NEET Chemistry: Organic Basics and Hydrocarbons · Chapter 1

Purification and Characterisation of Organic Compounds

What to remember

Choose the purification method from the nature of the impurity: sublimation for solids that sublime, crystallisation for solids with different solubility, distillation for liquids (simple, fractional, reduced pressure, steam), chromatography for small mixtures.
Lassaigne's test converts N, S and halogens into ionic salts (NaCN, Na₂S, NaX): N gives Prussian blue, S gives violet (nitroprusside) or black PbS, N + S together give blood red, halogens give silver halide precipitates after boiling with HNO₃.
Formulae for percentage composition: %C = (12/44) × (mass of CO₂/mass of compound) × 100; %H = (2/18) × (mass of H₂O/mass of compound) × 100; Dumas %N = (28 × V/22400 × m) × 100 with V in mL at STP; Kjeldahl %N = 1.4 × N × V/m.
Empirical formula comes from the simplest whole-number ratio of moles of atoms; molecular formula = n × empirical formula, where n = molar mass/empirical formula mass. Oxygen is found by difference.

Methods of purification

A pure compound is needed before its structure can be studied. The method used depends on the physical state and on the properties of the compound and of its impurities.

Method	Used for	Principle	Example
Sublimation	Solids that change directly to vapour on heating and back to solid on cooling	Impurities do not sublime	Camphor, naphthalene, anthracene
Crystallisation	Solids whose solubility in a solvent changes strongly with temperature	The impure solid is dissolved in the minimum hot solvent, the solution is cooled, pure crystals separate, and impurities stay in the mother liquor	Sugar from sugar-cane juice
Simple distillation	Liquids with a large difference in boiling point, or a liquid with non-volatile impurities	The lower-boiling liquid vaporises first and is condensed	Chloroform (334 K) and aniline (457 K)
Fractional distillation	Liquids with boiling points close to each other	A fractionating column gives repeated vaporisation and condensation	Separation of petroleum fractions
Distillation under reduced pressure	Liquids that decompose at or below their normal boiling point	Lowering the pressure lowers the boiling point	Glycerol
Steam distillation	Liquids that are steam volatile and immiscible with water	The liquid distils with steam at a temperature below 373 K, and the distillate is separated from water	Aniline
Differential extraction	Organic compounds present in water	A solvent immiscible with water (such as ether) dissolves the compound better than water does; the two layers are separated in a separating funnel	Extraction of an organic compound from its aqueous solution

In crystallisation, a mixture of compounds with different solubilities can also be separated by repeated crystallisation (fractional crystallisation). In differential extraction, the organic layer is then distilled to recover the compound.

Chromatography

The name comes from the Greek word chroma (colour), because the method was first used to separate coloured substances. A mixture is dissolved in a mobile phase, which passes over a stationary phase. The components move at different speeds and separate.

Type	Basis	Stationary phase	Mobile phase
Adsorption – column chromatography	Different components are adsorbed to different extents	Alumina or silica gel packed in a glass column	A suitable liquid (eluent)
Adsorption – thin layer chromatography (TLC)	Same as above	A thin layer of silica gel or alumina on a glass plate	A solvent that rises up the plate
Partition – paper chromatography	Continuous differential partitioning between two liquids	Water held in the pores of the paper	A suitable solvent that moves up the paper

In TLC, spots are placed on a base line. The plate is kept in a closed jar with the solvent below the base line. The solvent rises and carries the components to different heights. Colourless spots are made visible by iodine vapour, UV light or a spray reagent.

Retention factor: Rf = distance moved by the substance from the base line / distance moved by the solvent front from the base line. Rf has no unit and is always less than or equal to 1. Each compound has a characteristic Rf under fixed conditions.

Qualitative analysis

Carbon and hydrogen. The compound is heated strongly with dry copper(II) oxide. Carbon is oxidised to CO₂, which turns lime water milky. Hydrogen is oxidised to H₂O, which turns anhydrous copper(II) sulphate blue.

