8.4 Purification, Qualitative and Quantitative Analysis
Before a new compound can be named or used, it must be separated from everything else and its composition found. This section covers the purification methods — sublimation, crystallisation, the kinds of distillation, extraction and chromatography with Rf values — the tests that detect carbon, hydrogen, nitrogen, sulphur, halogens and phosphorus, and the classical methods that measure how much of each element a compound contains.
In 1972 the Chinese pharmacologist Tu Youyou and her team were searching old herbal texts for a cure for malaria. Extracts of sweet wormwood looked promising but kept failing — until Tu noticed that a 1,700-year-old recipe soaked the herb in cold water. Boiling destroyed the active compound. Extracting with ether at low temperature, then purifying the extract, gave pure artemisinin, now taken by millions of patients every year; in 2015 she received the Nobel Prize in Physiology or Medicine. The story is a reminder that most of a chemist's time is spent separating and purifying — and that the right method depends on the properties of the compound.
Once a compound is pure, the next questions are: which elements does it contain, and in what proportions? The answers give its empirical formula (Section 1.3), the first step towards a structure. This section describes the standard methods of purification, the classical tests for the elements, and the quantitative methods by which the percentage of each element is measured.
Purification: Sublimation, Crystallisation, Distillation and Extraction
📘 Definition 8.16 — Purity and the methods of purification
A pure organic compound has a sharp, constant melting point (solids) or boiling point (liquids), and gives a single spot in chromatography. The main methods of purification are:
- Sublimation — for solids that pass directly into vapour on heating (camphor, naphthalene, benzoic acid) from non-volatile impurities.
- Crystallisation — the impure solid is dissolved in a minimum of hot solvent in which it is sparingly soluble cold but readily soluble hot; on cooling the pure compound crystallises and the impurities stay in solution (activated charcoal removes coloured impurities). Repeated crystallisation separates compounds of similar solubility.
- Distillation — simple distillation for liquids with very different boiling points or a liquid from non-volatile impurities; fractional distillation, with a fractionating column, for liquids whose boiling points are close; distillation under reduced pressure for liquids that decompose at or below their normal boiling point; steam distillation for steam-volatile compounds that do not mix with water.
- Differential extraction — an organic compound dissolved in water is shaken with an immiscible organic solvent (ether, chloroform) in which it is more soluble; the layers are separated in a separating funnel and the solvent is evaporated.
- Chromatography — separation by differential movement through a stationary phase (next subsection).
🧭 Problem-solving strategy — Choosing a purification method
| Situation | Method | Typical example |
|---|---|---|
| solid that sublimes; impurity does not | sublimation | camphor from sand or common salt |
| solid; impurity more (or less) soluble in a chosen solvent | crystallisation | benzoic acid from hot water |
| liquids, boiling points far apart | simple distillation | chloroform (334 K) and aniline (457 K) |
| liquids, boiling points close together | fractional distillation | fractions of crude oil |
| liquid that decomposes at its boiling point | distillation under reduced pressure | glycerol from spent lye (soap industry) |
| steam-volatile, immiscible with water | steam distillation | aniline; essential oils from plants |
| compound in water, more soluble in another solvent | differential extraction | artemisinin into ether |
| small amounts, similar properties | chromatography | dyes in an ink, plant pigments |
📐 Rule 8.10 — Principle of distillation under reduced pressure and steam distillation
A liquid boils when its vapour pressure equals the pressure above it. Lowering the external pressure therefore lowers the boiling point, so a liquid can be distilled below the temperature at which it decomposes. In steam distillation, two immiscible liquids each exert their own vapour pressure, and the mixture boils when
$$p_{\text{organic}}+p_{\text{water}}=p_{\text{atmosphere}}$$
so it boils below $373\ \mathrm{K}$ (and far below the boiling point of the organic liquid); the organic compound distils over with the water and is then separated from it.
✏️ Example 1 — Choosing the method
Suggest a method for each and give a reason: (a) camphor contaminated with common salt; (b) benzoic acid containing a small amount of a coloured impurity; (c) a mixture of chloroform (b.p. $334\ \mathrm{K}$) and aniline (b.p. $457\ \mathrm{K}$); (d) separating petrol, kerosene and diesel from crude oil; (e) glycerol from the spent lye of soap making (glycerol decomposes at its boiling point, $563\ \mathrm{K}$); (f) an organic compound from its aqueous solution, when it is far more soluble in ether. (g) A water-insoluble liquid with b.p. $457\ \mathrm{K}$ steam-distils at $371\ \mathrm{K}$ under $760\ \mathrm{mmHg}$; the vapour pressure of water at $371\ \mathrm{K}$ is $707\ \mathrm{mmHg}$. What is the vapour pressure of the organic liquid at $371\ \mathrm{K}$?
💡 Show solution▶
(a), (b) Sublimation: camphor sublimes, salt does not. Crystallisation from hot water with a little activated charcoal (which adsorbs the coloured impurity); filter hot, then cool.
