JEE Mains Chemistry · Organic Chemistry - Some Basic Principles and Techniques

Estimation of Carbon, Hydrogen, Halogens, Sulphur and Phosphorus

Each element is turned into a product that can be weighed (CO₂, water, a silver halide, barium sulphate or magnesium pyrophosphate), and its percentage is the element's mass fraction in that product times the product's mass over the sample's mass.

All Organic Chemistry - Some Basic Principles and Techniques notes JEE Mains Chemistry strategy

Why this matters

Thirty-three PYQs, twenty-four of them asking for a number, and eleven from 2026, the most of any page in this chapter. Eleven are combustion analysis: carbon from CO₂, hydrogen from water, and oxygen by difference. Eleven are Carius estimations of chlorine, bromine or iodine as a silver halide. Eleven estimate sulphur as barium sulphate or phosphorus as magnesium pyrophosphate, or ask what the Carius tube is used for.

Concept 1 of 3: Combustion analysis for carbon, hydrogen and oxygen

Burn a weighed sample completely over copper(II) oxide. All its carbon becomes CO₂ and all its hydrogen becomes water. Two absorption tubes catch these, and each tube's gain in mass is the mass of the product it caught.

Definition

Water is caught first, in a U-tube of anhydrous $\mathrm{CaCl_2}$ ; $\mathrm{CO_2}$ next, in a U-tube of concentrated KOH (the potash tube).
Carbon is 12/44 of the mass of $\mathrm{CO_2}$ ; hydrogen is 2/18 of the mass of water.
Oxygen is found by difference: %O = 100 − (the sum of all the other percentages).
Empirical formula: divide each percentage by the atomic mass, then divide by the smallest result and round to whole numbers.
Many numerical questions ask for the answer as '× 10⁻¹' or '× 10⁻³ g': read the scale before you round.

Percentages of carbon and hydrogen

\%\,\mathrm{C} = \dfrac{12}{44} \times \dfrac{m_{\mathrm{CO_2}}}{m} \times 100 \qquad \%\,\mathrm{H} = \dfrac{2}{18} \times \dfrac{m_{\mathrm{H_2O}}}{m} \times 100

Worked example

0.30 g of a compound of carbon, hydrogen and oxygen gives 0.66 g of

\mathrm{CO_2}

and 0.27 g of water on complete combustion. Find the percentages of C, H and O.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q46Moderate

Example 1 · Organic Chemistry - Some Basic Principles and Techniques · Estimation of Carbon, Hydrogen, Halogens, Sulphur and Phosphorus

0.25 g of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of

{CaCl}_{2}

tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g, respectively. The percentage of oxygen in compound A is

\_\_\_\_

%. (Nearest integer) (Given: molar mass in

g{mol}^{- 1}H:1,C:12,O:16

)

Hydrogen is 2/18 of water, not 1/18

Each water molecule carries two hydrogen atoms. Using 1/18 halves the percentage of hydrogen and then throws off the oxygen found by difference.

Oxygen is never weighed directly

Combustion analysis gives only C and H. Oxygen comes from 100 minus everything else, so an error in carbon or hydrogen carries straight into it.

Match the tube to the gas

The

\mathrm{CaCl_2}

tube gains the mass of water and the potash (KOH) tube the mass of

\mathrm{CO_2}

. Swapping them swaps the two fractions, 12/44 and 2/18.

Drill 10 more on combustion analysis for carbon, hydrogen and oxygen

Concept 2 of 3: Carius method for halogens

Heat the compound with fuming nitric acid and silver nitrate in a sealed tube. Carbon and hydrogen are oxidised away, and every halogen atom ends up in a silver halide precipitate, which is filtered, dried and weighed.

Definition

A known mass of the compound is heated with fuming $\mathrm{HNO_3}$ and $\mathrm{AgNO_3}$ in a sealed hard-glass tube, the Carius tube.
The halogen is weighed as AgCl (143.5), AgBr (188) or AgI (235), with Cl = 35.5, Br = 80 and I = 127.
Which halogen it is can be settled first by Lassaigne's test.
Combined with a known percentage of carbon, the halogen percentage gives the empirical formula.

Percentage of halogen (Carius)

\%\,\mathrm{X} = \dfrac{\text{atomic mass of X}}{\text{molar mass of AgX}} \times \dfrac{m_{\mathrm{AgX}}}{m} \times 100

Worked example

In a Carius estimation, 0.20 g of an organic compound gives 0.287 g of AgCl. Find the percentage of chlorine (AgCl = 143.5 g mol⁻¹, Cl = 35.5 g mol⁻¹).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q40Moderate

Example 2 · Organic Chemistry - Some Basic Principles and Techniques · Estimation of Carbon, Hydrogen, Halogens, Sulphur and Phosphorus

In Carius method 0.2425 g of an organic compounds gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is

Divide by the silver halide's molar mass

The fraction is the halogen's atomic mass over the molar mass of the WHOLE precipitate: 35.5/143.5 for AgCl, not 35.5/108.

Keep the three silver halides apart

AgCl is 143.5, AgBr 188 and AgI 235 g mol⁻¹. Pairing bromine with 143.5 is the commonest slip, and it gives an answer that is often among the options.