C + 2CuO → 2Cu + CO₂ ; 2H + CuO → Cu + H₂O

CO₂ + Ca(OH)₂ → CaCO₃ + H₂O ; 5H₂O + CuSO₄ → CuSO₄·5H₂O (blue)

Lassaigne's test (sodium fusion). The compound is fused with sodium metal. Nitrogen, sulphur and halogens change to the ionic salts NaCN, Na₂S and NaX. The fused mass is boiled with distilled water and filtered. The filtrate is called the sodium fusion extract.

Na + C + N → NaCN ; 2Na + S → Na₂S ; Na + X → NaX

Element	Test on the extract	Observation
Nitrogen	Boil with FeSO₄, then acidify with conc. H₂SO₄	Prussian blue colour of Fe₄[Fe(CN)₆]₃
Sulphur	Add sodium nitroprusside	Violet colour
Sulphur	Acidify with acetic acid, add lead acetate	Black precipitate of PbS
Nitrogen and sulphur together	Acidify, add FeCl₃ (Fe³⁺)	Blood red colour of [Fe(SCN)]²⁺ (NaSCN forms during fusion)
Halogens	Boil with dilute HNO₃, then add AgNO₃	AgCl white, soluble in NH₄OH; AgBr pale yellow, sparingly soluble in NH₄OH; AgI yellow, insoluble in NH₄OH
Phosphorus	Heat with an oxidising agent (Na₂O₂) to form phosphate, boil with HNO₃, add ammonium molybdate	Yellow colour or precipitate of ammonium phosphomolybdate

Steps for Prussian blue: 6CN⁻ + Fe²⁺ → [Fe(CN)₆]⁴⁻ ; 3[Fe(CN)₆]⁴⁻ + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃ . Sulphide: S²⁻ + Pb²⁺ → PbS. For the N and S test: Fe³⁺ + SCN⁻ → [Fe(SCN)]²⁺.

The extract is boiled with HNO₃ before the AgNO₃ test to decompose NaCN and Na₂S. Otherwise they would also give precipitates with silver nitrate (AgCN and Ag₂S) and spoil the test.

Quantitative analysis

Let m be the mass of the organic compound taken.

Carbon and hydrogen (Liebig's method). The compound is heated with CuO. The CO₂ formed is absorbed in a U-tube of concentrated KOH solution. The H₂O formed is absorbed in a tube of anhydrous CaCl₂. The increase in mass of each tube is measured.

%C = (12/44) × (mass of CO₂/m) × 100
%H = (2/18) × (mass of H₂O/m) × 100

Nitrogen.

Method	Principle	Formula
Dumas	The compound is heated with CuO in a CO₂ atmosphere. Nitrogen is set free as N₂ (oxides of nitrogen are reduced over hot copper gauze). The N₂ is collected over concentrated KOH solution and its volume is measured at STP.	%N = (28 × V/22400 × m) × 100, V = volume of N₂ in mL at STP
Kjeldahl	The compound is heated with concentrated H₂SO₄ and a little K₂SO₄ and CuSO₄ (catalyst). Nitrogen changes to (NH₄)₂SO₄. This is heated with excess NaOH to release NH₃, which is absorbed in a measured volume of standard H₂SO₄. The unused acid is found by titration.	%N = 1.4 × N × V/m, N = normality of the acid, V = volume of acid neutralised by NH₃ in mL

Kjeldahl's method does not give satisfactory results for compounds with nitrogen in nitro (–NO₂) or azo (–N=N–) groups, or with nitrogen in a ring (such as pyridine). Their nitrogen is not converted to ammonium sulphate under the conditions of digestion. The method is simple and is widely used for food and fertilisers.

Halogens (Carius method). The compound is heated with fuming HNO₃ and AgNO₃ in a sealed Carius tube. Carbon and hydrogen are oxidised to CO₂ and H₂O. The halogen forms the silver halide, which is filtered, washed, dried and weighed.