(c), (d) Simple distillation: the boiling points differ by $123\ \mathrm{K}$, so chloroform distils first. Fractional distillation: the components boil over overlapping ranges; the fractionating column gives many successive vaporisation–condensation steps.
(e), (f) Distillation under reduced pressure (glycerol then boils below its decomposition temperature). Differential extraction with ether in a separating funnel, repeated several times, then evaporation of the ether.
(g) $p_{\text{organic}}=760-707=53\ \mathrm{mmHg}$. The mixture boils when the two vapour pressures add up to $760\ \mathrm{mmHg}$ — at $371\ \mathrm{K}$, well below $457\ \mathrm{K}$, so the compound is distilled without overheating.
Chromatography
📘 Definition 8.17 — Chromatography and the retardation factor
In chromatography a mixture is carried by a moving mobile phase (a liquid or gas) over a fixed stationary phase; components that are held more strongly by the stationary phase move more slowly, so the components separate.
- Adsorption chromatography: the stationary phase is an adsorbent, usually silica gel or alumina. In column chromatography it is packed in a glass tube and the components are washed out (eluted) one after another; in thin-layer chromatography (TLC) it is a thin layer on a glass or plastic plate.
- Partition chromatography: components distribute themselves between two liquids. In paper chromatography the stationary phase is the water held in the fibres of special paper and the mobile phase is a solvent that rises through it.
📐 Rule 8.11 — Properties of $R_{f}$
- $R_{f}$ is a ratio of two lengths: it has no unit and always lies between 0 and 1.
- For a given stationary phase, solvent and temperature, $R_{f}$ is characteristic of a compound, so spots are identified by comparison with standards run on the same plate.
- On a polar adsorbent (silica, alumina) the more polar compound is adsorbed more strongly and has the smaller $R_{f}$; a more polar solvent moves all the spots further.
✏️ Example 2 — Calculating and using Rf values
(a) On a TLC plate (silica gel) the solvent front moved $8.0\ \mathrm{cm}$ from the base line; spot A moved $6.0\ \mathrm{cm}$ and spot B $2.4\ \mathrm{cm}$. Find their $R_{f}$ values. Which compound is more polar? (b) An unknown run on the same plate travelled $2.5\ \mathrm{cm}$. Which of A and B could it be? (c) A paper chromatogram of an ink shows three spots at $1.2$, $3.6$ and $5.4\ \mathrm{cm}$ with the front at $7.2\ \mathrm{cm}$. Find the $R_{f}$ values. (d) How far would the middle spot move if the front were allowed to travel $10.0\ \mathrm{cm}$?
💡 Show solution▶
(a) $R_{f}(\mathrm{A})=\dfrac{6.0}{8.0}=0.75$, $R_{f}(\mathrm{B})=\dfrac{2.4}{8.0}=0.30$. B is held more strongly by the polar silica: B is the more polar compound.
(b) $R_{f}=\dfrac{2.5}{8.0}=0.31\approx0.30$: probably B. (Confirm by running the unknown and B side by side, and with a second solvent.)
(c) $\dfrac{1.2}{7.2}=0.17$, $\dfrac{3.6}{7.2}=0.50$, $\dfrac{5.4}{7.2}=0.75$.
(d) $R_{f}$ stays the same: $0.50\times10.0=5.0\ \mathrm{cm}$.
Check Every $R_{f}$ lies between 0 and 1 ✓; the order of the spots does not depend on how far the front runs ✓.
Qualitative Analysis: Which Elements Are Present?
Organic compounds are covalent, so their elements cannot be detected directly with the ionic tests of inorganic analysis. The trick is to convert each element into an ion or a simple compound first.
📘 Definition 8.18 — Tests for the elements
- Carbon and hydrogen. The compound is heated with copper(II) oxide. Carbon is oxidised to $\mathrm{CO_{2}}$, which turns lime water milky; hydrogen gives water, which turns white anhydrous copper(II) sulphate blue: $$\mathrm{CO_{2}(g)+Ca(OH)_{2}(aq)\rightarrow CaCO_{3}(s)+H_{2}O(l)}$$ $$\mathrm{CuSO_{4}(s)+5H_{2}O(l)\rightarrow CuSO_{4}{\cdot}5H_{2}O(s)}$$
- Lassaigne's test (sodium fusion) for N, S and halogens. The compound is fused with sodium metal, which converts covalent N, S and X into ionic $\mathrm{NaCN}$, $\mathrm{Na_{2}S}$ and $\mathrm{NaX}$; the melt is boiled with water and filtered to give the sodium fusion extract: $$\mathrm{Na+C+N\xrightarrow{\Delta}NaCN};\quad \mathrm{2Na+S\xrightarrow{\Delta}Na_{2}S};\quad \mathrm{Na+X\xrightarrow{\Delta}NaX}$$
- Phosphorus. Heating with sodium peroxide converts P into phosphate; boiling with nitric acid and adding ammonium molybdate gives a canary-yellow precipitate of ammonium phosphomolybdate, $\mathrm{(NH_{4})_{3}PO_{4}{\cdot}12MoO_{3}}$.
