Quantitative analysis of organic compounds

How is the composition of an organic compound measured?

In plain words: you cannot weigh the carbon inside a molecule, but you can burn the compound and weigh the carbon dioxide it makes. Since 44 g of CO₂ always contains 12 g of carbon, the mass of carbon follows. Every method on this page uses the same idea: convert the element into a compound of known formula, measure that, and scale back by the fraction of it that is the element.

Carbon and hydrogen

A known mass of the compound is burnt in excess oxygen with copper(II) oxide, which oxidises carbon to carbon dioxide and hydrogen to water (NCERT Section 8.10.1, pages 285–286):

C_xH_y + (x + y/4)O₂ → xCO₂ + (y/2)H₂O

The gases pass through a weighed U-tube of anhydrous calcium chloride, which absorbs the water, and then a U-tube of concentrated potassium hydroxide solution, which absorbs the carbon dioxide (NCERT Fig. 8.14). The order matters: KOH solution would also take up water. The gains in mass give the masses of water (m₁) and carbon dioxide (m₂) from m g of compound.

Here 12/44 is the fraction of carbon in CO₂ and 2/18 the fraction of hydrogen in H₂O (see mass per cent).

Nitrogen

Dumas method. The compound is heated with copper(II) oxide in an atmosphere of carbon dioxide; its nitrogen is set free as N₂, and carbon and hydrogen form CO₂ and H₂O (NCERT Section 8.10.2, pages 286–287):

Any oxides of nitrogen are reduced to N₂ over heated copper gauze. The gases are collected over potassium hydroxide solution, which absorbs the CO₂, and the volume V₁ of nitrogen is read in a graduated tube (a nitrometer, NCERT Fig. 8.15) at room temperature T₁. Because the gas is collected over an aqueous solution (KOH), it is saturated with water vapour, so its pressure is the atmospheric pressure minus the aqueous tension (the water-vapour pressure over the collecting solution): p₁ = atmospheric pressure − aqueous tension (with gas and external pressures equalised). Concentrated KOH lowers water vapour pressure, so use the stated collecting-solution correction rather than automatically substituting a pure-water table. The volume is converted to STP (273 K, 760 mm Hg), using the school ideal-gas approximation of 22 400 mL per mole and 28 g mol⁻¹ for N₂. This is the older 1-atm STP convention; the 1-bar convention has a different molar volume:

Kjeldahl’s method. The compound is heated with concentrated sulphuric acid (with a little CuSO₄, NCERT Fig. 8.16, as catalyst), which turns its nitrogen into ammonium sulphate. The mixture is then heated with excess sodium hydroxide, and the ammonia set free is absorbed in a known excess of standard sulphuric acid. The acid left over is titrated with standard alkali; the difference gives the acid used by the ammonia (NCERT pages 287–288; see acid–base titrations):

• nitrogen in the compound → (NH₄)₂SO₄ during acid digestion (schematic, not a balanced overall equation);
• (NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2NH₃ + 2H₂O;
• 2NH₃ + H₂SO₄ → (NH₄)₂SO₄.

If V mL of H₂SO₄ of molarity M is taken and V₁ mL of NaOH of the same molarity M neutralises what is left, the leftover acid is V₁/2 mL (each H₂SO₄ needs two NaOH). The acid used by ammonia is (V − V₁/2) mL, which took 2(V − V₁/2) mL of ammonia of molarity M, and 1000 mL of 1 M NH₃ contains 14 g of nitrogen. Mass of nitrogen = 14 × M × 2(V − V₁/2)/1000 g; multiplying by 100/m (and 14 × 100/1000 = 1.4):

NCERT’s simplified line on page 288 prints 2(V − V/2); the subscript 1 belongs on the second V, as in the line before it.

The unmodified Kjeldahl procedure described here does not work for nitrogen in nitro and azo groups or in a ring (as in pyridine), because that nitrogen is not converted to ammonium sulphate under these conditions (page 288).

