Purification and Characterisation of Organic Compounds MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Determine the molecular formula:}\) The empirical-formula mass is: \[ M_{\mathrm{emp}} = 2(12)+5+14 \] \[ M_{\mathrm{emp}} = 43\,\mathrm{g\,mol^{-1}} \] The independently measured molecular molar mass is \(86\,\mathrm{g\,mol^{-1}}\). Therefore: \[ n=\frac{86}{43} \] \[ n=2 \] The molecular formula is: \[ \mathrm{C_4H_{10}N_2} \] \(\textbf{Determine the original molecular amount:}\) Each analytical portion weighs: \[ m_{\mathrm{sample}}=0.0860\,\mathrm{g} \] Thus: \[ n_{\mathrm{compound}} = \frac{0.0860}{86} \] \[ n_{\mathrm{compound}} = 0.00100\,\mathrm{mol} \] \(\textbf{Predict complete Dumas nitrogen:}\) Each molecule contains two nitrogen atoms. Therefore: \[ n_{\mathrm{N,total}} = 2(0.00100) \] \[ n_{\mathrm{N,total}} = 0.00200\,\mathrm{mol} \] These nitrogen atoms form: \[ n_{\mathrm{N_2}} = \frac{0.00200}{2} \] \[ n_{\mathrm{N_2}} = 0.00100\,\mathrm{mol} \] The corresponding STP volume is: \[ V_{\mathrm{N_2}} = 0.00100\times22400 \] \[ V_{\mathrm{N_2}} = 22.4\,\mathrm{mL} \] This matches the Dumas observation. \(\textbf{Predict ordinary Kjeldahl ammonia:}\) The structure contains one accessible nitrogen atom per molecule. Therefore: \[ n_{\mathrm{N,accessible}} = 0.00100\,\mathrm{mol} \] Each accessible nitrogen atom gives one ammonia molecule under the stipulated quantitative conditions. Thus: \[ n_{\mathrm{NH_3}} = 0.00100\,\mathrm{mol} \] This also agrees with the observation. \(\textbf{Interpret the apparent analytical difference:}\) Complete Dumas estimation represents two nitrogen atoms per compound molecule. Ordinary Kjeldahl estimation represents only one under the specified conditions. The methods consequently need not report equal nitrogen percentages. The numerical difference is explained by the stated nitrogen environments, not by changing the molecular formula. \(\textbf{Interpretation:}\) An empirical formula provides the simplest nitrogen-containing ratio, while molecular mass determines the actual nitrogen count. A complete nitrogen method and a chemically selective nitrogen method may give different results on the same pure compound. Here those results are quantitatively consistent with the independently established structural information.

