The percentage formulae for carbon and hydrogen look interchangeable and are not. Carbon is weighed as CO₂, whose molar mass is 44; hydrogen is weighed as H₂O, molar mass 18. But hydrogen appears twice in a water molecule, so the mass of H in 18 g of H₂O is 2, not 1. Swapping 44 and 18 — or writing 1/18 instead of 2/18 — is the single most productive source of wrong options in this topic.
NCERT Class 11 Chemistry, Chapter 8 (Organic Chemistry — Some Basic Principles and Techniques), pages 285–286, set out the Liebig combustion method: a known mass of the compound is burnt in excess oxygen, the water is absorbed in anhydrous CaCl₂ and the carbon dioxide in KOH, and each absorber is weighed. From those two masses,
% C = (12 / 44) × (mass of CO₂ / mass of compound) × 100 % H = (2 / 18) × (mass of H₂O / mass of compound) × 100
Nitrogen has two routes, on pages 286–288. Dumas' method burns the compound with CuO, converts any oxides of nitrogen back to N₂ over hot copper gauze, and collects the N₂ over an aqueous alkali solution. The collected gas is moist, so the measured pressure must have the aqueous tension subtracted before the volume is reduced to STP. Kjeldahl's method digests the compound with concentrated H₂SO₄, converting nitrogen to ammonium sulphate, then liberates NH₃ with alkali and traps it in a known excess of standard acid; the unreacted acid is back-titrated. Kjeldahl's is not applicable to nitrogen in a ring (pyridine), or to nitro and azo compounds — that nitrogen does not convert to ammonium sulphate.
Watch-out: oxygen is never estimated directly. It is found by difference, after every other element has been determined.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In the Liebig combustion method for carbon and hydrogen, the water produced and the carbon dioxide produced are absorbed respectively in:
Answer: B. B is correct. Anhydrous CaCl₂ is the drying agent that absorbs water; KOH absorbs the acidic gas CO₂. This pairing is stated in NCERT Class 11 Chemistry, Chapter 8, page 286.
Why A is wrong: A is wrong because the absorbers are reversed: KOH is alkaline and absorbs the acidic CO₂, while anhydrous CaCl₂ is the desiccant for water.
Why C is wrong: C is wrong because copper(II) oxide is the oxidising agent used inside the combustion tube, not an absorber; the absorbers are weighed after combustion.
Why D is wrong: D is wrong because sodium metal belongs to Lassaigne's qualitative test, and lime water is a detection reagent for CO₂, not a weighable quantitative absorber.
On complete combustion, 0.246 g of an organic compound gave 0.198 g of carbon dioxide. The percentage of carbon in the compound is (atomic masses: C = 12, O = 16):
Answer: D. D is correct. % C = (12/44) × (0.198/0.246) × 100 = 0.2727 × 0.8049 × 100 = 21.95 ≈ 22.0%. The formula is given in NCERT Class 11 Chemistry, Chapter 8, page 286.
Why A is wrong: A is wrong because 80.5% is simply (0.198/0.246) × 100 — the whole mass of CO₂ treated as if it were carbon, with the 12/44 conversion omitted.
Why B is wrong: B is wrong because 43.9% is twice the correct value: it divides by 22 instead of 44 for CO₂ (12/22 × 0.198/0.246 × 100 = 43.9).
Why C is wrong: C is wrong because 12.0% is the result of dividing by 100 instead of by the molar mass of CO₂; 12/44 is a ratio, not a percentage in itself.
0.20 g of an organic compound on combustion gave 0.12 g of water. The percentage of hydrogen in the compound is (atomic masses: H = 1, O = 16):
Answer: B. B is correct. % H = (2/18) × (0.12/0.20) × 100 = 0.1111 × 0.60 × 100 = 6.7%. The factor is 2/18, not 1/18, because each H₂O molecule carries two hydrogen atoms — see NCERT Class 11 Chemistry, Chapter 8, page 286.
Why A is wrong: A is wrong because 3.3% uses 1/18 instead of 2/18, counting only one hydrogen atom per water molecule.
