Why purify, and how do we check purity?
Before studying a compound we must make it pure. All purification methods use some difference in properties between the compound and the impurity. A pure solid has a sharp melting point; a pure liquid has a fixed boiling point. Impurities lower and widen the melting range. Chromatography and spectroscopy are also used to test purity.
Sublimation and crystallisation
Sublimation
Some solids turn straight into vapour on heating without melting (camphor, naphthalene, iodine, benzoic acid). The vapour is cooled back to a solid, leaving non-volatile impurities behind.
Crystallisation
Uses a difference in solubility. The compound is dissolved in the smallest amount of a hot solvent in which it is very soluble when hot but little soluble when cold. The hot solution is filtered, then cooled slowly. Pure crystals form and are filtered off. Coloured impurities are removed with activated charcoal. If the compound and impurity have similar solubility, repeated crystallisation is used (fractional crystallisation).
Distillation and its types
- Simple distillation: liquids with very different boiling points (difference > 20–25 K), or a liquid from non-volatile impurities. Example: chloroform (334 K) and aniline (457 K).
- Fractional distillation: boiling points are close. A fractionating column gives many surfaces for repeated boiling and condensing, so the vapour at the top is richer in the more volatile liquid. Used to separate crude oil fractions.
- Distillation under reduced pressure (vacuum): for liquids that decompose at or below their normal boiling point. Lower pressure means lower boiling point. Used to get glycerol from spent lye in soap industry.
- Steam distillation: for substances that are steam-volatile and do not mix with water (aniline, essential oils). The mixture boils when p(water) + p(substance) = atmospheric pressure, so it boils below 373 K and the compound does not break down.
Differential extraction
An organic compound dissolved in water is shaken with an organic solvent (like ether) that does not mix with water and in which the compound dissolves more. The two layers separate in a separating funnel; the organic layer is run off and the solvent evaporated. For compounds that are only a little soluble, continuous extraction is used.
Chromatography
Chromatography (Greek chroma = colour) separates a mixture using two phases: a stationary phase (a solid or a liquid held on a solid) and a mobile phase (a liquid or gas) that moves over it. Components that stick more to the stationary phase move slowly.
Adsorption chromatography
- Column chromatography: a glass column packed with alumina or silica gel. The mixture is added at the top and an eluant (solvent) flows down. The least adsorbed part comes out first.
- Thin layer chromatography (TLC): a thin layer of silica gel on a glass plate. Spots are seen under UV light or with iodine vapour or ninhydrin (for amino acids).
Partition chromatography
Paper chromatography: water held in the paper fibres is the stationary phase; the solvent moving up is the mobile phase. The components divide (partition) between the two liquids.
Retardation factor
Rf = distance moved by the substance ÷ distance moved by the solvent front (both from the base line). Rf is always less than 1 and is fixed for a substance in a given solvent.
Qualitative analysis: detecting elements
Carbon and hydrogen
Heat the compound with copper(II) oxide. C becomes CO₂ (turns lime water milky); H becomes H₂O (turns white anhydrous copper sulphate blue).
Lassaigne's test (sodium fusion extract)
The compound is fused with sodium metal so covalent N, S, X turn into ionic salts: Na + C + N → NaCN; 2Na + S → Na₂S; Na + X → NaX. The hot tube is broken in water, boiled and filtered to get the sodium fusion extract.
- Nitrogen: boil extract with FeSO₄, then acidify with conc. H₂SO₄ → Prussian blue colour (iron(III) hexacyanidoferrate(II)).
- Sulphur: (a) add acetic acid and lead acetate → black PbS; (b) add sodium nitroprusside → violet colour.
- N and S together: NaSCN forms → blood-red colour with Fe³⁺. (Excess sodium breaks it into NaCN and Na₂S.)
- Halogens: acidify with HNO₃, add AgNO₃: white ppt soluble in NH₃ = Cl; pale yellow, sparingly soluble = Br; yellow, insoluble = I. Boiling with HNO₃ first removes CN⁻ and S²⁻, which would also give a precipitate.
- Phosphorus: oxidise with Na₂O₂ to phosphate; heat with HNO₃ and ammonium molybdate → yellow precipitate.
Quantitative analysis: how much of each element
Carbon and hydrogen (Liebig's method)
A weighed sample (m) is burnt in dry O₂ over CuO. Water is absorbed in a U-tube of anhydrous CaCl₂; CO₂ in a tube of concentrated KOH. The mass gains give:
%C = (12/44) × m(CO₂)/m × 100; %H = (2/18) × m(H₂O)/m × 100
Nitrogen: Dumas method
The compound is heated with CuO in CO₂; N becomes N₂ gas, collected over KOH solution (which absorbs CO₂). The N₂ volume is changed to STP (use dry pressure = p − aqueous tension).
%N = (28/22400) × V(N₂ at STP, mL)/m × 100
Nitrogen: Kjeldahl method
The compound is heated with conc. H₂SO₄ (and K₂SO₄, CuSO₄) so N becomes (NH₄)₂SO₄. NaOH releases NH₃, which is passed into a known volume of standard acid. The leftover acid is titrated to find how much acid the NH₃ used.
moles NH₃ = moles of H⁺ used; %N = 1.4 × M × (acidity) × V(acid used, mL)/m. Kjeldahl does not work for nitro, azo compounds or N in a ring (pyridine), because that N does not turn into ammonium sulphate.