Halogens: Carius method

A known mass of the compound is heated with fuming nitric acid and silver nitrate in a sealed hard-glass Carius tube in a furnace (NCERT Section 8.10.3, page 289, Fig. 8.17). Carbon and hydrogen are oxidised to CO₂ and H₂O, and the halogen forms silver halide, AgX, which is filtered, washed, dried and weighed. Here X means Cl, Br or I; this silver-halide method does not estimate fluorine because AgF is soluble. One mole of AgX contains one mole of X:

with m₁ g of AgX from m g of compound. For chlorine the factor is 35.5/143.5 and for bromine 80/188. NCERT’s line labelled percentage of halogen (page 289) gives only the mass of halogen in m₁ g of AgX; multiply it by 100/m for the percentage, as above.

Sulphur

The compound is heated in a Carius tube with sodium peroxide or fuming nitric acid, which oxidises sulphur to sulphuric acid. Excess barium chloride solution precipitates it as barium sulphate, which is filtered, washed, dried and weighed (NCERT Section 8.10.4, page 289). 233 g of BaSO₄ contains 32 g of sulphur:

Phosphorus

Heating with fuming nitric acid oxidises phosphorus to phosphoric acid, which is precipitated with ammonia and ammonium molybdate as ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃ (molar mass 1877 g mol⁻¹). Alternatively, magnesia mixture precipitates MgNH₄PO₄, which on ignition gives Mg₂P₂O₇ (magnesium pyrophosphate, 222 g mol⁻¹, with 62 g of phosphorus) (NCERT Section 8.10.5, page 290):

NCERT’s second formula on page 290 is printed with a stray × in place of × m in the denominator.

Oxygen

Oxygen may be estimated by difference: 100 minus the sum of the percentages of all the other elements (NCERT Section 8.10.6, page 290). That works only if every other component has been accounted for; missing elements, moisture, ash and analytical errors otherwise get wrongly assigned to oxygen (reasoned). It can also be estimated directly. The compound is decomposed by heating in a stream of nitrogen, and the gases are passed over red-hot coke, which turns all the oxygen into carbon monoxide; warm iodine pentoxide then oxidises the CO to CO₂ and sets free iodine:

• 2C + O₂ → 2CO (× 5);
• I₂O₅ + 5CO → I₂ + 5CO₂ (× 2).

These are stoichiometric bookkeeping equations; free O₂ need not be released from the sample. Each two moles of oxygen atoms from the compound ultimately correspond to two moles of CO₂: 32 g of oxygen gives 88 g of CO₂.

with m₁ g of CO₂. The iodine produced can be used instead. Today, carbon, hydrogen and nitrogen are measured on 1–3 mg samples by an automatic CHN elemental analyser (page 290).

Solved examples

Common mistakes to avoid

Quick revision summary

• C and H: burn with CuO; H₂O in anhydrous CaCl₂, then CO₂ in KOH; % C = 12m₂ × 100/(44m), % H = 2m₁ × 100/(18m).
• N (Dumas): N₂ volume over KOH, corrected for aqueous tension and to STP; 28 g per 22 400 mL.
• N (Kjeldahl): N → (NH₄)₂SO₄ → NH₃ into standard acid; not for nitro, azo or ring N.
• Halogens (Carius): AgX; S: BaSO₄ (32/233); P: phosphomolybdate (31/1877) or Mg₂P₂O₇ (62/222).
• O: by difference, or via CO → CO₂ with I₂O₅ (32 g O₂ → 88 g CO₂).

Questions students ask

How is the Kjeldahl method used to find the protein in food?

The Kjeldahl method finds protein indirectly: it measures the per cent of nitrogen in a food, and since most proteins contain about 16% nitrogen, the protein content is taken as % N × 6.25 (that is, 100/16). A lentil flour found to contain 3.8% nitrogen would be reported as about 24% protein, while some foods use their own factors, such as 6.38 for milk and 5.70 for wheat endosperm flour (whole-wheat factors differ). Because the method counts all the nitrogen that ends up as ammonium sulphate, not only that of protein, an adulterant such as urea in milk raises the apparent protein content. The Dumas method, now run in automatic combustion analysers, is also widely used for protein because it is faster and needs no concentrated acid.

