JEE Mains Chemistry · Organic Chemistry - Some Basic Principles and Techniques

Lassaigne's Test and Estimation of Nitrogen

Fusing an organic compound with sodium turns its nitrogen, sulphur and halogens into cyanide, sulphide and halide ions that give coloured tests; nitrogen is then measured either as N₂ gas (Dumas' method) or as ammonia (Kjeldahl's method).

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Why this matters

Thirty-eight PYQs, eleven of them asking for a number, and three from 2026. Eighteen are about Lassaigne's test: which elements it detects, the colour and formula of each product, why the extract is boiled with nitric acid, and which nitrogen compounds fail it. Twelve are about Dumas' method, mostly a volume of N₂ turned into a percentage. Eight are about Kjeldahl's method, where the traps are the dibasic sulphuric acid and the compounds the method cannot handle.

Concept 1 of 3: Lassaigne's test for nitrogen, sulphur, halogens and phosphorus

Nitrogen, sulphur and halogens bonded covalently in an organic molecule give none of the usual ionic tests. Fusing the compound with sodium metal turns them into ionic NaCN, Na₂S and NaX. The fused mass is boiled with water, and this sodium fusion extract is then tested for each ion.

Definition

Sodium fusion: $\mathrm{Na + C + N \to NaCN}$ ; $\mathrm{2Na + S \to Na_2S}$ ; $\mathrm{Na + X \to NaX}$ . Phosphorus is detected as phosphate after oxidation.
NaCN forms only when the compound contains carbon as well as nitrogen. Hydrazine, hydroxylamine and ammonium salts have no carbon and fail the test; urea, glycine and phenylhydrazine pass.
With both N and S present, the fusion can give NaSCN, which turns blood red with $\mathrm{Fe^{3+}}$ and gives no Prussian blue. With excess sodium it breaks down: $\mathrm{NaSCN + 2Na \to NaCN + Na_2S}$ .
Before the halogen test, boil the extract with dilute $\mathrm{HNO_3}$ to decompose NaCN and $\mathrm{Na_2S}$ ; otherwise they give their own precipitates with $\mathrm{AgNO_3}$ . Never use HCl, which adds chloride.
Iodine can also be shown by adding $\mathrm{CHCl_3}$ and chlorine water: the $\mathrm{CHCl_3}$ layer turns violet.
Carbon and hydrogen are detected differently, by heating with CuO: $\mathrm{CO_2}$ turns lime water milky and water turns anhydrous $\mathrm{CuSO_4}$ blue. Oxygen is not detected by Lassaigne's test.

Element	In the extract as	Reagent	Positive result
Nitrogen	NaCN	$\mathrm{FeSO_4}$ , boil, then conc. $\mathrm{H_2SO_4}$	Prussian blue, $\mathrm{Fe_4[Fe(CN)_6]_3}$
Sulphur	$\mathrm{Na_2S}$	Sodium nitroprusside	Violet, $\mathrm{Na_4[Fe(CN)_5NOS]}$
Sulphur	$\mathrm{Na_2S}$	Ethanoic acid and lead acetate	Black precipitate of PbS
Nitrogen and sulphur together	NaSCN	$\mathrm{Fe^{3+}}$ (iron(III) chloride)	Blood red, $\mathrm{[Fe(SCN)]^{2+}}$ With excess sodium the thiocyanate breaks down, and the separate cyanide and sulphide tests work instead.
Chlorine	NaCl	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	White AgCl, soluble in ammonia
Bromine	NaBr	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	Pale yellow AgBr, sparingly soluble in ammonia
Iodine	NaI	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	Yellow AgI, insoluble in ammonia
Phosphorus	$\mathrm{Na_3PO_4}$ (after oxidation with $\mathrm{Na_2O_2}$ )	$\mathrm{HNO_3}$ and ammonium molybdate	Yellow $\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}$

The fusion extract detects nitrogen, sulphur, the halogens and phosphorus, and nothing else.

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The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q43Moderate

Example 1 · Organic Chemistry - Some Basic Principles and Techniques · Lassaigne's Test and Estimation of Nitrogen

Given below are two statements: Statement I: In 'Lassaigne's Test, when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed. Statement II: If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give

NaCN

and

Na_{2}\text{ }S

. In the light of the above statements, choose the most appropriate answer from the options given below:

No carbon, no cyanide

The nitrogen test depends on NaCN, which needs carbon from the compound. Hydrazine and hydroxylamine contain nitrogen but no carbon, so they give no Prussian blue.

