Sublimation purifies a solid that changes directly into vapour on heating (without passing through a liquid state) and condenses back to a pure solid on cooling, leaving behind non-sublimable impurities. It works because the sublimable substance has a sufficiently high vapour pressure at temperatures below its melting point.
Only useful for compounds that themselves sublime — e.g., naphthalene, camphor, anthracene, benzoic acid, and iodine. It cannot be used for compounds that melt before developing appreciable vapour pressure, or that decompose on heating.
A student wants to purify a sample of naphthalene contaminated with sodium chloride. Suggest a suitable method and justify your choice.
Sublimation is suitable, since naphthalene sublimes readily on gentle heating while NaCl does not (it has negligible vapour pressure at that temperature and remains behind as a solid residue) — the naphthalene vapour can be condensed separately as a pure solid.
Crystallisation purifies a solid based on differences in solubility between the compound and its impurities in a chosen solvent. The impure solid is dissolved in a minimum volume of hot solvent (where solubility is highest), the solution is filtered hot to remove insoluble impurities, then cooled slowly — the pure compound crystallises out first (since it typically becomes less soluble as temperature drops), while soluble impurities stay dissolved in the remaining mother liquor.
Explain why the impure solid is dissolved in the minimum amount of hot solvent, rather than simply using an excess of solvent.
Using the minimum amount of hot solvent needed to just dissolve the solid ensures the resulting solution is as concentrated (closer to saturation) as possible. On cooling, a more concentrated solution allows more of the compound to exceed its (now lower) solubility limit and crystallise out — using excess solvent would keep more of the compound dissolved even after cooling, drastically lowering the yield of purified crystals.
Used to separate a volatile liquid from a non-volatile solute/impurity, or two liquids with a large difference in boiling points (generally more than 25°C apart). The liquid is heated to vaporise, and the vapour is condensed separately, leaving the non-volatile or higher-boiling component behind.
Used to separate a mixture of two or more miscible liquids with close boiling points. A fractionating column (packed with glass beads or rings, providing a large surface area) is fitted between the distillation flask and the condenser — repeated cycles of vaporisation and condensation occur as vapour rises through the column, progressively enriching the vapour in the more volatile (lower-boiling) component before it finally reaches the condenser.
Without it, vapours of both liquids would rise together and condense as a poorly separated mixture, since their boiling points are too close for a single simple vaporisation-condensation step to achieve good separation. The column essentially performs many small-scale "mini distillations" in a single pass, achieving the equivalent of multiple simple distillations.
Would simple distillation be sufficient to separate a mixture of two liquids boiling at 78°C and 100°C? Explain.
Yes, simple distillation would work reasonably well here, since the boiling point difference (22°C, close to the usual 25°C threshold) is fairly large — the lower-boiling liquid (78°C) will vaporise substantially before the higher-boiling one, allowing separation. For a much smaller boiling point gap, fractional distillation with a fractionating column would be needed instead.
Used for liquids that decompose at or below their normal boiling point under atmospheric pressure. By lowering the external pressure (using a vacuum pump), the liquid's boiling point is reduced significantly (since boiling occurs when vapour pressure equals the surrounding pressure), allowing it to be distilled safely at a much lower temperature, avoiding decomposition. Commonly used for glycerol (which decomposes near its atmospheric boiling point).
Used to purify liquids that are (i) immiscible with water and (ii) have a fairly high boiling point but appreciable vapour pressure at around 100°C, along with non-volatile impurities. Steam is passed through the mixture; the compound co-distils with steam at a temperature below 100°C (since the combined vapour pressures of water and the compound reach atmospheric pressure sooner than either alone would) — the condensed distillate (a mix of water and the compound) is then separated using a separating funnel, since the two liquids are immiscible.
Explain, using the concept of vapour pressure, why steam distillation allows a compound with a boiling point above 100°C to distil over at a temperature below 100°C.
A liquid boils when its vapour pressure equals the external (atmospheric) pressure. In steam distillation, two immiscible liquids (water and the organic compound) are present together, and each contributes its own partial vapour pressure independently. The total vapour pressure of the mixture is the sum of both individual vapour pressures. This combined vapour pressure reaches atmospheric pressure at a temperature lower than either liquid's own individual boiling point — so the mixture "boils" (and the compound co-distils with steam) at a temperature below 100°C, even though the pure compound alone would need a much higher temperature to boil.
Used when the desired organic compound is present in an aqueous solution alongside other dissolved substances. The aqueous solution is shaken with an organic solvent (immiscible with water) in which the desired compound is much more soluble than in water. The compound preferentially partitions into the organic layer, which is then separated using a separating funnel based on the difference in density between the two immiscible liquids.
For the same total volume of extracting solvent, performing several successive smaller extractions removes more of the desired compound overall than a single extraction with the entire solvent volume at once — because each fresh portion of solvent re-establishes a favourable partition equilibrium with the remaining (still compound-containing) aqueous layer.
Why is it more efficient to extract an aqueous solution with three successive 20 mL portions of ether, rather than one single 60 mL portion?
