Qualitative and Quantitative Analysis

Q: What is the difference between the Dumas and Kjeldahl methods?

In the Dumas method nitrogen is released as N₂ gas by heating with copper oxide, and its volume is measured and corrected to STP. In Kjeldahl's method nitrogen is converted to ammonium sulphate, released as ammonia with NaOH, absorbed in standard acid and estimated by back titration. Dumas works for all nitrogen compounds.

ChemistryPurification and Characterisation of Organic CompoundsFor JEE aspirants

Qualitative and quantitative analysis of an organic compound tells you which elements it contains and in what proportion. Carbon and hydrogen are detected with copper(II) oxide; nitrogen, sulphur, halogens and phosphorus through Lassaigne's sodium fusion test; functional groups through simple colour and precipitate tests. Each element is then estimated as something you can weigh or measure: $C O_{2}$ and $H_{2} O$ , $N_{2}$ or $N H_{3}$ , $AgX$ , $BaS O_{4}$ . Qualitative and quantitative analysis numericals, ending in an empirical and molecular formula, are regulars in NEET and JEE Main.

On this page1Detecting C and H2Lassaigne's test3Functional group tests4Estimating C and H5Estimating N6X, S, P and O7Percentages to formula

Key Formulas - Quick Reference

★ Must learn Liebig: $% C = \frac{12}{44} \times \frac{m _{CO_{2}}}{m} \times 100$ and $% H = \frac{2}{18} \times \frac{m _{H_{2} O}}{m} \times 100$
★ Must learn Dumas: $% N = \frac{28}{22400} \times \frac{V _{STP} ( mL )}{m} \times 100$ , with $V_{STP} = \frac{p _{1} V _{1} \times 273}{760 \times T _{1}}$ and $p_{1} = p_{atm} - aqueous tension$
★ Must learn Kjeldahl: $% N = \frac{1.4 \times M \times 2 ( V - \frac{V _{1}}{2} )}{m}$ (acid $V$ mL of molarity $M$ , NaOH $V_{1}$ mL of the same molarity)
★ Must learn Carius: $% X = \frac{at. mass of X}{M _{AgX}} \times \frac{m _{AgX}}{m} \times 100$ and $% S = \frac{32}{233} \times \frac{m _{BaSO_{4}}}{m} \times 100$
Phosphorus: $% P = \frac{62}{222} \times \frac{m _{Mg_{2} P_{2} O_{7}}}{m} \times 100$ or $\frac{31}{1877} \times \frac{m _{phosphomolybdate}}{m} \times 100$
Oxygen: $% O = 100 - (sum of the other percentages)$ ; directly, $% O = \frac{32}{88} \times \frac{m _{CO_{2}}}{m} \times 100$
★ Must learn Formula: moles $= %/ atomic mass$ , divide by the smallest to get the EF; $n = \frac{M}{EF mass}$ with $M = 2 \times vapour density$
Lassaigne colours: N Prussian blue; S black PbS or violet; N + S blood red; Cl, Br, I give white, pale yellow and yellow AgX

1. Detecting Carbon and Hydrogen

The compound is heated with dry copper(II) oxide. Carbon is oxidised to $C O_{2}$ , which turns lime water milky; hydrogen is oxidised to water, which turns white anhydrous copper sulphate blue.

C + 2CuO Δ 2Cu + C O_{2}

2H + CuO Δ Cu + H_{2} O

C O_{2} + Ca (OH)_{2} CaC O_{3} ↓ + H_{2} O

CuS O_{4} + 5 H_{2} O CuS O_{4} \cdot 5 H_{2} O

White anhydrous $CuS O_{4}$ becomes blue $CuS O_{4} \cdot 5 H_{2} O$ . Oxygen has no simple direct test; it is usually found by difference.

Figure 1: Detecting C and H. CuO oxidises carbon to

C O_{2}

(lime water turns milky) and hydrogen to

H_{2} O

(white anhydrous

CuS O_{4}

turns blue).

2. Lassaigne's Test: N, S, Halogens and P

Lassaigne's test: the compound is fused with metallic sodium so that nitrogen, sulphur and halogen, which are covalently bonded in the compound, become ionic ( $C N^{-}$ , $S^{2 -}$ , $X^{-}$ ). The fused mass is boiled with distilled water and filtered; the filtrate is the sodium fusion extract (SFE).

Figure 2: Sodium fusion. Sodium converts covalent N, S and halogen into

NaCN

N a_{2} S

and

NaX

, which dissolve in water as ions and can be tested like inorganic ions.

Na + C + N Δ NaCN

2Na + S Δ N a_{2} S

Na + X Δ NaX

Here C, N, S and X come from the organic compound. The extract is alkaline (excess sodium forms NaOH).