Carius does not estimate nitrogen

The Carius tube is used for halogens, sulphur and phosphorus. Nitrogen is estimated by Dumas' or Kjeldahl's method.

Drill 10 more on carius method for halogens

Concept 3 of 3: Estimation of sulphur and phosphorus

The route is the same as for halogens: oxidise the compound in a sealed tube and trap the element in a precipitate that can be weighed. Sulphur becomes sulphuric acid and is weighed as barium sulphate; phosphorus becomes phosphoric acid and is weighed as magnesium pyrophosphate or ammonium phosphomolybdate.

Definition

Sulphur: heat with fuming $\mathrm{HNO_3}$ (or sodium peroxide) in a Carius tube to give $\mathrm{H_2SO_4}$ ; add $\mathrm{BaCl_2}$ to precipitate $\mathrm{BaSO_4}$ (233 g mol⁻¹). Each $\mathrm{BaSO_4}$ carries one S.
Phosphorus: heat with fuming $\mathrm{HNO_3}$ to give $\mathrm{H_3PO_4}$ . Precipitate it as ammonium phosphomolybdate, $\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}$ , or as $\mathrm{MgNH_4PO_4}$ , which gives $\mathrm{Mg_2P_2O_7}$ (222 g mol⁻¹) on ignition. Each $\mathrm{Mg_2P_2O_7}$ carries TWO P, so the fraction is 62/222.
Methionine and cysteine are the amino acids that contain sulphur.
When a question gives its own molar mass for the precipitate, use it.

Percentages of sulphur and phosphorus

\%\,\mathrm{S} = \dfrac{32}{233} \times \dfrac{m_{\mathrm{BaSO_4}}}{m} \times 100 \qquad \%\,\mathrm{P} = \dfrac{62}{222} \times \dfrac{m_{\mathrm{Mg_2P_2O_7}}}{m} \times 100

Worked example

In a Carius estimation, 0.40 g of an organic compound gives 0.466 g of

\mathrm{BaSO_4}

. Find the percentage of sulphur (

\mathrm{BaSO_4}

= 233 g mol⁻¹, S = 32 g mol⁻¹).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q33Moderate

Example 3 · Organic Chemistry - Some Basic Principles and Techniques · Estimation of Carbon, Hydrogen, Halogens, Sulphur and Phosphorus

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass

32\text{ }g{\text{ }mol}^{- 1}

). Molar mass of barium sulphate is

233\text{ }g{\text{ }mol}^{- 1}

Two phosphorus atoms per pyrophosphate

\mathrm{Mg_2P_2O_7}

holds two P atoms, so phosphorus is 62/222 of its mass. Using 31/222 halves the answer.

Use the molar mass the question gives

Some papers print an unusual molar mass for

\mathrm{BaSO_4}

or for the pyrophosphate. The key is worked with the printed value, so use it even if it differs from 233 or 222.

Sulphur is weighed as a barium salt

Silver nitrate traps halogens; barium chloride traps sulphate. A question that adds

\mathrm{BaCl_2}

is estimating sulphur, and the precipitate is

\mathrm{BaSO_4}

Drill 10 more on estimation of sulphur and phosphorus

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Combustion analysis for carbon, hydrogen and oxygen
Percentages of carbon and hydrogen
$\%\,\mathrm{C} = \dfrac{12}{44} \times \dfrac{m_{\mathrm{CO_2}}}{m} \times 100 \qquad \%\,\mathrm{H} = \dfrac{2}{18} \times \dfrac{m_{\mathrm{H_2O}}}{m} \times 100$
Carius method for halogens
Percentage of halogen (Carius)
$\%\,\mathrm{X} = \dfrac{\text{atomic mass of X}}{\text{molar mass of AgX}} \times \dfrac{m_{\mathrm{AgX}}}{m} \times 100$
Estimation of sulphur and phosphorus
Percentages of sulphur and phosphorus
$\%\,\mathrm{S} = \dfrac{32}{233} \times \dfrac{m_{\mathrm{BaSO_4}}}{m} \times 100 \qquad \%\,\mathrm{P} = \dfrac{62}{222} \times \dfrac{m_{\mathrm{Mg_2P_2O_7}}}{m} \times 100$

Watch out for (9)

Hydrogen is 2/18 of water, not 1/18
→ Combustion analysis for carbon, hydrogen and oxygen
Oxygen is never weighed directly
→ Combustion analysis for carbon, hydrogen and oxygen
Match the tube to the gas
→ Combustion analysis for carbon, hydrogen and oxygen
Divide by the silver halide's molar mass
→ Carius method for halogens
Keep the three silver halides apart
→ Carius method for halogens
Carius does not estimate nitrogen
→ Carius method for halogens
Two phosphorus atoms per pyrophosphate
→ Estimation of sulphur and phosphorus
Use the molar mass the question gives
→ Estimation of sulphur and phosphorus
Sulphur is weighed as a barium salt
→ Estimation of sulphur and phosphorus

Test yourself on Organic Chemistry - Some Basic Principles and Techniques

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Organic Chemistry - Some Basic Principles and Techniques chapter test Drill all 33 questions