Organic analysis determines what elements are present in a compound and in what amounts. CBSE Class 11 covers the basics. NEET rarely tests this directly but the techniques underlie all organic chemistry.
Core Concepts
Detection of elements
Carbon and hydrogen — burn compound with CuO. CO2 turns limewater milky; H2O condenses. Nitrogen, sulphur, halogens — Lassaigne’s test (fuse with Na, then chemical tests).
The CuO test works because CuO is a mild oxidising agent. It oxidises carbon to CO2 and hydrogen to H2O. The CO2 is detected by passing it through limewater (), which turns milky due to formation of . The water is detected by anhydrous copper sulphate, which turns from white to blue.
Lassaigne’s test
Organic compound is fused with sodium, converting N, S and halogens to NaCN, Na2S and NaX respectively. These ions are then detected by standard inorganic tests.
The detailed steps:
Heat a small piece of sodium in a fusion tube until it melts. Add the organic compound. Heat to red hot. Plunge the hot tube into distilled water. The tube cracks, and the sodium extract dissolves.
Add FeSO4 to the sodium extract, boil, then add dilute H2SO4. A Prussian blue colour confirms nitrogen. The chemistry: , then with Fe3+, Prussian blue () forms.
Add sodium nitroprusside to the extract. A violet colour confirms sulphur (, violet). Alternatively, acidify with acetic acid and add lead acetate — black PbS precipitate confirms S.
Acidify with HNO3, then add AgNO3. White precipitate (soluble in NH3) = Cl. Pale yellow precipitate (partially soluble) = Br. Yellow precipitate (insoluble in NH3) = I.
If both N and S are present, we get NaSCN instead of separate NaCN and Na2S. This gives a blood-red colour with Fe3+ (ferric thiocyanate) instead of Prussian blue. Always test for N and S separately in this case.
Estimation of carbon and hydrogen
Burn known mass of compound in oxygen. CO2 is absorbed in KOH solution; H2O in anhydrous CaCl2. Masses of absorbed products give % of C and H.
The ratio 12/44 converts mass of CO2 to mass of carbon (since each CO2 molecule contains one C atom with mass 12 out of total 44).
The ratio 2/18 converts mass of water to mass of hydrogen (each H2O has 2 H atoms with total mass 2 out of 18).
Estimation of nitrogen
Dumas method — convert all N to N2 by combustion with CuO. Measure volume at known T and P. Kjeldahl method — convert N to ammonia with concentrated H2SO4, then titrate.
Dumas method in detail: The compound is heated with CuO in a stream of CO2. All carbon becomes CO2, hydrogen becomes H2O, and nitrogen becomes N2 (plus some oxides of nitrogen). The gas mixture is passed over hot copper gauze to reduce any NOx to N2. The N2 is collected over KOH (which absorbs CO2) and measured.
where = volume of N2 at STP (in mL), = mass of compound (in g), 28 = molar mass of N2, 22400 = molar volume at STP in mL.
Kjeldahl method: The compound is heated with conc. H2SO4 and a catalyst (K2SO4 + CuSO4). All nitrogen converts to (NH4)2SO4. This is then treated with excess NaOH to liberate NH3, which is distilled into a known volume of standard acid. Back-titration gives the nitrogen content.
Kjeldahl method does not work for nitrogen in nitro groups (-NO2), azo groups (-N=N-), or ring nitrogen (like in pyridine). These forms of nitrogen are not easily converted to ammonium sulphate by H2SO4. This is a classic NEET/JEE question.
Estimation of halogens and sulphur
Carius method — heat compound with fuming HNO3 and AgNO3 in a sealed tube. Halogens give AgX precipitate; sulphur gives BaSO4 precipitate. Weigh the precipitate.
For chlorine:
Worked Examples
Proteins contain about 16% nitrogen. Estimate N, multiply by 6.25, get protein content. Used in food labelling for decades.
Organic compounds do not dissociate in water. Fusing with sodium converts elements to soluble ionic forms that can be tested.
0.50 g of an organic compound on combustion gives 1.10 g of CO2 and 0.45 g of H2O. Find % of C and H.
Remaining = 100 - 60 - 10 = 30% (could be oxygen).
0.30 g of an organic compound was Kjeldahl analysed. The NH3 evolved was absorbed in 50 mL of 0.1 M H2SO4. The excess acid required 20 mL of 0.1 M NaOH for neutralisation.