2.1 Nitrogen: Prussian Blue

The extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. Cyanide first forms hexacyanidoferrate(II); some $F e^{2 +}$ is oxidised to $F e^{3 +}$ on heating, and the two combine into Prussian blue, iron(III) hexacyanidoferrate(II):

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

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

No carbon, no test: hydrazine ( $N_{2} H_{4}$ ) and hydroxylamine cannot form NaCN, so they give no Prussian blue. Fusing them with a little starch or sugar supplies the carbon.
Diazonium salts lose nitrogen as $N_{2}$ on heating before fusion is complete and often give a negative test.

2.2 Sulphur

(a) Lead acetate test: acidify the extract with acetic acid and add lead acetate; a black precipitate of lead sulphide shows sulphur. (b) Sodium nitroprusside test: a violet colour with the alkaline extract.

S^{2 -} + P b^{2 +} black PbS ↓

S^{2 -} + [Fe (CN)_{5} NO]^{2 -} violet [Fe (CN)_{5} NOS]^{4 -}

Acetic acid is used, not sulphuric acid, because sulphuric acid would itself precipitate white $PbS O_{4}$ with lead acetate and hide the result.

2.3 Nitrogen and Sulphur Together

When both are present, fusion with a limited amount of sodium gives sodium thiocyanate. There is no free cyanide, so no Prussian blue; instead $F e^{3 +}$ gives a blood red colour.

Na + C + N + S Δ NaSCN

F e^{3 +} + SC N^{-} blood red [Fe (SCN)]^{2 +}

With excess sodium the thiocyanate splits, and the extract then gives the separate tests for N and S:

NaSCN + 2Na NaCN + N a_{2} S

2.4 Halogens

The extract is acidified with nitric acid, boiled, and treated with silver nitrate. Boiling with $HN O_{3}$ is essential when N or S is present: it drives off cyanide and sulphide as HCN and $H_{2} S$ , which would otherwise give AgCN (white) or $A g_{2} S$ (black) and spoil the test.

NaCN + HN O_{3} NaN O_{3} + HCN ↑

N a_{2} S + 2HN O_{3} 2NaN O_{3} + H_{2} S ↑

X^{-} + A g^{+} AgX ↓

Halogen	Precipitate	In $N H_{3} (aq)$
Chlorine	AgCl, white (curdy)	soluble: forms $[Ag (N H_{3})_{2}]^{+}$
Bromine	AgBr, pale yellow	sparingly soluble
Iodine	AgI, yellow	insoluble
Fluorine	no precipitate (AgF is soluble)	not applicable

Beilstein test (quick check): a copper wire heated with the compound gives a green or blue-green flame (volatile copper halide). It does not detect F, does not say which halogen, and gives false positives with urea and thiourea.
Layer test for Br and I: add chlorine water and $CC l_{4}$ (or $C S_{2}$ ) to the acidified extract; $B r_{2}$ colours the organic layer orange-brown, $I_{2}$ colours it violet.

2.5 Phosphorus

The compound is heated with an oxidising agent, sodium peroxide, which converts P into phosphate. The solution is boiled with nitric acid and ammonium molybdate is added: a yellow colour or precipitate of ammonium phosphomolybdate shows phosphorus.

N a_{3} P O_{4} + 3HN O_{3} H_{3} P O_{4} + 3NaN O_{3}

H_{3} P O_{4} + 12 (N H_{4})_{2} Mo O_{4} + 21HN O_{3} ⟶ (N H_{4})_{3} P O_{4} \cdot 12Mo O_{3} ↓ + 21N H_{4} N O_{3} + 12 H_{2} O

Figure 3: Reading Lassaigne's test. Each ion formed in the fusion has its own colour test; for halogens, the colour and the solubility of

AgX

in ammonia identify Cl, Br or I.

Figure 4: The colour code examiners use. Blue for N, black or violet for S, blood red for N and S together; the silver halides darken from white to yellow as their solubility in ammonia falls.

Exam Trick

Blue for N, Black for S, Blood-red for both. For halides remember "white, pale yellow, yellow" for Cl, Br, I: the colour deepens and the solubility in ammonia falls down the group.

Key idea

Sodium fusion turns covalent N, S and X into ions; after that, every test is an ordinary inorganic ion test on the extract.

Quick Recall: tap to check

Why is the compound fused with metallic sodium?

To convert covalently bonded N, S and halogens into ionic NaCN,

N a_{2} S

and NaX, which dissolve in water and give ionic tests.

Why is nitric acid added to the extract before silver nitrate?

To decompose NaCN and

N a_{2} S

(as HCN and

H_{2} S

); otherwise AgCN and

A g_{2} S

would precipitate and interfere.