Explanation: \(\textbf{Determine the actual mass of compound P:}\) The crude sample contains two organic substances, both of which contribute to the combustion products. Therefore, the carbon dioxide and water produced by ethanol must be subtracted before the elemental composition of P can be calculated. The mass of P is: \[ m_{\mathrm P}=0.194-0.0460 \] \[ m_{\mathrm P}=0.148\,\mathrm{g} \] \(\textbf{Calculate the combustion products contributed by ethanol:}\) The molar mass of ethanol is: \[ M_{\mathrm{C_2H_6O}}=2(12)+6(1)+16 \] \[ M_{\mathrm{C_2H_6O}}=46\,\mathrm{g\,mol^{-1}} \] The amount of ethanol present is: \[ n_{\mathrm{ethanol}}=\frac{0.0460}{46} \] \[ n_{\mathrm{ethanol}}=0.00100\,\mathrm{mol} \] The balanced combustion reaction is: \[ \mathrm{C_2H_6O+3O_2\rightarrow2CO_2+3H_2O} \] Thus, one mole of ethanol produces two moles of carbon dioxide and three moles of water. Carbon dioxide contributed by ethanol: \[ n_{\mathrm{CO_2,ethanol}}=2(0.00100) \] \[ n_{\mathrm{CO_2,ethanol}}=0.00200\,\mathrm{mol} \] \[ m_{\mathrm{CO_2,ethanol}}=0.00200\times44 \] \[ m_{\mathrm{CO_2,ethanol}}=0.0880\,\mathrm{g} \] Water contributed by ethanol: \[ n_{\mathrm{H_2O,ethanol}}=3(0.00100) \] \[ n_{\mathrm{H_2O,ethanol}}=0.00300\,\mathrm{mol} \] \[ m_{\mathrm{H_2O,ethanol}}=0.00300\times18 \] \[ m_{\mathrm{H_2O,ethanol}}=0.0540\,\mathrm{g} \] \(\textbf{Determine the combustion products formed from P:}\) The total measured carbon dioxide contains contributions from both substances. Therefore: \[ m_{\mathrm{CO_2,P}}=0.352-0.0880 \] \[ m_{\mathrm{CO_2,P}}=0.264\,\mathrm{g} \] Similarly, the water produced by P is: \[ m_{\mathrm{H_2O,P}}=0.162-0.0540 \] \[ m_{\mathrm{H_2O,P}}=0.108\,\mathrm{g} \] \(\textbf{Calculate the carbon and hydrogen masses in P:}\) Carbon constitutes \(12/44\) of the mass of carbon dioxide. \[ m_{\mathrm C}=0.264\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0720\,\mathrm{g} \] Hydrogen constitutes \(2/18\) of the mass of water. \[ m_{\mathrm H}=0.108\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0120\,\mathrm{g} \] \(\textbf{Determine oxygen by mass difference:}\) Compound P contains only carbon, hydrogen and oxygen. The oxygen mass is therefore obtained by subtracting the calculated carbon and hydrogen masses from the actual mass of P. \[ m_{\mathrm O}=m_{\mathrm P}-m_{\mathrm C}-m_{\mathrm H} \] \[ m_{\mathrm O}=0.148-0.0720-0.0120 \] \[ m_{\mathrm O}=0.0640\,\mathrm{g} \] \(\textbf{Construct the empirical formula:}\) Convert the elemental masses into amounts of atoms. For carbon: \[ n_{\mathrm C}=\frac{0.0720}{12} \] \[ n_{\mathrm C}=0.00600\,\mathrm{mol} \] For hydrogen: \[ n_{\mathrm H}=\frac{0.0120}{1} \] \[ n_{\mathrm H}=0.0120\,\mathrm{mol} \] For oxygen: \[ n_{\mathrm O}=\frac{0.0640}{16} \] \[ n_{\mathrm O}=0.00400\,\mathrm{mol} \] The elemental mole ratio is: \[ C:H:O=0.00600:0.0120:0.00400 \] Dividing every term by \(0.00200\): \[ C:H:O=3:6:2 \] Therefore, the empirical formula of P is: \[ \mathrm{C_3H_6O_2} \] \(\textbf{Verify the molecular formula:}\) The empirical-formula mass is: \[ M_{\mathrm{emp}}=3(12)+6(1)+2(16) \] \[ M_{\mathrm{emp}}=74\,\mathrm{g\,mol^{-1}} \] The independently measured molecular molar mass is also \(74\,\mathrm{g\,mol^{-1}}\). Therefore, the molecular multiplier is: \[ k=\frac{M_{\mathrm{molecular}}}{M_{\mathrm{emp}}} \] \[ k=\frac{74}{74}=1 \] Hence, the molecular formula is identical to the empirical formula. \(\textbf{Interpretation:}\) The crucial step is recognising that the combustion products originate from two different organic compounds. Directly using the total carbon dioxide and water masses to determine the composition of P would incorrectly attribute ethanol-derived carbon and hydrogen to the unknown compound. The known mass and molecular formula of ethanol allow its contributions to be calculated independently. Subtracting these contributions isolates the combustion data belonging to P. The remaining elemental masses establish the simplest atomic ratio, while the molecular-mass measurement confirms that this ratio represents one complete molecule. \(\textbf{Final answer:}\) The molecular formula of compound P is \(\mathrm{C_3H_6O_2}\).