🧭 Problem-solving strategy — Reading Lassaigne's test
| Element | Test on the sodium fusion extract | Positive result |
|---|---|---|
| N | boil with $\mathrm{FeSO_{4}}$, acidify with conc. $\mathrm{H_{2}SO_{4}}$ | Prussian blue, $\mathrm{Fe_{4}[Fe(CN)_{6}]_{3}{\cdot}xH_{2}O}$ |
| S | (i) acetic acid + lead acetate; (ii) sodium nitroprusside | (i) black $\mathrm{PbS}$; (ii) violet colour |
| N and S together | add $\mathrm{FeCl_{3}}$ (the extract contains $\mathrm{SCN^{-}}$) | blood-red $\mathrm{[Fe(SCN)]^{2+}}$ (no Prussian blue) |
| halogens | boil with conc. $\mathrm{HNO_{3}}$, then add $\mathrm{AgNO_{3}}$ | white $\mathrm{AgCl}$ (soluble in $\mathrm{NH_{3}(aq)}$), pale yellow $\mathrm{AgBr}$ (sparingly soluble), yellow $\mathrm{AgI}$ (insoluble) |
✏️ Example 3 — Interpreting tests
What do these observations show? (a) The sodium fusion extract, boiled with $\mathrm{FeSO_{4}}$ and acidified, turns Prussian blue. (b) The extract gives a blood-red colour with $\mathrm{FeCl_{3}}$ but no Prussian blue. (c) After boiling with $\mathrm{HNO_{3}}$, the extract gives a yellow precipitate with $\mathrm{AgNO_{3}}$ that does not dissolve in ammonia. (d) Why must the extract be boiled with $\mathrm{HNO_{3}}$ before adding $\mathrm{AgNO_{3}}$? (e) Hydrazine, $\mathrm{NH_{2}NH_{2}}$, contains nitrogen but gives no Prussian blue. Why? (f) Why is sodium, not a sodium salt, used in the fusion?
💡 Show solution▶
(a), (b) Nitrogen present ($\mathrm{CN^{-}}$ → ferrocyanide → Prussian blue). Both N and S present: they ended up together as $\mathrm{NaSCN}$, which gives red $\mathrm{[Fe(SCN)]^{2+}}$; with excess sodium they would appear as $\mathrm{NaCN}$ and $\mathrm{Na_{2}S}$ instead.
(c) Iodine present ($\mathrm{AgI}$: yellow, insoluble in ammonia).
(d) To decompose $\mathrm{NaCN}$ and $\mathrm{Na_{2}S}$ (as HCN and $\mathrm{H_{2}S}$); otherwise white $\mathrm{AgCN}$ or black $\mathrm{Ag_{2}S}$ would interfere with the halide test.
(e) The test needs $\mathrm{CN^{-}}$, which can form only if carbon is present too. Hydrazine has no carbon, so no cyanide forms.
(f) The covalent N, S and X must be reduced to anions; sodium, a strong reducing agent, gives up its electrons to them and forms the ionic salts.
Quantitative Analysis: How Much of Each Element?
Each method converts one element completely into a substance of known formula, which is weighed (or, for nitrogen, measured as a gas volume or titrated). The mass of the element follows from the mass fraction of that element in the product.
🔧 Tool from earlier work — gas volumes
For a fixed amount of gas, $\dfrac{p_{1}V_{1}}{T_{1}}=\dfrac{p_{2}V_{2}}{T_{2}}$; at STP ($273\ \mathrm{K}$, $760\ \mathrm{mmHg}$) one mole of gas occupies $22\,400\ \mathrm{mL}$. A gas collected over water contains water vapour: $p_{\text{gas}}=p_{\text{atmosphere}}-\text{aqueous tension}$.
📐 Rule 8.12 — Percentage composition from the classical methods
For a sample of mass $m$ (all masses in the same unit):
- Carbon and hydrogen (Liebig) — burnt in oxygen over CuO; $\mathrm{CO_{2}}$ absorbed in KOH, $\mathrm{H_{2}O}$ in anhydrous $\mathrm{CaCl_{2}}$: $$\%\mathrm{C}=\frac{12}{44}\times\frac{m(\mathrm{CO_{2}})}{m}\times100,\qquad \%\mathrm{H}=\frac{2}{18}\times\frac{m(\mathrm{H_{2}O})}{m}\times100$$
- Nitrogen (Dumas) — heated with CuO; N is released as $\mathrm{N_{2}}$, collected over KOH solution; with $V_{0}$ the volume at STP in mL: $$\%\mathrm{N}=\frac{28}{22\,400}\times\frac{V_{0}}{m}\times100$$
- Nitrogen (Kjeldahl) — digested with conc. $\mathrm{H_{2}SO_{4}}$ (N → $\mathrm{(NH_{4})_{2}SO_{4}}$), then with NaOH; the $\mathrm{NH_{3}}$ is absorbed in a known excess of acid and the excess is titrated. With $n$ the millimoles of $\mathrm{NH_{3}}$ and $m$ in grams: $$\%\mathrm{N}=\frac{1.4\,n}{m}$$ For $V$ mL of $M$ molar $\mathrm{H_{2}SO_{4}}$ back-titrated with $V_{1}$ mL of $M$ molar NaOH this is $\%\mathrm{N}=\dfrac{1.4\times M\times2(V-V_{1}/2)}{m}$. Not applicable to nitro and azo compounds or to N in rings (pyridine).