Why is acetic acid, not sulphuric acid, used in the lead acetate test?

Sulphuric acid would precipitate white

PbS O_{4}

with lead acetate and mask the black PbS.

Will

CC l_{4}

give a white precipitate with silver nitrate?

No. The chlorine in

CC l_{4}

is covalently bonded, so there are no free chloride ions; it must first be fused with sodium.

3. Detecting Functional Groups

Once the elements are known, simple tests identify the functional group. Test in a sensible order: acids first, then phenols, carbonyl compounds, alcohols and amines.

Group	Test	Positive result
Carboxylic acid, –COOH	aqueous $NaHC O_{3}$	brisk effervescence of $C O_{2}$ (turns lime water milky)
Phenolic –OH	neutral $FeC l_{3}$ solution; bromine water	violet, blue or green colour; white precipitate of 2,4,6-tribromophenol
Alcoholic –OH	sodium metal; ceric ammonium nitrate	bubbles of $H_{2}$ ; red colour
Alcohol class	Lucas reagent (anhydrous $ZnC l_{2}$ + conc. HCl)	3°: turbid at once; 2°: in about 5 min; 1°: no turbidity at room temperature
Carbonyl, >C=O	2,4-dinitrophenylhydrazine (Brady's reagent)	yellow, orange or red precipitate
Aldehyde, –CHO	Tollens' reagent; Fehling's solution; Schiff's reagent	silver mirror; red $C u_{2} O$ (aliphatic aldehydes only); pink colour
Ketone	no Tollens' or Fehling's; sodium nitroprusside + NaOH	red colour (methyl ketones); iodoform test for $C H_{3} CO$ -
1° amine, – $N H_{2}$	$CHC l_{3}$ + alcoholic KOH (carbylamine test)	offensive smell of an isocyanide
Aromatic 1° amine	$NaN O_{2}$ + HCl at 273-278 K, then alkaline 2-naphthol	orange-red azo dye
Nitro, – $N O_{2}$	Zn dust + $N H_{4} Cl$ , then Tollens' (Mulliken-Barker test)	silver mirror

Key equations behind the tests:

RCOOH + NaHC O_{3} RCOONa + H_{2} O + C O_{2} ↑

C_{6} H_{5} OH + 3B r_{2} white ppt C_{6} H_{2} B r_{3} OH ↓ + 3HBr

RCHO + 2 [Ag (N H_{3})_{2}]^{+} + 3O H^{-} silver mirror RCO O^{-} + 2Ag ↓ + 4N H_{3} + 2 H_{2} O

RCHO + 2C u^{2 +} + 5O H^{-} red ppt RCO O^{-} + C u_{2} O ↓ + 3 H_{2} O

RN H_{2} + CHC l_{3} + 3KOH isocyanide RNC + 3KCl + 3 H_{2} O

RN O_{2} + 4 [H] Zn/ N H_{4} Cl RNHOH + H_{2} O

In the last test the hydroxylamine formed reduces Tollens' reagent, so a nitro compound finally gives a silver mirror. Hinsberg's reagent (benzenesulphonyl chloride) separates 1°, 2° and 3° amines.

Figure 5: Identify the group by testing in order of acidity first (

NaHC O_{3}

before

FeC l_{3}

), then carbonyl, then alcohol and amine; each positive result is confirmed by a second test.

Tollens' reagent

Ammoniacal $AgN O_{3}$ . All aldehydes, aliphatic and aromatic, give a silver mirror. Also given by formic acid and reducing sugars.

Fehling's solution

Alkaline $C u^{2 +}$ with tartrate. Aliphatic aldehydes give red $C u_{2} O$ ; aromatic aldehydes such as benzaldehyde do not.

Key idea

Each functional group has one quick test and one confirming test; examiners like pairs that separate look-alikes: phenol vs alcohol (

FeC l_{3}

), aldehyde vs ketone (Tollens'), aliphatic vs aromatic aldehyde (Fehling's).

Quick Recall: tap to check

How do you tell phenol from ethanol?

Phenol gives a violet colour with neutral

FeC l_{3}

and a white precipitate with bromine water; ethanol gives neither.

How do you tell an aldehyde from a ketone?

Aldehydes reduce Tollens' reagent (silver mirror); ketones do not.

Which amines give the carbylamine test?

Only primary amines, aliphatic and aromatic.

4. Estimating Carbon and Hydrogen (Liebig Method)

A weighed sample (m g) is burnt in a stream of dry oxygen over heated copper(II) oxide. Carbon becomes $C O_{2}$ and hydrogen becomes water:

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

The water is absorbed in a weighed U-tube of anhydrous $CaC l_{2}$ (gain $m_{1}$ ) and the $C O_{2}$ in a weighed U-tube of concentrated KOH (gain $m_{2}$ ), connected in that order.