%Cl = (35.5/143.5) × (mass of AgCl/m) × 100
%Br = (80/188) × (mass of AgBr/m) × 100
%I = (127/235) × (mass of AgI/m) × 100

Sulphur. The compound is heated with fuming HNO₃ (or Na₂O₂). Sulphur is oxidised to sulphuric acid, which is precipitated as BaSO₄ by adding BaCl₂.

%S = (32/233) × (mass of BaSO₄/m) × 100

Phosphorus. The compound is heated with fuming HNO₃. Phosphorus is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃, by adding ammonia and ammonium molybdate. Alternatively, magnesia mixture gives MgNH₄PO₄, which on ignition gives Mg₂P₂O₇.

%P = (62/222) × (mass of Mg₂P₂O₇/m) × 100

Oxygen. There is no simple direct test here. The percentage of oxygen is found by difference: %O = 100 − (sum of the percentages of all other elements).

Empirical and molecular formula

1. Write the percentage of each element (this equals grams in 100 g).
2. Divide each by its atomic mass to get moles of atoms.
3. Divide all by the smallest value to get the simplest ratio. Multiply by a small integer if a ratio is near x.5 or x.33.
4. The whole-number ratio gives the empirical formula.
5. n = molar mass/empirical formula mass. Molecular formula = (empirical formula) × n.

Atomic masses used: C = 12, H = 1, O = 16, N = 14, S = 32, Cl = 35.5, Br = 80, I = 127, P = 31.

Worked examples

Example 1 (C, H and formula). A compound of mass 0.30 g gave 0.44 g CO₂ and 0.18 g H₂O on combustion. Its molar mass is 180 g mol⁻¹. Find its molecular formula.

%C = (12/44) × (0.44/0.30) × 100 = 40 %
%H = (2/18) × (0.18/0.30) × 100 = 6.67 %
%O = 100 − 40 − 6.67 = 53.33 %
Moles: C = 40/12 = 3.33; H = 6.67/1 = 6.67; O = 53.33/16 = 3.33. Ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O (mass 30).
n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.

Example 2 (Dumas). 0.28 g of a compound gave 56 mL of nitrogen at STP. %N = (28 × 56)/(22400 × 0.28) × 100 = 1568/6272 × 100 = 25 %.

Example 3 (Kjeldahl). 0.35 g of a compound gave ammonia which neutralised 10 mL of 0.5 N H₂SO₄. %N = 1.4 × 0.5 × 10/0.35 = 20 %.

Example 4 (Carius). 0.20 g of a chlorine compound gave 0.287 g AgCl. %Cl = (35.5/143.5) × (0.287/0.20) × 100 = 35.5 %.

Example 5 (sulphur). 0.16 g of a compound gave 0.233 g BaSO₄. %S = (32/233) × (0.233/0.16) × 100 = 20 %.

Example 6 (phosphorus). 0.31 g of a compound gave 0.222 g Mg₂P₂O₇. %P = (62/222) × (0.222/0.31) × 100 = 20 %.

Example 7 (empirical formula from two elements). A hydrocarbon has 85.7 % C and 14.3 % H, and its molar mass is 56 g mol⁻¹. Moles: C = 85.7/12 = 7.14; H = 14.3. Ratio 1 : 2, so the empirical formula is CH₂ (mass 14). n = 56/14 = 4, so the molecular formula is C₄H₈.

