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Quantitative analysis questions look like a wall of decimals, but each one uses the same idea: turn an element into a compound you can weigh, then work out what fraction of that compound is the element. This post explains the combustion method for carbon and hydrogen, the Carius method for halogens and sulphur, and works one full example from weighings to formula.
Is this still on the syllabus?
Yes. Section 8.10 of the NCERT Class 11 chapter Organic Chemistry: Some Basic Principles and Techniques (Reprint 2026-27) covers carbon, hydrogen, nitrogen, halogens, sulphur and phosphorus. The CBSE Chemistry curriculum for 2026-27 lists "Quantitative Analysis" in Unit 8. The JEE (Main) 2026 syllabus asks for the basic principles of estimating these elements and for "numerical problems in organic quantitative analysis". The JEE (Advanced) 2026 brochure limits itself to empirical and molecular formulae "by combustion method only". Check the syllabus for your exam year.
How does combustion measure carbon and hydrogen?
A weighed sample is burnt in a stream of dry oxygen over copper(II) oxide. All the carbon becomes CO₂ and all the hydrogen becomes H₂O. The gases pass through a U-tube of anhydrous calcium chloride, which absorbs the water, and then through potassium hydroxide solution, which absorbs the CO₂. The gain in mass of each tube gives the mass of each product.
If m g of compound gives m₂ g of CO₂ and m₁ g of H₂O, NCERT writes:
% C = (12 × m₂ × 100) / (44 × m)
% H = (2 × m₁ × 100) / (18 × m)
The 12/44 is the mass of carbon in 44 g of CO₂. The 2/18 is the mass of hydrogen in 18 g of water. The NIOS Senior Secondary chemistry lesson calls this Liebig's combustion method, after Justus von Liebig (1803–1873) of the University of Giessen. A short history of his Kaliapparat, a small instrument, describes it as fundamental to organic analysis for several decades of the nineteenth century, when chemists used it to establish correct empirical formulas.
How does the Carius method measure halogens and sulphur?
For a halogen, the weighed compound is heated with fuming nitric acid and silver nitrate in a sealed hard-glass Carius tube. The halogen ends up as silver halide, AgX, which is filtered, dried and weighed:
% X = (atomic mass of X × m₁ × 100) / (molar mass of AgX × m)
For sulphur, the compound is heated in the Carius tube with fuming nitric acid or sodium peroxide. Sulphur is oxidised to sulphuric acid and precipitated as barium sulphate with barium chloride:
% S = (32 × m₁ × 100) / (233 × m)
Here m₁ is the mass of AgX or BaSO₄. The useful molar masses are AgCl 143.5, AgBr 188, AgI 235 and BaSO₄ 233 g mol⁻¹.
Worked example: from weighings to a formula
These are sample numbers for practice. A 0.218 g sample of a liquid containing C, H and Br gives 0.176 g of CO₂ and 0.090 g of H₂O on combustion. A second 0.218 g sample gives 0.376 g of AgBr in a Carius estimation. Find the percentages and the empirical formula.
Carbon: % C = (12 × 0.176 × 100) / (44 × 0.218) = 211.2 / 9.592 = 22.02%.
Hydrogen: % H = (2 × 0.090 × 100) / (18 × 0.218) = 18 / 3.924 = 4.59%.
Bromine: % Br = (80 × 0.376 × 100) / (188 × 0.218) = 3008 / 40.984 = 73.39%.
Check the total: 22.02 + 4.59 + 73.39 = 100.00%, so there is no oxygen. If the total had been short, the gap would be oxygen by difference.
Divide by atomic masses: C 22.02/12 = 1.835; H 4.59/1 = 4.59; Br 73.39/80 = 0.917.
Divide by the smallest (0.917): C 2.00, H 5.0, Br 1. The empirical formula is C₂H₅Br.
For more practice on step 6, our post on empirical and molecular formula problems works through cases that need multiplying by 2 or 3.
Check yourself against NCERT
NCERT's solved problems are a good test of your formulas. Problem 8.20: 0.246 g of compound gives 0.198 g CO₂ and 0.1014 g H₂O, which is 21.95% C and 4.58% H. Problem 8.23: 0.15 g gives 0.12 g AgBr, which is 34.04% Br. Problem 8.24: 0.157 g gives 0.4813 g BaSO₄, which is 42.10% S. If your numbers differ, find the step before reading on.
Which mistakes cost marks?
Using the wrong fraction. Hydrogen is 2/18 of water, not 1/18. Carbon is 12/44 of CO₂, not 12/28.
Using the wrong silver halide. Check whether the halogen is Cl, Br or I before choosing 143.5, 188 or 235.
Mixing up the two samples. Combustion and Carius use separate weighed samples; each percentage uses its own m.
Rounding too early. Keep four significant figures until the last step, or the empirical formula ratios drift from whole numbers.
Forgetting oxygen. It is never measured directly in these questions; it is 100 minus the rest.
A practice routine
Day 1: Write the four formulas from the one idea in the callout, without looking.
Day 2: Solve NCERT Problems 8.20, 8.23 and 8.24 and match the book's answers.
Day 3: Make up a compound, such as C₂H₄Cl₂, calculate the masses of CO₂, H₂O and AgCl it would give, then solve your own question backwards.
Day 4: Mixed set of six questions, including one where oxygen comes by difference.
Before these tests, make sure you can detect the elements in the first place; our post on Lassaigne's test covers that step. If the set-up keeps going wrong rather than the arithmetic, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see why each mass ratio is used.
Tonight, redo the worked example with 0.436 g of the same liquid and check that every percentage stays the same.
