Chapters

The laboratory part of NCERT Class 11 Unit 8 (Organic Chemistry: Some Basic Principles and Techniques): purification, qualitative analysis (Lassaigne's test) and quantitative analysis (sections 8.8–8.10), plus the functional-group tests and titration facts JEE Main files here from the Class 12 laboratory course (marked Beyond NCERT).

Formula sheet

Every formula in the chapter on one screen, for the last pass. m = mass of compound; m₁ = mass of what is weighed. Details and examples are in the sections below.

Quantity	Formula	Watch
% C	12/44 × m(CO₂)/m × 100	CO₂ caught in KOH
% H	2/18 × m(H₂O)/m × 100	H₂O caught in anhydrous CaCl₂
% O	100 − all others	Or 32/88 × m(CO₂ from I₂O₅)/m × 100
Dumas, N₂ at STP	V₀ = P₁V₁ × 273 / (760 × T₁)	P₁ = P(atm) − aqueous tension
Dumas, % N	28/22400 × V₀/m × 100	V₀ in mL at STP
Kjeldahl, % N	1.4 × (mmol H⁺ used by NH₃) / m	1 mol H₂SO₄ takes 2 NH₃
NCERT's Kjeldahl form	1.4 × M × 2(V − V₁/2) / m	V mL H₂SO₄ and V₁ mL NaOH, both molarity M
Carius, % X	(at. mass X / M of AgX) × m₁/m × 100	AgCl 143.5, AgBr 188, AgI 235
% S	32/233 × m(BaSO₄)/m × 100	BaSO₄ 233
% P (molybdate)	31/1877 × m₁/m × 100	(NH₄)₃PO₄·12MoO₃ 1877
% P (Mg₂P₂O₇)	62/222 × m₁/m × 100	two P per Mg₂P₂O₇
Rf	distance moved by substance / by solvent, from the base line	Always < 1; no unit
Steam distillation	P = P(organic) + P(water)	Boils below 373 K

0 · What JEE Main asks from this chapter

164 questions in JEE Main 2019–2026, about 1 per paper; 50 are numerical answers. Grouped by type (counts are approximate):

Type	Share	Typical question
Estimation numericals (C, H, O, N, X, S, P)	~33%	0.15 g gives 0.2397 g AgBr: % Br = 68 (2021)
Purification methods (which technique, matches)	~20%	Glycerol from spent-lye: distillation under reduced pressure (2020)
Chromatography and Rf	~15%	Spot 3.5 cm, solvent 5 cm: Rf = 7 × 10⁻¹ (2024)
Lassaigne's test and detection	~12%	Blood red with Fe³⁺: N and S both present (2024)
Functional-group tests (Class 12 lab)	~12%	Match: Baeyer's test → unsaturation (2024)
Titration and salt-preparation lab facts	~8%	Why dil. H₂SO₄ in Mohr's salt: stops FeSO₄ hydrolysis (2024)

The estimation questions repeat NCERT's own problems with new numbers. Learn the five formulas and the molar masses of AgCl, AgBr, AgI, BaSO₄ and Mg₂P₂O₇.

1 · Purification

The method depends on the nature of the compound and of the impurity. Purity is then checked by a sharp melting or boiling point; modern checks use chromatography and spectroscopy.

Method	Principle	Classic use
Sublimation	solid → vapour without melting	camphor, naphthalene, benzoic acid from non-sublimable solids (camphor from CaSO₄)
Crystallisation	different solubility of compound and impurity	most solid organic compounds
Simple distillation	volatile liquid from non-volatile impurity, or b.p.s far apart	chloroform (334 K) from aniline (457 K)
Fractional distillation	b.p.s close together	crude oil fractions; isomers a few °C apart
Reduced-pressure distillation	high b.p., or decomposes at or below b.p.	glycerol from spent-lye (soap industry)
Steam distillation	steam-volatile and immiscible with water	aniline from aniline–water; essential oils
Differential extraction	more soluble in an organic solvent than in water	organic compound from an aqueous layer
Chromatography	different adsorption or partition	separation, purification, purity test

Crystallisation: dissolve in a solvent in which the compound is sparingly soluble cold but well soluble hot; concentrate to near saturation; cool; filter. The filtrate (mother liquor) keeps the impurities. Use a mixture of two solvents if the compound is very soluble in one and hardly in the other. Activated charcoal removes coloured impurities. Impurities of similar solubility need repeated crystallisation.