Moles of H2SO4 taken = 0.05 × 0.1 = 0.005
Moles of NaOH used = 0.02 × 0.1 = 0.002
Moles of H2SO4 that reacted with NH3 = 0.005 - 0.001 = 0.004 (since H2SO4 is dibasic, 0.002 mol NaOH neutralises 0.001 mol H2SO4)
Moles of NH3 = 2 × 0.004 = 0.008
Mass of N = 0.008 × 14 = 0.112 g
Common Mistakes
Confusing Dumas and Kjeldahl. Dumas is combustion; Kjeldahl is chemical conversion to ammonia.
Saying Lassaigne test detects carbon. It detects N, S and halogens, not C.
Writing that estimation is qualitative. Estimation is quantitative; detection is qualitative.
Forgetting that Kjeldahl fails for nitro and azo groups. If the question specifies a nitro compound and asks for nitrogen estimation, the answer is Dumas method, not Kjeldahl.
Not accounting for the dibasic nature of H2SO4 in Kjeldahl calculations. One mole of H2SO4 reacts with two moles of NH3, and one mole of H2SO4 needs two moles of NaOH.
Exam Weightage and Revision
Organic analysis carries 1-2 questions in JEE Main (usually a Kjeldahl or Carius numerical). NEET rarely asks it as a standalone question but may include Lassaigne test identification. CBSE Class 11 boards allocate 3-5 marks.
| Question Type | Exam | Frequency |
|---|---|---|
| Lassaigne test colour/precipitate | NEET, CBSE | Occasional |
| Kjeldahl numerical | JEE Main | Most years |
| Carius method calculation | JEE Main | Every 2 years |
| Limitations of Kjeldahl | NEET, JEE | Common |
| C and H estimation numerical | JEE Main | Occasional |
For NEET, know the colours: Prussian blue for N, violet for S, white/pale yellow/yellow precipitates for Cl/Br/I. For JEE, be ready for a Kjeldahl back-titration numerical.
Practice Questions
Q1. An organic compound gives a blood-red colour with FeCl3 after Lassaigne’s test. What does this indicate?
Blood-red colour with FeCl3 indicates the presence of both nitrogen and sulphur in the compound. The sodium fusion produces NaSCN (sodium thiocyanate), which gives ferric thiocyanate () — a blood-red compound. If only nitrogen were present, we would get Prussian blue.
Q2. 0.40 g of an organic compound on combustion gave 0.88 g of CO2 and 0.36 g of H2O. Determine the empirical formula.
Mole ratio: C = 60/12 = 5, H = 10/1 = 10, O = 30/16 = 1.875
Dividing by smallest (1.875): C = 2.67, H = 5.33, O = 1
Multiply by 3: C = 8, H = 16, O = 3
Empirical formula:
Q3. Why cannot Kjeldahl method be used for estimating nitrogen in pyridine?
Pyridine has nitrogen in an aromatic ring. When heated with conc. H2SO4, the ring nitrogen is not easily converted to ammonium sulphate because the C-N bonds in the ring are strong and resistant to acid digestion. Dumas method works because it uses complete combustion at high temperature with CuO, which breaks all bonds.
FAQs
Why is sodium used in Lassaigne’s test and not any other metal?
Sodium is highly reactive, melts at a low temperature (98°C), and readily reduces non-metallic elements to their anionic forms (CN-, S2-, X-). Potassium would work too but is more dangerous to handle due to its higher reactivity. Sodium is the practical choice.
What is the difference between qualitative and quantitative analysis?
Qualitative analysis tells you which elements are present (detection). Quantitative analysis tells you how much of each element (estimation). Lassaigne’s test is qualitative; Kjeldahl, Dumas, and Carius are quantitative.
How has modern chemistry replaced these classical methods?
Modern labs use instrumental methods: mass spectrometry (molecular formula), NMR (structural connectivity), IR spectroscopy (functional groups), elemental analysers (C, H, N percentages by combustion + gas chromatography). These are faster, more accurate, and need smaller samples. But the principles are the same.
Three methods to know — Lassaigne for detection, Dumas/Kjeldahl for N, Carius for halogens and sulphur.
Organic analysis is how chemists first identify unknowns. It set the stage for modern spectroscopy.