- Halogens (Carius) — heated with fuming $\mathrm{HNO_{3}}$ and $\mathrm{AgNO_{3}}$; the silver halide is weighed: $$\%\mathrm{X}=\frac{A_{r}(\mathrm{X})}{M(\mathrm{AgX})}\times\frac{m(\mathrm{AgX})}{m}\times100\quad(\mathrm{AgCl}\ 143.5,\ \mathrm{AgBr}\ 188,\ \mathrm{AgI}\ 235)$$
- Sulphur (Carius) — oxidised by fuming $\mathrm{HNO_{3}}$ to $\mathrm{H_{2}SO_{4}}$, precipitated as $\mathrm{BaSO_{4}}$: $\ \%\mathrm{S}=\dfrac{32}{233}\times\dfrac{m(\mathrm{BaSO_{4}})}{m}\times100$.
- Phosphorus — oxidised to $\mathrm{H_{3}PO_{4}}$, weighed as $\mathrm{Mg_{2}P_{2}O_{7}}$ ($222$) or as ammonium phosphomolybdate ($1877$): $\ \%\mathrm{P}=\dfrac{62}{222}\times\dfrac{m(\mathrm{Mg_{2}P_{2}O_{7}})}{m}\times100$ or $\%\mathrm{P}=\dfrac{31}{1877}\times\dfrac{m(\text{phosphomolybdate})}{m}\times100$.
- Oxygen — usually by difference: $\%\mathrm{O}=100-(\text{sum of all the other percentages})$.
✏️ Example 4 — Carbon and hydrogen, then the empirical formula
$0.2475\ \mathrm{g}$ of a compound containing C, H and O gave $0.4950\ \mathrm{g}$ of $\mathrm{CO_{2}}$ and $0.2025\ \mathrm{g}$ of $\mathrm{H_{2}O}$ on complete combustion. Find the percentage composition and the empirical formula. If $M=88\ \mathrm{g\,mol^{-1}}$, what is the molecular formula?
💡 Show solution▶
%C $\dfrac{12}{44}\times\dfrac{0.4950}{0.2475}\times100=\dfrac{12}{44}\times200=54.55\%$.
%H, %O $\dfrac{2}{18}\times\dfrac{0.2025}{0.2475}\times100=9.09\%$; $\%\mathrm{O}=100-54.55-9.09=36.36\%$.
Ratio $\mathrm{C}:\mathrm{H}:\mathrm{O}=\dfrac{54.55}{12}:\dfrac{9.09}{1}:\dfrac{36.36}{16}=4.545:9.09:2.273=2:4:1$ → empirical formula $\mathrm{C_{2}H_{4}O}$ ($44\ \mathrm{g\,mol^{-1}}$).
Molecular $88/44=2$ → $\mathrm{C_{4}H_{8}O_{2}}$ (for example ethyl ethanoate).
Check $\mathrm{C_{4}H_{8}O_{2}}$: $\%\mathrm{C}=48/88=54.55\%$ ✓, $\%\mathrm{H}=8/88=9.09\%$ ✓.
✏️ Example 5 — Nitrogen by the Dumas and Kjeldahl methods
(a) $0.300\ \mathrm{g}$ of a compound gave $50.0\ \mathrm{mL}$ of $\mathrm{N_{2}}$ collected over water at $300\ \mathrm{K}$ and $715\ \mathrm{mmHg}$ (aqueous tension at $300\ \mathrm{K}$: $26.7\ \mathrm{mmHg}$). Find $\%\mathrm{N}$. (b) In a Kjeldahl determination, the ammonia from $0.500\ \mathrm{g}$ of a compound was passed into $50.0\ \mathrm{mL}$ of $0.250\ \mathrm{M}$ $\mathrm{H_{2}SO_{4}}$; the excess acid needed $35.0\ \mathrm{mL}$ of $0.500\ \mathrm{M}$ NaOH. Find $\%\mathrm{N}$. (c) Why can the Kjeldahl method not be used for nitrobenzene?
💡 Show solution▶
(a) STP volume $p=715-26.7=688.3\ \mathrm{mmHg}$; $V_{0}=\dfrac{688.3\times50.0\times273}{760\times300}=41.2\ \mathrm{mL}$.
(a) %N $\dfrac{28}{22\,400}\times\dfrac{41.2}{0.300}\times100=17.2\%$.