Fractional distillation: vapours pass up a fractionating column before the condenser. The higher-boiling vapour condenses first; the rising vapour gets richer in the more volatile liquid. The column gives many surfaces where rising vapour and falling liquid exchange heat; each condensation–vaporisation unit is a theoretical plate (columns with hundreds exist). The flask is left richer in the high boiler.

Distillation under reduced pressure: a liquid boils when its vapour pressure equals the external pressure, so lowering the pressure (water pump or vacuum pump) lowers the b.p.

Steam distillation: steam is blown through the hot mixture. It boils when p(organic) + p(water) = atmospheric pressure, so the organic liquid distils below its own b.p. and the mixture close to but below 373 K. Separate the water and compound afterwards in a separating funnel.

Differential extraction: shake the aqueous solution with an immiscible organic solvent in a separatory funnel; the layers separate; evaporate the solvent. If the compound is not very soluble in the solvent, use continuous extraction (the same solvent recycled).

Choosing a method · Beyond NCERT

• Simple distillation works for b.p.s about 20 °C or more apart (propanol and propanone, 2021). Close b.p.s (isohexane 60 °C, 3-methylpentane 63 °C) need fractional distillation; the lower boiler comes over first (2020).
• An azeotrope boils at a fixed composition and cannot be separated by fractional distillation (2026).
• Steam-volatile: o-nitrophenol and o-nitroaniline (intramolecular H-bond, weak attraction to water); the para isomers are not (intermolecular H-bonds) (2025).
• Density decides the layers: CH₂Cl₂ and CHCl₃ (denser) are the lower layer under water; ether and ethyl acetate float on top (2019).
• Acid–base extraction: from an organic solution, NaHCO₃ takes out a carboxylic acid, then NaOH takes a phenol, then dilute HCl takes an amine (2020).
• Sugar and NaCl: sugar dissolves in alcohol, salt doesn't (2026).

Chromatography

From the Greek chroma, colour: first used for plant pigments. The mixture is put on a stationary phase (solid or liquid); a mobile phase (solvent, solvent mixture or gas) moves over it, and the components separate.

Adsorption chromatography: components are adsorbed to different extents on an adsorbent, usually silica gel or alumina.

• Column chromatography: the adsorbent is packed in a glass tube with a stopcock; the mixture goes on top; the eluant flows down. The most strongly adsorbed component stays near the top; the least adsorbed comes out first.
• Thin layer chromatography (TLC): a 0.2 mm layer of silica gel or alumina on a glass plate (a chromaplate). The spot goes about 2 cm above the bottom; the plate stands in the eluant in a closed jar.
• Seeing colourless spots: UV light (if they fluoresce); a jar with iodine crystals (brown spots); a spray reagent (ninhydrin for amino acids).

Partition chromatography: continuous partitioning between a stationary and a mobile phase. In paper chromatography the stationary phase is the water held in the paper, not the paper itself (2023, 2025). The solvent rises by capillary action; the developed strip is the chromatogram.

Reading Rf · Beyond NCERT

• Rf is a ratio: no unit, always below 1. It depends on the solvent and the stationary phase, so it is not constant across solvents (2021).
• On silica, a more polar compound is held more strongly: lower Rf, and it leaves a column later. The least polar elutes first (2023). If A elutes first, B is the one with low Rf and stronger adsorption.
• Turning 2-methylpropene (Rf 0.42) into tert-butyl alcohol makes it more polar: its Rf drops (0.12 in the 2026 key).
• Small amounts of two close compounds (100 mg of p-nitrophenol and picric acid) are best separated by preparative TLC (2022).
• Purification by chromatography doesn't depend on the physical state of the pure compound (2021).