Why C is wrong: C is wrong because 13.3% doubles the correct answer, using 4/18 — as though water carried four hydrogen atoms.
Why D is wrong: D is wrong because 60.0% is (0.12/0.20) × 100, the full mass of water treated as if it were all hydrogen.
In Kjeldahl's method, the nitrogen of the organic compound is first converted to:
Answer: D. D is correct. Digestion with concentrated H₂SO₄ (with K₂SO₄ and CuSO₄) converts the nitrogen to ammonium sulphate, which is the whole basis of the subsequent liberation of ammonia. NCERT Class 11 Chemistry, Chapter 8, page 287.
Why A is wrong: A is wrong because collecting nitrogen gas over aqueous alkali is Dumas' method, a separate quantitative route on page 287.
Why B is wrong: B is wrong because sodium fusion to sodium cyanide is Lassaigne's qualitative detection, which tells you nitrogen is present but not how much.
Why C is wrong: C is wrong because oxides of nitrogen are an intermediate in Dumas' method, where they are then reduced back to N₂ over hot copper; Kjeldahl's has no such step.
Kjeldahl's method would give a falsely low nitrogen percentage for which of the following compounds?
Answer: A. A is correct. Pyridine's nitrogen is part of the aromatic ring, and ring nitrogen is not converted to ammonium sulphate during acid digestion — so the method under-reports it. NCERT Class 11 Chemistry, Chapter 8, page 288, lists ring nitrogen alongside nitro and azo compounds as the exclusions.
Why B is wrong: B is wrong because amide nitrogen is converted quantitatively to ammonium sulphate under Kjeldahl digestion — ethanamide is a standard Kjeldahl substrate.
Why C is wrong: C is wrong because the nitrogen in aniline is an amine nitrogen attached to the ring, not inside it; amine nitrogen digests normally to ammonium sulphate.
Why D is wrong: D is wrong because urea's two amide-type nitrogens digest completely; urea is in fact a common standard for checking Kjeldahl apparatus.
In Dumas' method, the nitrogen liberated is collected over an aqueous solution of potassium hydroxide. Before the volume is reduced to standard conditions, the measured pressure must be corrected by:
Answer: C. C is correct. The collected gas is saturated with water vapour, so the total measured pressure is the pressure of nitrogen plus the aqueous tension; subtracting the latter leaves the true nitrogen pressure. NCERT Class 11 Chemistry, Chapter 8, page 286.
Why A is wrong: A is wrong because it reverses the correction. Adding aqueous tension inflates the nitrogen pressure and gives a nitrogen percentage that is too high.
Why B is wrong: B is wrong because the CO₂ has been absorbed by the alkali and exerts no pressure in the collected gas; there is nothing to add back.
Why D is wrong: D is wrong because although KOH itself is non-volatile, the water of the aqueous solution is not — the gas above it is still saturated with water vapour.
0.50 g of an organic compound was digested by Kjeldahl's method. The ammonia liberated was absorbed in 5.0 × 10¹ cm³ of 0.10 M H₂SO₄, and the excess acid required 5.0 × 10¹ cm³ of 0.10 M NaOH for neutralisation. The percentage of nitrogen in the compound is (atomic mass: N = 14):
Answer: C. C is correct. Work in moles of H⁺ throughout, because H₂SO₄ is dibasic. Acid taken = 0.0500 dm³ × 0.10 M = 5.0 × 10⁻³ mol H₂SO₄ = 1.0 × 10⁻² mol H⁺. The excess is found from the alkali: 0.0500 dm³ × 0.10 M = 5.0 × 10⁻³ mol NaOH, so 5.0 × 10⁻³ mol H⁺ was left unreacted. Acid neutralised by the ammonia = 1.0 × 10⁻² − 5.0 × 10⁻³ = 5.0 × 10⁻³ mol H⁺, and since NH₃ + H⁺ → NH₄⁺ this is 5.0 × 10⁻³ mol NH₃, hence 5.0 × 10⁻³ mol N. Mass of N = 5.0 × 10⁻³ × 14 = 0.070 g, so % N = (0.070 / 0.50) × 100 = 14.0%. NCERT Class 11 Chemistry, Chapter 8, page 287.
