NCERT Solutions · Class 11 Chemistry Organic Chemistry – Some Basic Principles and Techniques

Dumas and Kjeldahl's methods both find the percentage of nitrogen in an organic compound, but they measure it in completely different ways — one measures a gas by volume, the other measures an acid by titration — and that difference decides which compounds each method can even be used on.(i) Dumas method — nitrogen is turned into \(\displaystyle N_2\) gas and its volume is measured.The organic compound is mixed with dry copper(II) oxide, \(\displaystyle CuO\), and heated strongly in a stream of carbon dioxide, \(\displaystyle CO_2\). The \(\displaystyle CuO\) oxidises the carbon and hydrogen of the compound to \(\displaystyle CO_2\) and water, while the nitrogen of the compound is set free mostly as nitrogen gas, \(\displaystyle N_2\), along with small amounts of nitrogen oxides:\[\text{Organic compound} + CuO \xrightarrow{\ \Delta\ } CO_2 + H_2O + N_2 \ (\text{+ some } N_xO_y) \]Any oxides of nitrogen formed are not left as they are — the gas stream is passed over hot copper gauze, which reduces them back to \(\displaystyle N_2\), so that all the nitrogen ends up as the free element:\[2NO + 2Cu \xrightarrow{\ \Delta\ } N_2 + 2CuO \]Check the balance: $\displaystyle 2$ N atoms on each side (in \(\displaystyle 2NO\) and in \(\displaystyle N_2\)), $\displaystyle 2$ O atoms on each side (in \(\displaystyle 2NO\) and in \(\displaystyle 2CuO\)), and $\displaystyle 2$ Cu atoms on each side — the copper is the reducing agent and is itself oxidised.The mixed gas stream (\(\displaystyle CO_2 + N_2\)) is then passed into a graduated tube standing over concentrated \(\displaystyle KOH\) solution. \(\displaystyle KOH\) absorbs \(\displaystyle CO_2\) completely (\(\displaystyle CO_2 + 2KOH \rightarrow K_2CO_3 + H_2O\)) but does not touch \(\displaystyle N_2\), so the gas that collects and can be read off the graduations is pure \(\displaystyle N_2\), measured at known temperature and pressure and then converted to STP.This is a measurement of moles of gas, not of mass — so the working goes through the gas volume, not through weighing anything. From the volume \(\displaystyle V\) (in mL, at STP) of \(\displaystyle N_2\) collected:\[\text{moles of } N_2 = \frac{V}{22400} \]using \(\displaystyle 22400\ \text{mL}\) as the volume occupied by $\displaystyle 1$ mole of any gas at STP. The mass of nitrogen this represents uses the molar mass of \(\displaystyle N_2\), which is \(\displaystyle 28\ \text{g mol}^{-1}\) (two nitrogen atoms, each of atomic mass $\displaystyle 14$):\[\text{mass of } N = \frac{V}{22400}\times 28 \]Dividing by the mass \(\displaystyle m\) (in g) of the organic compound taken and multiplying by $\displaystyle 100$ gives the percentage:\[\%\,N = \frac{28 \times V \times 100}{22400 \times m} \](ii) Kjeldahl's method — nitrogen is turned into ammonium sulfate, then \(\displaystyle NH_3\), and the \(\displaystyle NH_3\) is found by titration.The organic compound is heated with concentrated sulfuric acid, \(\displaystyle H_2SO_4\) (a small amount of \(\displaystyle K_2SO_4\) is added to raise the boiling point, and \(\displaystyle CuSO_4\) or selenium acts as a catalyst). This digestion converts the nitrogen present into ammonium sulfate, while the carbon and hydrogen are oxidised away as \(\displaystyle CO_2\) and water:\[\text{Organic compound } (N) + H_2SO_4 \xrightarrow{\ \Delta,\ \text{catalyst}\ } (NH_4)_2SO_4 + CO_2 + H_2O \]The digested mixture is then made strongly alkaline with excess sodium hydroxide, which displaces ammonia gas from the ammonium sulfate:\[(NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3\uparrow + 2H_2O \]Balance check: $\displaystyle 2$ N atoms, $\displaystyle 8$ H atoms ($\displaystyle 4$ in each \(\displaystyle NH_4^+\)) matched by $\displaystyle 2$ in \(\displaystyle NaOH\) and $\displaystyle 2$ more released as \(\displaystyle 2NH_3\) plus the water, $\displaystyle 1$ S and the O's all accounted for in the two sulfate salts — the equation is balanced.This liberated ammonia is distilled over into a known volume of a standard (accurately known strength) acid taken in excess, where it is absorbed and neutralised. The acid left over after neutralising the ammonia is then found by back-titrating it against standard \(\displaystyle NaOH\); the volume of acid actually consumed by the ammonia is obtained by subtracting the leftover from the total taken. This is the step people get wrong — you are not titrating the ammonia directly, you are finding how much of a known excess of acid it used up.From the volume \(\displaystyle V\) (mL) of acid of normality \(\displaystyle N\) that reacted with the ammonia: $\displaystyle 1000$ mL of a $\displaystyle 1$ N acid is chemically equivalent to $\displaystyle 14$ g of nitrogen (one gram-equivalent of \(\displaystyle N\), since the acid neutralises the ammonia mole-for-mole of nitrogen). So the mass of nitrogen present is:\[\text{mass of } N = \frac{V \times N \times 14}{1000} \]and the percentage of nitrogen is:\[\%\,N = \frac{\text{mass of } N}{m}\times 100 = \frac{V \times N \times 14 \times 100}{1000 \times m} = \frac{1.4\times N \times V}{m} \]where \(\displaystyle m\) is the mass (g) of the