Example (2022) · Solvent 3.25 cm, A 2.08 cm, B 1.05 cm. Rf(A) = 0.64, Rf(B) = 0.32; the ratio A : B = 2.

2 · Qualitative analysis

Carbon and hydrogen: heat with copper(II) oxide. C becomes CO₂ (lime water turns milky); H becomes H₂O (white anhydrous CuSO₄ turns blue).

Lassaigne's test (N, S, halogens, P): fuse the compound with sodium metal to turn covalent N, S and X into ionic NaCN, Na₂S, NaX; boil the fused mass with distilled water: the sodium fusion extract. (P is tested separately with an oxidising agent.)

Nitrogen: boil the extract with FeSO₄, then acidify with conc. H₂SO₄. Cyanide forms sodium hexacyanidoferrate(II); some Fe²⁺ is oxidised to Fe³⁺, which gives Prussian blue, iron(III) hexacyanidoferrate(II) (2021 asked which formula).

Sulphur:

• (a) Acidify with acetic acid, add lead acetate: black PbS. Not H₂SO₄: it would precipitate white PbSO₄ (NCERT exercise 8.31).
• (b) Sodium nitroprusside: violet S²⁻ + [Fe(CN)₅NO]²⁻ → [Fe(CN)₅NOS]⁴⁻, written Na₄[Fe(CN)₅NOS] (2022).

N and S together: fusion gives NaSCN, which gives blood red [Fe(SCN)]²⁺ with Fe³⁺, and no Prussian blue (no free CN⁻). With excess sodium, NaSCN + 2Na → NaCN + Na₂S, and the usual separate tests work.

Halogens: acidify with HNO₃, add AgNO₃. If N or S is present, first boil with conc. HNO₃ to decompose NaCN and Na₂S, which would otherwise give AgCN or Ag₂S precipitates (NCERT exercise 8.25; asked 2021, 2022, 2024).

Phosphorus: heat with sodium peroxide (P → phosphate), boil with HNO₃, add ammonium molybdate: yellow ammonium phosphomolybdate.

When Lassaigne's test fails or misleads · Beyond NCERT

• No carbon, no NaCN: hydrazine (N₂H₄), NH₂OH and NH₄Cl give no test for N (2024). CH₃NH₂·HCl gives both N and Cl (2023).
• Sulphanilic acid (N and S) gives blood red, not blue (2023).
• CCl₄ gives no AgCl with AgNO₃: its Cl is covalent; there is no Cl⁻ until sodium fusion (NCERT exercise 8.29).
• Layer test for Br and I: add CHCl₃ and chlorine water to the acidified extract; violet CHCl₃ layer means iodine, orange-brown means bromine (2026).

3 · Quantitative analysis

Mass per cent of each element gives the empirical formula (Mole Concept). Today a CHN elemental analyser does this on 1–3 mg automatically.

Carbon and hydrogen

Burn a known mass in excess O₂ with CuO. Absorb the water in a weighed U-tube of anhydrous CaCl₂ and the CO₂ in a U-tube of concentrated KOH, in that order; the gains in mass give the amounts. KOH is used because CO₂ is acidic (NCERT exercise 8.30).

Example (NCERT Problem 8.20) · 0.246 g gives 0.198 g CO₂ and 0.1014 g H₂O. C = 12/44 × 0.198 = 0.0540 g → 21.95%. H = 2/18 × 0.1014 = 0.01127 g → 4.58%.

Example (2021) · 1.80 g of CₓHᵧO_z gives 2.64 g CO₂ and 1.08 g H₂O. C = 0.72 g, H = 0.12 g, O = 1.80 − 0.84 = 0.96 g → 53.33% O.

