Part 3 · Chapter 33

Purification and Characterisation of Organic Compounds

Getting a compound clean, then asking what it is made of — the physical methods that separate and the analytical tests that reveal the elements and their amounts

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • How to choose a purification method — sublimation, crystallisation, distillation, extraction or chromatography — from a compound's properties.
  • The four kinds of distillation and when each one applies.
  • Chromatography by adsorption and partition, and the meaning of the Rf value.
  • How to detect C, H, N, S and the halogens, including Lassaigne's test.
  • How to estimate C and H (Liebig) and nitrogen (Dumas, Kjeldahl).
  • The Carius method for halogens and sulphur, and the percentage formulae for every element.
Section 33-1

Why Purify and Characterise?

An organic compound straight from a reaction is rarely pure. Before its identity can be settled, it must be purified — freed from unreacted starting material, by-products and solvent — and then characterised: its elements detected (qualitative analysis) and their amounts measured (quantitative analysis). Purification rests on physical differences such as volatility and solubility; characterisation rests on chemistry that converts hidden covalent elements into things we can weigh or titrate.

Section 33-2

Sublimation & Crystallisation

Sublimation purifies solids that pass directly from solid to vapour on heating, leaving non-volatile impurities behind. It suits compounds such as camphor, naphthalene and anthracene.

Crystallisation is the workhorse for solids and exploits differences in solubility. The impure solid is dissolved in the minimum of hot solvent (one that dissolves the compound well when hot, poorly when cold), the hot solution is filtered to remove insoluble impurities, and on cooling the pure compound crystallises while soluble impurities stay in the mother liquor. Coloured impurities are removed with a little activated charcoal.

Choosing the solvent is everything. A good crystallisation solvent dissolves the compound generously when hot and grudgingly when cold, dissolves the impurities either always or never, and does not react with the compound. Get the solvent right and the crystals fall out pure; get it wrong and nothing separates.
Section 33-3

Distillation & Its Variants

Distillation separates liquids by boiling point. Four variants cover the cases you meet.

MethodUsed whenExample
Simple distillationliquid + non-volatile impurity, or boiling points far apartchloroform from aniline; water from salt
Fractional distillationboiling points close togetherpetroleum refining; acetone–water
Distillation under reduced pressureliquid decomposes at its normal boiling pointglycerol; sugar-cane juice concentration
Steam distillationcompound steam-volatile and immiscible with wateraniline; bromobenzene; essential oils
💨
Steam distillation — the key idea
the mixture boils when \( p_{\text{compound}} + p_{\text{water}} = p_{\text{atmospheric}} \)

Because the two immiscible liquids contribute their vapour pressures independently, the mixture boils below 100 °C — so a compound that would otherwise decompose at its own boiling point distils over gently with the steam. It must be steam-volatile and immiscible with, and unreactive toward, water.

Section 33-4

Extraction & Chromatography

Differential (solvent) extraction recovers a compound from an aqueous solution by shaking it in a separating funnel with an immiscible organic solvent in which the compound is far more soluble. The compound moves into the organic layer, which is run off and evaporated. Several small extractions remove more than one large one.

Chromatography separates a mixture by the differing affinities of its components for a stationary and a mobile phase. Two mechanisms appear in the syllabus.

TypeSeparates byStationary / mobile phase
Adsorption (column, TLC)differential adsorption on a solidsilica gel / alumina · a liquid eluent
Partition (paper)differential partition between two liquidswater held on paper · a moving solvent
baseline solvent front d (spot) D (front)
Rf = distance moved by spot ÷ distance moved by solvent
The Rf value. For each component, \( R_f = \dfrac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent}} \). It lies between 0 and 1, is characteristic of a compound in a given system, and lets you identify and compare substances on a single plate.
Section 33-5

Detecting C and H (Qualitative)

Carbon and hydrogen are detected together by heating the compound with dry copper(II) oxide. Carbon is oxidised to carbon dioxide and hydrogen to water.

Test reactions
\[ \ce{C + 2CuO -> CO2 ^ + 2Cu};\qquad \ce{2H + CuO -> H2O + Cu} \]

The carbon dioxide turns lime water milky; the water turns anhydrous white copper(II) sulphate blue (or pink cobalt chloride paper). Both observations together confirm carbon and hydrogen.

Section 33-6

Lassaigne's Test: N, S, Halogens

Nitrogen, sulphur and halogens are held in covalent form and give no ionic tests directly. Lassaigne's test first fuses the compound with metallic sodium, converting these elements into ionic sodium salts that dissolve in water as the sodium fusion extract.