Figure 6: Estimating C and H. The gain in mass of the

CaC l_{2}

tube is the water (

m_{1}

) and of the KOH tube the

C O_{2}

(

m_{2}

). The

CaC l_{2}

tube must come first, because KOH solution would also absorb water.

44 g of $C O_{2}$ contains 12 g of C and 18 g of water contains 2 g of H, so

% C = \frac{12 \times m _{2} \times 100}{44 \times m} % H = \frac{2 \times m _{1} \times 100}{18 \times m}

KOH is used for $C O_{2}$ because $C O_{2}$ is acidic: $2KOH + C O_{2} ⟶ K_{2} C O_{3} + H_{2} O$ .

Exam Trick

C is 3/11 of $C O_{2}$ , H is 1/9 of water. These two fractions save a step in every combustion numerical:

% C = \frac{3}{11} \times \frac{m _{2}}{m} \times 100 % H = \frac{1}{9} \times \frac{m _{1}}{m} \times 100

5. Estimating Nitrogen

5.1 Dumas Method

The compound is heated with copper(II) oxide in an atmosphere of $C O_{2}$ . Nitrogen is set free as $N_{2}$ :

C_{x} H_{y} N_{z} + (2 x + \frac{y}{2}) CuO ⟶ x C O_{2} + \frac{y}{2} H_{2} O + \frac{z}{2} N_{2} + (2 x + \frac{y}{2}) Cu

Any oxides of nitrogen are reduced to $N_{2}$ over a heated copper gauze. The gases are collected over KOH solution in a nitrometer: KOH absorbs $C O_{2}$ , so only $N_{2}$ collects.

Figure 7: Dumas method. All the nitrogen is freed as

N_{2}

; KOH removes the

C O_{2}

, so the gas volume read in the nitrometer is nitrogen (saturated with water vapour, so subtract the aqueous tension).

Pressure of dry $N_{2}$ : $p_{1} = p_{atm} - aqueous tension$ (the gas is collected over an aqueous solution).
Convert to STP: $V = \frac{p _{1} V _{1} \times 273}{760 \times T _{1}}$ mL.
Mass of N: 22400 mL of $N_{2}$ at STP weighs 28 g, so $V$ mL weighs $\frac{28 V}{22400}$ g.
$% N = \frac{28 \times V \times 100}{22400 \times m}$ .

5.2 Kjeldahl's Method

The compound is digested with concentrated $H_{2} S O_{4}$ (with $K_{2} S O_{4}$ to raise the boiling point and $CuS O_{4}$ as catalyst), which turns its nitrogen into ammonium sulphate. The digest is heated with excess NaOH, and the ammonia is absorbed in a known excess of standard $H_{2} S O_{4}$ . The acid left over is titrated with standard NaOH.

Organic compound + H_{2} S O_{4} (N H_{4})_{2} S O_{4}

(N H_{4})_{2} S O_{4} + 2NaOH N a_{2} S O_{4} + 2N H_{3} + 2 H_{2} O

2N H_{3} + H_{2} S O_{4} (N H_{4})_{2} S O_{4}

Figure 8: Kjeldahl's method. N becomes

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

, the

N H_{3}

released by NaOH is caught in a known excess of acid, and back titration shows how much acid the ammonia used.

Let $V$ mL of $H_{2} S O_{4}$ of molarity $M$ be taken and $V_{1}$ mL of NaOH of the same molarity be needed for the excess acid. Since NaOH is monoacidic and $H_{2} S O_{4}$ dibasic, $V_{1}$ mL of NaOH neutralise $V_{1} /2$ mL of the acid, so $(V - V_{1} /2)$ mL of acid reacted with ammonia. This acid neutralises $2 (V - V_{1} /2)$ mL of $N H_{3}$ solution of molarity $M$ , and 1000 mL of 1 M $N H_{3}$ contains 14 g N:

% N = \frac{14 \times M \times 2 ( V - \frac{V _{1}}{2} )}{1000} \times \frac{100}{m} = \frac{1.4 \times M \times 2 ( V - \frac{V _{1}}{2} )}{m}

Figure 9: Back titration in numbers (Solved Example 8). Acid taken

=

acid used by

N H_{3}

+

acid left; NaOH counts the acid left, and each

H_{2} S O_{4}

takes two

N H_{3}

, giving

46.7%

Where Kjeldahl fails: nitrogen in nitro and azo groups and in rings (pyridine, quinoline) is not converted into ammonium sulphate under these conditions, so these compounds need the Dumas method.