Example (2025) · 1.0 g gives 1.46 g CO₂ and 0.567 g H₂O: C 0.398 g (0.0332 mol), H 0.063 g (0.063 mol), O 0.539 g (0.0337 mol) → C : H : O = 1 : 2 : 1 → CH₂O, empirical formula mass 30.

Example (NCERT exercise 8.32) · 0.20 g with 69% C and 4.8% H: C 0.138 g → CO₂ = 0.138 × 44/12 = 0.506 g; H 0.0096 g → H₂O = 0.0096 × 9 = 0.0864 g.

Nitrogen: Dumas method

Heat with CuO in an atmosphere of CO₂: N becomes free N₂. Pass the gases over heated copper gauze to reduce any nitrogen oxides to N₂ (2023). Collect over KOH solution, which absorbs the CO₂; N₂ collects at the top of a graduated tube (a nitrometer).

Example (NCERT Problem 8.21) · 0.3 g gives 50 mL N₂ at 300 K and 715 mm; aqueous tension 15 mm. P₁ = 700 mm. V₀ = 700 × 50 × 273 / (760 × 300) = 41.9 mL. N = 28 × 41.9 / 22400 = 0.0524 g → 17.46%.

Example (2025, 2026) · The same numbers with 0.4 g and 60 mL: V₀ = 50.3 mL → 15.71%. With 0.50 g and 70 mL: V₀ = 58.7 mL → 14.67%, nearest integer 15.

Example (2021) · CuO for 57.5 g N,N-dimethylpentan-1-amine, C₇H₁₇N (115): 0.5 mol × (2 × 7 + 17/2) = 0.5 × 22.5 = 11.25 mol → 1125 × 10⁻².

Nitrogen: Kjeldahl's method

Heat with conc. H₂SO₄ (CuSO₄ as catalyst, K₂SO₄ to raise the b.p.): N becomes (NH₄)₂SO₄. Heat with excess NaOH; absorb the NH₃ in a known excess of standard H₂SO₄; titrate the leftover acid with standard alkali. The difference is the acid used by NH₃.

(1 mmol NH₃ = 14 mg N, so % N = 14 × mmol / (1000 m) × 100 = 1.4 × mmol / m.)

Kjeldahl fails for N in nitro and azo groups and N in a ring (pyridine): that N doesn't become (NH₄)₂SO₄. Diazonium salts fail too (they lose N₂). Amines, amides and urea work.

Example (NCERT Problem 8.22) · NH₃ from 0.5 g neutralises 10 mL of 1 M H₂SO₄ = 10 mmol H₂SO₄ = 20 mmol NH₃ = 0.28 g N → 56%.

Example (NCERT exercise 8.33) · 0.50 g; NH₃ into 50 mL of 0.5 M H₂SO₄ (50 mmol H⁺); the excess takes 60 mL of 0.5 M NaOH (30 mmol). H⁺ used = 20 mmol → 20 mmol N = 0.28 g → 56%.

Example (2021, the reverse) · 0.8 g at 42% N = 0.336 g N = 24 mmol NH₃ → 12 mmol H₂SO₄ → 12 mL of 1 M.

Halogens: Carius method

Heat a known mass with fuming HNO₃ and AgNO₃ in a sealed hard-glass Carius tube in a furnace. C and H oxidise; the halogen forms AgX, which is filtered, washed, dried and weighed. 1 mol AgX holds 1 mol X.

Example (NCERT Problem 8.23) · 0.15 g gives 0.12 g AgBr: Br = 0.12 × 80/188 = 0.0511 g → 34.04%. (Asked again in 2022 as 0.45 g and 0.36 g: the same 34.04%.)

Example (NCERT exercise 8.34) · 0.3780 g gives 0.5740 g AgCl: Cl = 0.5740 × 35.5/143.5 = 0.1420 g → 37.57%.

Example (2026) · Iodine by the layer test; 0.15 g gives 0.12 g AgI: I = 0.12 × 127/235 = 0.0649 g → 43%.