(b) $\mathrm{H^{+}}$ taken: $2\times0.250\times50.0=25.0\ \mathrm{mmol}$; left over (neutralised by NaOH): $0.500\times35.0=17.5\ \mathrm{mmol}$; used by $\mathrm{NH_{3}}$: $7.5\ \mathrm{mmol}$, so $n(\mathrm{NH_{3}})=7.5\ \mathrm{mmol}$. $\%\mathrm{N}=\dfrac{1.4\times7.5}{0.500}=21.0\%$.
(c) The nitrogen of a nitro group is not converted into ammonium sulphate under the digestion conditions, so the method gives a low result; nitro, azo and ring nitrogen need the Dumas method.
Check (b) $7.5\ \mathrm{mmol}$ N = $0.105\ \mathrm{g}$, and $0.105/0.500=21.0\%$ ✓.
✏️ Example 6 — Halogen, sulphur, phosphorus and oxygen
(a) In a Carius determination $0.250\ \mathrm{g}$ of a compound gave $0.376\ \mathrm{g}$ of AgBr. Find $\%\mathrm{Br}$. (b) $0.160\ \mathrm{g}$ of a compound gave $0.466\ \mathrm{g}$ of $\mathrm{BaSO_{4}}$. Find $\%\mathrm{S}$. (c) $0.150\ \mathrm{g}$ of a compound gave $0.111\ \mathrm{g}$ of $\mathrm{Mg_{2}P_{2}O_{7}}$. Find $\%\mathrm{P}$. (d) A compound of C, H and O contains $40.0\%$ C and $6.7\%$ H. Find $\%\mathrm{O}$ and the empirical formula.
💡 Show solution▶
(a) $\dfrac{80}{188}\times\dfrac{0.376}{0.250}\times100=64.0\%$.
(b) $\dfrac{32}{233}\times\dfrac{0.466}{0.160}\times100=40.0\%$.
(c) $\mathrm{Mg_{2}P_{2}O_{7}}$ contains 2 P ($62$) in $222$: $\dfrac{62}{222}\times\dfrac{0.111}{0.150}\times100=20.7\%$.
(d) $\%\mathrm{O}=100-40.0-6.7=53.3\%$. Ratio $\dfrac{40.0}{12}:\dfrac{6.7}{1}:\dfrac{53.3}{16}=3.33:6.7:3.33=1:2:1$ → $\mathrm{CH_{2}O}$.
Exercises
I
Purification.
- 1 Give two criteria of purity of an organic compound.
- 2 State the principle of crystallisation. Why is a little activated charcoal sometimes added?
- 3 How would you obtain pure naphthalene from a mixture with sand?
- 4 Ethanol (b.p. $351\ \mathrm{K}$) and water (b.p. $373\ \mathrm{K}$) are miscible. Which kind of distillation separates them better, and why?
- 5 Why is glycerol purified by distillation under reduced pressure?
- 6 State the conditions a compound must satisfy to be purified by steam distillation.
- 7 $o$-Nitrophenol and $p$-nitrophenol are separated by steam distillation. Which one distils, and why?
- 8 In differential extraction, why are several extractions with small portions of ether better than one extraction with the same total volume?
💡 Show solutions▶
1 A sharp, constant melting point (solid) or boiling point (liquid); a single spot in TLC or paper chromatography.
2 The compound and the impurities differ in solubility in the chosen solvent: the compound is much more soluble hot than cold, so it crystallises on cooling while the impurities stay dissolved (or are filtered off hot). Charcoal adsorbs coloured impurities.
3 Sublimation: on gentle heating naphthalene vaporises and condenses as pure crystals on a cool surface; sand remains.
4 Fractional distillation: the boiling points are only $22\ \mathrm{K}$ apart, and the column provides repeated vaporisation–condensation. (It gives about 95.6% ethanol at best, because the two form a constant-boiling mixture.)
5 Glycerol decomposes at or below its normal boiling point ($563\ \mathrm{K}$); at reduced pressure it boils at a much lower temperature, without decomposition.
6 It must be immiscible with water, volatile in steam (appreciable vapour pressure near $373\ \mathrm{K}$), and not decomposed by steam.
7 $o$-Nitrophenol: its $\mathrm{OH}$ forms an intramolecular hydrogen bond with the neighbouring $\mathrm{NO_{2}}$, so its molecules do not hold each other (or water) strongly and it is steam-volatile. $p$-Nitrophenol forms intermolecular hydrogen bonds (Section 4.4) and stays behind.
8 Each extraction removes a fixed fraction of what is left in the water, so repeated small extractions remove more in total than a single large one.
II
Chromatography.
- 1 Name the stationary and mobile phases in TLC and in paper chromatography.
- 2 A spot moved $3.0\ \mathrm{cm}$ while the solvent moved $12.0\ \mathrm{cm}$. Find $R_{f}$.
- 3 A compound has $R_{f}=0.65$. How far will it move if the solvent front travels $8.0\ \mathrm{cm}$?
- 4 Why can $R_{f}$ never be greater than 1?
- 5 On silica gel, which has the larger $R_{f}$ in the same solvent: a hydrocarbon or an alcohol? Explain.
- 6 Give three ways of making colourless spots visible.
