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Class 11 Organic Chemistry — Basic Principles and Techniques

By Aniket Bhardwaj · 4 October 2026 · CBSE Class 11 Chemistry

Before organic chemistry gets into naming compounds and drawing mechanisms, NCERT's "Organic Chemistry — Some Basic Principles and Techniques" chapter covers something more practical: how an organic chemist actually purifies an unknown compound and finds out what elements it contains and in what proportion. This is a separate chapter from nomenclature and isomerism, and it is the one place in Class 11 where organic chemistry becomes genuinely numerical. Every method here comes with a real worked calculation.

Purification methods — which technique fits which situation

MethodWhen it is usedExample
CrystallisationSolids whose solubility differs sharply with temperaturePurifying impure alum or benzoic acid
SublimationSolids that pass directly from solid to vapour, leaving a non-sublimable impurity behindNaphthalene, camphor, anthracene
Simple distillationLiquids that do not decompose on heating and boil well below 300°C, with no other liquid close in boiling pointSeparating water from a salt solution's solvent
Fractional distillationLiquids whose boiling points are close togetherSeparating different fractions of crude petroleum
Steam distillationSubstances insoluble in water that have some appreciable vapour pressure at 373 KIsolating essential oils, aniline
Distillation under reduced pressureLiquids that decompose at or near their normal boiling pointPurifying glycerol
Differential extractionExtracting an organic compound from an aqueous mixture using an immiscible solventUsing a separating funnel with ether or another organic solvent
ChromatographySeparating very small amounts of closely related compoundsColumn (adsorption) and paper (partition) chromatography

Qualitative analysis — detecting which elements are present

Nitrogen, sulphur, halogens and phosphorus cannot be detected directly on an organic compound; the compound is first fused with sodium metal to convert these elements into water-soluble ionic compounds — this is Lassaigne's test, and the resulting solution is called the sodium fusion extract (Lassaigne's extract).

Na + C, N, S (in the compound) → NaCN, Na₂S, NaCNS (as relevant) — the extract is then tested with specific reagents
ElementTest on the extractPositive result
NitrogenFeSO₄, then acidified with dilute H₂SO₄Prussian blue colouration
SulphurSodium nitroprussideViolet/purple colouration
Sulphur (alternative)Lead acetateBlack precipitate of PbS
HalogensDilute HNO₃, then AgNO₃White (Cl), pale yellow (Br), yellow (I) precipitate
PhosphorusBoiled with conc. HNO₃, then ammonium molybdateYellow precipitate

Quantitative analysis — Liebig's method for carbon and hydrogen

A known mass of an organic compound is burnt completely in a stream of oxygen; the CO₂ and H₂O produced are absorbed separately (in KOH and anhydrous CaCl₂ respectively) and weighed.

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

Worked example 1 — percentage of carbon and hydrogen

0.24 g of an organic compound gave 0.352 g of CO₂ and 0.144 g of H₂O on complete combustion. Find the percentage of carbon and hydrogen.

%C = (12 ÷ 44) × (0.352 ÷ 0.24) × 100 = 0.2727 × 1.4667 × 100 = 40.0%
%H = (2 ÷ 18) × (0.144 ÷ 0.24) × 100 = 0.1111 × 0.6 × 100 = 6.67%

If the compound is known to contain only carbon, hydrogen and oxygen, the remaining percentage is found by difference: 100 − 40.0 − 6.67 = 53.33% oxygen.

Duma's method and Kjeldahl's method for nitrogen

In Duma's method, the compound is heated with copper oxide in a stream of CO₂; the nitrogen released is collected over concentrated KOH solution (which absorbs the CO₂, leaving pure N₂) and its volume is measured at STP.

%N = (28 × volume of N₂ at STP in mL) ÷ (22,400 × mass of compound in g) × 100

Kjeldahl's method instead converts the nitrogen to ammonium sulphate by heating with concentrated H₂SO₄; the liberated ammonia is absorbed in a known volume of a standard acid, and the unreacted acid is back-titrated.

%N = (1.4 × normality of acid × volume of acid used in mL) ÷ mass of compound in g

Kjeldahl's method fails for compounds where nitrogen is present as a ring nitrogen (e.g. pyridine) or in a nitro or azo group, because that nitrogen does not convert to ammonium sulphate under the reaction conditions — Duma's method must be used for those instead.

Worked example 2 — percentage of nitrogen, both methods

Duma's method: 0.15 g of an organic compound gave 20 mL of N₂ gas collected at STP. Find %N.

%N = (28 × 20) ÷ (22,400 × 0.15) × 100 = 560 ÷ 3360 × 100 = 16.67%

Kjeldahl's method: 0.5 g of an organic compound required 20 mL of 0.1 N H₂SO₄ to neutralise the ammonia liberated. Find %N.

%N = (1.4 × 0.1 × 20) ÷ 0.5 = 2.8 ÷ 0.5 = 5.6%

Carius method for halogens, sulphur and phosphorus

The compound is heated with fuming nitric acid in the presence of silver nitrate (for halogens) inside a sealed Carius tube. The precipitate formed (AgX, BaSO₄ or Mg₂P₂O₇) is filtered, dried and weighed.

%halogen = (atomic mass of X × mass of AgX formed) ÷ (molar mass of AgX × mass of compound) × 100

Worked example 3 — percentage of chlorine by Carius method

0.3 g of an organic compound gave 0.5 g of AgCl by the Carius method. Find the percentage of chlorine. (Atomic mass Cl = 35.5, molar mass AgCl = 143.5)

%Cl = (35.5 × 0.5) ÷ (143.5 × 0.3) × 100 = 17.75 ÷ 43.05 × 100 = 41.23%

Common mistakes that cost marks

  • Testing an organic compound directly with AgNO₃ or FeSO₄ instead of first preparing the sodium fusion extract — the elements in the original compound are covalently bound and give no test until converted to ionic form.
  • Mixing up the constants in the %N formulas. 28 = molar mass of N₂ (Duma's), 22,400 = molar volume at STP in mL, and 1.4 = 14/10 (Kjeldahl's) — writing the wrong constant is a common slip.
  • Applying Kjeldahl's method to ring, nitro or azo nitrogen compounds, where it systematically underestimates %N.
  • Forgetting the molar mass of the precipitate (AgCl = 143.5, BaSO₄ = 233, Mg₂P₂O₇ = 222) — these must be calculated correctly before the percentage formula can be applied.

Where this chapter appears in exams

ExamTypical use
CBSE Class 11Percentage composition numericals, qualitative test identification
JEE/NEETEmpirical formula problems built directly on Liebig's/Duma's/Carius results
IIT-JAM / CUET-PGApplied as a routine step before structure-determination questions

Check your elemental-analysis calculations. Use the free calculator suite's Molar Mass tool to verify the molar masses of AgCl, BaSO₄ and Mg₂P₂O₇ used in these formulas.

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