organic compound taken. The constant is $\displaystyle 1.4$, not $\displaystyle 14$ — the extra factor of $\displaystyle 10$ comes purely from converting the $\displaystyle 1000$ mL basis into a percentage (÷$\displaystyle 1000$ × $\displaystyle 100$); it is not a separate chemical fact to memorise.The difference that actually matters — which compounds each method can be used on.Dumas' method converts all the nitrogen in the compound to \(\displaystyle N_2\) gas by oxidation–reduction, so it works for every nitrogen-containing compound, including nitro compounds, azo compounds, and compounds with nitrogen in a ring (such as pyridine).Kjeldahl's method depends on the nitrogen being convertible to ammonium sulfate by hot concentrated \(\displaystyle H_2SO_4\). Nitrogen that is directly bonded to another nitrogen or to oxygen (as in nitro, azo, or diazo groups) or is locked in an aromatic ring (as in pyridine) does not get reduced to ammonium sulfate under these conditions, so Kjeldahl's method fails for these compounds and gives a low result, while Dumas' method still succeeds.Answer: Dumas' method oxidises the compound with \(\displaystyle CuO\) in a \(\displaystyle CO_2\) atmosphere, reduces any nitrogen oxides back to \(\displaystyle N_2\) over hot copper, and measures the volume \(\displaystyle V\) of \(\displaystyle N_2\) collected over \(\displaystyle KOH\): \(\displaystyle \%N = \dfrac{28\times V\times100}{22400\times m}\). Kjeldahl's method digests the compound with conc. \(\displaystyle H_2SO_4\) to \(\displaystyle (NH_4)_2SO_4\), liberates \(\displaystyle NH_3\) with excess \(\displaystyle NaOH\), absorbs it in a known excess of standard acid, and finds the amount consumed by back-titration: \(\displaystyle \%N = \dfrac{1.4\times N\times V}{m}\). Dumas measures nitrogen as a gas and works for every nitrogen compound (nitro, azo, pyridine-type included); Kjeldahl measures nitrogen by acid–base titration and fails for compounds whose nitrogen is not converted to ammonium sulfate, i.e. nitro, azo, diazo compounds and ring nitrogen such as in pyridine.

Halogens, sulphur and phosphorus are all estimated the same way: convert the element into one particular insoluble salt, weigh that salt, and scale back to find what fraction of the original compound was that element. This is the Carius method (for halogens and sulphur) and a closely related oxidation-and-precipitation method for phosphorus. In every case you need two masses — the mass of organic compound you started with, and the mass of the final weighable salt — because the percentage formula is built entirely from that ratio and the known molar mass of the salt.1. Estimation of halogens (Carius method)A known mass of the organic compound is heated with fuming nitric acid, \(\displaystyle \text{HNO}_3 \), in the presence of silver nitrate, \(\displaystyle \text{AgNO}_3 \), inside a sealed hard-glass tube (the Carius tube). The carbon and hydrogen in the compound are oxidised away as carbon dioxide and water, while the halogen present is converted straight into the insoluble silver halide:\[\text{Organic compound} + \text{HNO}_3 \xrightarrow{\text{AgNO}_3,\ \Delta} \text{CO}_2 + \text{H}_2\text{O} + \text{AgX}\ (\text{precipitate}) \]Here \(\displaystyle X \) stands for the halogen (Cl, Br or I) present in the compound. This equation is not meant to be balanced atom-for-atom the way a normal reaction is, because "organic compound" is not one fixed formula — it is a bookkeeping statement of what happens to each type of atom: carbon and hydrogen end up oxidised, and the halogen ends up locked into \(\displaystyle \text{AgX} \), a solid that will not redissolve in the acidic solution.The silver halide precipitate is filtered, washed, dried and weighed. Let \(\displaystyle m \) be the mass of organic compound taken and \(\displaystyle m_1 \) be the mass of \(\displaystyle \text{AgX} \) obtained. Since every mole of \(\displaystyle \text{AgX} \) contains exactly one mole of the halogen, the mass of halogen inside \(\displaystyle m_1 \) grams of \(\displaystyle \text{AgX} \) is found from the fraction of \(\displaystyle \text{AgX} \)'s molar mass that belongs to \(\displaystyle X \):\[\%\ \text{of halogen} = \frac{\text{atomic mass of } X}{\text{molar mass of AgX}} \times \frac{m_1}{m} \times 100 \]The step people get wrong here is treating \(\displaystyle m_1 \), the mass of AgX, as if it were the mass of halogen itself. It is not — \(\displaystyle \text{AgX} \) also contains the mass of silver, so you must always multiply by the atomic-mass fraction \(\displaystyle \dfrac{\text{atomic mass of }X}{\text{molar mass of AgX}} \) before you can call the number "mass of halogen."Using atomic masses \(\displaystyle \text{Ag} = 108,\ \text{Cl} = 35.5,\ \text{Br} = 80,\ \text{I} = 127 \):\[\%\ \text{Cl} = \frac{35.5}{143.5}\times\frac{m_1}{m}\times100,\quad \%\ \text{Br} = \frac{80}{188}\times\frac{m_1}{m}\times100,\quad \%\ \text{I} = \frac{127}{235}\times\frac{m_1}{m}\times100 \](the denominators $\displaystyle 143.5$, $\displaystyle 188$ and $\displaystyle 235$ are the molar masses of \(\displaystyle \text{AgCl} \), \(\displaystyle \text{AgBr} \) and \(\displaystyle \text{AgI} \) respectively).2. Estimation of sulphur (Carius method)The same Carius-tube procedure is used: a known mass of the compound is heated with fuming nitric acid, but this time the sulphur present is oxidised all the way to sulphuric acid:\[\text{S (in compound)} + \text{HNO}_3 \xrightarrow{\Delta} \text{H}_2\text{SO}_4 \]The resulting sulphuric acid is then precipitated as barium sulphate by adding excess barium chloride solution:\[\text{H}_2\text{SO}_4 + \text{BaCl}_2 \longrightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \]Checking this balances: $\displaystyle 1$ Ba, $\displaystyle 1$ S and $\displaystyle 4$ O appear on each side, and the $\displaystyle 2$ chlorines on the left reappear as $\displaystyle 2$ HCl on the right — so the coefficient $\displaystyle 2$ in front of HCl is what makes it balance, not 1.The white \(\displaystyle \text{BaSO}_4 \) precipitate is filtered, washed, dried and weighed. With \(\displaystyle m \) = mass of compound and \(\displaystyle m_1 \) = mass of \(\displaystyle \text{BaSO}_4 \) formed, and using \(\displaystyle \text{Ba}=137,\ \text{S}=32,\ \text{O}=16 \) so that the molar mass of \(\displaystyle \text{BaSO}_4 = 137+32+64 = 233 \):\[\%\ \text{of sulphur} = \frac{32}{233}\times\frac{m_1}{m}\times100 \]The same aside applies again: \(\displaystyle m_1 \) is the mass of the whole \(\displaystyle \text{BaSO}_4 \) salt, not the mass of sulphur — the factor \(\displaystyle \dfrac{32}{233} \) is what strips out the barium and oxygen to leave just the sulphur's share.3. Estimation of phosphorusA known mass of the organic compound is heated with fuming nitric acid, which oxidises the phosphorus present to phosphoric acid:\[\text{P (in compound)} + \text{HNO}_3 \xrightarrow{\Delta} \text{H}_3\text{PO}_4 \]This phosphoric acid is precipitated as magnesium ammonium phosphate by adding magnesia mixture (a solution of \(\displaystyle \text{MgCl}_2 \) and \(\displaystyle \text{NH}_4\text{OH} \)):\[\text{MgCl}_2 + \text{NH}_3 + \text{H}_3\text{PO}_4 \longrightarrow \text{MgNH}_4\text{PO}_4\downarrow + 2\text{HCl} \](Balance check: $\displaystyle 1$ Mg and $\displaystyle 1$ P on each side; hydrogens total \(\displaystyle 3+3=6 \) on the left and \(\displaystyle 4+2=6 \) on the right; the $\displaystyle 2$ chlorines on the left match the $\displaystyle 2$ in \(\displaystyle 2\text{HCl} \).)The \(\displaystyle \text{MgNH}_4\text{PO}_4 \) precipitate, once filtered and washed, is ignited (strongly heated), which converts it to magnesium pyrophosphate, \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) — this is the compound actually weighed, because \(\displaystyle \text{MgNH}_4\text{PO}_4 \) itself is not stable enough on heating in air to give a reliable mass:\[2\text{MgNH}_4\text{PO}_4 \xrightarrow{\Delta} \text{Mg}_2\text{P}_2\text{O}_7 + 2\text{NH}_3\uparrow + \text{H}_2\text{O} \](Balance check: Mg \(\displaystyle 2=2\); N \(\displaystyle 2=2\); P \(\displaystyle 2=2\); H on the left \(\displaystyle 2\times4=8\), on the right \(\displaystyle 2\times3 + 2 = 8\); O on the left \(\displaystyle 2\times4=8\), on the right \(\displaystyle 7+1=8\).)With \(\displaystyle m \) = mass of compound and \(\displaystyle m_1 \) = mass of \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) finally weighed, and using \(\displaystyle \text{Mg}=24,\ \text{P}=31,\ \text{O}=16 \), so molar mass of \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 = 2(24)+2(31)+7(16) = 48+62+112 = 222 \), and noting that each mole of \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) carries two phosphorus atoms:\[\%\ \text{of phosphorus} = \frac{2\times31}{222}\times\frac{m_1}{m}\times100 = \frac{62}{222}\times\frac{m_1}{m}\times100 \]Here the mistake to avoid is using only one atomic mass of phosphorus ($\displaystyle 31$) in the numerator instead of \(\displaystyle 2\times31=62\) — the weighed salt \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) has two phosphorus atoms per formula unit, so leaving out that factor of $\displaystyle 2$ would understate the phosphorus content by half.Answer: In all three estimations, a known mass \(\displaystyle m \) of the organic compound is oxidised (heated with fuming \(\displaystyle \text{HNO}_3 \), for halogens in the presence of \(\displaystyle \text{AgNO}_3 \)) so that the element of interest is converted into one specific insoluble, weighable salt of mass \(\displaystyle m_1 \): the halogen as \(\displaystyle \text{AgX} \), sulphur as \(\displaystyle \text{BaSO}_4 \) (via \(\displaystyle \text{H}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \)), and phosphorus as \(\displaystyle \text{Mg}_2\text{P}_2\text{O}_7 \) (via \(\displaystyle \text{H}_3\text{PO}_4 \rightarrow \text{MgNH}_4\text{PO}_4\downarrow \xrightarrow{\Delta} \text{Mg}_2\text{P}_2\text{O}_7 \)). The percentage of each element is then \(\displaystyle \%\ \text{element} = \dfrac{\text{mass of element in one mole of the salt}}{\text{molar mass of the salt}} \times \dfrac{m_1}{m} \times 100 \), giving \(\displaystyle \%\text{X} = \dfrac{\text{at. mass of X}}{\text{molar mass of AgX}}\times\dfrac{m_1}{m}\times100 \), \(\displaystyle \%\text{S} = \dfrac{32}{233}\times\dfrac{m_1}{m}\times100 \), and \(\displaystyle \%\text{P} = \dfrac{62}{222}\times\dfrac{m_1}{m}\times100 \).