Dumas method

N is measured as a volume of $N_{2}$ gas. Works for all nitrogen compounds. Needs pressure, temperature and aqueous-tension corrections.

Kjeldahl's method

N is measured as $N H_{3}$ by titration. Fast, suits many samples (soil, fertilisers, food proteins). Fails for nitro, azo and ring nitrogen.

Exam Trick

Kjeldahl in one line: $% N = \frac{1.4 \times meq}{m}$ , where meq is the milliequivalents of acid used up by $N H_{3}$ . With normalities, $meq = N_{a} V_{a} - N_{b} V_{b}$ (acid taken minus base used in the back titration).

Key idea

Dumas counts nitrogen molecules as gas; Kjeldahl counts them as ammonia by titration. In both, the numerical is moles of N times 14 over the sample mass.

Quick Recall: tap to check

Why is a KOH solution used to absorb carbon dioxide?

C O_{2}

is acidic and reacts completely with KOH to form

K_{2} C O_{3}

Why must the

CaC l_{2}

tube come before the KOH tube in the Liebig method?

KOH solution would absorb water as well as

C O_{2}

, making %C too high and %H too low.

Why is the aqueous tension subtracted in the Dumas method?

The

N_{2}

is collected over an aqueous solution, so the gas is saturated with water vapour; only

p_{atm} - aqueous tension

is due to

N_{2}

Why can nitrobenzene not be analysed by Kjeldahl's method?

Nitro nitrogen is not converted into ammonium sulphate on digestion, so the result would be too low.

6. Estimating Halogens, Sulphur, Phosphorus and Oxygen

6.1 Halogens: Carius Method

A weighed compound is heated with fuming nitric acid and silver nitrate in a sealed hard-glass Carius tube in a furnace. C and H are oxidised to $C O_{2}$ and water; the halogen forms AgX, which is filtered, washed, dried and weighed. One mole of AgX holds one mole of X:

% X = \frac{atomic mass of X \times m _{1} \times 100}{molar mass of AgX \times m}

Figure 10: Carius method. Fuming

HN O_{3}

destroys the organic part; the halogen is weighed as

AgX

and sulphur as

BaS O_{4}

6.2 Sulphur

The compound is heated in a Carius tube with fuming nitric acid (or sodium peroxide). Sulphur is oxidised to sulphuric acid, which is precipitated as barium sulphate with excess barium chloride:

H_{2} S O_{4} + BaC l_{2} BaS O_{4} ↓ + 2HCl

233 g of $BaS O_{4}$ contains 32 g of S, so $% S = \frac{32 \times m _{1} \times 100}{233 \times m}$ .

6.3 Phosphorus

Heating with fuming nitric acid oxidises P to phosphoric acid. It is weighed either as ammonium phosphomolybdate, $(N H_{4})_{3} P O_{4} \cdot 12Mo O_{3}$ (molar mass 1877 g/mol), after adding ammonia and ammonium molybdate, or as $M g_{2} P_{2} O_{7}$ (222 g/mol) after precipitating $MgN H_{4} P O_{4}$ with magnesia mixture and igniting it:

2MgN H_{4} P O_{4} Δ M g_{2} P_{2} O_{7} + 2N H_{3} + H_{2} O

% P = \frac{31 \times m _{1} \times 100}{1877 \times m} or % P = \frac{62 \times m _{1} \times 100}{222 \times m}

(62 g is the mass of the two P atoms in one mole of $M g_{2} P_{2} O_{7}$ .)

6.4 Oxygen

Usually by difference: $% O = 100 - (sum of all other percentages)$ . It can also be found directly. The compound is decomposed by heating in a stream of $N_{2}$ ; the gaseous products are passed over red-hot coke (1373 K), which turns all the oxygen into CO; the CO is then passed over warm iodine pentoxide:

2C + O_{2} 1373 K 2CO

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

Multiplying the first equation by 5 and the second by 2 shows that each mole of $O_{2}$ gives two moles of $C O_{2}$ : 32 g of oxygen gives 88 g of $C O_{2}$ , so $\%\text{O} = \dfrac{32 \times m_1 \times 100}{88 \times m}$ (the liberated iodine can also be titrated). Today C, H and N are measured together on 1-3 mg of sample in an automatic CHN elemental analyser.

Figure 11: Every gravimetric formula is (element fraction of the weighed product)

\times m_{1} \times 100/ m

. Note how little P there is in ammonium phosphomolybdate (1.65%), which makes that weighing very sensitive.

7. From Percentages to Empirical and Molecular Formula

Find every percentage (oxygen by difference if it is not measured).
Divide each percentage by the atomic mass: this gives the relative number of moles.
Divide by the smallest value; if a ratio ends near .5, .33 or .25, multiply all by 2, 3 or 4 to get whole numbers. This is the empirical formula (EF).
Find the molar mass M (for a vapour, $M = 2 \times$ vapour density).
$n = \frac{M}{EF mass}$ ; molecular formula $= (EF)_{n}$ .