Sodium fusion
\[ \ce{Na + C + N -> NaCN};\quad \ce{2Na + S -> Na2S};\quad \ce{Na + X -> NaX} \]
ElementReagent on the extractPositive result
Nitrogen\(\ce{FeSO4}\), then \(\ce{Fe^3+}\)/acidPrussian-blue colour (ferric ferrocyanide)
Sulphursodium nitroprussideviolet colour (also \(\ce{Pb(OAc)2}\) → black \(\ce{PbS}\))
Halogenboil with \(\ce{HNO3}\), add \(\ce{AgNO3}\)white \(\ce{AgCl}\) / pale-yellow \(\ce{AgBr}\) / yellow \(\ce{AgI}\)
Two traps in the halogen test. First, if nitrogen or sulphur is present, the extract must be boiled with nitric acid first to drive off \(\ce{HCN}\)/\(\ce{H2S}\), which would otherwise give false silver precipitates. Second, if both N and S are present, sodium thiocyanate \(\ce{NaSCN}\) forms instead, giving a blood-red colour with \(\ce{Fe^3+}\) — not Prussian blue. Read the colour carefully.
Section 33-7

Estimating C and H (Liebig)

In Liebig's method a known mass of compound is burnt in excess oxygen over hot copper oxide. The water formed is absorbed in anhydrous calcium chloride and the carbon dioxide in concentrated potassium hydroxide; both absorbers are weighed before and after.

🔥
Carbon & hydrogen percentages
from the masses of \(\ce{CO2}\) and \(\ce{H2O}\) collected

$$ \%\,\ce{C} = \frac{12}{44}\times\frac{m_{\ce{CO2}}}{m_{\text{compound}}}\times 100 $$

$$ \%\,\ce{H} = \frac{2}{18}\times\frac{m_{\ce{H2O}}}{m_{\text{compound}}}\times 100 $$

Section 33-8

Estimating Nitrogen

Two methods estimate nitrogen, and the choice depends on how the nitrogen is bound.

Dumas method. The compound is burnt with copper oxide; all the nitrogen comes off as \(\ce{N2}\) gas, whose volume is measured over potassium hydroxide (which absorbs the \(\ce{CO2}\)). It works for all nitrogen compounds.

Dumas — percentage of nitrogen
\[ \%\,\ce{N} = \frac{28}{22400}\times\frac{V_{\ce{N2}}\,(\text{mL at STP})}{m_{\text{compound}}}\times 100 \]

Kjeldahl method. The compound is heated with concentrated sulphuric acid, converting its nitrogen to ammonium sulphate. Adding alkali liberates ammonia, which is distilled into a known excess of standard acid; back-titration gives the acid that reacted with the ammonia. It is quick and accurate but fails for nitro, azo and ring nitrogen.

Kjeldahl — percentage of nitrogen
\[ \%\,\ce{N} = \frac{1.4\times N_{\text{acid}}\times V_{\text{acid}}\,(\text{mL})}{m_{\text{compound}}} \]
Section 33-9

Halogens, Sulphur, Phosphorus & Oxygen

The Carius method handles halogens and sulphur: the compound is heated with fuming nitric acid in a sealed Carius tube, which oxidises and frees the element to be precipitated and weighed. Phosphorus is treated similarly and oxygen is usually found by difference.

ElementWeighed asPercentage
Halogen (X)silver halide \(\ce{AgX}\)\( \dfrac{\text{at. mass } X}{M_{\ce{AgX}}}\times\dfrac{m_{\ce{AgX}}}{m}\times 100 \)
Sulphurbarium sulphate \(\ce{BaSO4}\)\( \dfrac{32}{233}\times\dfrac{m_{\ce{BaSO4}}}{m}\times 100 \)
Phosphorusmagnesium pyrophosphate \(\ce{Mg2P2O7}\)\( \dfrac{62}{222}\times\dfrac{m_{\ce{Mg2P2O7}}}{m}\times 100 \)
Oxygenby difference: \(100 - \sum(\text{others})\)
The pattern behind every formula. Each percentage is the same idea: the fraction (mass of element inside the weighed product) ÷ (molar mass of that product), multiplied by the mass ratio of product to original sample, times 100. Learn the principle, not six separate formulae.
Worked Examples

Putting It to Work

1 Choose the method

Problem. Aniline (b.p. 184 °C, steam-volatile, immiscible with water) is to be purified. Which method, and why?

