Mission 1: How many water molecules are in a single drop of water?

The mole is a count

• Amount of substance, n, is measured in moles (mol).
• One mole contains 6.02 × 10²³ entities: the Avogadro constant, N_A = 6.02 × 10²³ mol⁻¹ (data booklet).
• Number of entities: N = n × N_A.
• Always name the entity: atom, molecule, ion, electron or formula unit.
• 1 mol CO₂ = 6.02 × 10²³ molecules, but 3 × 6.02 × 10²³ = 1.81 × 10²⁴ atoms.
• 1 mol Al₂(SO₄)₃ contains 5 mol of ions: 2 Al³⁺ and 3 SO₄²⁻.

Relative masses on the carbon-12 scale

• Atoms are compared with one twelfth of the mass of a ¹²C atom, so ¹²C = 12 exactly.
• Relative atomic mass, A_r: use the two-decimal values in the data booklet.
• Relative formula mass, M_r: add the A_r of every atom in the formula.
• A_r and M_r are ratios, so they have no units.
• A bracket multiplies all inside it: M_r(Mg(OH)₂) = 24.31 + 2(16.00 + 1.01) = 58.33.
• Water of crystallization counts: M_r(CuSO₄·5H₂O) = 159.62 + 5 × 18.02 = 249.72.

Molar mass and n = m/M

• Molar mass, M: the mass of one mole, in g mol⁻¹. Same number as M_r.
• n = m/M (data booklet). Units check: g ÷ g mol⁻¹ = mol.
• Use the formula of the real entity: nitrogen gas is N₂, M = 28.02 g mol⁻¹.
• Worked example: 11.0 g of CO₂. n = 11.0 ÷ 44.01 = 0.250 mol.
• Molecules = 0.250 × 6.02 × 10²³ = 1.50 × 10²³.

Mission 2: Methanal, ethanoic acid and glucose all boil down to CH₂O. How can three different substances share one formula?

Empirical and molecular formulae

• Empirical formula: the simplest whole-number ratio of atoms of each element.
• Molecular formula: how many atoms of each element one molecule really contains.
• Molecular formula = (empirical formula)ₓ, where x is a whole number.
• H₂O₂ → HO. C₂H₄ → CH₂. H₂O is already its own empirical formula.
• Neither formula shows which atoms are bonded to which.

Percentage by mass

• % by mass of an element = (number of its atoms × A_r) ÷ M_r × 100.
• Worked example: ammonium nitrate, NH₄NO₃. M_r = 2(14.01) + 4(1.01) + 3(16.00) = 80.06.
• % N = 2 × 14.01 ÷ 80.06 × 100 = 28.02 ÷ 80.06 × 100 = 35.00%.
• Weight every atom by its A_r: a share of atoms is not a share of mass.

From percentages to empirical formula

• Treat percentages as grams in a 100 g sample. Divide each by its A_r.
• Divide every result by the smallest.
• Scale to whole numbers: ×2 for .50, ×3 for .33 or .67, ×4 for .25 or .75. Never round 1.50.
• Worked example: 43.64% P, 56.36% O. P: 43.64 ÷ 30.97 = 1.409; O: 56.36 ÷ 16.00 = 3.523.
• Ratio 1 : 2.50, × 2 = 2 : 5. Empirical formula P₂O₅.

From empirical to molecular formula

• You also need the molar mass, M, of the compound.
• x = M ÷ mass of one empirical formula unit.
• Worked example: P₂O₅ has empirical formula mass 2(30.97) + 5(16.00) = 141.94.
• The compound's M is 283.88 g mol⁻¹, so x = 283.88 ÷ 141.94 = 2.
• Molecular formula = (P₂O₅)₂ = P₄O₁₀.

Mission 3: A 500 cm³ hospital drip bag holds 9.00 g of salt per dm³. How many moles of salt are in the bag?

Molar concentration

• Molar concentration, C = n ÷ V: amount of solute per dm³ of solution, in mol dm⁻³.
• Square brackets mean concentration in mol dm⁻³: [NaCl] = 0.100 mol dm⁻³.
• [NaCl] is never an amount in moles.
• V is the volume of the whole solution, not of the water added.
• A standard solution is made up to the mark in a volumetric flask.

n = CV and volume units

• n = CV (data booklet), with V in dm³.
• 1 dm³ = 1000 cm³, so divide cm³ by 1000.
• Worked example: 40.0 cm³ of KOH(aq), [KOH] = 0.150 mol dm⁻³.
• n = 0.150 × 0.0400 = 0.00600 mol.
• Rearranged: 0.0125 mol in 250 cm³ gives C = 0.0125 ÷ 0.250 = 0.0500 mol dm⁻³.

g dm⁻³ and mol dm⁻³

• Concentration can also be a mass per volume: g dm⁻³.
• g dm⁻³ ÷ M = mol dm⁻³. Units check: g dm⁻³ ÷ g mol⁻¹ = mol dm⁻³.
• mol dm⁻³ × M = g dm⁻³.
• Worked example: [NaOH] = 0.250 mol dm⁻³, M = 40.00 g mol⁻¹: 0.250 × 40.00 = 10.0 g dm⁻³.