Paper chromatography separates a mixture because its components split differently between two phases that do not mix — a film of water held inside the paper (stationary phase) and a solvent that flows through the paper (mobile phase).Chromatography as a family of techniques works because different components of a mixture get distributed unequally between a stationary phase and a mobile phase: a component that clings more to the stationary phase lags behind, and a component that dissolves more readily in the mobile phase races ahead. Paper chromatography is the specific technique where this distribution happens by partition, not adsorption — the mixture keeps re-dissolving into and out of two liquid phases rather than sticking to a solid surface.The two phases in paper chromatography How the separation is set upA small spot of the sample solution is applied near one end of a paper strip, on a line called the origin, and allowed to dry. The paper is then suspended vertically inside a closed chamber with the lower edge dipping into the solvent, keeping the origin spot above the solvent surface so the spot itself never touches the liquid directly. As the solvent rises up the dry paper by capillary action, it reaches the spot and carries the dissolved components upward along with it. This whole process of letting the solvent travel up the paper is called developing the chromatogram.Why the components separateAs the mobile phase moves through the region of the spot, each component of the mixture continuously partitions itself between the stationary water film and the moving solvent — dissolving into the mobile phase, being carried a little further, then re-partitioning back toward the stationary phase, over and over as the solvent front advances. A component that is more soluble in the mobile solvent (relative to how strongly it is held by the stationary water) spends more of its time in the moving phase and travels further up the paper in a given time; a component that partitions more strongly into the stationary water film is held back and travels only a short distance. Because different substances have different relative solubilities in the two phases, they end up travelling different distances in the same time — this is what separates one spot on the origin line into several distinct spots at different heights.After the solvent front has moved a convenient distance, the paper is removed and dried, and the spots are located (directly if coloured, or by spraying with a developing reagent, or under UV light if fluorescent). Each spot's position is described using the retardation factor, \(\displaystyle R_f \):\[R_f = \dfrac{\text{distance moved by the substance from the origin}}{\text{distance moved by the solvent front from the origin}} \]Here "distance moved by the substance" is measured to the centre of its spot, and "distance moved by the solvent front" is measured to the farthest point the mobile phase reached — both measured from the same origin line, along the same straight path up the paper. Because \(\displaystyle R_f \) is a ratio of two distances measured under the same run, it is a fixed, characteristic property of a given substance for a given solvent system, temperature, and type of paper — it does not depend on how far that particular run happened to develop. This is the property that lets an unknown spot be identified: its \(\displaystyle R_f \) is compared against the \(\displaystyle R_f \) of a known reference substance run under the same conditions.A step people get wrong: paper chromatography is a partition technique, so the separating agent is the invisible water trapped in the cellulose, not the visible paper fibres — describing the paper itself as the "stationary phase" misses the actual principle being tested.Answer: Paper chromatography works on the principle of partition (differential distribution) of the components of a mixture between two phases — a stationary phase of water held in the pores of cellulose paper, and a mobile phase of an organic solvent that flows past it by capillary action. A component that is more soluble in the mobile phase moves further; one that is held more strongly by the stationary water film moves less. This unequal movement separates the original mixed spot into distinct spots, each located by its retardation factor \(\displaystyle R_f = \dfrac{\text{distance moved by substance}}{\text{distance moved by solvent}} \), which is characteristic of that substance under fixed conditions and is used to identify it.