Summary table of tests and reagents

To detect or estimate	Method or reagent	Result
C and H	Heat with CuO	CO₂ (lime water milky), H₂O (anhydrous CuSO₄ blue)
N (qualitative)	Lassaigne, FeSO₄, conc. H₂SO₄	Prussian blue
S (qualitative)	Nitroprusside or lead acetate	Violet or black
N and S together	Fe³⁺	Blood red
Halogens	HNO₃, AgNO₃	AgCl white, AgBr pale yellow, AgI yellow
P	Ammonium molybdate	Yellow
N (quantitative)	Dumas or Kjeldahl	% from N₂ volume or acid used
Halogens, S, P (quantitative)	Carius	AgX, BaSO₄, Mg₂P₂O₇

NEET traps

Kjeldahl's method fails for nitro, azo and ring nitrogen (pyridine). Dumas works for all nitrogen compounds.
In Kjeldahl's method the N₂ gas is not collected. The nitrogen is converted to NH₃ and measured by acid. In Dumas method N₂ is collected over KOH.
Prussian blue is Fe₄[Fe(CN)₆]₃. Blood red is [Fe(SCN)]²⁺ and needs both N and S.
Boiling the extract with HNO₃ is done before adding AgNO₃, to remove CN⁻ and S²⁻. It is not done for the sulphur or nitrogen test.
AgCl dissolves in NH₄OH, AgBr only sparingly, AgI does not dissolve.
Steam distillation is for steam-volatile liquids that do not mix with water; reduced pressure distillation is for liquids that decompose near their boiling point. Do not swap them.
Column chromatography and TLC are adsorption methods. Paper chromatography is a partition method.
Rf is never greater than 1 and has no unit.
In Liebig's method the H₂O is absorbed in CaCl₂ and the CO₂ in KOH. Do not reverse the absorbents.
Oxygen is found by difference, not by a direct test in this syllabus.

One-liners

Camphor and naphthalene are purified by sublimation.
Sugar is crystallised from its aqueous solution.
Chloroform and aniline are separated by simple distillation.
Petroleum is separated into fractions by fractional distillation.
Glycerol is distilled under reduced pressure.
Aniline is purified by steam distillation.
Differential extraction uses two immiscible liquids.
Column chromatography uses alumina or silica gel as stationary phase.
In paper chromatography the stationary phase is water held in the paper.
Sodium fusion converts N, S and halogens into NaCN, Na₂S and NaX.
Sodium nitroprusside gives violet colour with sulphide.
Lead acetate gives black PbS with sulphide in acetic acid.
Yellow precipitate with ammonium molybdate shows phosphorus.
Carius method estimates halogens, sulphur and phosphorus.
Empirical formula mass × n = molar mass.