Solution. Steam-volatile and water-immiscible points to one method:

Working
\[ \textbf{steam distillation} \Rightarrow \text{distils below 100 °C, avoiding decomposition} \]
2 Carbon & hydrogen

Problem. Burning \(0.20\,\text{g}\) of a compound gives \(0.44\,\text{g}\) \(\ce{CO2}\) and \(0.18\,\text{g}\) \(\ce{H2O}\). Find % C and % H.

Solution. Apply the Liebig formulae:

Working
\[ \%\ce{C} = \tfrac{12}{44}\cdot\tfrac{0.44}{0.20}\cdot100 = \textbf{60\%} \]
\[ \%\ce{H} = \tfrac{2}{18}\cdot\tfrac{0.18}{0.20}\cdot100 = \textbf{10\%} \]
3 Nitrogen by Kjeldahl

Problem. \(0.50\,\text{g}\) of a compound needs \(20\,\text{mL}\) of \(0.5\,\text{N}\) acid to neutralise the ammonia. Find % N.

Solution. Use the Kjeldahl formula:

Working
\[ \%\ce{N} = \frac{1.4\times 0.5\times 20}{0.50} = \textbf{28\%} \]
4 Halogen by Carius

Problem. \(0.30\,\text{g}\) of a chloro-compound gives \(0.287\,\text{g}\) \(\ce{AgCl}\) (\(M=143.5\)). Find % Cl. (\(\ce{Cl}=35.5\))

Solution. Fraction of Cl in AgCl times the mass ratio:

Working
\[ \%\ce{Cl} = \frac{35.5}{143.5}\cdot\frac{0.287}{0.30}\cdot100 \approx \textbf{23.7\%} \]
5 Read the colour

Problem. A Lassaigne extract from a compound containing both N and S gives a blood-red colour with \(\ce{Fe^3+}\), not Prussian blue. Why?

Solution. N and S together form thiocyanate:

Working
\[ \ce{Na + C + N + S -> NaSCN} \Rightarrow \ce{[Fe(SCN)]^2+}\ \text{(blood-red)} \]
6 Rf value

Problem. On a TLC plate a spot moves \(3.0\,\text{cm}\) while the solvent front moves \(5.0\,\text{cm}\). Find Rf.

Solution. Divide the two distances:

Working
\[ R_f = \frac{3.0}{5.0} = \textbf{0.60} \]
Review

Chapter Summary

Solid methods

Sublimation (volatile solids) and crystallisation (solubility difference) purify solids.

Distillation

Simple, fractional, reduced-pressure and steam — chosen by boiling point and stability.

Chromatography

Adsorption and partition separate mixtures; Rf identifies each component.

Detect elements

CuO for C and H; Lassaigne's sodium fusion for N, S and halogens.

Estimate C, H, N

Liebig (C, H); Dumas and Kjeldahl (N) — each with its percentage formula.

Estimate X, S, P, O

Carius weighs AgX or BaSO₄; P as Mg₂P₂O₇; oxygen by difference.

Practice

Problems

For each item, first decide whether it is purification, detection or estimation, then apply the matching principle. Difficulty rises down the list.

  1. State the principle of crystallisation and the properties of an ideal solvent.
  2. Name the four types of distillation and give one application of each.
  3. Explain why steam distillation lets a compound distil below 100 °C.
  4. Define the Rf value and state its range. How is it useful?
  5. How are carbon and hydrogen detected in an organic compound?
  6. Describe Lassaigne's test and why sodium fusion is necessary.
  7. Why must the extract be boiled with nitric acid before testing for halogens?
  8. A compound contains both N and S. What colour does the nitrogen test give, and why?
  9. Burning \(0.246\,\text{g}\) of a compound gives \(0.198\,\text{g}\) \(\ce{CO2}\) and \(0.1014\,\text{g}\) \(\ce{H2O}\). Find % C and % H.
  10. State the Dumas and Kjeldahl percentage formulae and the limitation of Kjeldahl.
  11. \(0.35\,\text{g}\) of a compound gives \(0.466\,\text{g}\) \(\ce{BaSO4}\). Find % S.
  12. Explain how oxygen is usually estimated and why a direct method is harder.
Tip: read every technique as a physical or chemical difference put to work. Volatility drives sublimation and distillation; solubility drives crystallisation and extraction; differing affinity drives chromatography; and every estimation is the same accounting — convert the hidden element into a weighable compound, then take its mass fraction. Ask "what difference is being exploited?" and the method, and its formula, follow.