Silver nitrate does not only test for halogens — sulphur and nitrogen in the same fusion mixture also give precipitates with it, and nitric acid's job is to destroy those before the test even starts.When an organic compound is fused with sodium metal to make the sodium extract (Lassaigne's test), every heteroatom in the molecule gets converted to a sodium salt in the same solution: All of these can sit together in the one extract. If silver nitrate, \(\displaystyle \text{AgNO}_3\), were added straight to this mixture, it would not react with the halide ion alone — sulphide and cyanide ions are also precipitated by \(\displaystyle \text{Ag}^+\):\[\text{Na}_2\text{S} + 2\text{AgNO}_3 \rightarrow \text{Ag}_2\text{S}\!\downarrow + 2\text{NaNO}_3 \] (silver sulphide, a black precipitate)\[\text{NaCN} + \text{AgNO}_3 \rightarrow \text{AgCN}\!\downarrow + \text{NaNO}_3 \] (silver cyanide, a white precipitate, and it can also form soluble complexes with excess CN⁻)Both equations are already balanced: count sodium, silver, nitrate, and the S or CN group on each side and they match one-for-one. Notice the black \(\displaystyle \text{Ag}_2\text{S}\) and the white \(\displaystyle \text{AgCN}\) would appear right alongside the halide precipitate — you would have no way to tell, by colour, whether the precipitate you see is telling you about a halogen or about leftover sulphur/nitrogen from the fusion.This is the step people skip: they treat "add dilute HNO₃" as a generic acidifying habit, when its real, specific job here is to remove sulphide and cyanide before silver nitrate ever touches the solution.Boiling the sodium extract with dilute nitric acid decomposes both interfering ions and drives the products off as gases, so they leave the test tube instead of sitting in solution:\[\text{Na}_2\text{S} + 2\text{HNO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{S}\!\uparrow \] (hydrogen sulphide gas escapes)\[\text{NaCN} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{HCN}\!\uparrow \] (hydrogen cyanide gas escapes)Check the balance on each: sodium, nitrogen, oxygen and hydrogen atoms match on both sides in the first equation ($\displaystyle 2$ Na, $\displaystyle 2$ N, $\displaystyle 6$ O, $\displaystyle 2$ H, $\displaystyle 1$ S throughout), and in the second ($\displaystyle 1$ Na, $\displaystyle 2$ N, $\displaystyle 3$ O, $\displaystyle 1$ H, $\displaystyle 1$ C throughout) — one \(\displaystyle \text{HNO}_3\) is exactly enough to protonate one \(\displaystyle \text{CN}^-\) and free one \(\displaystyle \text{NO}_3^-\). Any carbonate picked up from atmospheric \(\displaystyle \text{CO}_2\) during fusion is destroyed the same way, \[\text{Na}_2\text{CO}_3 + 2\text{HNO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{O} + \text{CO}_2\!\uparrow, \] so that carbonate cannot form a silver precipitate either.The halide ion itself is untouched by this treatment — \(\displaystyle \text{NaX}\) and \(\displaystyle \text{HNO}_3\) do not react to remove \(\displaystyle X^-\), because the hydrohalic acids stay fully ionised in water rather than escaping as gas under gentle boiling. So after boiling, the solution still carries all its original halide ion, but no more sulphide, cyanide, or carbonate. Only now is silver nitrate added:\[\text{NaX} + \text{AgNO}_3 \rightarrow \text{AgX}\!\downarrow + \text{NaNO}_3 \]giving a precipitate that is unambiguously the silver halide — white and curdy for \(\displaystyle \text{AgCl}\) (chloride), pale yellow for \(\displaystyle \text{AgBr}\) (bromide), or yellow for \(\displaystyle \text{AgI}\) (iodide).A second reason sits underneath the same acidification, and it is worth naming separately: silver halides are insoluble even in dilute nitric acid, while silver carbonate, silver phosphate and similar silver salts of other anions are not. So working in acidic medium is a built-in filter — even if some other anion had survived, any precipitate it formed would simply redissolve in the acid, while a true halide precipitate would not. That is why the medium for this test is kept acidic with \(\displaystyle \text{HNO}_3\) throughout, not just at the start.**Answer: Nitric acid is added first because the sodium extract can also contain sodium sulphide and sodium cyanide (from sulphur or nitrogen in the compound), and these give their own precipitates with silver nitrate (black \(\displaystyle \text{Ag}_2\text{S}\), white \(\displaystyle \text{AgCN}\)) that would be mistaken for, or mask, the halide test. Boiling with dilute \(\displaystyle \text{HNO}_3\) decomposes \(\displaystyle \text{Na}_2\text{S}\) and \(\displaystyle \text{NaCN}\) into gaseous \(\displaystyle \text{H}_2\text{S}\) and \(\displaystyle \text{HCN}\) (and any carbonate into \(\displaystyle \text{CO}_2\)), which escape from solution, while the halide ion is unaffected. Adding \(\displaystyle \text{AgNO}_3\) after this leaves the silver halide, \(\displaystyle \text{AgX}\), as the only precipitate formed — and because \(\displaystyle \text{AgX}\) (unlike silver salts of carbonate or phosphate) does not dissolve back into dilute nitric acid, the acidic medium also guards against any other silver precipitate being wrongly read as a positive halogen test.

