Carius Method: Halogen and Sulphur Calculations

By Aniket Bhardwaj · 29 October 2026 · Organic

Quantitative elemental analysis is an essential part of organic chemistry. In entrance and board examinations, questions on determining the percentage composition of halogens and sulphur in organic compounds appear regularly. The Carius method is the classical analytical procedure taught in Class 11 and Class 12 chemistry for this purpose. Many students find these questions easy to score on, but arithmetic slips, wrong molar masses, or swapping atomic weights often lead to lost marks. Understanding the chemistry behind the method and learning a steady step-by-step calculation pattern ensures complete accuracy under exam pressure.

Principles of the Carius Method

In the Carius method, a known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate inside a sealed hard-glass tube known as a Carius tube. Fuming nitric acid acts as a powerful oxidising agent. It oxidises carbon to carbon dioxide and hydrogen to water.

When the compound contains a halogen (chlorine, bromine, or iodine), the halogen reacts with silver nitrate to form a precipitate of silver halide. Silver chloride forms as a white precipitate, silver bromide as a pale yellow precipitate, and silver iodide as a yellow precipitate. The precipitate is filtered, washed thoroughly with dilute nitric acid and distilled water, dried, and weighed accurately on an analytical balance.

When the organic compound contains sulphur, it is heated in the Carius tube with fuming nitric acid alone, or with sodium peroxide. The sulphur is quantitatively oxidised to sulphuric acid. Barium chloride solution is then added to precipitate the sulphate ions as insoluble barium sulphate. The barium sulphate precipitate is filtered, washed, ignited, and weighed. From the mass of the precipitate obtained, the mass and percentage of sulphur in the original organic sample are determined.

Standard IUPAC Atomic Masses

Always use the standard atomic masses recommended by IUPAC and standard textbooks:

Carbon (C): 12.011 g/mol (commonly taken as 12.0 g/mol)
Hydrogen (H): 1.008 g/mol (commonly taken as 1.0 g/mol)
Nitrogen (N): 14.007 g/mol (commonly taken as 14.0 g/mol)
Oxygen (O): 15.999 g/mol (commonly taken as 16.0 g/mol)
Sulphur (S): 32.06 g/mol (commonly taken as 32.0 g/mol)
Chlorine (Cl): 35.45 g/mol (commonly taken as 35.5 g/mol)
Bromine (Br): 79.904 g/mol (commonly taken as 80.0 g/mol)
Iodine (I): 126.90 g/mol (commonly taken as 127.0 g/mol)
Silver (Ag): 107.87 g/mol (commonly taken as 108.0 g/mol)
Barium (Ba): 137.33 g/mol (commonly taken as 137.3 g/mol or 137.0 g/mol)

Molar masses of precipitates used in standard school calculations:

Silver chloride (AgCl): 108 + 35.5 = 143.5 g/mol
Silver bromide (AgBr): 108 + 80 = 188.0 g/mol
Silver iodide (AgI): 108 + 127 = 235.0 g/mol
Barium sulphate (BaSO4): 137 + 32 + 4(16) = 233.0 g/mol (using Ba = 137.3 gives 233.3 g/mol; textbooks commonly use 233 g/mol)

Key Formulas for the Carius Method:

Percentage of Chlorine = [Atomic mass of Cl / Molar mass of AgCl] × [Mass of AgCl / Mass of organic compound] × 100
Percentage of Chlorine = [35.5 / 143.5] × [m(AgCl) / m(compound)] × 100

Percentage of Bromine = [Atomic mass of Br / Molar mass of AgBr] × [Mass of AgBr / Mass of organic compound] × 100
Percentage of Bromine = [80 / 188] × [m(AgBr) / m(compound)] × 100

Percentage of Iodine = [Atomic mass of I / Molar mass of AgI] × [Mass of AgI / Mass of organic compound] × 100
Percentage of Iodine = [127 / 235] × [m(AgI) / m(compound)] × 100

Percentage of Sulphur = [Atomic mass of S / Molar mass of BaSO4] × [Mass of BaSO4 / Mass of organic compound] × 100
Percentage of Sulphur = [32 / 233] × [m(BaSO4) / m(compound)] × 100

Variables and units: m(compound) is the initial sample mass in grams (g); m(AgX) and m(BaSO4) are the precipitate masses in grams (g); result is expressed in percentage (%).

Exam details

Source: practice problem written for this article (not a past-paper question)
Relevant exams: CBSE Class 11, JEE Main, NEET, CUET-UG, State Board Examinations
Topic: Quantitative Organic Analysis (Halogen and Sulphur Estimation)
Difficulty: School to Foundation Competitive Level

Worked Numerical Examples

Question: In a Carius estimation of halogen, 0.240 g of an organic compound gave 0.287 g of silver chloride (AgCl). Calculate the percentage of chlorine present in the organic compound.

