CBSE Class 11 & JEE Mains • Module 04 of 20 • Knowing What You Have and How Pure It Is
Topics Covered: Criteria of Purity (BP, MP, Rf) · Methods of Purification (crystallisation, distillation, sublimation, chromatography) · Qualitative Analysis — Lassaigne's Sodium Fusion Test · Quantitative Analysis — Liebig's (C & H), Dumas (N), Kjeldahl (N), Carius (S, X), Victor Meyer (mol. mass)
A pure compound has a sharp, well-defined melting point (for solids) and boiling point (for liquids). Any impurity causes:
Chromatographic purity: A single spot on TLC (thin-layer chromatography) with a single, consistent Rf value confirms purity.
Principle: Dissolve the impure solid in minimum hot solvent → cool slowly → the desired compound crystallises out (impurities remain in solution).
Solvent selection: The compound must be soluble in the hot solvent but sparingly soluble when cold. Common solvents: water, ethanol, acetone.
Fractional crystallisation: Used when two compounds have similar solubility — repeated crystallisation separates them.
Example: Purification of crude naphthalene from coal tar using hot ethanol. Cooling → pure naphthalene crystals form.
Simple Distillation: Separates liquids with significantly different boiling points (>25°C difference) or separates a liquid from dissolved solids.
Fractional Distillation: Liquids with close boiling points (e.g., ethanol BP 78°C and water BP 100°C). Uses a fractionating column. Petroleum refining uses fractional distillation.
Steam Distillation: For compounds that decompose at their boiling point but are immiscible with water. The compound distils with steam at a temperature lower than 100°C. Example: Aniline (BP 184°C) is steam-distilled at ~98°C because it is slightly volatile and immiscible with water. Principle: total vapour pressure = Pcompound + Pwater = atmospheric pressure → mixture boils below 100°C.
Distillation under reduced pressure (vacuum distillation): For high-boiling or heat-sensitive compounds. Lowering pressure lowers boiling point.
Some solids convert directly to vapour (no liquid phase) when heated — they sublime. On cooling, the pure vapour solidifies.
Used for: Camphor, iodine, naphthalene, anthracene, benzoic acid, ammonium chloride.
Apparatus: A porcelain dish with the impure substance. Covered with an inverted funnel. Gentle heating → pure sublimate collects on the funnel walls.
Principle: Separation based on differential distribution between two phases — a stationary phase (silica, alumina, paper) and a moving phase (solvent, mobile phase). Different compounds travel different distances based on their affinity to each phase.
Types of Chromatography:
Purpose: To detect the presence of N, S, and halogens (Cl, Br, I) in an organic compound.
Principle: Alkali metals (Na) when fused with the organic compound at high temperature convert the covalently bonded elements into their ionic (water-soluble) forms that can then be tested.
Why must we acidify with HNO₃ before Halogen test?
SFE may contain CN⁻ and S²⁻ (from N and S in the compound). These also precipitate with Ag⁺ forming AgCN (white) and Ag₂S (black) — which would give false positives for the halogen test. Acidifying with HNO₃ first destroys CN⁻ and S²⁻ ions.
HCN gas and H₂S gas are evolved → removed → then AgNO₃ test is reliable for halogens only.
Principle: The organic compound is burned completely in excess O₂. C → CO₂ (absorbed in KOH), H → H₂O (absorbed in anhydrous CaCl₂ / Mg(ClO₄)₂). The increase in weights of the absorbents gives mass of CO₂ and H₂O formed.
Carbon:
Hydrogen:
Derivation logic: 44g CO₂ contains 12g C. So (12/44) converts mass of CO₂ to mass of C. Similarly, 18g H₂O contains 2g H.
Principle: Organic compound + CuO (oxidising) → heated → all N → N₂ gas (collected over KOH solution which absorbs CO₂). Volume of N₂ measured at room temperature and pressure.
Where V = volume of N₂ at STP (in mL), P = corrected pressure, W = mass of compound in grams. At STP: 22400 mL N₂ = 28g N.
Principle: Compound + conc. H₂SO₄ → digested → N → (NH₄)₂SO₄. Boil with NaOH → NH₃ released. NH₃ collected in excess H₂SO₄ (standard). Back-titrate excess H₂SO₄ with standard NaOH → volume used → calculate N%.