Fusion converts covalently bound N, S and halogens into free ions, because only ions give the simple precipitation and colour tests of qualitative analysis.In an organic compound, nitrogen, sulphur and the halogens are not present as separate \(\displaystyle \text{N}^{3-} \), \(\displaystyle \text{S}^{2-} \) or \(\displaystyle \text{X}^- \) ions — they are covalently bonded to carbon (as in \(\displaystyle \text{C-N} \), \(\displaystyle \text{C-S} \), \(\displaystyle \text{C-X} \) bonds). A covalent bond does not ionise in solution, so none of the standard inorganic spot tests can see these elements directly: This is the step students get wrong: testing the original organic compound straight away with \(\displaystyle \text{AgNO}_3 \) or lead acetate and concluding "no halogen/sulphur present" when in fact the element is present but covalently bound, so it simply cannot react in that form.Why sodium specifically. Metallic sodium is an extremely reactive reducing agent. When the organic compound is fused (strongly heated) with sodium metal, the sodium is reactive enough to break the covalent \(\displaystyle \text{C-N} \), \(\displaystyle \text{C-S} \) and \(\displaystyle \text{C-X} \) bonds and combine directly with these atoms, converting them into simple, water-soluble, ionic sodium salts. This molten mixture, plunged into distilled water, gives the "sodium fusion extract" (Lassaigne's extract), which now contains free ions that behave exactly like an ordinary inorganic salt solution.The fusion reactions (elements combine one-to-one, so each is already balanced by atom count on both sides):For nitrogen, sodium combines with the carbon and nitrogen of the compound to give sodium cyanide: \[\text{Na} + \text{C} + \text{N} \longrightarrow \text{NaCN} \]For sulphur, sodium combines directly with sulphur to give sodium sulphide: \[2\text{Na} + \text{S} \longrightarrow \text{Na}_2\text{S} \]For a halogen (\(\displaystyle \text{X} = \text{Cl}, \text{Br}, \text{I} \)), sodium combines directly with the halogen to give the sodium halide: \[\text{Na} + \text{X} \longrightarrow \text{NaX} \]If both nitrogen and sulphur are present in the same compound, sodium preferentially forms sodium thiocyanate instead of separate cyanide and sulphide: \[\text{Na} + \text{C} + \text{N} + \text{S} \longrightarrow \text{NaSCN} \]This is another common pitfall: when a compound has both N and S, the ordinary Prussian-blue test for nitrogen can fail (because the \(\displaystyle \text{CN}^- \) has been tied up as \(\displaystyle \text{SCN}^- \) instead), which is why a separate confirmatory test is needed for thiocyanate in such cases.Once the fusion extract is boiled and filtered, it contains free \(\displaystyle \text{Na}^+ \), \(\displaystyle \text{CN}^- \), \(\displaystyle \text{S}^{2-} \), \(\displaystyle \text{X}^- \) or \(\displaystyle \text{SCN}^- \) ions in aqueous solution — ordinary inorganic ions that now respond normally to \(\displaystyle \text{AgNO}_3 \) (halides), sodium nitroprusside or lead acetate (sulphide), and the iron(II)/iron(III) Prussian-blue test (cyanide).Answer: Fusion with metallic sodium is done because N, S and halogens exist in an organic compound as covalent bonds, not as free ions, so they cannot give the usual precipitation/colour tests of qualitative inorganic analysis. Reactive sodium metal, on strong heating, breaks these covalent bonds and combines with the atoms to form water-soluble, ionic sodium salts — \(\displaystyle \text{NaCN} \) (\(\displaystyle \text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN} \)) for nitrogen, \(\displaystyle \text{Na}_2\text{S} \) (\(\displaystyle 2\text{Na} + \text{S} \rightarrow \text{Na}_2\text{S} \)) for sulphur, \(\displaystyle \text{NaX} \) (\(\displaystyle \text{Na} + \text{X} \rightarrow \text{NaX} \)) for a halogen, or \(\displaystyle \text{NaSCN} \) when both N and S are present. These ionic salts, dissolved in the sodium fusion extract, now give the standard inorganic tests directly.

Sublimation separates a solid that vaporises directly (without melting) from one that does not — and camphor is exactly that kind of solid.To pick a separation technique, ask what physical property differs between the two components. Here that property is volatility of the solid itself: Aside — the mistake to avoid: it is tempting to reach for filtration or a solvent-based separation (like solvent extraction) because the two solids "look" like they should be separated by dissolving one away. But nothing in the question dissolves either component in a liquid; the property being exploited is volatility on heating, which is what makes sublimation the right technique — not solubility.Procedure. Heat the mixture gently in a china dish covered with an inverted funnel (its stem plugged with cotton wool). Camphor sublimes, rises as vapour, and re-solidifies as pure crystals on the cooler inner wall of the funnel, where it can be scraped off. Calcium sulphate, being non-volatile, is left behind as the residue in the dish.Answer: Sublimation. Camphor sublimes on gentle heating (solid → vapour → solid, collected on a cold surface) while the non-volatile calcium sulphate is left behind as residue, so the two are separated.