Explanation: \(\textbf{Distinguish adhering impurity from incorporated impurity:}\) The wet cake contains Q in two different locations. Some Q is present in the adhering mother liquor, while another portion is incorporated within the collected crystals. The stated washing procedure removes the mother liquor completely. However, it does not remove Q incorporated within the crystals. The effects of washing on purity and recovery must therefore be evaluated separately. \(\textbf{Determine the initial amount of P:}\) P is present in both the crystals and adhering mother liquor. Therefore: \[ m_{\mathrm{P,initial}}=4.00+0.0200 \] \[ m_{\mathrm{P,initial}}=4.0200\,\mathrm{g} \] \(\textbf{Determine P remaining after washing:}\) The wash removes all adhering mother liquor. It also dissolves \(0.0800\,\mathrm{g}\) of crystalline P. Therefore: \[ m_{\mathrm{P,final}}=4.00-0.0800 \] \[ m_{\mathrm{P,final}}=3.9200\,\mathrm{g} \] \(\textbf{Determine Q remaining in the crystals:}\) The \(0.0800\,\mathrm{g}\) of Q in the adhering mother liquor is removed. However, the incorporated Q remains. Thus: \[ m_{\mathrm{Q,final}}=0.0600\,\mathrm{g} \] \(\textbf{Calculate the final dried-product purity:}\) The final dry mass is: \[ m_{\mathrm{final}}=3.9200+0.0600 \] \[ m_{\mathrm{final}}=3.9800\,\mathrm{g} \] Therefore: \[ \%\mathrm{purity_P}=\frac{3.9200}{3.9800}\times100 \] \[ \%\mathrm{purity_P}=98.4925\ldots\% \] \(\textbf{Calculate retention relative to the initial P inventory:}\) \[ \%\mathrm{P\ retained}=\frac{3.9200}{4.0200}\times100 \] \[ \%\mathrm{P\ retained}=97.5124\ldots\% \] \(\textbf{Check the total P lost:}\) The wash removes P from two sources: \[ m_{\mathrm{P,lost}}=0.0200+0.0800 \] \[ m_{\mathrm{P,lost}}=0.1000\,\mathrm{g} \] This equals: \[ 4.0200-3.9200=0.1000\,\mathrm{g} \] \(\textbf{Interpretation:}\) Washing removes soluble impurities present in adhering mother liquor, but it need not eliminate impurities incorporated within crystals. It can also dissolve a portion of the desired product. The final purity is therefore less than \(100\%\), even though all adhering mother liquor has been removed. The percentage retained must be calculated from the complete initial P inventory, including P originally dissolved in the adhering liquid. \(\textbf{Final answer:}\) \[ \boxed{\left(98.49\%\ \mathrm{purity},\,97.51\%\ \mathrm{retention}\right)} \]

Explanation: \(\textbf{Relate chromatographic migration to the solvent front:}\) The retention factor is: \[ R_f=\frac{d_{\mathrm{solute}}}{d_{\mathrm{solvent}}} \] Therefore: \[ d_{\mathrm{solute}}=R_fd_{\mathrm{solvent}} \] The absolute distance between two spots depends on both their \(R_f\) difference and the solvent-front migration. \(\textbf{Calculate the spot positions in Solvent X:}\) For P: \[ d_{P,X}=0.32(5.00) \] \[ d_{P,X}=1.60\,\mathrm{cm} \] For Q: \[ d_{Q,X}=0.56(5.00) \] \[ d_{Q,X}=2.80\,\mathrm{cm} \] Their separation is: \[ \Delta d_X=2.80-1.60 \] \[ \Delta d_X=1.20\,\mathrm{cm} \] \(\textbf{Calculate the corresponding positions in Solvent Y:}\) For P: \[ d_{P,Y}=0.80(10.0) \] \[ d_{P,Y}=8.00\,\mathrm{cm} \] For Q: \[ d_{Q,Y}=0.84(10.0) \] \[ d_{Q,Y}=8.40\,\mathrm{cm} \] Their separation is: \[ \Delta d_Y=8.40-8.00 \] \[ \Delta d_Y=0.400\,\mathrm{cm} \] \(\textbf{Compare migration with separation:}\) Both substances travel farther in Y. However: \[ \Delta d_X\gt \Delta d_Y \] Under the stated compact-spot assumption, X gives a greater distance between the two spots. \(\textbf{Interpretation:}\) A chromatographic solvent that makes both substances migrate farther need not separate them better. The difference in their relative migration is more important than their individual migration distances alone. In Y, both spots approach the solvent front and remain close together. In X, they travel shorter distances but become more widely separated. The appropriate solvent must therefore be judged by the resulting separation of the components. \(\textbf{Final answer:}\) \[ \left(\Delta d_X,\Delta d_Y\right)=\left(1.20\,\mathrm{cm},0.400\,\mathrm{cm}\right) \]

Explanation: The mixture presents two different separation difficulties. R is much more volatile than P and Q at atmospheric pressure. Consequently, the first stage can exploit this relatively large boiling-point difference. Controlled removal of R by simple distillation can be carried out near its \(65^\circ\mathrm{C}\) boiling temperature without requiring P to boil. This temperature is below the stated decomposition threshold of P. After R has been removed, P and Q remain. Their normal boiling points differ by only \(3^\circ\mathrm{C}\). This close-boiling pair is poorly suited to simple distillation when meaningful purification is required. Furthermore, boiling P at its normal boiling point would expose it to a temperature substantially above its decomposition threshold. Reducing the pressure to \(200\,\mathrm{mmHg}\) lowers the boiling temperatures of P and Q to \(79^\circ\mathrm{C}\) and \(82^\circ\mathrm{C}\), respectively. Both temperatures lie below the decomposition threshold of P. However, reduced pressure alone does not eliminate the problem of their close boiling points. A fractionating column remains useful because repeated vaporisation and condensation improve separation of the two volatile components. The appropriate sequence therefore combines an initial simple-distillation stage with fractional distillation under reduced pressure. The method selection depends jointly on volatility differences and chemical stability rather than either property alone.

