Lesson 2: Empirical Formulas and Solutions

An empirical formula shows the simplest whole-number ratio of atoms in a compound; the molecular formula shows the actual numbers, and is a whole-number multiple of the empirical formula. Glucose is C₆H₁₂O₆ molecularly but CH₂O empirically — six compounds share that empirical formula, which is why analysis must pair composition with molar mass. This lesson (Structure 1.4–1.5) completes the analytical toolkit: percentage composition, combustion analysis, hydrates, and the solution chemistry — concentration, dilution and standard solutions — that underpins every titration you will ever do.

From percentage composition to formula

The routine is mechanical and must become automatic. Assume 100 g of compound, so percentages become grams. Convert each mass to moles (÷ A_r). Divide every mole value by the smallest to get the ratio; if a ratio lands on .5, .33 or .25, multiply the whole set (×2, ×3, ×4) to clear fractions. Worked through: a compound is 40.0% C, 6.7% H, 53.3% O. Moles: C = 40.0 ÷ 12.01 = 3.33; H = 6.7 ÷ 1.01 = 6.63; O = 53.3 ÷ 16.00 = 3.33. Dividing by 3.33 gives 1 : 2 : 1 — empirical formula CH₂O (mass 30.03). If the compound's molar mass is 180.18 g mol⁻¹, the multiplier is 180.18 ÷ 30.03 = 6, so the molecular formula is C₆H₁₂O₆. The same three steps solve combustion analysis: all the carbon in the sample ends up in the CO₂ (count it there), all the hydrogen in the H₂O (remember: two H per water), and oxygen — if present — is found by difference from the sample mass.

Solutions: concentration and dilution

Concentration in IB is amount per volume: c = n ÷ V, with V in dm³ (1 dm³ = 1000 cm³ — the conversion that decides more marks than any other in this topic). A standard solution is one of accurately known concentration, prepared by dissolving a weighed primary standard and making up to the mark in a volumetric flask. Dilution adds solvent only, so moles of solute are conserved: c₁V₁ = c₂V₂. Serial dilution (repeated tenfold steps) is how very low concentrations are reached accurately, and each step multiplies the uncertainty — a favourite evaluation point.

Worked example 1: combustion analysis in full

A 0.920 g sample of a compound containing only C, H and O burns completely to give 1.76 g of CO₂ and 1.08 g of H₂O. Its molar mass is 46.08 g mol⁻¹. Deduce its molecular formula. Carbon: n(CO₂) = 1.76 ÷ 44.01 = 0.0400 mol, so n(C) = 0.0400 mol and m(C) = 0.0400 × 12.01 = 0.480 g. Hydrogen: n(H₂O) = 1.08 ÷ 18.02 = 0.0599 mol, so n(H) = 2 × 0.0599 = 0.120 mol and m(H) = 0.120 × 1.01 = 0.121 g. Oxygen by difference: m(O) = 0.920 − 0.480 − 0.121 = 0.319 g, so n(O) = 0.319 ÷ 16.00 = 0.0199 mol. Ratio: divide by 0.0199 → C : H : O = 2.01 : 6.03 : 1.00 ≈ 2 : 6 : 1, empirical formula C₂H₆O (mass 46.08). The empirical mass already equals the molar mass, so the molecular formula is C₂H₆O — ethanol or methoxymethane; analysis alone cannot distinguish structural isomers, which is why spectroscopy exists.

Worked example 2: preparing a standard solution — the practical core

Prepare 250.0 cm³ of 0.0500 mol dm⁻³ sodium carbonate solution. n = c × V = 0.0500 × 0.2500 = 0.01250 mol. M(Na₂CO₃) = 2 × 22.99 + 12.01 + 3 × 16.00 = 105.99 g mol⁻¹. Mass required = 0.01250 × 105.99 = 1.325 g. Technique, step by step: weigh the solid by difference (container + solid, tip into a beaker, reweigh) on a balance reading to ±0.01 g; dissolve in distilled water, stir, warm if needed; transfer through a funnel into the volumetric flask, rinsing the beaker and funnel three times into the flask so no solute is lost; fill to just below the mark, then add dropwise until the meniscus sits on the mark at eye level; stopper and invert ten times to homogenise. Now the uncertainty budget: the balance gives 1.325 ± 0.01 g across two weighings (±0.02 g worst case, 1.5%), while the 250.0 cm³ flask is ±0.3 cm³ (0.12%) — the weighing dominates the uncertainty, so a more precise balance improves this preparation more than anything else. That is the kind of quantified evaluation the IA rewards.

Nature of Science — and the real world

Empirical formulas were nineteenth-century chemistry's great detective tool: Liebig's combustion train let analysts deduce formulas decades before anyone could see a molecule, and the method's name is a philosophy — *empirical*, from evidence. The same logic now runs automated CHN analysers in forensic and pharmaceutical laboratories, verifying the purity of every batch of medicine. Standard solutions underpin water-quality monitoring worldwide: the concentration of lead or nitrate in drinking water is, at root, a comparison against a standard someone prepared in a volumetric flask.

Common mistakes

• Forgetting that each H₂O from combustion carries two hydrogen atoms — the classic combustion-analysis slip.
• Clearing fractional ratios incorrectly: 1 : 1.5 : 1 must be doubled to 2 : 3 : 2, not rounded.
• Leaving V in cm³ in c = n ÷ V. Convert to dm³ first, every time.
• Treating the molecular formula as reachable from composition alone; without a molar mass, only the empirical formula is known.

IB Exam Experience

Question 1. What is the empirical formula of a compound with composition by mass 52.2% C, 13.0% H and 34.8% O?

