Purification & Characterisation of Organic Compounds Formula Sheet — NEET Chemistry

Every key Purification & Characterisation of Organic Compounds formula, definition and fact for NEET Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNCERT · 4 sub-topics

Purification — crystallisation, sublimation, distillation, differential extraction, chromatography
Qualitative analysis — detection of nitrogen, sulphur, phosphorus and halogens
Quantitative analysis — estimation of carbon, hydrogen and nitrogen
Calculation of empirical and molecular formulae

Purification I — Sublimation, Crystallisation, Distillation

Purification of organic compounds: Once isolated or synthesised, a compound is purified by a method matched to its

physical state and the impurity

. Purity is finally judged by a

sharp melting or boiling point

— a pure solid/liquid melts/boils over a narrow range.

Situation	Method	Example / note
Solid sublimes on heating	Sublimation	camphor, $N H_{4} Cl$
Solid, solubility differs with $T$	Crystallisation	cool hot satd. soln.; charcoal
Liquid + non-volatile impurity	Simple distillation	or two liquids, $Δ$ b.p. large
Two liquids, close b.p.	Fractional distillation	column; crude-oil refining
Liquid, high b.p. / decomposes	Reduced-pressure distillation	boils below b.p.; glycerol
Steam-volatile + water-immiscible	Steam distillation	$p = p_{1} + p_{2}$ ; aniline
Compound dissolved in water	Differential extraction	separating funnel; solvent

★ Remember · Sublimation

A solid that turns directly to

vapour

on heating (no liquid stage) is separated from a non-sublimable impurity — e.g. camphor, naphthalene, anthracene,

N H_{4} Cl

🎯 Exam · Crystallisation

Relies on the

difference in solubility

of compound vs impurity. Dissolve in a hot solvent (sparingly soluble cold, freely soluble hot), concentrate, then cool — pure crystals form while impurities stay in the

mother liquor

. Coloured impurities are removed by

activated charcoal

; close solubilities need

fractional crystallisation

Simple distillation

▸Separates a volatile liquid from a $non-volatile$ impurity, or two liquids with a $large b.p. difference$ .
▸The lower-boiling component vaporises first, condenses, and is collected in the receiver.
▸Example: chloroform (b.p. 334 K) from aniline (b.p. 457 K).

⚠️ Watch out · Sharp m.p./b.p. ≠ identity

A sharp melting point confirms

purity

, not which compound it is — identity needs the elemental/spectral analysis on the later pages.

🚫 Examiner Trap · Examiner traps

(1) A SHARP melting/boiling point proves

purity

, NOT identity (that needs elemental/spectral analysis). (2) Sublimation: solid

\to

vapour directly (camphor, N

H_{4}

Cl). (3) Crystallisation exploits a solubility difference; charcoal removes coloured impurities; impurities stay in the mother liquor. (4) Simple distillation needs a LARGE b.p. difference (or a non-volatile impurity).

Distillation Variants & Differential Extraction

Schematic of three distillation set-ups: simple distillation with flask, condenser and receiver; fractional distillation with a fractionating column; and steam distillation with a steam inlet feeding the heated flask — Simple, fractional and steam distillation set-ups.

★ Remember · Fractional distillation

For liquids whose boiling points are

close

. A

fractionating column

gives repeated condensation–vaporisation cycles (each = one 'theoretical plate'), enriching the rising vapour in the more volatile component. Industrial use:

crude-oil refining

in the petroleum industry.

🎯 Exam · Reduced-pressure distillation

For liquids with

very high b.p.

or that

decompose

on heating. Lowering the external pressure makes the liquid boil at a

lower temperature

(it boils when vapour pressure

=

external pressure). Example: glycerol from spent-lye in the soap industry.

Steam distillation condition

p = p_{1} + p_{2}

Liquid distils when its vapour pressure

p_{1}

plus that of water

p_{2}

equals atmospheric p — so it comes over

below

its own b.p. (and below 373 K).