A mixture boils when the sum of the vapour pressures of its components equals the atmospheric pressure — not when either liquid alone reaches its own boiling point.In steam distillation you have two liquids that do not mix into a true solution and do not react with each other: water, and the organic liquid (which must be steam-volatile and virtually insoluble in water). Because the two are immiscible, each one is free to build up its own vapour pressure above the mixture exactly as if the other were not there. Water contributes its vapour pressure \(\displaystyle p_{\text{water}} \) and the organic liquid contributes its vapour pressure \(\displaystyle p_{\text{organic}} \), and — this is the step that is easy to get wrong — these are not weighted by mole fraction the way they would be in an ideal solution (Raoult's law). Since the two liquids are immiscible, each behaves as a pure substance occupying its own vapour space, so the total vapour pressure above the mixture is simply the sum of the two separate vapour pressures:\[p_{\text{total}} = p_{\text{water}}^{\circ} + p_{\text{organic}}^{\circ} \]Here \(\displaystyle p_{\text{water}}^{\circ} \) is the vapour pressure that pure water would exert at that temperature, and \(\displaystyle p_{\text{organic}}^{\circ} \) is the vapour pressure that the pure organic liquid would exert at that same temperature.A liquid (or a liquid mixture) boils at the temperature where its total vapour pressure becomes equal to the atmospheric pressure:\[p_{\text{total}} = p_{\text{atm}} \]Now compare this with what each liquid would need on its own. Pure water boils when \(\displaystyle p_{\text{water}}^{\circ} \) alone reaches \(\displaystyle p_{\text{atm}} \) (at \(\displaystyle 373\ \text{K} \), i.e. \(\displaystyle 100\,^{\circ}\text{C} \), at $\displaystyle 1$ atm). Pure organic liquid boils when \(\displaystyle p_{\text{organic}}^{\circ} \) alone reaches \(\displaystyle p_{\text{atm}} \), which for most steam-volatile organic liquids happens at a temperature well above \(\displaystyle 100\,^{\circ}\text{C} \).But in the mixture, water is already contributing part of the total pressure. So the organic liquid does not have to supply the full \(\displaystyle p_{\text{atm}} \) by itself — it only has to supply the leftover amount:\[p_{\text{organic}}^{\circ} = p_{\text{atm}} - p_{\text{water}}^{\circ} \]Because \(\displaystyle p_{\text{organic}}^{\circ} \) needed to reach boiling is now smaller than \(\displaystyle p_{\text{atm}} \), it is reached at a lower temperature than the organic liquid's own normal boiling point. In fact, since \(\displaystyle p_{\text{water}}^{\circ} \) is also less than \(\displaystyle p_{\text{atm}} \) below \(\displaystyle 100\,^{\circ}\text{C} \), the two partial contributions add up to \(\displaystyle p_{\text{atm}} \) at some temperature that is below \(\displaystyle 100\,^{\circ}\text{C} \) as well — so the mixture as a whole boils below the boiling point of either pure component. This is exactly why steam distillation is used: it lets a high-boiling organic liquid (or one that would decompose if heated all the way to its own boiling point) be carried over and vaporised at a much lower, safer temperature, together with steam.Answer: In steam distillation the organic liquid and water are immiscible, so each exerts its own vapour pressure independently and the mixture's total vapour pressure is the sum \(\displaystyle p_{\text{total}} = p_{\text{water}}^{\circ} + p_{\text{organic}}^{\circ} \) rather than a mole-fraction-weighted average. The mixture boils as soon as this sum equals the atmospheric pressure, which happens well before either pure liquid's own vapour pressure alone could reach atmospheric pressure. Hence the organic liquid vaporises (distils over with the steam) at a temperature below its normal boiling point.

A white AgCl precipitate needs free Cl⁻ ions available to combine with Ag⁺ — and CCl₄ has none to give up, before or after heating.The silver nitrate test detects halide ions already present in solution: \[\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)\downarrow \] Here \(\displaystyle \text{Ag}^+ \) is the silver ion (supplied by silver nitrate, \(\displaystyle \text{AgNO}_3 \)) and \(\displaystyle \text{Cl}^- \) is a chloride ion free in solution. The equation is already balanced — one silver ion combines with one chloride ion to give one formula unit of silver chloride, \(\displaystyle \text{AgCl} \), a white insoluble solid, and the charges balance too: \(\displaystyle +1\) and \(\displaystyle -1\) cancel to a neutral precipitate.This test only works when chlorine is present as a free ion. That happens for an ionic chloride such as \(\displaystyle \text{NaCl} \), which dissociates completely in water, or for a compound that hydrolyses in water to release \(\displaystyle \text{HCl} \), which then ionises: \[\text{HCl}(aq) \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq) \]In tetrachloromethane, \(\displaystyle \text{CCl}_4 \) (carbon tetrachloride), neither route is open. All four chlorine atoms are joined to the central carbon by ordinary covalent bonds — one C–Cl bond, repeated four times — so there is no ionic \(\displaystyle \text{Cl}^- \) sitting free in the liquid to react with \(\displaystyle \text{Ag}^+ \) directly.The remaining question is whether heating with water could still break those C–Cl bonds and generate \(\displaystyle \text{HCl} \) on the spot. Compare \(\displaystyle \text{CCl}_4 \) with its close relative \(\displaystyle \text{SiCl}_4 \) (silicon tetrachloride), which hydrolyses readily in moist air: \[\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl} \] Checking the balance: Si, $\displaystyle 1$ on each side; Cl, $\displaystyle 4$ on each side; H, \(\displaystyle 4\times2=8\) on the left matches \(\displaystyle 4\) (in \(\displaystyle \text{Si(OH)}_4 \)) \(\displaystyle +\ 4\) (in \(\displaystyle 4\,\text{HCl}\)) \(\displaystyle =8\) on the right; O, $\displaystyle 4$ on the left matches $\displaystyle 4$ (in \(\displaystyle \text{Si(OH)}_4 \)) on the right. In this reaction, water's oxygen lone pair attacks the silicon atom directly, because silicon (period $\displaystyle 3$) has empty valence \(\displaystyle 3d\) orbitals available to briefly hold five groups at once, in a five-coordinate transition state, before a chloride ion leaves and \(\displaystyle \text{HCl} \) is expelled.The trap is assuming carbon can do the same thing just because it carries the same four chlorine atoms as silicon — it cannot. Carbon is a period-$\displaystyle 2$ element: its valence shell is only \(\displaystyle 2s\) and \(\displaystyle 2p\), with no \(\displaystyle 3d\), or any other empty, orbital available. There is no orbital on the carbon atom for water's lone pair to attack in forming that five-coordinate intermediate, so the C–Cl bond in \(\displaystyle \text{CCl}_4 \) has no pathway to break by hydrolysis, no matter how strongly the mixture is heated. (The four bulky Cl atoms crowding the small carbon add steric protection on top of this, but the decisive reason is the missing d-orbital.)Since \(\displaystyle \text{CCl}_4 \) neither ionises nor hydrolyses, heating it with aqueous \(\displaystyle \text{AgNO}_3 \) never produces a free \(\displaystyle \text{Cl}^- \) ion for \(\displaystyle \text{Ag}^+ \) to capture, so the characteristic white precipitate of \(\displaystyle \text{AgCl} \) never forms.Answer: No. \(\displaystyle \text{CCl}_4 \) does not give a white precipitate of \(\displaystyle \text{AgCl} \) with silver nitrate, even on heating, because its C–Cl bonds are covalent, not ionic, and — unlike \(\displaystyle \text{SiCl}_4 \) — carbon has no vacant d-orbitals for water to attack, so \(\displaystyle \text{CCl}_4 \) cannot be hydrolysed and no free \(\displaystyle \text{Cl}^- \) ion is ever released to react with \(\displaystyle \text{Ag}^+ \).

