Qualitative & Quantitative Analysis of Organic Compounds

Why analysis matters

Establishing the structure of an organic compound is a staircase, and elemental analysis is its first step. Before a molecular formula can be proposed, the chemist must know which elements are present and the mass percentage of each. From the percentages an empirical formula is computed, and combined with a molar mass it yields the molecular formula — the foundation on which spectroscopy and degradation studies later build the full structure.

Carbon and hydrogen are common to almost every organic compound; oxygen, nitrogen, sulphur, the halogens and phosphorus appear selectively. Because these elements are held in covalent linkages, they do not respond to ordinary ionic tests directly. The whole craft of qualitative organic analysis therefore turns on a single idea: convert the covalently bound elements into ionic species, then apply familiar inorganic reactions.

A note on sources

Source-limitation caveat. The standard Class 11 NCERT file for this unit (Unit 8, §8.10) was supplied in an encoding-corrupted (mojibake) form and could not be read reliably. The detection and estimation content below is therefore grounded in the parallel NIOS Chemistry treatment (Chapter 23, §23.5–§23.6), which covers the full NEET syllabus. The estimation formulae are presented exactly as standardised; the lone numerical illustration in the worked-example box uses figures that are explicitly labelled as illustrative, and no experiment-specific data has been fabricated.

Detecting carbon and hydrogen

Almost every organic compound chars or burns with a sooty flame, which is a crude indication of carbon. The definitive test heats the compound with dry copper(II) oxide, a powerful oxidant. Carbon is oxidised to carbon dioxide and hydrogen to water:

$$\ce{C + 2CuO ->[\Delta] 2Cu + CO2 ^}$$ $$\ce{2H + CuO ->[\Delta] Cu + H2O}$$

The two products are then identified by their classic confirmatory reactions. Carbon dioxide turns lime water milky through precipitation of calcium carbonate, while the water vapour turns white anhydrous copper sulphate blue:

$$\ce{CO2 + Ca(OH)2 -> CaCO3 v + H2O}$$ $$\ce{CuSO4 + 5H2O -> CuSO4.5H2O}$$

Lassaigne's test — the principle

Nitrogen, sulphur, halogens and phosphorus in an organic compound are bound covalently, so they cannot be detected as ions directly. Lassaigne's test solves this by fusing the compound with metallic sodium. Sodium is a strong reducing agent that, at fusion temperature, converts these covalent elements into their ionic sodium salts:

$$\ce{Na + C + N ->[\Delta] NaCN}$$ $$\ce{2Na + S ->[\Delta] Na2S}$$ $$\ce{Na + X ->[\Delta] NaX}\quad(\ce{X = Cl, Br, I})$$

The hot fused mass is plunged into distilled water and boiled. The soluble salts — sodium cyanide, sodium sulphide and sodium halide — pass into solution to give the sodium fusion extract, also called Lassaigne's extract (L.E.). Every subsequent identification is performed on this aqueous extract.

Detecting nitrogen

The extract is boiled with freshly prepared ferrous sulphate and then acidified with concentrated sulphuric acid. Cyanide ions combine with ferrous ions to form hexacyanoferrate(II); some ferrous ions are oxidised to ferric, and the ferric ions react with the hexacyanoferrate(II) to deposit the deep blue pigment Prussian blue:

$$\ce{Fe^2+ + 6CN^- -> [Fe(CN)6]^4-}$$ $$\ce{3[Fe(CN)6]^4- + 4Fe^3+ -> Fe4[Fe(CN)6]3 v}$$

The blue $\ce{Fe4[Fe(CN)6]3}$ confirms nitrogen. If both nitrogen and sulphur are present, fusion instead yields sodium thiocyanate; with ferric ions this gives a blood-red ferric thiocyanate rather than the blue pigment, because no free cyanide remains:

$$\ce{Na + C + N + S ->[\Delta] NaSCN}$$ $$\ce{Fe^3+ + 3SCN^- -> Fe(SCN)3}\;(\text{blood red})$$

Fusion with a large excess of sodium decomposes the thiocyanate back to cyanide and sulphide, $\ce{NaSCN + 2Na -> NaCN + Na2S}$, restoring the separate Prussian-blue and sulphide tests.

Sulphur, halogens and phosphorus

Sulphur is confirmed in two complementary ways. Treating the extract with sodium nitroprusside gives a characteristic violet colour; alternatively, acidifying with acetic acid and adding lead acetate precipitates black lead sulphide:

$$\ce{S^2- + [Fe(CN)5NO]^2- -> [Fe(CN)5NOS]^4-}\;(\text{violet})$$ $$\ce{Pb^2+ + S^2- -> PbS v}\;(\text{black})$$

For halogens, the extract is first acidified with dilute nitric acid and then treated with silver nitrate. The colour and solubility of the silver halide identify the halogen:

$$\ce{NaCl + AgNO3 -> AgCl v + NaNO3}$$ $$\ce{NaBr + AgNO3 -> AgBr v + NaNO3}$$ $$\ce{NaI + AgNO3 -> AgI v + NaNO3}$$