What is the percentage of nitrogen in urea, and how is it estimated?

Urea, NH₂CONH₂ (molar mass 60 g mol⁻¹), contains 28 × 100/60 = 46.7% nitrogen (see Example 2 of the introduction), and because its nitrogen is in amide groups, which concentrated sulphuric acid turns into ammonium sulphate, it can be estimated by Kjeldahl’s method. For example, 0.18 g of urea (0.0030 mol) gives 0.0060 mol of NH₃, which neutralises 0.0030 mol of H₂SO₄, or 30 mL of 0.1 M acid; then % N = 14 × 0.0060 × 100/0.18 = 46.7%. The Dumas method gives the same result when combustion, reduction and gas collection are quantitative.

Are the Kjeldahl, Dumas and Carius methods in the JEE Main 2027 and NEET (UG) 2026 syllabuses?

Yes, the Kjeldahl, Dumas and Carius methods fall within both syllabuses: JEE Main 2027 (NTA) and NEET (UG) 2026 (NMC) list ‘Quantitative analysis (basic principles only) – estimation of carbon, hydrogen, nitrogen, halogens, sulphur and phosphorus’ and ‘numerical problems in organic quantitative analysis’ under Unit 13, Purification and Characterisation of Organic Compounds. The methods are not named there, but they are how NCERT Section 8.10 estimates nitrogen, halogens and sulphur, so percentage calculations like the solved examples here are to be expected. Oxygen is not in the list, although NCERT describes its estimation too.

Which elements are estimated by the Carius method?

The Carius method estimates halogens and sulphur: the compound is heated with fuming nitric acid in a sealed hard-glass Carius tube, which oxidises its carbon and hydrogen and turns a halogen, with silver nitrate present, into silver halide, or sulphur into sulphuric acid, weighed as barium sulphate. Sealing the tube lets the acid be heated far above its boiling point, allowing vigorous oxidation; quantitative recovery still has to be established for the sample. For example, 0.40 g of an iodo compound giving 0.470 g of AgI (235 g mol⁻¹) contains (127 × 0.470 × 100)/(235 × 0.40) = 63.5% iodine.

How do you pronounce Kjeldahl?

Kjeldahl is usually pronounced KEL-dahl in English, with the j silent: the method is named after the Danish chemist Johan Kjeldahl, who introduced it in 1883 while working at the Carlsberg Laboratory in Copenhagen. It is spelt with kj at the start and dahl at the end, and NCERT calls it Kjeldahl’s method.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A compound contains 69% carbon and 4.8% hydrogen. What masses of CO₂ and H₂O are produced when 0.20 g of it is burnt completely (NCERT Exercise 8.32)?

   Show answerHide answer

   Carbon = 0.20 × 0.69 = 0.138 g, so CO₂ = 0.138 × 44/12 = 0.506 g. Hydrogen = 0.20 × 0.048 = 0.0096 g, so H₂O = 0.0096 × 18/2 = 0.0864 g.

2. Q2. 0.3780 g of an organic chloro compound gave 0.5740 g of AgCl in a Carius estimation. Find the percentage of chlorine (NCERT Exercise 8.34).

   Show answerHide answer

   AgCl = 108 + 35.5 = 143.5. % Cl = (35.5 × 0.5740 × 100)/(143.5 × 0.3780) = 37.57%.

3. Q3. 0.468 g of an organic sulphur compound gave 0.668 g of BaSO₄ by the Carius method. Find the percentage of sulphur (NCERT Exercise 8.35).

   Show answerHide answer

   % S = (32 × 0.668 × 100)/(233 × 0.468) = 19.60%.