Mission 4: Why does helium behave almost exactly as the gas model predicts, but ammonia doesn't?

The ideal gas model

• Particles are in constant, random motion.
• Particle volume is negligible compared with the volume of the gas.
• There are no intermolecular forces, attractive or repulsive.
• All collisions are elastic: total kinetic energy is unchanged.
• Pressure comes from particles colliding with the walls, equally in all directions.

Why real gases deviate

• Real particles have a finite volume and do attract each other.
• High pressure: particles are close, so attractions act more and their own volume is no longer negligible.
• Low temperature: slow particles are affected more by the attractions.
• Stronger intermolecular forces mean larger deviation: NH₃ (hydrogen bonding) more than He.
• Near room temperature and atmospheric pressure, most gases behave almost ideally.

Seeing deviation on a graph

• For an ideal gas, PV/nRT = 1 at every pressure.
• Attractions soften and reduce wall collisions, so P is lower than ideal: PV/nRT below 1.
• At very high pressure the particles' own volume cuts the free space, so PV/nRT rises above 1.
• You explain these in words; no calculation is needed.

Mission 5: An identical balloon of hydrogen and one of carbon dioxide: which holds more molecules?

Avogadro's law

• Avogadro's law: at the same T and p, equal gas volumes hold equal numbers of molecules.
• It works because particle volume is negligible: size doesn't change how many fit.
• It counts molecules, not atoms.
• Equal volumes hold equal amounts, so their masses are in the ratio of their molar masses.

Reacting gas volumes

• At the same T and p, gas volume ∝ amount in moles.
• So gases react in the volume ratio of their coefficients.
• Worked example: 2CO(g) + O₂(g) → 2CO₂(g).
• 40 cm³ of CO reacts with 20 cm³ of O₂ to make 40 cm³ of CO₂.
• Only gases count: a liquid or solid product has almost no volume.

Molar volume

• Molar volume, V_m: the volume of one mole of gas.
• It is the same for every ideal gas at a given T and p.
• At STP (273.15 K, 100 kPa): V_m = 22.7 dm³ mol⁻¹ (data booklet).
• V = n × V_m, so n = V ÷ V_m.
• Worked example: 7.00 g of N₂. n = 7.00 ÷ 28.02 = 0.250 mol; V = 0.250 × 22.7 = 5.68 dm³.
• Only for gases, and only at STP. Other conditions: use PV = nRT (Mission 6).

Mission 6: Why should drivers check tyre pressure before a long drive, not after?

Pressure and volume

• Fixed amount of gas, constant temperature: p ∝ 1/V.
• Halve the volume and the pressure doubles: particles hit the walls twice as often.
• p against V: a falling curve.
• p against 1/V: a straight line through the origin.

Temperature, in kelvin

• Constant pressure: V ∝ T. Constant volume: p ∝ T. T must be in kelvin.
• Hotter particles move faster and hit the walls harder and more often. They do not expand.
• Plotted against T in K: straight lines through the origin.
• Plotted against °C: straight lines that extend back to zero at −273 °C (0 K).
• Investigate p–V: gas syringe with a pressure sensor, at constant T.
• Investigate V–T: syringe in a water bath. p–T: sealed flask with a pressure sensor.

The ideal gas equation

• PV = nRT (data booklet). R = 8.31 J K⁻¹ mol⁻¹ (data booklet).
• SI units only: P in Pa, V in m³, T in K.
• 1 kPa = 10³ Pa. 1 dm³ = 10⁻³ m³. 1 cm³ = 10⁻⁶ m³.
• Convert first, then substitute.

Worked example: finding a molar mass

• 0.880 g of a gas occupies 4.72 × 10⁻⁴ m³ at 1.05 × 10⁵ Pa and 25 °C (298 K).
• n = PV ÷ RT = (1.05 × 10⁵ × 4.72 × 10⁻⁴) ÷ (8.31 × 298) = 49.6 ÷ 2476 = 0.0200 mol.
• M = m ÷ n = 0.880 ÷ 0.0200 = 44.0 g mol⁻¹.
• M ≈ 44 g mol⁻¹ is consistent with CO₂ (44.01 g mol⁻¹).

The combined gas law

• For a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂ (data booklet). T in kelvin.
• Leave out any quantity that stays constant.
• Worked example: 0.0150 m³ of gas at 1.01 × 10⁵ Pa and 290 K moves to 5.05 × 10⁴ Pa and 250 K.
• V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 0.0150 × 2 × (250/290) = 0.0259 m³.
• Check direction: lower pressure → bigger; cooler → smaller.

The big ideas

• A mole is 6.02 × 10²³ of a named entity: N = n × N_A.
• A_r and M_r have no units; molar mass is in g mol⁻¹; n = m/M.
• Empirical formula: simplest ratio. Molecular formula = (empirical)ₓ, with x = M ÷ empirical formula mass.
• Concentration [X] is in mol dm⁻³; n = CV with V in dm³; divide g dm⁻³ by M to get mol dm⁻³.
• Ideal gas: negligible particle volume, no forces, elastic collisions. Real gases deviate when cold or compressed.
• Equal gas volumes (same T, p) hold equal numbers of molecules. PV = nRT in Pa, m³ and K.