- 7 Two compounds give spots at $R_{f}=0.42$ and $0.44$. Suggest how to separate them better.
- 8 Applied. A sweet's colouring gives spots at $2.9\ \mathrm{cm}$ and $5.7\ \mathrm{cm}$, with the front at $8.0\ \mathrm{cm}$. Standards on the same paper: yellow dye Y ($R_{f}=0.36$), red dye R ($0.52$), blue dye B ($0.71$). Which dyes are present?
💡 Show solutions▶
1 TLC: silica gel or alumina layer (stationary), a solvent or solvent mixture (mobile). Paper: water held in the paper fibres (stationary), a solvent rising through the paper (mobile).
2 $R_{f}=3.0/12.0=0.25$.
3 $0.65\times8.0=5.2\ \mathrm{cm}$.
4 A substance is carried by the solvent and cannot move further than the solvent front itself.
5 The hydrocarbon: it is non-polar and only weakly adsorbed by the polar silica; the alcohol is held by hydrogen bonding to the surface.
6 Ultraviolet light (spots that fluoresce or quench the plate's fluorescence); a jar of iodine crystals (brown spots); spraying with a reagent (e.g. ninhydrin for amino acids).
7 Change the solvent (different polarity or mixture), use a longer plate, or use column chromatography; any change that makes the two $R_{f}$ values differ more.
8 $2.9/8.0=0.36$ (Y) and $5.7/8.0=0.71$ (B): the yellow and blue dyes (a green sweet); no red dye.
III
Qualitative analysis.
- 1 Why are organic compounds fused with sodium before testing for N, S and halogens?
- 2 Write the equation for the formation of sodium cyanide during the fusion.
- 3 Name and give the formula of the coloured product in the test for nitrogen.
- 4 Describe two tests for sulphur in the sodium fusion extract.
- 5 What does a blood-red colour with iron(III) chloride indicate?
- 6 How do you tell whether the halogen present is Cl, Br or I?
- 7 How are carbon and hydrogen detected in an organic compound?
- 8 Describe the test for phosphorus and its positive result.
💡 Show solutions▶
1 In organic compounds N, S and X are covalently bonded and give no ionic tests; fusion with sodium converts them into the ionic $\mathrm{NaCN}$, $\mathrm{Na_{2}S}$ and $\mathrm{NaX}$, which dissolve in water.
2 $\mathrm{Na+C+N\xrightarrow{\Delta}NaCN}$.
3 Prussian blue, iron(III) hexacyanidoferrate(II), $\mathrm{Fe_{4}[Fe(CN)_{6}]_{3}{\cdot}xH_{2}O}$.
4 (i) Acidify with acetic acid and add lead acetate: black $\mathrm{PbS}$. (ii) Add sodium nitroprusside: violet colour.
5 Both N and S are present (as thiocyanate, giving $\mathrm{[Fe(SCN)]^{2+}}$).
6 Boil the extract with conc. $\mathrm{HNO_{3}}$, add $\mathrm{AgNO_{3}}$: white precipitate soluble in ammonia → Cl; pale yellow, sparingly soluble → Br; yellow, insoluble → I.
7 Heat with dry CuO: the $\mathrm{CO_{2}}$ formed turns lime water milky (carbon), and the water formed turns anhydrous $\mathrm{CuSO_{4}}$ blue (hydrogen).
8 Heat with $\mathrm{Na_{2}O_{2}}$ (P → phosphate), boil with $\mathrm{HNO_{3}}$, add ammonium molybdate: canary-yellow ammonium phosphomolybdate.
IV
Quantitative analysis ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{N}=14$, $\mathrm{O}=16$, $\mathrm{S}=32$, $\mathrm{Cl}=35.5$, $\mathrm{I}=127$, $\mathrm{Ag}=108$, $\mathrm{Ba}=137$).
- 1 $0.30\ \mathrm{g}$ of a compound of C, H and O gave $0.44\ \mathrm{g}$ $\mathrm{CO_{2}}$ and $0.18\ \mathrm{g}$ $\mathrm{H_{2}O}$. Find $\%\mathrm{C}$, $\%\mathrm{H}$ and $\%\mathrm{O}$.
- 2 Find the empirical formula of the compound in 1, and its molecular formula if $M=180\ \mathrm{g\,mol^{-1}}$.
- 3 $0.150\ \mathrm{g}$ of a compound gave $0.287\ \mathrm{g}$ AgCl in a Carius determination. Find $\%\mathrm{Cl}$.
- 4 $0.200\ \mathrm{g}$ of a compound gave $0.235\ \mathrm{g}$ AgI. Find $\%\mathrm{I}$.
- 5 The ammonia from $0.60\ \mathrm{g}$ of urea, $\mathrm{CO(NH_{2})_{2}}$, exactly neutralised $10.0\ \mathrm{mL}$ of $1.00\ \mathrm{M}$ $\mathrm{H_{2}SO_{4}}$. Find $\%\mathrm{N}$ and compare with the value from the formula.