Figure 12: The formula pipeline (Solved Example 6). Divide each percentage by the atomic mass, scale to the smallest, round to whole numbers for the EF, then multiply by

n = M / EF mass

(with

M = 2 \times VD

JEE Advanced

Molar mass of an acid from its silver salt. The acid is converted into its silver salt, which is ignited to leave pure silver ( $RCOOAg ⟶ Ag$ ). If $w$ g of silver salt leaves $x$ g of Ag, the equivalent mass of the salt is $\frac{108 w}{x}$ ; replacing Ag (108) by H (1) gives the equivalent mass of the acid:

E_{acid} = \frac{108 w}{x} - 107, M = n \times E_{acid} (n = basicity)

Example: 0.458 g of the silver salt of a monobasic acid leaves 0.216 g of silver, so

E = \frac{108 \times 0.458}{0.216} - 107 = 229 - 107 = 122

which fits benzoic acid, M = 122. Bases are handled the same way through their chloroplatinates $(BH)_{2} PtC l_{6}$ , which leave Pt (195) on ignition; for a monoacidic base

M = \frac{1}{2} (\frac{195 w}{x} - 410)

Key idea

Percent, divide by atomic mass, divide by the smallest, round, then scale by

n = M / EF mass

: the same five steps finish every quantitative-analysis numerical.

Figure 13: The whole topic in one view: detect the elements, identify the groups, estimate each element as a weighable product, then convert percentages into a formula.

8. Solved Examples

Solved Example 1

A 0.246 g sample of an organic compound burns completely to give 0.198 g of

C O_{2}

and 0.1014 g of water. Find the percentages of carbon and hydrogen.

Solution:

% C = \frac{12 \times 0.198 \times 100}{44 \times 0.246} = 21.95%

% H = \frac{2 \times 0.1014 \times 100}{18 \times 0.246} = 4.58%

Solved Example 2

In a Dumas estimation, 0.3 g of a compound gave 50 mL of nitrogen collected at 300 K and 715 mm Hg. The aqueous tension at 300 K is 15 mm. Find the percentage of nitrogen.

Solution:

Pressure of dry $N_{2}$ $= 715 - 15 = 700$ mm.

V_{STP} = \frac{700 \times 50 \times 273}{760 \times 300} = 41.9 mL

% N = \frac{28 \times 41.9 \times 100}{22400 \times 0.3} = 17.46%

Solved Example 3

In a Kjeldahl estimation, the ammonia evolved from 0.5 g of a compound neutralised 10 mL of 1 M

H_{2} S O_{4}

. Find the percentage of nitrogen.

Solution:

10 mL of 1 M $H_{2} S O_{4}$ neutralises 20 mL of 1 M $N H_{3}$ (two $N H_{3}$ per $H_{2} S O_{4}$ ). 1000 mL of 1 M $N H_{3}$ contains 14 g N, so 20 mL contains $\dfrac{14 \times 20}{1000} = 0.28$ g N.

% N = \frac{0.28 \times 100}{0.5} = 56.0%

Solved Example 4

In a Carius estimation, 0.15 g of an organic compound gave 0.12 g of AgBr. Find the percentage of bromine.

Solution:

Molar mass of AgBr $= 108 + 80 = 188$ g/mol, so 188 g AgBr contains 80 g Br.

% Br = \frac{80 \times 0.12 \times 100}{188 \times 0.15} = 34.04%

Solved Example 5

In a sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. Find the percentage of sulphur.

Solution:

Molar mass of $BaS O_{4} = 137 + 32 + 64 = 233$ g/mol, containing 32 g S.

% S = \frac{32 \times 0.4813 \times 100}{233 \times 0.157} = 42.10%

Solved Example 6

A compound contains C 40.0%, H 6.67% and O 53.33%. Its vapour density is 30. Find its empirical and molecular formula.

Solution:

Moles: C $= 40.0/12 = 3.33$ ; H $= 6.67/1 = 6.67$ ; O $= 53.33/16 = 3.33$ . Dividing by 3.33 gives C : H : O $= 1 : 2 : 1$ , so EF $= C H_{2} O$ (EF mass 30).

$M = 2 \times 30 = 60$ , so $n = 60/30 = 2$ and the molecular formula is $C_{2} H_{4} O_{2}$ (for example, acetic acid). Figure 12 shows the same working.

Solved Example 7

0.2475 g of a compound containing C, H and O gave 0.495 g of

C O_{2}

and 0.2025 g of water on combustion. Its vapour density is 44. Find its molecular formula.