Sulphur

Heat in a Carius tube with sodium peroxide or fuming HNO₃: S becomes H₂SO₄. Add excess BaCl₂ solution; weigh the BaSO₄. 1 mol BaSO₄ (233 g) holds 32 g S.

Example (NCERT Problem 8.24) · 0.157 g gives 0.4813 g BaSO₄: S = 0.4813 × 32/233 = 0.0661 g → 42.10%. (2026 used 0.314 g with the same 0.4813 g: 21.05%.)

Example (NCERT exercise 8.35) · 0.468 g gives 0.668 g BaSO₄: S = 0.0917 g → 19.60%.

Example (2025) · 160 mg gives 466 mg BaSO₄ = 2 mmol → 64 mg S → 40%.

Phosphorus

Heat with fuming HNO₃: P becomes H₃PO₄. Precipitate it as ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃ (1877 g mol⁻¹), with ammonia and ammonium molybdate; or as MgNH₄PO₄ with magnesia mixture, which on ignition gives Mg₂P₂O₇ (222 g mol⁻¹, 62 g P).

Example (2026) · 1.00 g gives 1.79 g Mg₂P₂O₇: P = 1.79 × 62/222 = 0.500 g → 50%.

Oxygen

Usually by difference: 100 − (sum of the others). Directly: heat the compound in a stream of N₂; pass the gases over red-hot coke (all O → CO); pass through warm I₂O₅ (CO → CO₂, I₂ freed).

So 1 mol O₂ from the compound gives 2 mol CO₂: 32 g O per 88 g CO₂.

O can also be found from the iodine set free.

4 · Functional-group tests · Beyond NCERT

From the Class 12 laboratory course; JEE files them here and asks them as matches.

Group	Test	Positive result
Unsaturation	Br₂ in CCl₄ or Br₂ water	red-orange colour discharged
Unsaturation	Baeyer's (cold dil. alkaline KMnO₄)	pink colour discharged (brown MnO₂)
Alcohol	ceric ammonium nitrate (CAN)	red colour
Alcohol 1°/2°/3°	Lucas (conc. HCl + ZnCl₂)	3° turbid at once, 2° in ~5 min, 1° not at room temperature
Phenol	neutral FeCl₃	violet (blue, green) colour
Phenol	phthalein dye (phthalic anhydride, conc. H₂SO₄, then NaOH)	pink-red
Phenol, aniline	bromine water	white precipitate
Aldehyde, ketone	2,4-DNP	yellow-orange or orange-red precipitate
Aldehyde	Tollens'	silver mirror
Aliphatic aldehyde	Fehling's	red Cu₂O precipitate
Aldehyde	Schiff's reagent	pink-magenta colour
CH₃CO– or CH₃CH(OH)–	iodoform (I₂ + NaOH)	yellow CHI₃
Carboxylic acid	NaHCO₃	CO₂ effervescence
Carboxylic acid	ester test (alcohol + conc. H₂SO₄)	fruity smell
1° amine	carbylamine (CHCl₃ + alc. KOH)	foul smell of isocyanide
1°/2°/3° amine	Hinsberg's (C₆H₅SO₂Cl)	1° sulphonamide dissolves in alkali; 2° doesn't; 3° no reaction
1° aromatic amine	azo dye (HNO₂ at 0–5 °C, then β-naphthol)	orange-red dye
Carbohydrate	Molisch's (α-naphthol, conc. H₂SO₄)	violet ring
Ketose	Seliwanoff's (resorcinol, HCl)	red
Reducing sugar	Benedict's, Barfoed's (copper reagents)	red Cu₂O
Protein, peptide	biuret (alkaline CuSO₄)	violet
Amino acid	ninhydrin	blue-violet

Example (2021) · CAN tests an alcohol; CHCl₃ + alc. KOH (carbylamine) tests a 1° amine. Of Seliwanoff's, biuret, Barfoed's and Benedict's tests, only Seliwanoff's uses no copper reagent.