- 6 $0.250\ \mathrm{g}$ of a compound gave $40.0\ \mathrm{mL}$ of $\mathrm{N_{2}}$ at $300\ \mathrm{K}$ and $725\ \mathrm{mmHg}$ (aqueous tension $26.7\ \mathrm{mmHg}$). Find $\%\mathrm{N}$.
- 7 $0.200\ \mathrm{g}$ of a compound gave $0.466\ \mathrm{g}$ $\mathrm{BaSO_{4}}$. Find $\%\mathrm{S}$.
- 8 Applied. A urea fertiliser is checked by the Kjeldahl method: the ammonia from $0.250\ \mathrm{g}$ was absorbed in $25.0\ \mathrm{mL}$ of $0.250\ \mathrm{M}$ $\mathrm{H_{2}SO_{4}}$, and the excess acid needed $9.8\ \mathrm{mL}$ of $0.500\ \mathrm{M}$ NaOH. Find $\%\mathrm{N}$. Is the fertiliser pure urea?
💡 Show solutions▶
1 $\%\mathrm{C}=\tfrac{12}{44}\times\tfrac{0.44}{0.30}\times100=40.0\%$; $\%\mathrm{H}=\tfrac{2}{18}\times\tfrac{0.18}{0.30}\times100=6.67\%$; $\%\mathrm{O}=53.3\%$.
2 $\tfrac{40.0}{12}:\tfrac{6.67}{1}:\tfrac{53.3}{16}=3.33:6.67:3.33=1:2:1$ → $\mathrm{CH_{2}O}$ ($30$); $180/30=6$ → $\mathrm{C_{6}H_{12}O_{6}}$.
3 $\tfrac{35.5}{143.5}\times\tfrac{0.287}{0.150}\times100=47.3\%$.
4 $\tfrac{127}{235}\times\tfrac{0.235}{0.200}\times100=63.5\%$.
5 $n(\mathrm{NH_{3}})=2\times1.00\times10.0=20.0\ \mathrm{mmol}$; $\%\mathrm{N}=\tfrac{1.4\times20.0}{0.60}=46.7\%$. From the formula: $\tfrac{28}{60}\times100=46.7\%$ ✓.
6 $p=698.3\ \mathrm{mmHg}$; $V_{0}=\tfrac{698.3\times40.0\times273}{760\times300}=33.4\ \mathrm{mL}$; $\%\mathrm{N}=\tfrac{28}{22\,400}\times\tfrac{33.4}{0.250}\times100=16.7\%$.
7 $\tfrac{32}{233}\times\tfrac{0.466}{0.200}\times100=32.0\%$.
8 $\mathrm{H^{+}}$: $2\times0.250\times25.0=12.5\ \mathrm{mmol}$; NaOH: $0.500\times9.8=4.9\ \mathrm{mmol}$; $n(\mathrm{NH_{3}})=7.6\ \mathrm{mmol}$; $\%\mathrm{N}=\tfrac{1.4\times7.6}{0.250}=42.6\%$. Pure urea has $46.7\%$, so the fertiliser is about $91\%$ urea.
🎯 Exam corner
E1. NEET In Lassaigne's test, a blood-red colour on adding $\mathrm{FeCl_{3}}$ to the sodium fusion extract shows the presence of (A) N only (B) S only (C) both N and S (D) a halogen
💡 Show solution▶
Idea N and S together give $\mathrm{NaSCN}$; $\mathrm{SCN^{-}}$ forms red $\mathrm{[Fe(SCN)]^{2+}}$ with $\mathrm{Fe^{3+}}$.
Answer (C).
E2. JEE Main Numerical answer. In a Carius determination, $0.250\ \mathrm{g}$ of an organic compound gave $0.404\ \mathrm{g}$ of AgCl. Find the percentage of chlorine (nearest integer). ($\mathrm{Ag}=108$, $\mathrm{Cl}=35.5$)
💡 Show solution▶
Formula $\%\mathrm{Cl}=\dfrac{35.5}{143.5}\times\dfrac{0.404}{0.250}\times100=0.2474\times1.616\times100=39.98$.
Answer 40.
E3. JEE Main The Kjeldahl method cannot be used to estimate nitrogen in (A) urea (B) ethanamide (C) nitrobenzene (D) aniline
💡 Show solution▶
Idea Nitro and azo nitrogen, and nitrogen in a ring (pyridine), is not converted into ammonium sulphate during digestion; amines and amides are.
Answer (C).
E4. Board · 3 marks Explain the principle of steam distillation. Why can $o$-nitrophenol be separated from $p$-nitrophenol by this method?
💡 Show solution▶
1 mark A mixture of water and an immiscible, steam-volatile liquid boils when the sum of their vapour pressures equals atmospheric pressure, $p_{1}+p_{2}=p_{\text{atm}}$.
1 mark So it boils below $373\ \mathrm{K}$, below the boiling point of either component, and the organic compound distils with the steam without decomposing.
1 mark $o$-Nitrophenol has an intramolecular hydrogen bond, so it is volatile in steam; $p$-nitrophenol is held by intermolecular hydrogen bonds and is not.
