QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 8.6NCERT Class 11 · Chemistry · Chapter 8

Quantitative Elemental Analysis of Organic Compounds

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 9 min read

A complete mathematical and experimental masterclass for estimating the percentage composition of carbon, hydrogen, nitrogen (Dumas and Kjeldahl methods), halogens, sulphur, and phosphorus in organic compounds.

1. Estimation of Carbon and Hydrogen (Liebig's Combustion Method)

A known mass (m grams) of the organic compound is heated in excess of pure dry dioxygen in the presence of pure copper(II) oxide (CuO). Carbon is quantitatively oxidized to carbon dioxide, and hydrogen is oxidized to water:

CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O
  • Water vapor is absorbed in a pre-weighed U-tube containing anhydrous calcium chloride (CaCl2). Increase in mass = mH2O.
  • Carbon dioxide is absorbed in a pre-weighed bulb containing concentrated potassium hydroxide (KOH) solution. Increase in mass = mCO2.

Mathematical Formulas:

% Carbon = [ 12 / 44 ] × [ mCO2 / m ] × 100
% Hydrogen = [ 2 / 18 ] × [ mH2O / m ] × 100

where m = mass of organic compound taken (g), mCO2 = mass of CO2 produced (g), and mH2O = mass of H2O produced (g).

2. Estimation of Nitrogen

A. Dumas Method

The nitrogen-containing organic compound is heated with excess cupric oxide (CuO) in a combustion tube in an atmosphere of pure CO2 gas. Nitrogen is converted into dinitrogen (N2) gas. Any traces of nitrogen oxides formed are reduced back to N2 by passing over a heated copper gauze:

CxHyNz + (2x + y/2) CuO → x CO2 + (y/2) H2O + (z/2) N2 + (2x + y/2) Cu

The gaseous mixture is collected over concentrated aqueous KOH in a nitrometer. KOH absorbs CO2 completely, leaving moist N2 gas collected at the top.

Dumas Calculation Steps:

  1. Correct observed atmospheric pressure (P1) for aqueous tension (f) at room temperature T1:
    PN2 = P1 − f
  2. Reduce measured volume V1 (mL) of N2 to Standard Temperature and Pressure (STP: P0 = 760 mm Hg, T0 = 273 K):
    VSTP = [ (P1 − f) × V1 × 273 ] / [ 760 × T1 ]
  3. Since 22,400 mL of N2 gas at STP weighs 28 g:
    % Nitrogen = [ 28 / 22400 ] × [ VSTP / m ] × 100

B. Kjeldahl's Method

The organic compound is digested with concentrated H2SO4 in the presence of K2SO4 (boiling point elevator) and CuSO4 (catalyst). Nitrogen is converted quantitatively into ammonium sulphate [(NH4)2SO4]:

Organic compound + Conc. H2SO4 →[CuSO4, K2SO4, Δ] (NH4)2SO4
(NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3↑ + 2H2O

The liberated ammonia gas is distilled into a known volume (V mL) of standard acid (e.g., M molar H2SO4 or HCl). The unreacted residual acid is determined by back-titration against standard NaOH solution.

Kjeldahl Calculation Formula:

Let Vacid = volume of standard acid neutralized strictly by liberated NH3 (in mL) of molarity M (for mono-basic acid like HCl, or normality N):

% Nitrogen = [ 1.4 × N × Vacid (neutralized by NH3) ] / m

where N = normality of the acid and m = mass of organic compound (g).

Limitations of Kjeldahl's Method:
Kjeldahl's method is inapplicable to:
  • Nitro compounds (−NO2, e.g., nitrobenzene)
  • Azo compounds (−N=N−, e.g., azobenzene)
  • Compounds with nitrogen present in aromatic or heterocyclic rings (e.g., pyridine, quinoline)
In these substances, the nitrogen is not converted into ammonium sulphate under standard digestion conditions.

3. Estimation of Halogens (Carius Method)

A known mass (m grams) of organic compound is heated with fuming nitric acid (HNO3) and solid silver nitrate (AgNO3) in a sealed hard glass tube (Carius tube). Carbon and hydrogen are oxidized to CO2 and H2O, while halogen is converted into insoluble silver halide (AgX):

X + AgNO3 →[Fuming HNO3, Δ] AgX↓ (Precipitate)

The precipitate is filtered, washed, dried, and weighed accurately (m1 grams).