Steam distillation

▸For substances that are $steam-volatile$ and $immiscible$ with water.
▸Steam is passed through the heated liquid; the mixture distils and is separated in a separating funnel.
▸Classic example: purifying $aniline$ from an aniline–water mixture.

★ Remember · Differential (solvent) extraction

A compound in

aqueous

solution is shaken with an

immiscible

organic solvent in a

separating funnel

; it moves into the layer in which it is more soluble. If solubility is low,

continuous extraction

reuses the same solvent.

🚫 Examiner Trap · Examiner traps

(1)

Fractional

distillation is for CLOSE boiling points (column

=

many theoretical plates). (2)

Reduced-pressure

for high-b.p. or heat-sensitive liquids (boils when vapour pressure

=

reduced external pressure). (3)

Steam

distillation:

p = p_{1} + p_{2}

, for steam-volatile + water-IMMISCIBLE substances (aniline) — distils below 10

0^{\circ}

C. (4) Differential extraction uses an immiscible solvent in a separating funnel.

Chromatography

Chromatography: A technique to

separate, purify and test the purity

of mixtures (Greek chroma = colour). Components distribute differently between a

stationary phase

and a moving

mobile phase

, so they migrate at different rates and separate.

Column chromatography and a TLC plate, with the retardation factor R_f=x/y. — Column chromatography and a TLC plate, with the retardation factor $R_{f} = x / y$ .

★ Remember · Adsorption chromatography

Components are

adsorbed

to different degrees on a solid (silica gel or alumina).

Column

: mixture flows down an adsorbent column with an eluent; the most strongly adsorbed stays near the top.

TLC

: a thin adsorbent layer on glass; spots rise with the solvent.

Retardation factor (TLC)

R_{f} = \frac{distance moved by substance}{distance moved by solvent}

Each component has a characteristic

R_{f}

(between 0 and 1) in a given system.

🎯 Exam · Partition chromatography

Components

partition

continuously between two liquid phases.

Paper chromatography

is the common example — water trapped in the chromatography paper is the

stationary phase

and the developed strip is a

chromatogram

Detecting colourless spots

▸ $UV light$ — fluorescent spots glow.
▸ $Iodine vapour$ — many compounds give brown spots.
▸ $Ninhydrin$ spray — detects amino acids.

🚫 Examiner Trap · Examiner traps

(1)

R_{f} = \frac{substance distance}{solvent distance}

, always between 0 and 1. (2) In COLUMN chromatography the most-strongly-adsorbed component stays at the TOP (elutes last). (3) Paper chromatography is

partition

(trapped water

=

stationary phase), not adsorption. (4) Detect colourless spots with UV,

I_{2}

vapour, or ninhydrin (amino acids).

Qualitative Analysis I — C, H & the Lassaigne Test

Carbon

C + 2 CuO \to 2 Cu + C O_{2}

(lime-water milky)

Hydrogen

2 H + CuO \to Cu + H_{2} O

(anhyd.

CuS O_{4}

blue)

Lassaigne's (sodium fusion) test: N, S, halogens and P are covalently bound, so they are first made

ionic

by fusing the compound with sodium metal. The fused mass is boiled with water to give the

sodium fusion extract

, then tested.

Nitrogen

Na + C + N \to NaCN

Sulphur

2Na + S \to N a_{2} S

Halogen

Na + X \to NaX

=

Cl, Br, I)

🎯 Exam · Test for nitrogen → Prussian blue

Boil the extract with

FeS O_{4}

, then acidify with conc.

H_{2} S O_{4}

6 C N^{-} + F e^{2 +} \to [Fe (CN)_{6}]^{4 -}

, then

3 [Fe (CN)_{6}]^{4 -} + 4 F e^{3 +} \to F e_{4} [Fe (CN)_{6}]_{3}

—

Prussian blue

⚠️ Watch out · Fuse with excess sodium

Enough Na must be used so that N and S end up as separate

C N^{-}

and

S^{2 -}

ions; otherwise an

SC N^{-}

forms and gives a blood-red (not Prussian blue) nitrogen test.