Solution:

% C = \frac{12 \times 0.495 \times 100}{44 \times 0.2475} = 54.55% % H = \frac{2 \times 0.2025 \times 100}{18 \times 0.2475} = 9.09%

$% O = 100 - (54.55 + 9.09) = 36.36%$ .

Moles: C $= 4.545$ , H $= 9.09$ , O $= 2.273$ ; ratio $2 : 4 : 1$ , so EF $= C_{2} H_{4} O$ (mass 44).

$M = 2 \times 44 = 88$ , $n = 88/44 = 2$ : molecular formula $C_{4} H_{8} O_{2}$ .

Solved Example 8

0.30 g of a compound was digested by Kjeldahl's method. The ammonia was absorbed in 50 mL of 0.25 M

H_{2} S O_{4}

, and the excess acid needed 30 mL of 0.5 M NaOH. Find %N and decide whether the compound is urea,

N H_{2} CON H_{2}

, or acetamide,

C H_{3} CON H_{2}

Solution:

Acid taken $= 50 \times 0.25 = 12.5$ mmol. NaOH $= 30 \times 0.5 = 15$ mmol, which neutralises 7.5 mmol of $H_{2} S O_{4}$ . Acid used by $N H_{3}$ $= 12.5 - 7.5 = 5$ mmol, so $N H_{3} = 10$ mmol and N $= 10 \times 14 = 140$ mg.

% N = \frac{0.140 \times 100}{0.30} = 46.7%

Urea has $\frac{28}{60} \times 100 = 46.7%$ N; acetamide has 23.7%. The compound is urea (Figure 9).

Solved Example 9

The sodium fusion extract of a compound gives a blood red colour with

FeC l_{3}

but no Prussian blue. The compound contains
(A) nitrogen only
(B) sulphur only
(C) both nitrogen and sulphur
(D) a halogen

Solution:

Answer: (C). N and S together form NaSCN, which gives blood red $[Fe (SCN)]^{2 +}$ . There is no free cyanide, so no Prussian blue. Thiourea, $N H_{2} CSN H_{2}$ , behaves this way.

Solved Example 10

Kjeldahl's method can be used to estimate nitrogen in
(A) nitrobenzene
(B) pyridine
(C) azobenzene
(D) acetamide

Solution:

Answer: (D). The amide nitrogen of acetamide becomes ammonium sulphate on digestion. Nitro (A), ring (B) and azo (C) nitrogen do not, so those need the Dumas method.

Solved Example 11

A compound gives an orange precipitate with 2,4-DNP and a silver mirror with Tollens' reagent, but no red precipitate with Fehling's solution. It is
(A) acetone
(B) acetaldehyde
(C) benzaldehyde
(D) acetic acid

Solution:

Answer: (C). 2,4-DNP shows a carbonyl group and Tollens' shows an aldehyde. Aromatic aldehydes do not reduce Fehling's solution, so it is benzaldehyde. Acetaldehyde would also give Fehling's test; acetone fails Tollens'; acetic acid fails 2,4-DNP.

Practice Questions

Differentiate between the principle of estimating nitrogen by the Dumas method and by Kjeldahl's method.Answer: Dumas: N is released as $N_{2}$ gas and its volume is measured. Kjeldahl: N is converted to $(N H_{4})_{2} S O_{4}$ , released as $N H_{3}$ and estimated by acid-base titration.
State the principle of estimating halogens, sulphur and phosphorus in an organic compound.Answer: Oxidise with fuming $HN O_{3}$ (Carius): X is weighed as AgX, S as $BaS O_{4}$ , P as ammonium phosphomolybdate or $M g_{2} P_{2} O_{7}$ .
A compound contains 69% C and 4.8% H, the rest being O. What masses of $C O_{2}$ and water are formed when 0.20 g of it burns completely?Answer: C = 0.138 g gives 0.506 g $C O_{2}$ ; H = 0.0096 g gives 0.0864 g $H_{2} O$ .
0.50 g of a compound was treated by Kjeldahl's method. The ammonia was absorbed in 50 mL of 0.5 M $H_{2} S O_{4}$ , and the residual acid needed 60 mL of 0.5 M NaOH. Find %N.Answer: Acid used by $N H_{3}$ = 25 - 15 = 10 mmol, so $N H_{3}$ = 20 mmol and N = 0.28 g: 56.0%.
0.3780 g of an organic chloro compound gave 0.5740 g of AgCl in a Carius estimation. Find %Cl.Answer: $% Cl = \frac{35.5 \times 0.5740 \times 100}{143.5 \times 0.3780} = 37.57%$ .
In a Carius estimation of sulphur, 0.468 g of a compound gave 0.668 g of $BaS O_{4}$ . Find %S.Answer: $% S = \frac{32 \times 0.668 \times 100}{233 \times 0.468} = 19.60%$ .
In Lassaigne's test for nitrogen, the Prussian blue colour is due to (A) $N a_{4} [Fe (CN)_{6}]$ (B) $F e_{4} [Fe (CN)_{6}]_{3}$ (C) $F e_{2} [Fe (CN)_{6}]$ (D) $F e_{3} [Fe (CN)_{6}]_{4}$ .Answer: (B) $F e_{4} [Fe (CN)_{6}]_{3}$ , iron(III) hexacyanidoferrate(II).