NOT applicable for: Compounds with N in nitro (–NO₂), azo (–N=N–), pyridine ring (N in non-amine forms). These don't convert to (NH₄)₂SO₄ quantitatively.
Where M = molarity of H₂SO₄ used, V = volume of H₂SO₄ (mL) neutralised by NH₃, W = mass of compound (g). (Factor 1.4 = 14/10 from unit conversions.)
Principle: Compound + fuming HNO₃ → oxidised in sealed Carius tube (heated to 250°C in a furnace) → S → BaSO₄ (if Ba²⁺ is added) → Halogen → AgX precipitate.
Derivation: 233g BaSO₄ contains 32g S.
Oxygen is NOT directly estimated by any standard method. It is calculated by difference:
Step 1 — Find Empirical Formula: Divide % by atomic mass of each element. Divide all by the smallest quotient → gives simple whole number ratio → Empirical Formula.
Step 2 — Find Molecular Formula: Determine molecular mass (by Victor Meyer method, mass spectrometry etc.). Molecular Formula = n × Empirical Formula where n = Molecular mass / Empirical formula mass.
Example: A compound gives %C = 40%, %H = 6.7%, %O = 53.3%.
Moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33
Divide by smallest (3.33): C:H:O = 1:2:1 → Empirical formula = CH₂O
If mol. mass = 60g/mol → n = 60/30 = 2 → Molecular formula = C₂H₄O₂ (acetic acid or glycolaldehyde)
Ex 1 M: 0.30g of an organic compound on Liebig's combustion gives 0.44g CO₂ and 0.18g H₂O. Find %C and %H.
Solution:
%C = (12/44) × (0.44/0.30) × 100 = (0.2727) × (1.467) × 100 = 40%
%H = (2/18) × (0.18/0.30) × 100 = (0.1111) × (0.60) × 100 = 6.67%
Ex 2 M: In Kjeldahl's method, 0.50g compound liberates NH₃ that neutralises 25 mL of 0.1M H₂SO₄. Find %N.
Solution:
%N = (1.4 × M × V) / W = (1.4 × 0.1 × 25) / 0.50 = 3.5/0.50 = 7%
Ex 3 H: A compound has %C = 75%, %H = 25%, mol. mass = 16. Find molecular formula.
Solution: Moles: C = 75/12 = 6.25, H = 25/1 = 25. Divide by 6.25: C:H = 1:4. Empirical formula = CH₄. Empirical mass = 16. n = 16/16 = 1. Molecular formula: CH₄ (methane).
E Q1. What colour precipitate forms in Lassaigne's test when compound contains nitrogen? Name the compound formed.
E Q2. Why can't Kjeldahl's method be used for compounds containing −NO₂ groups?
M Q3. 0.20g compound on combustion gives 0.33g CO₂ and 0.135g H₂O. Find %C and %H. If %O = 53.3%, find empirical formula.
M Q4. An organic compound (0.40g) on Carius method gives 0.58g AgCl. Find %Cl.
H Q5. A compound contains C, H, and S. On analysis: %C = 24.4, %H = 4.1, %S = remaining. Mol. mass = 98. Find empirical and molecular formula. (Assume remaining = %S). [H₂SO₄ has mol. mass 98 and contains S = 32.7%]
| Estimation | Method | Formula |
|---|---|---|
| %C | Liebig combustion | 12/44 × mass CO₂ / mass compound × 100 |
| %H | Liebig combustion | 2/18 × mass H₂O / mass compound × 100 |
| %N | Dumas | 28 × V(N₂ at STP) / 22400 × mass compound × 100 |
| %N | Kjeldahl | 1.4 × M × V / W |
| %S | Carius | 32/233 × mass BaSO₄ / mass compound × 100 |
| %Cl | Carius | 35.5/143.5 × mass AgCl / mass compound × 100 |
| %O | By difference | 100 − (%C + %H + %N + %S + %X) |
| Purification Method | Used When |
|---|---|
| Crystallisation | Solid + impurities; solubility difference with temperature |
| Simple distillation | Liquids with BP difference >25°C or liquid + solid |
| Fractional distillation | Liquids with close boiling points |
| Steam distillation | High-BP compounds immiscible with water; heat-sensitive |
| Sublimation | Camphor, iodine, naphthalene; sublime directly |
| Chromatography (TLC) | Small quantities; Rf for ID; monitor reactions |