Why A is wrong: A is wrong because 7.0% works in moles of H₂SO₄ instead of moles of H⁺, halving the nitrogen. Each mole of the dibasic acid supplies two moles of H⁺ and so neutralises two moles of ammonia.
Why B is wrong: B is wrong because 28.0% uses the whole 1.0 × 10⁻² mol of H⁺ taken and never subtracts the excess found by the back-titration. Only the acid the ammonia actually consumed counts.
Why D is wrong: D is wrong because 42.0% adds the excess to the acid taken instead of subtracting it. The back-titration measures what was left over, so it must come off the total.
An organic compound contains carbon, hydrogen and oxygen only. Combustion of 0.100 g of it gave 0.147 g of carbon dioxide and 0.0900 g of water. The percentage of oxygen in the compound is (atomic masses: C = 12, H = 1, O = 16):
Answer: A. A is correct. % C = (12/44) × (0.147/0.100) × 100 = 40.1%; % H = (2/18) × (0.0900/0.100) × 100 = 10.0%. Oxygen is not measured, it is found by difference: 100 − 40.1 − 10.0 = 49.9%. NCERT Class 11 Chemistry, Chapter 8, page 290, states that the percentage of oxygen is usually found by difference between the total percentage composition and the sum of the percentages of all other elements.
Why B is wrong: B is wrong because the oxygen being estimated is the oxygen inside the compound, obtained by difference from the other elements. The oxygen supplied from the air for the combustion does not enter the calculation at all.
Why C is wrong: C is wrong because 40.1% is the percentage of carbon. It is the right number answering the wrong question.
Why D is wrong: D is wrong because 59.9% subtracts only the carbon term, 100 − 40.1, and forgets the 10.0% hydrogen. Every other element must come off before the remainder is oxygen.
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Given
Mass of organic compound burnt = 0.2160 g
Mass of carbon dioxide formed = 0.3300 g
Mass of water formed = 0.1350 g
Atomic masses (exact, by definition of the scale used here): C = 12, H = 1, O = 16
Required
The percentage by mass of carbon and of hydrogen in the compound, and the percentage of oxygen if the compound contains only C, H and O.
Concept
In Liebig's combustion method every carbon atom of the compound ends up in one molecule of CO₂ and every two hydrogen atoms end up in one molecule of H₂O. So the mass of carbon in the sample equals the carbon fraction of the CO₂ collected, and the mass of hydrogen equals the hydrogen fraction of the H₂O collected. Oxygen cannot be measured this way — combustion supplies oxygen from the air — so it is obtained by difference.
Formula
% C = (12 / 44) × (mass of CO₂ / mass of compound) × 100
% H = (2 / 18) × (mass of H₂O / mass of compound) × 100
% O = 100 − (% C + % H)
Substitution
% C = (12 / 44) × (0.3300 / 0.2160) × 100
% H = (2 / 18) × (0.1350 / 0.2160) × 100
Calculation
12/44 = 0.27273 and 0.3300/0.2160 = 1.5278, so % C = 0.27273 × 1.5278 × 100 = 41.67%
2/18 = 0.11111 and 0.1350/0.2160 = 0.62500, so % H = 0.11111 × 0.62500 × 100 = 6.944%
% O = 100 − (41.67 + 6.94) = 51.39%
The molar masses 44 and 18 and the atomic masses 12, 1 and 16 are exact defined values on the relative-mass scale, as is the factor 2 counting hydrogen atoms per water molecule and the factor 100 converting to per cent. None of them limits the significant figures. Only the three weighed masses do, and each carries four significant figures.
Final answer
% C = 41.67%, % H = 6.944%, % O = 51.39% (each to four significant figures, set by the four-significant-figure weighings).
Common trap
The trap that costs most marks here is putting 18 under the carbon dioxide or 44 under the water — the two formulae have the same shape and differ only in the constants. A second, subtler one is writing 1/18 for hydrogen. Both give a plausible-looking percentage, and both appear as options. Before substituting, say aloud which gas you weighed: CO₂ goes with 12/44, H₂O goes with 2/18.