• (A) C₂H₆O
• (B) C₂H₅O
• (C) CH₃O
• (D) C₄H₁₂O₂

Question 2. 5.85 g of sodium chloride is dissolved and made up to 500 cm³ of solution. What is its concentration in mol dm⁻³?

• (A) 0.100
• (B) 0.200
• (C) 0.0200
• (D) 2.00

Question 3. 25.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid is diluted to a total volume of 250 cm³. What is the new concentration?

• (A) 0.100 mol dm⁻³
• (B) 0.0200 mol dm⁻³
• (C) 0.200 mol dm⁻³
• (D) 20.0 mol dm⁻³

Answer key. 1. (A) — moles: C = 52.2 ÷ 12.01 = 4.35; H = 13.0 ÷ 1.01 = 12.9; O = 34.8 ÷ 16.00 = 2.18. Divide by 2.18 → 2 : 6 : 1. 2. (B) — n = 5.85 ÷ 58.44 = 0.100 mol; c = 0.100 ÷ 0.500 = 0.200 mol dm⁻³. 3. (C) — c₂ = c₁V₁ ÷ V₂ = 2.00 × 25.0 ÷ 250 = 0.200 mol dm⁻³.

Structured question (Paper 2 style). A 2.50 g sample of hydrated magnesium sulfate, MgSO₄·xH₂O, is heated to constant mass. The mass of anhydrous salt remaining is 1.22 g. (a) Calculate the amount, in mol, of anhydrous MgSO₄. [1] (b) Determine the mass and amount, in mol, of water lost. [2] (c) Deduce the value of x in MgSO₄·xH₂O. [1] (d) State one assumption made in this method, and the effect on the calculated x if the sample was not heated for long enough. [2] Marking notes. (a) n = 1.22 ÷ 120.37 = 0.0101 mol. (b) m(H₂O) = 2.50 − 1.22 = 1.28 g [1]; n = 1.28 ÷ 18.02 = 0.0710 mol [1]. (c) x = 0.0710 ÷ 0.0101 = 7.02 ≈ 7. (d) Assumption: all water of crystallisation was driven off (and no decomposition of the salt occurred) [1]; insufficient heating leaves water behind, so the water lost is underestimated and x comes out too small [1].

IA idea: Determine the water of crystallisation of a hydrated salt gravimetrically, investigating how the value of x depends on heating time and temperature.

Two more analytical workhorses: gravimetry and parts per million

Gravimetric analysis turns stoichiometry into a weighing exercise: convert the analyte into an insoluble compound of known composition, filter, dry and weigh it. To find the sulfate content of a fertiliser, dissolve a known mass, add excess barium chloride, and collect the BaSO₄ precipitate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). If 0.500 g of fertiliser yields 0.583 g of BaSO₄ (M = 233.4 g mol⁻¹), then n(SO₄²⁻) = 0.583 ÷ 233.4 = 2.50 × 10⁻³ mol, so m(SO₄²⁻) = 2.50 × 10⁻³ × 96.06 = 0.240 g, and the fertiliser is 48.0% sulfate by mass. The technique's marks live in its precautions: precipitate from hot solution for larger, purer crystals; wash the precipitate with distilled water to remove coprecipitated ions; dry to constant mass (heat, cool, weigh, repeat until two readings agree) so the mass is truly water-free; and use a sintered-glass crucible or ashless filter paper so the collection medium adds no mass of its own. Each precaution answers a specific error — that pairing of precaution to error is precisely what an evaluation section should look like.

Very dilute solutions are described in parts per million (ppm): 1 ppm = 1 mg of solute per dm³ of solution (for water, 1 kg ≈ 1 dm³). Concentration in mol dm⁻³ converts via the molar mass: 0.50 ppm of lead (Pb, A_r = 207) is 0.50 mg dm⁻³ = 5.0 × 10⁻⁴ g dm⁻³, which is 5.0 × 10⁻⁴ ÷ 207 = 2.4 × 10⁻⁶ mol dm⁻³. The WHO guideline limit for lead in drinking water is 0.01 ppm — about 5 × 10⁻⁸ mol dm⁻³ — a concentration reachable only by the serial dilutions described in this lesson, and measured by instruments (atomic absorption, ICP-MS) rather than titration. Being able to move fluently between %, mol dm⁻³ and ppm is what "solution fluency" means, and Paper 2 data questions move between them without warning.

Standardising a base: the primary-standard technique

Sodium hydroxide solution cannot be weighed out accurately — the solid absorbs water and CO₂ from the air — so it is standardised against a primary standard: potassium hydrogen phthalate (KHP, M = 204.22 g mol⁻¹), a stable, pure, high-molar-mass solid acid that reacts 1 : 1 with NaOH. A student dissolves 0.510 g of KHP and titrates it with the NaOH solution: titre 24.60 cm³. n(KHP) = 0.510 ÷ 204.22 = 2.497 × 10⁻³ mol = n(NaOH). c(NaOH) = 2.497 × 10⁻³ ÷ 0.02460 = 0.1015 mol dm⁻³ — close to, but not, the intended 0.100, and every subsequent result using this base inherits the corrected value. The design choices are the teaching: KHP's large molar mass shrinks the weighing percentage uncertainty (0.01 g on 0.510 g is 2.0%... halved by taking 1.020 g and a ~49 cm³ titre — but that overflows a 50 cm³ burette, so the design balances weighing uncertainty against titre capacity, a genuine experimental trade-off). Phenolphthalein is the indicator (a weak acid titrated by a strong base — the Lesson 14 equivalence logic), and the standardisation is repeated to concordance before the base is trusted. Every quantitative analysis you will ever meet — including the IA's — stands on this pattern: an accurately known standard, a measured ratio, a corrected unknown.