Common Mistakes to Avoid

Watch out

Acidifying the extract with HCl or $H_{2} S O_{4}$ before adding $AgN O_{3}$ : HCl adds chloride itself. Use nitric acid, and boil to remove $C N^{-}$ and $S^{2 -}$ .
Using sulphuric acid in the lead acetate test: it precipitates white $PbS O_{4}$ . Use acetic acid.
Forgetting to subtract the aqueous tension, or to convert the Dumas volume to STP before using 22400 mL.
In Kjeldahl numericals, forgetting that one $H_{2} S O_{4}$ takes two $N H_{3}$ (or two NaOH): $V_{1}$ mL of NaOH cancels only $V_{1} /2$ mL of acid of the same molarity.
Placing the KOH tube before the $CaC l_{2}$ tube: KOH solution absorbs water too, so %C comes out high and %H low.
Expecting Prussian blue from hydrazine or hydroxylamine: without carbon no NaCN forms.
Using Fehling's test to show benzaldehyde: aromatic aldehydes give Tollens' test but not Fehling's.
Applying Kjeldahl's method to nitro, azo or ring nitrogen (nitrobenzene, azobenzene, pyridine): use Dumas.

Frequently Asked Questions

What is Lassaigne's test and why is sodium used?

Lassaigne's test detects nitrogen, sulphur, halogens and phosphorus. The compound is fused with sodium, which converts covalently bonded N, S and X into ionic sodium cyanide, sodium sulphide and sodium halide. These dissolve in water as the sodium fusion extract and give ordinary ionic tests.

Why is the sodium fusion extract boiled with nitric acid before adding silver nitrate?

If nitrogen or sulphur is present, the extract contains cyanide and sulphide ions, which would precipitate as silver cyanide and silver sulphide and be mistaken for silver halide. Boiling with nitric acid drives them off as HCN and $H_{2} S$ . Nitric acid is used because it adds no chloride.

Why does a compound with both nitrogen and sulphur give a blood red colour?

With a limited amount of sodium, N and S together form sodium thiocyanate instead of separate cyanide and sulphide. Thiocyanate gives a blood red complex $[Fe (SCN)]^{2 +}$ with iron(III) ions, and because no free cyanide is present no Prussian blue appears. Excess sodium splits thiocyanate into cyanide and sulphide.

What is the difference between the Dumas and Kjeldahl methods?

In the Dumas method nitrogen is released as $N_{2}$ gas by heating with copper oxide, and its volume is measured and corrected to STP. In Kjeldahl's method nitrogen is converted to ammonium sulphate, released as ammonia with NaOH, absorbed in standard acid and estimated by back titration. Dumas works for all nitrogen compounds.

Why is Kjeldahl's method not used for nitrobenzene or pyridine?

Digestion with concentrated sulphuric acid does not convert nitrogen of nitro and azo groups, or nitrogen held in an aromatic ring such as pyridine, into ammonium sulphate. That nitrogen escapes estimation, so the result is too low. The Dumas method is used for such compounds.

How is the percentage of oxygen found in an organic compound?

Usually by difference: 100 minus the sum of all other percentages. It can also be estimated directly by converting all oxygen to carbon monoxide over red-hot coke and oxidising the CO with iodine pentoxide; 32 g of oxygen gives 88 g of carbon dioxide.

What type of questions come from this topic in NEET?

NEET asks straightforward numericals on percentage of C, H, N, halogen or sulphur using the standard formulas, followed by empirical and molecular formula, plus one-liners on Lassaigne's test colours, Prussian blue, the use of nitric acid before silver nitrate and the compounds for which Kjeldahl's method fails.

How is qualitative and quantitative analysis tested in JEE Main and JEE Advanced?

JEE Main favours Kjeldahl back-titration and Dumas numericals, blood red versus Prussian blue, and functional group tests from the practical chemistry unit. JEE Advanced adds identification chains using 2,4-DNP, Tollens', Fehling's, iodoform and carbylamine tests, and molar mass by the silver salt method.

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