Practice questions

Camphor containing non-volatile impurities is best purified by:
1. Sublimation
2. Differential extraction
3. Distillation under reduced pressure
4. Fractional distillation
Answer
A. Sublimation
Camphor changes directly from solid to vapour on heating, and the non-volatile impurities stay behind.
Glycerol is purified by distillation under reduced pressure because it:
1. sublimes on heating
2. decomposes at or near its normal boiling point
3. has a boiling point very close to that of water
4. is immiscible with water
Answer
B. decomposes at or near its normal boiling point
Lowering the pressure lowers the boiling point, so glycerol distils without decomposing.
A liquid that is steam volatile and immiscible with water, such as aniline, is purified by:
1. column chromatography
2. fractional crystallisation
3. sublimation
4. steam distillation
Answer
D. steam distillation
Steam distillation is used for steam-volatile, water-immiscible liquids.
In a TLC experiment the spot moved 3 cm and the solvent front moved 6 cm from the base line. The Rf value is:
1. 1.5
2. 2.0
3. 0.5
4. 0.3
Answer
C. 0.5
Rf = 3/6 = 0.5.
When an organic compound containing nitrogen, sulphur and a halogen is fused with sodium metal, the products are:
1. NH₃, H₂S and HX
2. NaCN, Na₂S and NaX
3. NaCNO, Na₂SO₃ and NaXO₃
4. NaNO₃, Na₂SO₄ and NaX
Answer
B. NaCN, Na₂S and NaX
Sodium fusion converts N, S and halogens to the ionic salts NaCN, Na₂S and NaX.
The Prussian blue colour obtained in the test for nitrogen is due to:
1. Na₄[Fe(CN)₆]
2. Fe₂(SO₄)₃
3. [Fe(SCN)]²⁺
4. Fe₄[Fe(CN)₆]₃
Answer
D. Fe₄[Fe(CN)₆]₃
Fe³⁺ reacts with [Fe(CN)₆]⁴⁻ to give Fe₄[Fe(CN)₆]₃.
The sodium fusion extract is boiled with dilute HNO₃ before adding AgNO₃ in order to:
1. dissolve the silver halide formed
2. convert halides to halogens
3. decompose NaCN and Na₂S
4. oxidise sodium to its oxide
Answer
C. decompose NaCN and Na₂S
CN⁻ and S²⁻ would also give precipitates with Ag⁺; boiling with HNO₃ removes them.
0.70 g of an organic compound gave ammonia in Kjeldahl's method which neutralised 25 mL of 0.2 N H₂SO₄. The percentage of nitrogen is:
1. 10 %
2. 20 %
3. 7 %
4. 14 %
Answer
A. 10 %
%N = 1.4 × N × V/m = 1.4 × 0.2 × 25/0.70 = 7/0.70 = 10 %.
0.20 g of an organic compound gave 22.4 mL of N₂ at STP in Dumas method. The percentage of nitrogen is:
1. 7 %
2. 28 %
3. 10 %
4. 14 %
Answer
D. 14 %
%N = 28 × 22.4/(22400 × 0.20) × 100 = 627.2/4480 × 100 = 14 %.
0.50 g of a bromine compound gave 0.94 g AgBr in the Carius method (Br = 80, Ag = 108). The percentage of bromine is:
1. 94 %
2. 80 %
3. 40 %
4. 47 %
Answer
B. 80 %
%Br = (80/188) × (0.94/0.50) × 100 = 80 × 0.005/0.50 × 100 = 80 %.
0.20 g of an organic compound gave 0.466 g BaSO₄ (molar mass 233). The percentage of sulphur is:
1. 32 %
2. 16 %
3. 23 %
4. 46.6 %
Answer
A. 32 %
%S = (32/233) × (0.466/0.20) × 100 = 32 × 0.002/0.20 × 100 = 32 %.
A compound contains 80 % carbon and 20 % hydrogen by mass. Its empirical formula is:
1. CH₂
2. C₂H₃
3. CH₃
4. CH₄
Answer
C. CH₃
Moles C = 80/12 = 6.67; moles H = 20. Ratio 1 : 3, so CH₃.
In thin layer chromatography the stationary phase is commonly:
1. moist filter paper
2. water held in cellulose
3. silica gel spread on a glass plate
4. an organic solvent
Answer
C. silica gel spread on a glass plate
TLC is an adsorption method with silica gel or alumina on a plate. Water in paper is the stationary phase of paper chromatography.
In the quantitative analysis of an organic compound containing C, H and O only, the percentage of oxygen is usually obtained:
1. from the volume of N₂ at STP
2. by subtracting the sum of the other percentages from 100
3. from the mass of AgCl formed
4. from the mass of BaSO₄ formed
Answer
B. by subtracting the sum of the other percentages from 100
Oxygen is found by difference.
The detection of phosphorus in an organic compound is done by forming phosphate and then adding ammonium molybdate. The positive result is a:
1. yellow colour or precipitate
2. violet colour
3. blood red colour
4. black precipitate
Answer
A. yellow colour or precipitate
Ammonium phosphomolybdate is yellow.
In paper chromatography the stationary phase is:
1. alumina on the paper
2. silica gel on the paper
3. the organic solvent used for development
4. water held in the pores of the paper
Answer
D. water held in the pores of the paper
Paper chromatography is a partition method; water on cellulose is the stationary phase.
Given below are two statements: Statement I: Camphor can be purified by sublimation. Statement II: In sublimation the solid passes through the liquid state before turning to vapour.
1. Statement I is correct but Statement II is incorrect
2. Both Statement I and Statement II are correct
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Statement I is correct but Statement II is incorrect
Sublimation is the direct change of solid to vapour, with no liquid stage.
Given below are two statements: Statement I: A black precipitate with lead acetate in acetic acid solution indicates the presence of sulphide. Statement II: Sodium nitroprusside gives a blood red colour with the sulphide ion.
1. Both Statement I and Statement II are correct
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Statement I is correct but Statement II is incorrect
Nitroprusside gives violet, not blood red.
Given below are two statements: Statement I: Phosphorus is detected by a blue colour with ammonium molybdate. Statement II: The extract is boiled with H₂SO₄ before adding AgNO₃ in the halogen test.
1. Both Statement I and Statement II are correct
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Both Statement I and Statement II are incorrect
Ammonium molybdate gives a yellow colour; HNO₃, not H₂SO₄, is used before AgNO₃.
Given below are two statements: Statement I: The Rf value of a compound can be greater than 1. Statement II: Rf has no unit.
1. Statement I is incorrect but Statement II is correct
2. Both Statement I and Statement II are correct
3. Both Statement I and Statement II are incorrect
4. Statement I is correct but Statement II is incorrect
Answer
A. Statement I is incorrect but Statement II is correct
The substance cannot travel farther than the solvent front, so Rf ≤ 1. It is a ratio of distances, so it has no unit.
Given below are two statements: Statement I: AgI is yellow and insoluble in NH₄OH. Statement II: AgCl is white and soluble in NH₄OH.
1. Both Statement I and Statement II are incorrect
2. Both Statement I and Statement II are correct
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are correct
Both statements describe the silver halide tests correctly.
Given below are two statements: Statement I: Fractional distillation is used to separate liquids with boiling points close to each other. Statement II: Simple distillation is enough to separate liquids whose boiling points differ widely.
1. Both Statement I and Statement II are incorrect
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are correct
4. Statement I is incorrect but Statement II is correct
Answer
C. Both Statement I and Statement II are correct
Both statements are correct.
Given below are two statements: Statement I: Differential extraction is based on the different solubility of a compound in two immiscible solvents. Statement II: The empirical formula of a compound gives the actual number of atoms in a molecule.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is incorrect but Statement II is correct
4. Statement I is correct but Statement II is incorrect
Answer
D. Statement I is correct but Statement II is incorrect
The molecular formula, not the empirical formula, gives the actual number of atoms.