🚫 Examiner Trap · Examiner traps

(1) C

\to

O_{2}

(lime water milky); H

\to

H_{2}

O (anhydrous CuS

O_{4}

turns blue). (2)

Lassaigne

: fuse with Na to convert covalent N/S/X/P into IONIC species first. (3) N test

=

Prussian blue (

F e_{4} [Fe (CN)_{6}]_{3}

). (4) Use EXCESS Na so N and S form separate CN

^{-}

and

S^{2 -}

(else SCN

^{-}

gives a blood-red, not blue, N test).

Qualitative Analysis II — Sulphur, Halogens & Phosphorus

★ Remember · Test for sulphur

(a) Acidify extract with acetic acid

+

lead acetate

\to

black

PbS

S^{2 -} + P b^{2 +} \to PbS

. (b) With sodium nitroprusside

\to

violet

[Fe (CN)_{5} NOS]^{4 -}

🎯 Exam · Nitrogen + sulphur together

When both are present,

Na + C + N + S \to NaSCN

; with

F e^{3 +}

it gives

blood-red

[Fe (SCN)]^{2 +}

and

no

Prussian blue (no free

C N^{-}

). Excess Na decomposes

SC N^{-}

C N^{-} + S^{2 -}

, restoring the usual tests.

Test for halogens (silver halide)

X^{-} + A g^{+} \to AgX

Acidify the extract with dilute

HN O_{3}

, then add

AgN O_{3}

Chlorine

white

AgCl

— soluble in

N H_{4} OH

Bromine

pale-yellow

AgBr

— sparingly soluble

Iodine

yellow

AgI

— insoluble in

N H_{4} OH

⚠️ Watch out · Remove N and S first

If N or S is present, boil the extract with conc.

HN O_{3}

before the silver test — this destroys

C N^{-}

and

S^{2 -}

, which would otherwise precipitate with

A g^{+}

and give a false halogen result.

Test for phosphorus

▸Heat with an oxidising agent (sodium peroxide) — P is oxidised to $phosphate$ .
▸Boil with $HN O_{3}$ $+$ ammonium molybdate $\to$ $yellow$ ammonium phosphomolybdate $(N H_{4})_{3} P O_{4} \cdot 12 Mo O_{3}$ .

🚫 Examiner Trap · Examiner traps

(1) S: black PbS (lead acetate) or violet (nitroprusside). (2) N

+

S together

\to

NaSCN

\to

blood-red

[Fe(SCN)

]^{2 +}

and NO Prussian blue. (3) AgX colours: AgCl white (soluble in N

H_{4}

OH), AgBr pale-yellow, AgI yellow (insoluble). (4) If N/S present, boil the extract with conc. HN

O_{3}

FIRST to expel CN

^{-}

S^{2 -}

before the silver halide test.

Quantitative — Carbon, Hydrogen & Nitrogen

Liebig combustion train: the sample is burnt with oxygen and copper oxide, the water formed is absorbed in an anhydrous calcium chloride U-tube and the carbon dioxide in a potassium hydroxide U-tube, with the percentage formulae — Liebig combustion: weigh the $H_{2} O$ and $C O_{2}$ absorbed.

Reaction

C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O

% Carbon

\frac{12 \times m _{C O_{2}} \times 100}{44 \times m}

% Hydrogen

\frac{2 \times m _{H_{2} O} \times 100}{18 \times m}

★ Remember · Two absorbents in series

H_{2} O

is absorbed in

anhydrous

CaC l_{2}

and

C O_{2}

conc. KOH

. The gain in mass of each U-tube gives

m_{H_{2} O}

and

m_{C O_{2}}

(

m =

mass of sample).

Dumas % N

\frac{28 \times V \times 100}{22400 \times m}

(

V =

vol.

N_{2}

at STP)

Kjeldahl % N

\frac{1.4 \times M \times 2 ( V - \frac{V _{1}}{2} )}{m}

⚠️ Watch out · Kjeldahl limitation

Kjeldahl fails for N in

nitro, azo

groups or a

ring

(e.g. pyridine) — that N does not convert to

(N H_{4})_{2} S O_{4}

. Use Dumas for those.