Halogen Molar Mass of AgX Percentage Formula
Chlorine (Cl) AgCl = 143.5 g/mol
% Cl = [ 35.5 / 143.5 ] × [ mAgCl / m ] × 100
Bromine (Br) AgBr = 188.0 g/mol
% Br = [ 80.0 / 188.0 ] × [ mAgBr / m ] × 100
Iodine (I) AgI = 235.0 g/mol
% I = [ 127.0 / 235.0 ] × [ mAgI / m ] × 100

4. Estimation of Sulphur and Phosphorus (Carius Method)

A. Sulphur

The organic compound is heated in a Carius tube with fuming HNO3 or sodium peroxide (Na2O2) to oxidize sulphur to sulphuric acid (H2SO4). Barium chloride (BaCl2) solution is added to precipitate barium sulphate (BaSO4, molar mass 233.3 g/mol):

S + fuming HNO3 → H2SO4
H2SO4 + BaCl2 → BaSO4↓ (White ppt) + 2HCl

% Sulphur = [ 32 / 233 ] × [ mBaSO4 / m ] × 100

B. Phosphorus

The compound is heated with fuming HNO3 to oxidize phosphorus into phosphoric acid (H3PO4). It can be precipitated either as:

  • Ammonium Phosphomolybdate: (NH4)3PO4•12MoO3 (molar mass 1877 g/mol):
    % P = [ 31 / 1877 ] × [ mprecipitate / m ] × 100
  • Magnesium Pyrophosphate: H3PO4 is treated with magnesia mixture (MgCl2 + NH4Cl + NH4OH) to precipitate MgNH4PO4, which upon ignition yields Mg2P2O7 (molar mass 222 g/mol):
    % P = [ 62 / 222 ] × [ mMg2P2O7 / m ] × 100

5. Estimation of Oxygen

Percentage of oxygen is usually determined by subtracting the sum of percentages of all other detected elements from 100:

% Oxygen = 100 − [ % C + % H + % N + % Halogens + % S + % P ]

Direct estimation (Unterzaucher method) involves pyrolyzing the compound over carbon at 1373 K to convert all oxygen into CO, which is subsequently oxidized to CO2 by I2O5 and quantified.

6. Interactive Solved JEE Problems

Q1. In a Liebig combustion analysis, 0.246 g of an organic compound gave 0.198 g of CO2 and 0.1014 g of H2O. Calculate the percentages of carbon and hydrogen.
Given: m = 0.246 g, mCO2 = 0.198 g, mH2O = 0.1014 g.
1. % Carbon:
% C = (12 / 44) × (0.198 / 0.246) × 100 = 0.2727 × 0.80488 × 100 = 21.95%.
2. % Hydrogen:
% H = (2 / 18) × (0.1014 / 0.246) × 100 = 0.1111 × 0.4122 × 100 = 4.58%.
Q2. In a Dumas nitrogen estimation, 0.30 g of an organic substance produced 50 mL of moist nitrogen gas at 300 K and 715 mm Hg pressure. Calculate the percentage of nitrogen (Aqueous tension at 300 K = 15 mm Hg).
1. Pressure of dry N2: P1 = 715 − 15 = 700 mm Hg.
2. Volume of N2 at STP:
VSTP = (P1 × V1 × 273) / (760 × T1) = (700 × 50 × 273) / (760 × 300) = 9,555,000 / 228,000 = 41.91 mL.
3. % Nitrogen:
% N = (28 / 22400) × (41.91 / 0.30) × 100 = (0.00125) × 139.7 × 100 = 17.46%.
Q3. During Kjeldahl's estimation of nitrogen, the ammonia evolved from 0.50 g of an organic compound completely neutralized 10 mL of 1 M H2SO4. Find the percentage of nitrogen in the compound.
1. Normality of H2SO4: N = Molarity × Basicity = 1 M × 2 = 2 N.
2. Milligram equivalents of acid neutralized: N × V = 2 N × 10 mL = 20 meq.
3. % Nitrogen:
% N = [ 1.4 × (N × V) ] / m = [ 1.4 × 20 ] / 0.50 = 28 / 0.50 = 56.0%.
Q4. In Carius method for halogen estimation, 0.15 g of an organic compound gave 0.12 g of silver bromide (AgBr). Find the percentage of bromine in the compound (Molar mass of AgBr = 188 g/mol).
Calculation:
% Br = (80 / 188) × (mAgBr / m) × 100
% Br = (80 / 188) × (0.12 / 0.15) × 100 = 0.4255 × 0.80 × 100 = 34.04%.
Q5. 0.32 g of an organic compound on heating with fuming nitric acid and barium chloride in a Carius tube gave 0.466 g of barium sulphate. Calculate the percentage of sulphur in the compound (Molar mass of BaSO4 = 233 g/mol).
Calculation:
% S = (32 / 233) × (mBaSO4 / m) × 100
% S = (32 / 233) × (0.466 / 0.32) × 100 = 0.13734 × 1.45625 × 100 = 20.0%.
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