Example (2025) · NaHCO₃ → carboxylic acid; neutral FeCl₃ → phenolic –OH; CAN → alcohol; alkaline KMnO₄ → unsaturation.

Solubility classes: insoluble in water but soluble in NaOH and not in NaHCO₃: a phenol (m-cresol, 2019). Soluble in NaHCO₃: a carboxylic acid. Soluble in dil. HCl: an amine. Dissolves only in hot NaOH: an ester or amide (hydrolysed).

5 · Titration and preparation facts · Beyond NCERT

• Primary standard: pure, dry, stable in air, not hygroscopic, high molar mass, reacts fast and exactly. Oxalic acid, potassium hydrogen phthalate (for NaOH, with phenolphthalein), K₂Cr₂O₇, Mohr's salt. Not NaOH, not Na₂Cr₂O₇ (hygroscopic) (2024, 2026).
• Phenolphthalein is a weak acid (colourless in acid, pink in base). Methyl orange is a weak base; its quinonoid (red) form is more deeply coloured than the benzenoid (yellow) form (2023).
• NaOH's strength is found best with NaOH in the burette and oxalic acid in the conical flask (2020). No Bunsen burner or measuring cylinder is needed for HCl vs NaOH (2020).
• KMnO₄ vs oxalic acid: heat to about 60 °C first (the reaction is slow until Mn²⁺ forms); KMnO₄ vs Mohr's salt (FAS) needs no heating (2025).
• Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O: add dil. H₂SO₄ to stop hydrolysis of FeSO₄; avoid prolonged heating, which oxidises Fe²⁺ to Fe³⁺ (2023, 2024).

6 · Traps

Trap	What goes wrong	Do this
Dumas pressure	Using the barometer reading	Subtract aqueous tension first
Kjeldahl, H₂SO₄	1 mol H₂SO₄ = 1 mol NH₃	It takes 2 NH₃; count H⁺ in mmol
Kjeldahl back-titration	Using the acid taken	Acid used = acid taken − acid left
Kjeldahl scope	Applying it to nitrobenzene, pyridine	Nitro, azo, ring N and diazonium fail
Carius molar masses	Using Br's mass for AgBr	AgCl 143.5, AgBr 188, AgI 235, BaSO₄ 233
Mg₂P₂O₇	Using 31 for P	Two P per formula: 62/222
Lead acetate test	Acidifying with H₂SO₄	Acetic acid, so PbSO₄ doesn't form
N + S together	Expecting Prussian blue	Blood red from SCN⁻; excess Na restores both
Halogen test	Skipping the HNO₃ boil	NaCN, Na₂S would precipitate with Ag⁺
Hydrazine, NH₄Cl	Expecting a N test	No carbon, so no NaCN
Rf	High Rf = strongly adsorbed	High Rf = weakly held, less polar
Paper chromatography	Paper is the stationary phase	The water in it is
Steam distillation	"Aniline is miscible with water"	It works because aniline is immiscible
Layers	Organic layer always on top	CH₂Cl₂ and CHCl₃ sink below water

7 · Numbers worth knowing

Substance	g mol⁻¹	Element fraction
CO₂	44	C 12/44
H₂O	18	H 2/18
AgCl	143.5	Cl 35.5/143.5
AgBr	188	Br 80/188
AgI	235	I 127/235
BaSO₄	233	S 32/233
Mg₂P₂O₇	222	P 62/222
(NH₄)₃PO₄·12MoO₃	1877	P 31/1877

• 22 400 mL of N₂ at STP weighs 28 g; 1 mmol NH₃ holds 14 mg N.
• b.p.: chloroform 334 K, aniline 457 K. TLC layer about 0.2 mm; spot about 2 cm above the edge.

8 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• NCERT exercise 8.39: the best and latest technique for isolation, purification and separation is chromatography.
• Fractional distillation separates the fractions of crude oil in the petroleum industry.
• The labelled apparatus figures (simple and fractional distillation, steam distillation, separating funnel, Dumas and Kjeldahl set-ups, Carius tube) are drawn in board answers.