🚫 Examiner Trap · Examiner traps

(1) Combustion:

H_{2}

O absorbed in anhydrous CaC

l_{2}

, C

O_{2}

in KOH (mass gain of each tube). (2)

%

= \frac{12 m _{C O_{2}} \times 100}{44 m}

%

= \frac{2 m _{H_{2} O} \times 100}{18 m}

— note the 12/44 and 2/18 factors. (3) Dumas measures

N_{2}

gas; Kjeldahl measures N

H_{3}

. (4)

Kjeldahl fails

for N in nitro, azo or ring (pyridine) — use Dumas there.

Quantitative — Halogens, S, P, O & Formula

Converting an empirical formula to the molecular formula (n= molar mass div empirical-formula mass). — Converting an empirical formula to the molecular formula ( $n =$ molar mass $\div$ empirical-formula mass).

Element	Method	% formula
Halogen (Carius)	fuming $HN O_{3} + AgN O_{3} \to$ AgX	$% X = \frac{at.mass X \times m _{1} \times 100}{mol.mass AgX \times m}$
Sulphur (Carius)	oxidise $\to H_{2} S O_{4} \to BaS O_{4}$	$% S = \frac{32 m _{1} \times 100}{233 m}$
Phosphorus	$\to M g_{2} P_{2} O_{7}$ (mass 222)	$% P = \frac{62 m _{1} \times 100}{222 m}$
Oxygen	by difference	$% O = 100 - \sum (others)$

★ Remember · Carius method

Heat the compound with fuming

HN O_{3}

in a sealed

Carius tube

. For halogens,

AgN O_{3}

gives

AgX

; for sulphur,

BaC l_{2}

gives

BaS O_{4}

— each weighed.

% Halogen

\frac{at. mass X \times m _{AgX} \times 100}{mol. mass AgX \times m}

% Sulphur

\frac{32 \times m _{BaS O_{4}} \times 100}{233 \times m}

% Phosphorus

\frac{62 \times m _{M g_{2} P_{2} O_{7}} \times 100}{222 \times m}

🎯 Exam · Oxygen — usually by difference

% O = 100 - \sum (other-element %)

. It can also be found directly: heat in

N_{2}

, pass products over red-hot coke (

O \to CO

), then

I_{2} O_{5}

I_{2} O_{5} + 5 CO \to I_{2} + 5 C O_{2}

Empirical → molecular formula

▸Divide each element's $%$ by its $atomic mass$ → relative moles.
▸Divide by the smallest → simplest whole-number ratio $=$ $empirical formula (EF)$ .
▸ $n = \frac{molar mass}{EF mass}$ ; then $molecular formula = (EF)_{n}$ .

★ Remember · Modern microanalysis

CHN elemental analyser

determines C, H and N together from just 1–3 mg of sample and displays the percentages directly.

🚫 Examiner Trap · Examiner traps

(1)

Carius

: heat with fuming HN

O_{3}

in a sealed tube; weigh AgX (halogen) or BaS

O_{4}

(sulphur). (2) Use the right ratio:

%

X uses (at.mass X)/(mol.mass AgX);

%

S uses 32/233. (3) Oxygen is usually found

by difference

(

100 - \sum

others). (4) Empirical

\to

molecular: divide

%

by atomic mass, then by smallest;

n =

molar mass / EF mass.

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What are the most important Purification & Characterisation of Organic Compounds formulas for NEET?

This Purification & Characterisation of Organic Compounds formula sheet covers all the high-yield Chemistry formulas, definitions and facts you need for NEET, across Purification — crystallisation, sublimation, distillation, differential extraction, chromatography, Qualitative analysis — detection of nitrogen, sulphur, phosphorus and halogens, Quantitative analysis — estimation of carbon, hydrogen and nitrogen, Calculation of empirical and molecular formulae — each shown with the key result and, where useful, a worked example.

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