Halogens: Carius method
Heat with fuming HNO₃ and AgNO₃ in a sealed Carius tube; halogen forms AgX, which is weighed.
%X = (atomic mass of X / molar mass of AgX) × m(AgX)/m × 100 (AgCl 143.5, AgBr 188, AgI 235)
Sulphur
Heat with fuming HNO₃ (or Na₂O₂); S becomes H₂SO₄, precipitated as BaSO₄. %S = (32/233) × m(BaSO₄)/m × 100
Phosphorus
P becomes phosphoric acid, precipitated as MgNH₄PO₄ and ignited to Mg₂P₂O₇. %P = (62/222) × m(Mg₂P₂O₇)/m × 100
Oxygen
Usually by difference: %O = 100 − (sum of all other %).
Try it
Paper chromatography of sketch-pen ink (see Try it above) and find Rf values. In the 3D free play, set the product mass yourself and predict the percentage before the readout shows it.
Key formulas and definitions
- Rf = distance moved by substance / distance moved by solvent front
- %C = (12/44) × m(CO₂)/m × 100; %H = (2/18) × m(H₂O)/m × 100
- Dumas: %N = (28/22400) × V(N₂, STP mL)/m × 100; V_STP = V × (p/760) × (273/T)
- Kjeldahl: moles N = moles NH₃ = moles H⁺ used; %N = 1.4 × M × acidity × V(mL)/m
- Carius: %Cl = (35.5/143.5) × m(AgCl)/m × 100; %Br uses 80/188
- %S = (32/233) × m(BaSO₄)/m × 100; %P = (62/222) × m(Mg₂P₂O₇)/m × 100; %O = 100 − Σ others
Worked examples
1. On a paper chromatogram the solvent front moved 7.5 cm and a dye moved 3.0 cm. Find the Rf value.
Line 1: Rf = distance of dye ÷ distance of solvent. Line 2: Rf = 3.0 ÷ 7.5. Line 3: Rf = 0.40.
2. 0.246 g of an organic compound gave 0.198 g CO₂ and 0.1014 g H₂O on burning. Find %C and %H.
Line 1: %C = (12/44) × (0.198/0.246) × 100. Line 2: = 0.2727 × 0.8049 × 100 = 21.95%. Line 3: %H = (2/18) × (0.1014/0.246) × 100 = 0.1111 × 0.4122 × 100 = 4.58%.
3. Dumas method: 0.20 g of a compound gave 30 mL of dry N₂ at 300 K and 750 mm Hg. Find %N.
Line 1: Change to STP: V = 30 × (750/760) × (273/300) = 26.94 mL. Line 2: Mass of N₂ = 28 × 26.94/22400 = 0.03368 g. Line 3: %N = 0.03368/0.20 × 100 = 16.84%.
4. Kjeldahl method: NH₃ from 0.50 g of a compound neutralised 8.0 mL of 0.5 M H₂SO₄. Find %N.
Line 1: Moles H⁺ used = 2 × 0.5 × 0.008 = 0.008 mol. Line 2: Moles NH₃ = moles N = 0.008 mol → mass N = 0.008 × 14 = 0.112 g. Line 3: %N = 0.112/0.50 × 100 = 22.4%.
5. Carius method: 0.15 g of an organic compound gave 0.287 g of AgCl. Find %Cl.
Line 1: Mass of Cl = (35.5/143.5) × 0.287 = 0.0710 g. Line 2: %Cl = 0.0710/0.15 × 100. Line 3: %Cl = 47.3%.
6. 0.16 g of a compound gave 0.466 g of BaSO₄ in the sulphur estimation. Find %S.
Line 1: Mass of S = (32/233) × 0.466 = 0.064 g. Line 2: %S = 0.064/0.16 × 100. Line 3: %S = 40%.
7. 0.20 g of a compound gave 0.111 g of Mg₂P₂O₇. Find %P.
Line 1: Mass of P = (62/222) × 0.111 = 0.031 g. Line 2: %P = 0.031/0.20 × 100. Line 3: %P = 15.5%.
8. A compound contains C 40.0% and H 6.67%; the rest is oxygen. Find %O and the empirical formula.
Line 1: %O = 100 − (40.0 + 6.67) = 53.33%. Line 2: Moles: C 40/12 = 3.33, H 6.67/1 = 6.67, O 53.33/16 = 3.33. Line 3: Divide by 3.33: C 1 : H 2 : O 1 → empirical formula CH₂O.
Common mistakes
- Using the wet gas volume in the Dumas method. Subtract aqueous tension if N₂ is collected over water, then convert to STP.
- Forgetting that H₂SO₄ gives 2 H⁺ per molecule in Kjeldahl calculations.
- Applying Kjeldahl to nitro compounds or pyridine. Their N does not change to ammonium sulphate.
- Measuring Rf from the bottom edge of the paper instead of from the base line where the spot was put.