Which of the following statements are correct? 1. Carbon is detected as CO₂, which turns lime water milky. 2. Hydrogen is detected as water, which turns anhydrous copper(II) sulphate blue. 3. Oxygen is detected by heating the compound with CuO.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
Oxygen is not detected by heating with CuO; it is found by difference.
Which of the following statements about Liebig's method are correct? 1. CO₂ is absorbed in anhydrous CaCl₂. 2. H₂O is absorbed in anhydrous CaCl₂. 3. The increase in mass of the KOH tube gives the carbon content.
1. 1 and 2 only
2. 1 and 3 only
3. 1, 2 and 3
4. 2 and 3 only
Answer
D. 2 and 3 only
CO₂ is absorbed in KOH and water in CaCl₂.
Assertion (A): The sodium fusion extract is boiled with HNO₃ before testing for halogens. Reason (R): HNO₃ decomposes NaCN and Na₂S that would otherwise give precipitates with AgNO₃.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Boiling removes CN⁻ and S²⁻ which interfere.
Assertion (A): Kjeldahl's method is not suitable for nitro compounds. Reason (R): The nitrogen of nitro compounds is not converted to ammonium sulphate during digestion.
1. Both A and R are true but R is not the correct explanation of A
2. Both A and R are true and R is the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true and R is the correct explanation of A
Nitro, azo and ring nitrogen are not changed to (NH₄)₂SO₄.
Assertion (A): Glycerol is purified by distillation under reduced pressure. Reason (R): Reducing the pressure raises the boiling point of a liquid.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Reduced pressure lowers the boiling point, so R is false.
Assertion (A): The Rf value of a substance can never exceed 1. Reason (R): A substance cannot move farther than the solvent front.
1. Both A and R are true but R is not the correct explanation of A
2. A is true but R is false
3. A is false but R is true
4. Both A and R are true and R is the correct explanation of A
Answer
D. Both A and R are true and R is the correct explanation of A
Rf is the ratio of the distance moved by the substance to that by the solvent front.
Assertion (A): Column chromatography is an adsorption method. Reason (R): Paper chromatography is a partition method.
1. Both A and R are true but R is not the correct explanation of A
2. Both A and R are true and R is the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true but R is not the correct explanation of A
Both are true, but one does not explain the other.
Assertion (A): Oxygen in an organic compound is always estimated by the Carius method. Reason (R): Oxygen percentage can be found by subtracting the sum of the percentages of all other elements from 100.
1. Both A and R are true and R is the correct explanation of A
2. A is false but R is true
3. Both A and R are true but R is not the correct explanation of A
4. A is true but R is false
Answer
B. A is false but R is true
Oxygen is found by difference, not by Carius method.
Match List I with List II. List I: a. Nitrogen b. Sulphur c. Nitrogen and sulphur together d. Phosphorus List II: i. Violet colour with sodium nitroprusside ii. Prussian blue iii. Blood red colour iv. Yellow colour with ammonium molybdate
1. a-i, b-ii, c-iii, d-iv
2. a-iii, b-i, c-ii, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
C. a-ii, b-i, c-iii, d-iv
N gives Prussian blue, S gives violet, N and S give blood red, P gives yellow.
Match List I with List II. List I: a. Sublimation b. Steam distillation c. Distillation under reduced pressure d. Fractional distillation List II: i. Glycerol ii. Aniline iii. Camphor iv. Petroleum fractions
1. a-iii, b-i, c-ii, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-iii, b-ii, c-i, d-iv
Answer
D. a-iii, b-ii, c-i, d-iv
Camphor sublimes; aniline is steam distilled; glycerol is distilled under reduced pressure; petroleum is fractionated.