Nitric acid, not hydrochloric acid

The extract is acidified with

\mathrm{HNO_3}

before adding

\mathrm{AgNO_3}

. HCl would add chloride and give white AgCl whatever the compound contained.

Lassaigne's test cannot detect oxygen

The fusion extract shows N, S, halogens and P. A set of elements that includes oxygen or carbon is not what the sodium fusion extract detects.

Drill 17 more on lassaigne's test for nitrogen, sulphur, halogens and phosphorus

Concept 2 of 3: Dumas method for nitrogen

Heat the compound with copper(II) oxide and all its nitrogen comes off as nitrogen gas. Collect the gas, correct its volume to STP, and its mass gives the percentage of nitrogen directly.

Definition

The compound is heated with CuO in an atmosphere of $\mathrm{CO_2}$ : carbon gives $\mathrm{CO_2}$ , hydrogen gives water and nitrogen gives $\mathrm{N_2}$ , with some oxides of nitrogen.
The gases pass over heated copper gauze, which reduces the oxides of nitrogen to $\mathrm{N_2}$ .
The $\mathrm{N_2}$ is collected over KOH solution, which absorbs the $\mathrm{CO_2}$ .
Gas collected over an aqueous solution is wet: its own pressure is the measured pressure minus the aqueous tension, f.
CuO needed for $\mathrm{C_xH_yN_z}$ : $(2x + y/2)$ mol per mole of compound.
Dumas' method works for every kind of nitrogen compound.

Percentage of nitrogen (Dumas)

V_{\mathrm{STP}} = \dfrac{(p - f)\,V}{T} \times \dfrac{273}{760} \qquad \%\,\mathrm{N} = \dfrac{28}{22400} \times \dfrac{V_{\mathrm{STP}}\,(\text{mL})}{m\,(\text{g})} \times 100

Worked example

In Dumas' method, 0.25 g of an organic compound gives 40 mL of nitrogen collected at 290 K and 740 mm Hg. The aqueous tension at 290 K is 20 mm Hg. Find the percentage of nitrogen.

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JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q49Moderate

Example 2 · Organic Chemistry - Some Basic Principles and Techniques · Lassaigne's Test and Estimation of Nitrogen

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is

\_\_\_\_

% (Aqueous tension at 300 K is 15 mm ).

Subtract the aqueous tension first

The collected nitrogen is saturated with water vapour. Use the measured pressure minus the aqueous tension before reducing the volume to STP; forgetting it raises the answer by a few per cent.

Nitrogen gas is 28, not 14

22400 mL of

\mathrm{N_2}

at STP weighs 28 g. Using 14 g halves the percentage.

Copper gauze, not copper oxide, reduces the oxides

Copper(II) oxide oxidises the compound; the heated copper gauze further along reduces any oxides of nitrogen back to

\mathrm{N_2}

Drill 11 more on dumas method for nitrogen

Concept 3 of 3: Kjeldahl method for nitrogen

Digest the compound so that its nitrogen becomes ammonium sulphate, release the ammonia with alkali, and catch it in a known amount of standard acid. The acid used up tells you the moles of ammonia, and each ammonia molecule carries one nitrogen atom.

Definition

The compound is heated with concentrated $\mathrm{H_2SO_4}$ and a little $\mathrm{CuSO_4}$ , which acts as a catalyst. Its nitrogen becomes $\mathrm{(NH_4)_2SO_4}$ .
Excess NaOH releases $\mathrm{NH_3}$ , which is distilled into a known volume of standard acid. The acid left over is titrated with alkali.
Moles of N = moles of $\mathrm{NH_3}$ = moles of acid used × basicity (2 for $\mathrm{H_2SO_4}$ , 1 for HCl).
The method fails for nitro and azo compounds and for nitrogen in a ring (pyridine): their nitrogen does not all turn into ammonium sulphate.
In the formula, V is the volume of acid actually neutralised by the ammonia, in mL, and M its molarity.