Similar NEET-style question
0.1580 g of an organic compound containing only carbon, hydrogen and oxygen gave 0.2320 g of carbon dioxide and 0.0950 g of water on complete combustion. Calculate the percentage of each of the three elements. *(Answer: C = (12/44) × (0.2320/0.1580) × 100 = 40.05%; H = (2/18) × (0.0950/0.1580) × 100 = 0.11111 × 0.60127 × 100 = 6.68%; O = 100 − 40.05 − 6.68 = 53.27%.)*
Quantitative analysis of compounds is very important in organic chemistry. It helps chemists in the determination of mass per cent of elements present in a compound. You have learnt in Unit-1 that mass per cent of elements is required for the determination of emperical and molecular formula. The percentage composition of elements present in an organic compound is determined by the following methods:
-- NCERT Class 11 Chemistry, Ch. 8, p. 285Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively. CxHy + (x + y/4) O2 -----> x CO2 + (y/2) H2O The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Fig. 8.14). The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated. Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m1 and m2 g respectively; Percentage of carbon = (12 x m2 x 100) / (44 x m) Percentage of hydrogen = (2 x m1 x 100) / (18 x m) Fig.8.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of substance are absorbed in anhydrous calcium chloride and potassium hydroxide solutions respectively contained in U tubes.
-- NCERT Class 11 Chemistry, Ch. 8, p. 286Problem 8.20 On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound. Solution Percentage of carbon = (12 x 0.198 x 100) / (44 x 0.246) = 21.95% Percentage of hydrogen = (2 x 0.1014 x 100) / (18 x 0.246) = 4.58%
-- NCERT Class 11 Chemistry, Ch. 8, p. 286There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl's method. (i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water. CxHyNz + (2x + y/2) CuO -----> x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig.8.15). Let the mass of organic compound = m g Volume of nitrogen collected = V1 mL Room temperature = T1 K Volume of nitrogen at STP = (P1V1 x 273) / (760 x T1) (Let it be V mL) Where p1 and V1 are the pressure and volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas is collected. The value of p1 is obtained by the relation; p1 = Atmospheric pressure - Aqueous tension 22400 mL N2 at STP weighs 28 g. V mL N2 at STP weighs = (28 x V) / 22400 g Percentage of nitrogen = (28 x V x 100) / (22400 x m) Fig. 8.15 Dumas method. The organic compound yields nitrogen gas on heating it with copper(II) oxide in the presence of CO2 gas. The mixture of gases is collected over potassium hydroxide solution in which CO2 is absorbed and volume of nitrogen gas is determined.
-- NCERT Class 11 Chemistry, Ch. 8, p. 287Problem 8.21 In Dumas' method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm) Solution Volume of nitrogen collected at 300K and 715mm pressure is 50 mL Actual pressure = 715-15 = 700 mm Volume of nitrogen at STP = (273 x 700 x 50) / (300 x 760) = 41.9 mL 22,400 mL of N2 at STP weighs = 28 g 41.9 mL of nitrogen weighs = (28 x 41.9) / 22400 g Percentage of nitrogen = (28 x 41.9 x 100) / (22400 x 0.3) = 17.46%
-- NCERT Class 11 Chemistry, Ch. 8, p. 287(ii) Kjeldahl's method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Fig. 8.16). The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia. Organic compound + H2SO4 -----> (NH4)2SO4 --(2NaOH)--> Na2SO4 + 2NH3 + 2H2O 2NH3 + H2SO4 -----> (NH4)2SO4 Let the mass of organic compound taken = m g Volume of H2SO4 of molarity, M, taken = V mL Volume of NaOH of molarity, M, used for titration of excess of H2SO4 = V1 mL V1 mL of NaOH of molarity M = V1/2 mL of H2SO4 of molarity M Volume of H2SO4 of molarity M unused = (V - V1/2) mL (V - V1/2) mL of H2SO4 of molarity M = 2(V-V1/2) mL of NH3 solution of molarity M. 1000 mL of 1 M NH3 solution contains 17g NH3 or 14 g of N 2(V-V1/2) mL of NH3 solution of molarity M contains: (14 x M x 2(V - V1/2)) / 1000 g N Percentage of N = [(14 x M x 2(V - V1/2)) / 1000] x (100 / m) = (1.4 x M x 2(V - V/2)) / m Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions. Fig.8.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO4 to get ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed in known volume of standard acid.