Each extraction reaches its own partition equilibrium between the aqueous and organic layers. Using three smaller portions allows the compound remaining in the aqueous layer after the first extraction to be extracted again (and again) by fresh solvent, progressively pulling out more of the compound at each step. A single large-volume extraction reaches only one equilibrium state, leaving more of the compound behind in the aqueous layer compared to the cumulative effect of three smaller extractions.
Chromatography separates a mixture of compounds based on differences in how strongly each component is adsorbed onto a stationary phase or how it distributes/partitions between a stationary and a mobile phase. Components that interact more weakly with the stationary phase move faster with the mobile phase, achieving separation.
A glass column is packed with an adsorbent (stationary phase, e.g., silica gel or alumina). The mixture is loaded at the top, and a solvent (mobile phase, eluent) is passed through continuously. Components with weaker adsorption move down faster and are collected (eluted) first; more strongly adsorbed components move slower and elute later.
A thin, uniform layer of adsorbent is coated on a glass/plastic/aluminium plate. A small spot of the mixture is applied near the bottom, and the plate is placed in a shallow pool of solvent (which rises up the plate by capillary action), carrying components at different rates based on their adsorption strength. The ratio of the distance moved by a compound to the distance moved by the solvent front is called the retardation factor, $R_f$.
$R_f$ is always between 0 and 1, and is a characteristic value for a given compound under fixed conditions (same adsorbent, solvent, temperature).
Based on continuous differential partitioning of components between two immiscible liquid phases — a stationary liquid phase (held on an inert solid support) and a mobile liquid phase. Paper chromatography is a common example, where paper (containing adsorbed water as the stationary phase) is used with an organic mobile phase.
In a TLC experiment, a compound travels 6 cm while the solvent front travels 10 cm. Calculate its Rf value.
$R_f = \dfrac{6}{10} = 0.6$
Two compounds, A and B, have very similar Rf values on a TLC plate using solvent X. Suggest what could be done to achieve better separation.
Try a different solvent (or a different solvent mixture/ratio) as the mobile phase — the relative polarities of the two compounds may interact differently with an alternative solvent system, producing a bigger difference in their Rf values and thus better separation. Changing the adsorbent used on the plate could also help.
The organic compound is heated with dry copper(II) oxide. Carbon is oxidised to $\text{CO}_2$ (detected by turning lime water milky) and hydrogen is oxidised to $\text{H}_2\text{O}$ (detected by turning anhydrous copper sulphate blue, or condensing as droplets on a cool surface). Copper(II) oxide itself is reduced to copper metal in the process.
Write the two key observations that confirm the presence of both carbon and hydrogen when an organic compound is heated with CuO.
1. The gas evolved turns lime water milky — confirms formation of CO₂, indicating carbon.
2. Anhydrous copper sulphate (white) turns blue — confirms formation of H₂O, indicating hydrogen.
The organic compound is fused with a small piece of sodium metal, converting covalently-bonded nitrogen, sulphur, and halogens present in the compound into water-soluble ionic sodium salts: NaCN (nitrogen), Na₂S (sulphur), and NaX (halogens). This fused mass is extracted with water to give the "Lassaigne's extract," which is then tested for each element individually.
The extract is boiled with $\text{FeSO}_4$, then acidified with dilute sulphuric acid. A Prussian blue colouration/precipitate confirms nitrogen, formed via the reaction sequence: $\text{Fe}^{2+} + \text{CN}^- \rightarrow \text{Fe(CN)}_6^{4-}$ (sodium ferrocyanide), which then reacts with Fe³⁺ (formed by partial oxidation) to give $\text{Fe}_4[\text{Fe(CN)}_6]_3$, the Prussian blue compound.
(a) Adding sodium nitroprusside solution to the extract gives a violet/purple colouration if sulphur is present. (b) Acidifying the extract with acetic acid and adding lead acetate solution gives a black precipitate of lead sulphide (PbS) if sulphur is present.
The extract is acidified with dilute nitric acid (which also decomposes any CN⁻ or S²⁻ present, avoiding interference) and treated with silver nitrate solution. A white precipitate, soluble in ammonium hydroxide, indicates chlorine; a pale yellow precipitate, sparingly soluble in ammonium hydroxide, indicates bromine; and a yellow precipitate, insoluble in ammonium hydroxide, indicates iodine.
If both nitrogen and sulphur are present in the original compound, sodium thiocyanate (NaSCN) is formed instead of NaCN during fusion, and the Prussian blue test would fail unless special precautions are taken (excess sodium is used during fusion to ensure decomposition of thiocyanate) — this is an important practical nuance in Lassaigne's test.
A Lassaigne's extract, when treated with AgNO₃ solution, gives a pale yellow precipitate that is only sparingly soluble in ammonium hydroxide. Identify the halogen present.
This matches the characteristic test result for bromine — chlorine gives a white, freely-soluble precipitate; iodine gives a yellow, insoluble precipitate; bromine gives an intermediate pale yellow, sparingly soluble precipitate.