E5. NEET In a TLC experiment a spot moved $3.2\ \mathrm{cm}$ and the solvent front $8.0\ \mathrm{cm}$. The $R_{f}$ value is (A) 0.25 (B) 0.40 (C) 2.5 (D) 4.8
💡 Show solution▶
Formula $R_{f}=3.2/8.0=0.40$; (C) and (D) are impossible because $R_{f}\le1$.
Answer (B).
E6. JEE Advanced Numerical answer. An organic compound with $M=60\ \mathrm{g\,mol^{-1}}$ contains $20.0\%$ C, $6.67\%$ H and $46.67\%$ N, the rest being oxygen. The ammonia obtained from $0.150\ \mathrm{g}$ of it by the Kjeldahl method is absorbed in $0.100\ \mathrm{M}$ $\mathrm{H_{2}SO_{4}}$. What volume (in mL) of this acid is exactly neutralised?
💡 Show solution▶
Formula $\%\mathrm{O}=26.66$. Ratio $\tfrac{20.0}{12}:\tfrac{6.67}{1}:\tfrac{46.67}{14}:\tfrac{26.66}{16}=1.667:6.67:3.333:1.666=1:4:2:1$ → $\mathrm{CH_{4}N_{2}O}$ ($M=60$, so this is the molecular formula: urea).
Ammonia Mass of N in $0.150\ \mathrm{g}$: $0.150\times0.4667=0.0700\ \mathrm{g}$ = $5.00\ \mathrm{mmol}$ N = $5.00\ \mathrm{mmol}$ $\mathrm{NH_{3}}$.
Acid $\mathrm{2NH_{3}(g)+H_{2}SO_{4}(aq)\rightarrow(NH_{4})_{2}SO_{4}(aq)}$: $2.50\ \mathrm{mmol}$ $\mathrm{H_{2}SO_{4}}$; $V=2.50/0.100=$ 25.0 mL.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Artemisinin, the antimalarial isolated by Tu Youyou's team, has the molecular formula $\mathrm{C_{15}H_{22}O_{5}}$. If you were handed a few milligrams of a white solid claimed to be artemisinin, how would you check its purity and confirm its composition with the methods of this section?
Why it matters. The World Health Organization estimates that malaria still causes well over half a million deaths a year, most of them young children in Africa. Fake and sub-standard antimalarial tablets are a serious problem in many countries, and regulators and non-profit laboratories test suspect tablets with exactly these tools: chromatography for identity and purity, elemental analysis and melting point for composition.
What to do.
- Explain why Tu Youyou's team extracted the herb with cold ether rather than boiling water. Which purification method of this section would you use to remove the ether afterwards without heating the extract strongly?
- Calculate the percentage composition of $\mathrm{C_{15}H_{22}O_{5}}$ ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$).
- Predict the masses of $\mathrm{CO_{2}}$ and $\mathrm{H_{2}O}$ that $10.0\ \mathrm{mg}$ of pure artemisinin would give on combustion. A sample gives $22.1\ \mathrm{mg}$ $\mathrm{CO_{2}}$ instead: what does that suggest?
- Design a TLC test (stationary phase, standard, how to see colourless spots) to compare a tablet extract with a genuine sample. What result would make you suspicious?
Tools: Rules 8.10–8.12, Interactives 1 and 2, the empirical-formula method of Section 1.3, and a WHO fact sheet on malaria.
Hand in: Your purification and analysis plan, the calculated composition and combustion masses, and a short paragraph on how these simple tests protect patients.
Going further. Elemental analysis cannot tell artemisinin from an isomer with the same formula. Which further measurement (one from this section) could, and why?
Instructor note: heat destroys artemisinin (its peroxide bridge is fragile), so evaporate the ether at reduced pressure (rotary evaporator) or gently. $M=282$: C $63.8\%$, H $7.8\%$, O $28.4\%$. From $10.0\ \mathrm{mg}$: $\mathrm{CO_{2}}=10.0\times\tfrac{15\times44}{282}=23.4\ \mathrm{mg}$, $\mathrm{H_{2}O}=10.0\times\tfrac{11\times18}{282}=7.0\ \mathrm{mg}$. $22.1\ \mathrm{mg}$ means only about $60\%$ C: an impurity or filler with less carbon (for example a starch-like binder or an inorganic solid). TLC: silica plate, genuine artemisinin as standard on the same plate, visualise with iodine vapour or a spray reagent and heat; extra spots, or no spot at the standard's $R_{f}$, are warning signs. Going further: melting point or $R_{f}$ in two different solvents distinguishes isomers.
🧪 Self-check quiz
0/5 correct
Q1. A spot moves $2.5\ \mathrm{cm}$ while the solvent front moves $10.0\ \mathrm{cm}$. Its $R_{f}$ is
Q2. Glycerol is recovered from the spent lye of the soap industry by
Q3. A Prussian blue colour in Lassaigne's test indicates
Q4. In the Carius method for halogens, the compound is heated with
Q5. The Kjeldahl method is not suitable for