$\ce{AgCl}$ is white and dissolves readily in ammonia; $\ce{AgBr}$ is pale yellow and only partly soluble; $\ce{AgI}$ is deep yellow and insoluble. When nitrogen or sulphur is also present, the extract must first be boiled with concentrated $\ce{HNO3}$ to expel $\ce{CN^-}$ and $\ce{S^2-}$, which would otherwise give $\ce{AgCN}$ or $\ce{Ag2S}$ and be mistaken for a halide. Phosphorus is detected by oxidising the compound with sodium peroxide to phosphate, then adding ammonium molybdate in nitric acid; a yellow precipitate of ammonium phosphomolybdate confirms phosphorus.

$$\ce{Na3PO4 + 3HNO3 -> H3PO4 + 3NaNO3}$$ $$\ce{H3PO4 + 12(NH4)2MoO4 + 21HNO3 -> (NH4)3PO4.12MoO3 v + 21NH4NO3 + 12H2O}$$

Master table — qualitative tests

Estimating carbon and hydrogen

Quantitative estimation puts numbers to the elements. The Liebig combustion method burns a known mass of compound in a stream of pure oxygen over heated copper oxide. All carbon ends as $\ce{CO2}$, all hydrogen as $\ce{H2O}$. The water vapour is absorbed in anhydrous calcium chloride and the carbon dioxide in concentrated potassium hydroxide; the gains in mass of the two absorbers give the masses of $\ce{H2O}$ and $\ce{CO2}$ produced.

Since 44 g of $\ce{CO2}$ contains 12 g of carbon and 18 g of $\ce{H2O}$ contains 2 g of hydrogen, the percentages follow directly:

$$\%\,\mathrm{C} = \frac{12}{44}\times\frac{\text{mass of }\ce{CO2}}{\text{mass of compound}}\times 100$$ $$\%\,\mathrm{H} = \frac{2}{18}\times\frac{\text{mass of }\ce{H2O}}{\text{mass of compound}}\times 100$$

Estimating nitrogen — Dumas & Kjeldahl

Nitrogen has two estimation routes. In the Dumas method a known mass is heated with copper oxide in a carbon-dioxide atmosphere; nitrogen is liberated as $\ce{N2}$ gas, any oxides of nitrogen being reduced over hot copper gauze. The gas is collected over potassium hydroxide, which absorbs $\ce{CO2}$, leaving the nitrogen to be measured volumetrically.

$$\ce{C_xH_yN_z + (2x + y/2)CuO -> x CO2 + (y/2)H2O + (z/2)N2 + (2x + y/2)Cu}$$

With the volume of $\ce{N2}$ reduced to STP (28 g of $\ce{N2}$ occupying 22400 mL), the percentage is:

$$\%\,\mathrm{N} = \frac{28}{22400}\times\frac{V_{\text{STP}}}{\text{mass of compound}}\times 100$$

The Kjeldahl method instead digests the compound (about 0.5 g) with concentrated $\ce{H2SO4}$, $\ce{K2SO4}$ to raise the boiling point, and a $\ce{CuSO4}$ or mercury catalyst. Organic nitrogen is converted quantitatively to ammonium sulphate; excess NaOH then liberates ammonia, which is absorbed in a measured excess of standard acid. The acid left unreacted is found by back-titration with standard alkali.

$$\ce{N ->[conc.\,H2SO4] (NH4)2SO4}$$ $$\ce{(NH4)2SO4 + 2NaOH -> Na2SO4 + 2NH3 ^ + 2H2O}$$ $$\ce{2NH3 + H2SO4 -> (NH4)2SO4}$$

Because 1000 mL of 1 M ($\equiv$ 1 N for the acid–NH₃ reaction) ammonia contains 14 g of nitrogen, the standard working formula is:

$$\%\,\mathrm{N} = \frac{1.4 \times N_{\text{acid}} \times V_{\text{acid used by }NH_3}}{\text{mass of compound}}$$

where $V_{\text{acid used by }NH_3}$ is the volume of standard acid (in mL, of normality $N_{\text{acid}}$) actually neutralised by the ammonia, obtained as (acid taken) − (acid back-titrated by alkali).

Halogens, sulphur, phosphorus, oxygen

Halogens, sulphur and phosphorus are estimated by the Carius method: a known mass of compound is heated in a sealed hard-glass Carius tube with fuming nitric acid. The targeted element is oxidised and precipitated in a weighable form. For a halogen, silver nitrate is included so that the halogen separates as silver halide:

$$\%\,\mathrm{X} = \frac{\text{atomic mass of X}}{\text{molecular mass of AgX}}\times\frac{\text{mass of AgX}}{\text{mass of compound}}\times 100$$

With the standard atomic and formula masses this reduces to the familiar working factors — $35.5/143.5$ for chlorine (as $\ce{AgCl}$), $80/188$ for bromine (as $\ce{AgBr}$) and $127/235$ for iodine (as $\ce{AgI}$). Sulphur is oxidised to sulphate and precipitated as barium sulphate; phosphorus is oxidised to phosphoric acid and weighed as ammonium phosphomolybdate or as magnesium pyrophosphate:

$$\%\,\mathrm{S} = \frac{32}{233}\times\frac{\text{mass of }\ce{BaSO4}}{\text{mass of compound}}\times 100$$ $$\%\,\mathrm{P} = \frac{\text{atomic mass of P}}{\text{mol. mass of ppt.}}\times\frac{\text{mass of ppt.}}{\text{mass of compound}}\times 100$$

Oxygen is normally not estimated directly. Once every other element present has been measured, the oxygen percentage is taken by difference: $\%\,\mathrm{O} = 100 - (\text{sum of the percentages of all other elements})$.

Master table — quantitative methods