Potassium hydroxide is chosen because it reacts with carbon dioxide completely and irreversibly — the "how much extra it weighs" trick for measuring carbon only works if every last molecule of \(\displaystyle \text{CO}_2 \) actually gets caught.The setup. In estimating carbon (Liebig's method), the organic compound is burnt in a stream of oxygen. Every carbon atom in the compound ends up as carbon dioxide gas, \(\displaystyle \text{CO}_2 \), and every hydrogen atom ends up as water vapour, \(\displaystyle \text{H}_2\text{O} \). These two gases are passed, one after the other, through separate weighed absorption tubes: water vapour is caught first (usually in anhydrous \(\displaystyle \text{CaCl}_2 \)), and the remaining gas stream — now carrying only \(\displaystyle \text{CO}_2 \) — is passed into a tube containing potassium hydroxide solution.The whole method depends on one idea: mass gained by the absorbing tube = mass of the gas it absorbed. So whatever absorbs \(\displaystyle \text{CO}_2 \) must take up all of it, not just some, or the measured mass undercounts the carbon in the sample.Why KOH specifically. Carbon dioxide is an acidic oxide — dissolved in water it behaves as a weak acid. Potassium hydroxide is a strong base. An acid meeting a strong base does not partially react and stop; it reacts until one of them is used up. Since KOH is kept present in excess in the tube, every molecule of \(\displaystyle \text{CO}_2 \) that passes through finds a hydroxide ion waiting for it, and the reaction runs to completion:\[2\text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]Here \(\displaystyle \text{KOH} \) is potassium hydroxide, \(\displaystyle \text{CO}_2 \) is carbon dioxide gas, \(\displaystyle \text{K}_2\text{CO}_3 \) is potassium carbonate (the salt formed), and \(\displaystyle \text{H}_2\text{O} \) is water.Balancing it: start by counting potassium. The product \(\displaystyle \text{K}_2\text{CO}_3 \) carries $\displaystyle 2$ potassium atoms, so the left side needs $\displaystyle 2$ KOH to match. Carbon: one \(\displaystyle \text{CO}_2 \) gives the one carbon in \(\displaystyle \text{K}_2\text{CO}_3 \) — already even. Hydrogen: $\displaystyle 2$ KOH supplies $\displaystyle 2$ hydrogen atoms, and the right side's \(\displaystyle \text{H}_2\text{O} \) also has $\displaystyle 2$ — matched. Oxygen as a check: left side has \(\displaystyle 2(1) \) from KOH plus \(\displaystyle 2 \) from \(\displaystyle \text{CO}_2 \), giving $\displaystyle 4$; right side has \(\displaystyle 3 \) from \(\displaystyle \text{K}_2\text{CO}_3 \) plus \(\displaystyle 1 \) from \(\displaystyle \text{H}_2\text{O} \), also 4. Every atom balances, so the equation is correct as written.Why this matters for the measurement, not just the chemistry. If a weaker or more volatile absorbent were used instead, some \(\displaystyle \text{CO}_2 \) could pass through unabsorbed, or the absorption could be an equilibrium that doesn't go to completion — either way the tube would gain less mass than the true amount of \(\displaystyle \text{CO}_2 \) evolved, and the calculated percentage of carbon would come out too low. Because the acid–base reaction between \(\displaystyle \text{CO}_2 \) and KOH goes essentially $\displaystyle 100$% to \(\displaystyle \text{K}_2\text{CO}_3 \), the mass gained by the KOH tube is trusted as exactly the mass of \(\displaystyle \text{CO}_2 \) produced, which is then used to back-calculate the mass, and hence percentage, of carbon in the original organic compound.Answer: A solution of KOH is used because it is a strong base that reacts completely and quantitatively with the acidic gas \(\displaystyle \text{CO}_2 \), via \(\displaystyle 2\text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \), so the increase in mass of the KOH tube gives exactly the mass of carbon dioxide evolved, which is needed to calculate the percentage of carbon in the compound.