-- NCERT Class 11 Chemistry, Ch. 8, p. 287Problem 8.22 During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.5 g of the compound in Kjeldahl's estimation of nitrogen, neutralized 10 mL of 1 M H2SO4. Find out the percentage of nitrogen in the compound. Solution 1 M of 10 mL H2SO4 = 1M of 20 mL NH3 1000 mL of 1M ammonia contains 14 g nitrogen 20 mL of 1M ammonia contains (14 x 20)/1000 g nitrogen Percentage of nitrogen = (14 x 20 x 100) / (1000 x 0.5) = 56.0%
-- NCERT Class 11 Chemistry, Ch. 8, p. 288Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.8.17) in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed. Let the mass of organic compound taken = m g Mass of AgX formed = m1 g 1 mol of AgX contains 1 mol of X Mass of halogen in m1 g of AgX = (atomic mass of X x m1 g) / (molecular mass of AgX) Percentage of halogen = [(atomic mass of X x m1 g) / (molecular mass of AgX)] x 100 / m Fig. 8.17 Carius method. Halogen containing organic compound is heated with fuming nitric acid in the presence of silver nitrate.
-- NCERT Class 11 Chemistry, Ch. 8, p. 289Problem 8.23 In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound. Solution Molar mass of AgBr = 108 + 80 = 188 g mol-1 188 g AgBr contains 80 g bromine 0.12 g AgBr contains (80 x 0.12)/188 g bromine Percentage of bromine = (80 x 0.12 x 100) / (188 x 0.15) = 34.04%
-- NCERT Class 11 Chemistry, Ch. 8, p. 289A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate. Let the mass of organic compound taken = m g and the mass of barium sulphate formed = m1 g 1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur m1 g BaSO4 contains (32 x m1)/233 g sulphur Percentage of sulphur = (32 x m1 x 100) / (233 x m)
-- NCERT Class 11 Chemistry, Ch. 8, p. 289Problem 8.24 In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound? Solution Molecular mass of BaSO4 = 137+32+64 = 233 g 233 g BaSO4 contains 32 g sulphur 0.4813 g BaSO4 contains (32 x 0.4813)/233 g sulphur Percentage of sulphur = (32 x 0.4813 x 100) / (233 x 0.157) = 42.10%
-- NCERT Class 11 Chemistry, Ch. 8, p. 290A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3PO4.12MoO3, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7. Let the mass of organic compound taken = m g and mass of ammonium phospho molydate = m1 g Molar mass of (NH4)3PO4.12MoO3 = 1877 g Percentage of phosphorus = (31 x m1 x 100) / (1877 x m) % If phosphorus is estimated as Mg2P2O7, Percentage of phosphorus = (62 x m1 x 100) / 222 x ... where, 222 u is the molar mass of Mg2P2O7, m, the mass of organic compound taken, m1, the mass of Mg2P2O7 formed and 62, the mass of two phosphorus atoms present in the compound Mg2P2O7.
-- NCERT Class 11 Chemistry, Ch. 8, p. 290The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows: A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine. Compound --(heat)--> O2 + other gaseous products 2C + O2 --(1373K)--> 2CO] x 5 (A) I2O5 + 5CO -----> I2 + 5CO2] x 2 (B) On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide. Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated. Let the mass of organic compound taken be m g Mass of carbon dioxide produced be m1 g m1 g carbon dioxide is obtained from (32 x m1)/88 g O2 Percentage of oxygen = (32 x m1 x 100) / (88 x m) % The percentage of oxygen can be derived from the amount of iodine produced also. Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as C,H,N elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time. A detailed discussion of such methods is beyond the scope of this book.
-- NCERT Class 11 Chemistry, Ch. 8, p. 2901 question from NEET 2022. Answers verified against NTA official keys.
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