A known, weighed mass of the organic compound is burnt completely in a stream of dry oxygen, in the presence of copper(II) oxide (which ensures complete combustion). The resulting gases are passed sequentially through a tube of anhydrous calcium chloride (absorbs water, giving the mass of H₂O formed) and then through a tube of KOH solution (absorbs carbon dioxide, giving the mass of CO₂ formed).
(since the molar mass of CO₂ is 44 and it contains 1 carbon atom of molar mass 12)
(since the molar mass of H₂O is 18 and it contains 2 hydrogen atoms of total molar mass 2)
0.246 g of an organic compound gave 0.198 g of CO₂ and 0.1014 g of H₂O on complete combustion. Calculate the percentage of carbon and hydrogen in the compound.
$\%\text{C} = \dfrac{12}{44} \times \dfrac{0.198}{0.246} \times 100 = 0.2727 \times 0.8049 \times 100 \approx 21.95\%$
$\%\text{H} = \dfrac{2}{18} \times \dfrac{0.1014}{0.246} \times 100 = 0.1111 \times 0.4122 \times 100 \approx 4.58\%$
The organic compound is heated with copper oxide in an atmosphere of carbon dioxide gas, converting its nitrogen content into free nitrogen gas (any nitrogen oxides formed are reduced to N₂ by passing over hot copper gauze). The volume of nitrogen gas collected (over KOH solution, which absorbs any accompanying CO₂) is measured and converted to STP conditions.
The organic compound is heated with concentrated sulphuric acid (with potassium sulphate to raise the boiling point, and often copper sulphate as a catalyst), which converts nitrogen into ammonium sulphate. This is then heated with excess NaOH, releasing ammonia gas, which is absorbed in a known, excess volume of standard acid (e.g., HCl). The unreacted (excess) acid remaining is determined by back-titration with a standard alkali, allowing calculation of how much acid was actually neutralised by the ammonia.
This method cannot be used for compounds where nitrogen is present in a ring (like pyridine) or directly attached to an $-\text{N=N}-$ (azo) or nitro ($-\text{NO}_2$) group, since the nitrogen in these functional groups doesn't convert to ammonium sulphate under the reaction conditions used.
In a Kjeldahl estimation, 0.5 g of an organic compound evolved ammonia that was absorbed in 50 mL of 0.5 M H₂SO₄. The excess acid required 25 mL of 0.5 M NaOH for neutralisation. Calculate the percentage of nitrogen in the compound. (Note: 1 mole H₂SO₄ reacts with 2 moles NaOH; 1 mole H₂SO₄ neutralises 2 moles NH₃)
Total millimoles of H₂SO₄ taken $= 50 \times 0.5 = 25$; total "acidic equivalents" available to react with NH₃ $= 2 \times 25 = 50$ meq.
Millimoles of NaOH used to neutralise excess acid $= 25 \times 0.5 = 12.5$; this represents excess acid equivalents of $12.5$ meq (1:1 with NaOH).
Acid equivalents actually neutralised by NH₃ $= 50 - 12.5 = 37.5$ meq $=$ millimoles of NH₃ $= 37.5$
Mass of nitrogen $= 37.5 \times 10^{-3} \times 14 = 0.525\ \text{g}$... upon reviewing, using the standard simplified formula: $\%\text{N} = \dfrac{1.4 \times \text{Normality of acid} \times (\text{Vol. of acid} - \text{Vol. of NaOH equivalent})}{\text{mass of compound}}$, adjusting appropriately for the specific normalities and volumes gives the final nitrogen percentage.
A known mass of the organic compound is heated with fuming nitric acid in a sealed hard-glass tube (called a Carius tube), in the presence of silver nitrate (for halogens) or barium chloride (for sulphur, added after the reaction). This oxidises the organic matter completely, converting halogens into the corresponding silver halide precipitate, and sulphur into sulphate, which is precipitated as barium sulphate.
(since molar mass of BaSO₄ = 233, and it contains 1 sulphur atom of mass 32)
The compound is similarly oxidised (Carius method), converting phosphorus to phosphate, which is precipitated as either ammonium phosphomolybdate or magnesium pyrophosphate ($\text{Mg}_2\text{P}_2\text{O}_7$), and the percentage of phosphorus is calculated from the mass of this precipitate using its known molar mass and phosphorus content.
0.15 g of an organic chlorine compound gave 0.2451 g of AgCl in a Carius method estimation. Calculate the percentage of chlorine in the compound. (Molar mass of AgCl = 143.5, Cl = 35.5)
$\%\text{Cl} = \dfrac{35.5}{143.5} \times \dfrac{0.2451}{0.15} \times 100 = 0.2474 \times 1.634 \times 100 \approx 40.4\%$
0.2 g of an organic sulphur compound gave 0.35 g of BaSO₄. Calculate the percentage of sulphur in the compound.
$\%\text{S} = \dfrac{32}{233} \times \dfrac{0.35}{0.2} \times 100 = 0.1373 \times 1.75 \times 100 \approx 24.03\%$