Solution:
Step 1: Identify given values.
Mass of organic compound = 0.240 g
Mass of silver chloride (AgCl) obtained = 0.287 g

Step 2: Find the molar masses.
Atomic mass of chlorine (Cl) = 35.5 g/mol
Molar mass of silver chloride (AgCl) = 108 + 35.5 = 143.5 g/mol

Step 3: Calculate the mass of chlorine in the precipitate.
143.5 g of AgCl contains 35.5 g of chlorine.
Mass of chlorine = (35.5 / 143.5) × 0.287 g
Notice that 0.287 / 143.5 = 0.002 mol.
Mass of chlorine = 35.5 × 0.002 = 0.071 g.

Step 4: Calculate the percentage of chlorine in the compound.
Percentage of chlorine = (Mass of chlorine / Mass of organic compound) × 100
Percentage of chlorine = (0.071 g / 0.240 g) × 100
0.071 / 0.240 = 0.295833...
0.295833... × 100 = 29.58%

Double-check verification via direct formula:
% Cl = (35.5 × 0.287 × 100) / (143.5 × 0.240) = 1018.85 / 34.44 = 29.58%

Answer: 29.58%

Question: During a Carius estimation of bromine, 0.376 g of an organic compound yielded 0.376 g of silver bromide (AgBr). Determine the percentage of bromine in the compound.

Solution:
Step 1: Identify given values.
Mass of organic compound = 0.376 g
Mass of silver bromide (AgBr) obtained = 0.376 g

Step 2: Find the molar masses.
Atomic mass of bromine (Br) = 80 g/mol
Molar mass of silver bromide (AgBr) = 108 + 80 = 188 g/mol

Step 3: Calculate the mass of bromine in the precipitate.
188 g of AgBr contains 80 g of bromine.
Mass of bromine = (80 / 188) × 0.376 g
Notice that 0.376 / 188 = 0.002 mol.
Mass of bromine = 80 × 0.002 = 0.160 g.

Step 4: Compute the percentage of bromine in the sample.
Percentage of bromine = (0.160 g / 0.376 g) × 100
0.160 / 0.376 = 160 / 376 = 20 / 47 ≈ 0.42553
0.42553 × 100 = 42.55%

Double-check verification via alternative route:
% Br = [80 × 0.376 / (188 × 0.376)] × 100
Here the mass of compound and mass of AgBr cancel out directly!
% Br = (80 / 188) × 100 = 8000 / 188 = 42.55%

Answer: 42.55%

Question: In the estimation of sulphur by the Carius method, 0.300 g of an organic substance gave 0.699 g of barium sulphate (BaSO4). Calculate the percentage of sulphur in the substance.

Solution:
Step 1: Identify given values.
Mass of organic compound = 0.300 g
Mass of barium sulphate (BaSO4) = 0.699 g

Step 2: Find the molar masses.
Atomic mass of sulphur (S) = 32 g/mol
Molar mass of BaSO4 = 137 + 32 + (4 × 16) = 233 g/mol

Step 3: Calculate the mass of sulphur present in 0.699 g of BaSO4.
233 g of BaSO4 contains 32 g of sulphur.
Mass of sulphur = (32 / 233) × 0.699 g
Notice that 0.699 / 233 = 0.003 mol.
Mass of sulphur = 32 × 0.003 = 0.096 g.

Step 4: Calculate the percentage of sulphur in the organic substance.
Percentage of sulphur = (0.096 g / 0.300 g) × 100
0.096 / 0.300 = 96 / 300 = 0.32
0.32 × 100 = 32.00%

Double-check verification using the comprehensive formula:
% S = (32 × 0.699 × 100) / (233 × 0.300) = 2236.8 / 69.9 = 32.00%

Answer: 32.00%

Common Mistakes Students Make

Inverting the molar mass fraction: Writing 143.5 / 35.5 instead of 35.5 / 143.5. Remember that the mass of the element must always be smaller than the mass of the precipitate.
Confusing precipitate formulas: Taking AgCl2 or Ba(SO4)2 instead of AgCl and BaSO4. Halogens are univalent in silver halides (AgX), and barium sulphate has the formula BaSO4.
Arithmetic rounding too early: Rounding intermediate values like moles to one decimal place before multiplying by 100 leads to incorrect final digits in numerical value questions.
Mixing up sample mass and precipitate mass: Placing the precipitate mass in the denominator and the sample mass in the numerator. Always divide by the mass of the original organic compound.
Using molecular mass instead of atomic mass: Using 71 for Cl2 instead of 35.5 for atomic Cl, or 64 for S2 instead of 32 for S. The percentage refers to the element, not a diatomic gas.

Exam Relevance and Applications

Examination Type	Question Format	Core Skills Evaluated
CBSE & State Boards (Class 11)	Short answer theory and two-mark numericals	Reactions in Carius tube, direct percentage formula application
JEE Main	Single correct options and integer-type numericals	Precise arithmetic calculation, percentage of halogen/sulphur, empirical formula derivation
NEET	Direct formula single-choice objective questions	Rapid identification of precipitate type and quick molar ratio calculations
CUET-UG	Multiple-choice conceptual and calculation questions	Identification of precipitates, colours, and stoichiometric conversions

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