Percentage of nitrogen (Kjeldahl)

\%\,\mathrm{N} = \dfrac{1.4 \times M \times V \times b}{m} \qquad (b = \text{basicity of the acid},\ V \text{ in mL},\ m \text{ in g})

Worked example

In Kjeldahl's method, the ammonia from 0.60 g of an organic compound neutralises 10 mL of 0.5 M

\mathrm{H_2SO_4}

. Find the percentage of nitrogen.

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The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q31Moderate

Example 3 · Organic Chemistry - Some Basic Principles and Techniques · Lassaigne's Test and Estimation of Nitrogen

When 1 g of compound ( X ) is subjected to Kjeldahl's method for estimation of nitrogen, 15

mL,1MH_{2}SO_{4}

was neutralized by ammonia evolved. The percentage of nitrogen in compound

(X)

is :

Sulphuric acid is dibasic

One mole of

\mathrm{H_2SO_4}

neutralises two moles of

\mathrm{NH_3}

. Forgetting the factor of 2 gives exactly half the right percentage, and that value is always among the options.

Not for nitro, azo or ring nitrogen

Kjeldahl's method does not work for nitro and azo compounds or for pyridine, because their nitrogen is not fully converted into ammonium sulphate. Use Dumas' method for them.

Use the acid actually neutralised

When a question gives the total acid and the excess titrated back with alkali, the ammonia used only the difference. Subtract before you multiply by the basicity.

Drill 7 more on kjeldahl method for nitrogen

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Dumas method for nitrogen
Percentage of nitrogen (Dumas)
$V_{\mathrm{STP}} = \dfrac{(p - f)\,V}{T} \times \dfrac{273}{760} \qquad \%\,\mathrm{N} = \dfrac{28}{22400} \times \dfrac{V_{\mathrm{STP}}\,(\text{mL})}{m\,(\text{g})} \times 100$
Kjeldahl method for nitrogen
Percentage of nitrogen (Kjeldahl)
$\%\,\mathrm{N} = \dfrac{1.4 \times M \times V \times b}{m} \qquad (b = \text{basicity of the acid},\ V \text{ in mL},\ m \text{ in g})$

Reference tables (1)

Lassaigne's test for nitrogen, sulphur, halogens and phosphorus8 rows

Element	In the extract as	Reagent	Positive result
Nitrogen	NaCN	$\mathrm{FeSO_4}$ , boil, then conc. $\mathrm{H_2SO_4}$	Prussian blue, $\mathrm{Fe_4[Fe(CN)_6]_3}$
Sulphur	$\mathrm{Na_2S}$	Sodium nitroprusside	Violet, $\mathrm{Na_4[Fe(CN)_5NOS]}$
Sulphur	$\mathrm{Na_2S}$	Ethanoic acid and lead acetate	Black precipitate of PbS
Nitrogen and sulphur together	NaSCN	$\mathrm{Fe^{3+}}$ (iron(III) chloride)	Blood red, $\mathrm{[Fe(SCN)]^{2+}}$ With excess sodium the thiocyanate breaks down, and the separate cyanide and sulphide tests work instead.
Chlorine	NaCl	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	White AgCl, soluble in ammonia
Bromine	NaBr	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	Pale yellow AgBr, sparingly soluble in ammonia
Iodine	NaI	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	Yellow AgI, insoluble in ammonia
Phosphorus	$\mathrm{Na_3PO_4}$ (after oxidation with $\mathrm{Na_2O_2}$ )	$\mathrm{HNO_3}$ and ammonium molybdate	Yellow $\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}$

The fusion extract detects nitrogen, sulphur, the halogens and phosphorus, and nothing else.

Watch out for (9)

No carbon, no cyanide
→ Lassaigne's test for nitrogen, sulphur, halogens and phosphorus
Nitric acid, not hydrochloric acid
→ Lassaigne's test for nitrogen, sulphur, halogens and phosphorus
Lassaigne's test cannot detect oxygen
→ Lassaigne's test for nitrogen, sulphur, halogens and phosphorus
Subtract the aqueous tension first
→ Dumas method for nitrogen
Nitrogen gas is 28, not 14
→ Dumas method for nitrogen
Copper gauze, not copper oxide, reduces the oxides
→ Dumas method for nitrogen
Sulphuric acid is dibasic
→ Kjeldahl method for nitrogen
Not for nitro, azo or ring nitrogen
→ Kjeldahl method for nitrogen
Use the acid actually neutralised
→ Kjeldahl method for nitrogen

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