Match List I with List II. List I: a. Percentage of carbon b. Percentage of hydrogen c. Percentage of nitrogen (Dumas) d. Percentage of sulphur List II: i. (32/233) × (mass of BaSO₄/m) × 100 ii. (28 × V/22400 × m) × 100 iii. (12/44) × (mass of CO₂/m) × 100 iv. (2/18) × (mass of H₂O/m) × 100
1. a-iv, b-iii, c-ii, d-i
2. a-iii, b-iv, c-ii, d-i
3. a-i, b-iv, c-ii, d-iii
4. a-iii, b-iv, c-i, d-ii
Answer
B. a-iii, b-iv, c-ii, d-i
Carbon uses 12/44 with CO₂, hydrogen uses 2/18 with H₂O, Dumas uses 28V/22400, sulphur uses 32/233.
Match List I with List II. List I: a. AgCl b. AgBr c. AgI d. PbS List II: i. Yellow, insoluble in NH₄OH ii. Black iii. White, soluble in NH₄OH iv. Pale yellow, sparingly soluble in NH₄OH
1. a-i, b-iv, c-iii, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-iv, b-iii, c-i, d-ii
Answer
C. a-iii, b-iv, c-i, d-ii
AgCl white soluble; AgBr pale yellow sparingly soluble; AgI yellow insoluble; PbS black.
Match List I with List II. List I: a. Crystallisation b. Differential extraction c. Adsorption chromatography d. Partition chromatography List II: i. Differential adsorption on a solid ii. Continuous partitioning between two liquid phases iii. Different solubility in hot and cold solvent iv. Different solubility in two immiscible liquids
1. a-iii, b-i, c-iv, d-ii
2. a-iii, b-iv, c-i, d-ii
3. a-iv, b-iii, c-i, d-ii
4. a-iii, b-iv, c-ii, d-i
Answer
B. a-iii, b-iv, c-i, d-ii
Match each technique with its principle.
Match List I with List II. List I: a. Carius method b. Liebig's method c. Kjeldahl's method d. Dumas method List II: i. Nitrogen collected as N₂ over KOH ii. Carbon and hydrogen as CO₂ and H₂O iii. Nitrogen converted to (NH₄)₂SO₄ iv. Halogens as silver halides
1. a-iii, b-ii, c-iv, d-i
2. a-ii, b-iv, c-iii, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-iv, b-ii, c-iii, d-i
Answer
D. a-iv, b-ii, c-iii, d-i
Carius gives AgX; Liebig gives CO₂ and H₂O; Kjeldahl gives (NH₄)₂SO₄; Dumas gives N₂.
0.60 g of an organic compound on complete combustion gave 0.88 g CO₂ and 0.36 g H₂O. The percentage of hydrogen in the compound is about:
1. 4.00 %
2. 10.0 %
3. 6.67 %
4. 12.0 %
Answer
C. 6.67 %
%H = (2/18) × (0.36/0.60) × 100 = 0.04/0.60 × 100 = 6.67 %.
A compound has 40 % C, 6.67 % H and 53.33 % O by mass and a molar mass of 60 g mol⁻¹. Its molecular formula is:
1. C₂H₄O₂
2. CH₂O
3. C₂H₂O₂
4. C₃H₆O₃
Answer
A. C₂H₄O₂
Empirical formula CH₂O (mass 30). n = 60/30 = 2, so C₂H₄O₂.
0.25 g of a chlorine compound gave 0.287 g AgCl in the Carius method (Ag = 108, Cl = 35.5). The percentage of chlorine is about:
1. 28.4 %
2. 35.5 %
3. 14.2 %
4. 56.8 %
Answer
A. 28.4 %
%Cl = (35.5/143.5) × (0.287/0.25) × 100 = 0.071/0.25 × 100 = 28.4 %.
A sodium fusion extract gives a violet colour with sodium nitroprusside and a black precipitate with lead acetate in acetic acid. No Prussian blue forms and no precipitate forms with AgNO₃ after boiling with HNO₃. The compound contains:
1. nitrogen only
2. sulphur and halogen
3. nitrogen and sulphur
4. sulphur only among N, S and halogens
Answer
D. sulphur only among N, S and halogens
Violet and black PbS show S; the other tests are negative, so N and halogens are absent.
0.40 g of an organic compound gave 56 mL of N₂ at STP in the Dumas method. The percentage of nitrogen is:
1. 35 %
2. 14 %
3. 17.5 %
4. 25 %
Answer
C. 17.5 %
%N = 28 × 56/(22400 × 0.40) × 100 = 1568/8960 × 100 = 17.5 %.
In Kjeldahl's method 0.40 g of a compound gave ammonia that neutralised 20 mL of 0.1 N H₂SO₄. The percentage of nitrogen is:
1. 28 %
2. 7 %
3. 3.5 %
4. 14 %
Answer
B. 7 %
%N = 1.4 × 0.1 × 20/0.40 = 2.8/0.40 = 7 %. Check: moles NH₃ = 0.1 × 20/1000 = 0.002; N = 0.028 g; 0.028/0.40 = 7 %.
In a TLC experiment two spots are at 2.4 cm and 3.6 cm from the base line, and the solvent front is at 6.0 cm. The Rf values of the two compounds are:
1. 0.40 and 0.60
2. 0.60 and 0.40
3. 2.5 and 1.7
4. 0.24 and 0.36
Answer
A. 0.40 and 0.60
Rf = 2.4/6.0 = 0.40 and 3.6/6.0 = 0.60.

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