⁻¹
Gas at room conditions (1.8)	V = n × 24.0	dm³
Ideal gas (1.8)	pV = nRT	Pa, m³, K
Molar mass of a vapour (1.8)	M = mRT / pV	g, Pa, m³, K
ppm by mass (1.4)	(mass of part / total mass) × 10⁶	same mass units
ppm from a percentage (1.4)	% × 10⁴
Percentage yield (1.9)	actual / theoretical × 100%	same mass units
Atom economy (1.9)	M(desired product) / Σ M(all products) × 100%	g mol⁻¹

Unit conversions: cm³ ÷ 1000 = dm³; dm³ ÷ 1000 = m³; cm³ × 10⁻⁶ = m³; kPa × 1000 = Pa; °C + 273 = K.

Method boxes

Empirical formula (1.6)

1. Write the mass or percentage of each element. If oxygen is not given, find it by difference.
2. Divide each by its Ar.
3. Scale the results so the smallest becomes 1.
4. If you get a value such as 1.5 or 1.33, multiply all values by 2 or 3.
5. For the molecular formula, divide Mr by the empirical formula mass.

Reacting masses (1.7)

1. Balanced equation.
2. Moles of the substance you know.
3. Mole ratio from the coefficients.
4. Moles × M for the substance you want.

Gas volumes (1.8)

1. At room temperature and pressure: V = n × 24.0 dm³.
2. At any other conditions: pV = nRT, after converting to Pa, m³ and K.
3. For gas-to-gas questions, volumes are in the same ratio as the equation’s coefficients, provided all gases are measured at the same conditions.

Ionic equations (1.3, 1.12)

1. Write the full equation with state symbols.
2. Split every aqueous ionic substance into its ions. Keep solids, liquids, gases and covalent molecules together.
3. Cross out ions that are identical on both sides.
4. Check atoms and total charge balance.

Small worked reminders

Concentration. 1.06 g of sodium carbonate in 100 cm³ of solution: n = 1.06 / 106.0 = 0.0100 mol; c = 0.0100 / 0.100 = 0.100 mol dm⁻³; mass concentration = 10.6 g dm⁻³.

Gas ratio. For N₂(g) + 3H₂(g) → 2NH₃(g), 30 cm³ of hydrogen reacts with 10 cm³ of nitrogen to give 20 cm³ of ammonia, all at the same conditions.

Atom economy can reach 100%. When the equation has only one product, as in an addition reaction, every reactant atom ends up in the desired product.

Core Practical 1 in brief (1.11)

• Aim: measure the volume occupied by one mole of gas.
• Measure: mass of solid reactant (by difference), final gas volume, room temperature and pressure.
• Calculation: moles of gas from the equation and the mass used; molar volume = gas volume / moles.
• Typical sources of low results: gas escaping before the bung is fitted, gas dissolving in water if collected over water, an oxide layer on a metal, leaks at joints.
• Improve: use a larger mass so the percentage uncertainty in the mass is smaller; add the solid in a small tube that is tipped over after the bung is fitted.
• Percentage uncertainty = (uncertainty / measured value) × 100%. For a value found by difference, add the uncertainties of both readings.

Test-tube reactions to know (1.12)

Reaction type	Observation	Ionic equation
Metal displacement, e.g. Mg in Cu²⁺(aq)	Pink-brown solid; blue colour fades	Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Acid + metal	Fizzing; solid disappears	Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)
Acid + carbonate	Fizzing; gas turns limewater cloudy	CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l)
Acid + alkali	No visible change; mixture warms	H⁺(aq) + OH⁻(aq) → H₂O(l)
Precipitation, e.g. Ag⁺ + Cl⁻	White precipitate	Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Precipitation, e.g. Ba²⁺ + SO₄²⁻	White precipitate	Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Must-know distinctions

• Empirical vs molecular formula: ratio vs actual count. They can be the same (H₂O) or different (C₂H₄ and CH₂).
• Relative molecular mass vs relative formula mass: molecules vs giant structures.
• Percentage yield vs atom economy: yield depends on the experiment; atom economy depends only on the equation.
• ppm by volume vs ppm by mass: gases in air vs dissolved substances.
• mol dm⁻³ vs g dm⁻³: convert using molar mass.

Quick self-test

1. Work out the relative formula mass of calcium hydrogencarbonate, Ca(HCO₃)₂.
2. Calculate the amount in moles in 9.45 g of nitric acid, HNO₃.
3. How many molecules are in 0.0500 mol of oxygen gas?
4. 0.0400 mol of solute is dissolved in 160 cm³ of solution. Give the concentration.
5. Give the mass concentration of 0.150 mol dm⁻³ sodium hydroxide, NaOH.
6. Find the amount of gas in 450 cm³ at 35.0 °C and 98.0 kPa.
7. By mass, a hydrocarbon contains 82.8% C and 17.2% H. Find its empirical formula.
8. A 2.0 kg soil sample contains 0.0036 g of cadmium. Give this in ppm by mass.
9. Calculate the atom economy for making calcium oxide by CaCO₃ → CaO + CO₂.
10. Write the ionic equation, with state symbols, for any strong acid reacting with any alkali.
11. Calculate the volume at room temperature and pressure of 0.88 g of carbon dioxide.
12. A theoretical yield is 4.00 g and 3.10 g is obtained. Give the percentage yield.

Answers

1. 40.1 + 2 × (1.0 + 12.0 + 3 × 16.0) = 162.1
2. M = 63.0 g mol⁻¹; n = 9.45 / 63.0 = 0.150 mol
3. 0.0500 × 6.02 × 10²³ = 3.01 × 10²² molecules
4. 0.0400 / 0.160 = 0.250 mol dm⁻³
5. 0.150 × 40.0 = 6.00 g dm⁻³
6. n = pV / RT = (98 000 × 450 × 10⁻⁶) / (8.31 × 308) = 0.0172 mol
7. C: 82.8 / 12.0 = 6.90; H: 17.2 / 1.0 = 17.2; ratio 1 : 2.49, i.e. 2 : 5, so C₂H₅
8. (0.0036 / 2000) × 10⁶ = 1.8 ppm
9. 56.1 / 100.1 × 100 = 56.0%
10. H⁺(aq) + OH⁻(aq) → H₂O(l)
11. n = 0.88 / 44.0 = 0.0200 mol; V = 0.0200 × 24.0 = 0.480 dm³
12. 3.10 / 4.00 × 100 = 77.5%

Where marks are usually lost

• Writing Mr for an ionic compound when the question asks about relative formula mass, or giving Ar or Mr a unit.
• Using 25 cm³ instead of 0.025 dm³ in c = n / V.
• Leaving kPa, cm³ or °C in pV = nRT; each one gives a wrong answer.
• Rounding a mole ratio too early: 0.0499 : 0.0748 is 1 : 1.5, which means 2 : 3, not 1 : 1.
• Ionic equations with spectator ions left in, or with charges that do not balance.
• No state symbols where they are requested, especially (s) on a precipitate.
• Using the coefficient of the wrong substance when converting moles across an equation.
• Calculating atom economy with only one of the waste products in the denominator.
• Describing an observation with the name of a product (“hydrogen is made”) instead of what is seen (“fizzing”).

Official syllabus

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