Explanation: For two immiscible liquids with both liquid phases present, the total vapour pressure is the sum of their respective saturated vapour pressures at the common temperature. At \(90^\circ\mathrm{C}\), the combined pressure is: \[ P_{\mathrm{total}}=620+105 \] \[ P_{\mathrm{total}}=725\,\mathrm{mmHg} \] Statement I is correct because this total is smaller than \(760\,\mathrm{mmHg}\). At that external pressure, the mixture has not reached its boiling condition at \(90^\circ\mathrm{C}\). Since the component vapour pressures increase with temperature, a higher temperature would be required to reach \(760\,\mathrm{mmHg}\). Statement II is correct because the combined vapour pressure equals \(725\,\mathrm{mmHg}\) at the specified temperature. Statement III is incorrect because reducing the external pressure to \(700\,\mathrm{mmHg}\) allows the equality between total vapour pressure and external pressure to be reached at a lower temperature, under the stipulated continuous behaviour. A pressure below the value already attained at \(90^\circ\mathrm{C}\) does not require further heating above that temperature. Statement IV is correct because two related but different quantities govern the process. The sum of the component vapour pressures determines whether boiling occurs. Their ratio determines the relative molecular amounts contributed to the vapour under the ideal immiscible-vapour relation. A valid steam-distillation prediction must therefore distinguish the boiling condition from the vapour-composition relation.

Explanation: The original compound contains both nitrogen and sulphur, so thiocyanate formation is possible during suitable sodium fusion. In Fusion I, the blood-red iron(III) reaction is consistent with thiocyanate in the extract. Formation of thiocyanate can reduce the amount of cyanide and sulphide available for their separate conventional detection reactions. Therefore, the absence of a clear Prussian-blue or nitroprusside response in Fusion I does not establish the absence of either element. Fusion II uses sufficient excess sodium. Under suitable conditions, sodium thiocyanate can undergo conversion represented by: \[ \mathrm{NaSCN+2Na\rightarrow NaCN+Na_2S} \] The resulting cyanide can participate in the prescribed iron-complex nitrogen test. The resulting sulphide can produce the characteristic sodium nitroprusside response. The disappearance of the blood-red thiocyanate observation is therefore compatible with the changed distribution of analytical species. The halogen test addresses a different converted ion, chloride. Because suitable nitric-acid pretreatment is performed before silver nitrate addition, the chloride observation can be interpreted without relying on the earlier nitrogen and sulphur tests. Both fusion experiments give the same characteristic chloride result. The complete pattern is therefore explained by condition-dependent nitrogen and sulphur fusion chemistry together with independently detectable chloride. There is no basis for proposing conversion of chlorine into sulphur or rejecting the original nitrogen–sulphur composition.

Explanation: The prescribed organic phosphorus test involves two chemically distinct requirements. First, phosphorus originally bound within the organic compound must be converted into the phosphate form needed for detection. Second, phosphate must react with ammonium molybdate under suitable acidic conditions to produce the characteristic analytical response. The unspiked unknown gives no yellow precipitate. That result could reflect absence of phosphorus or failure of the preceding conversion when phosphorus is present. Portion II tests an additional question because authentic phosphate is introduced after the oxidation stage. Its positive result shows that the final detecting reaction can respond to phosphate in the prepared sample medium. It does not prove that the earlier oxidation stage converted any original organic phosphorus successfully. The separate authentic phosphate reference also tests the detecting chemistry rather than the complete organic-to-phosphate pathway. Thus, the added phosphate is not evidence that the unknown originally contained phosphorus. The observation also does not establish that the unknown is phosphorus-free. A phosphorus-containing organic reference processed through the complete oxidation and detection procedure would supply more direct evidence about the reliability of the conversion stage. The decisive distinction is between checking a final detecting reaction and validating every preceding step needed to make the original analyte detectable.

Explanation: \(\textbf{Represent the unknown oxide amounts:}\) Let: \[ x=n_{\mathrm{NO}} \] \[ y=n_{\mathrm{NO_2}} \] The two reactions produce molecular nitrogen but consume different amounts of copper. The measured nitrogen gas and consumed copper therefore provide independent stoichiometric constraints. \(\textbf{Convert the nitrogen volume into moles:}\) \[ n_{\mathrm{N_2}}=\frac{67.2}{22400} \] \[ n_{\mathrm{N_2}}=0.00300\,\mathrm{mol} \] Each NO molecule contains one nitrogen atom. Each \(\mathrm{NO_2}\) molecule also contains one nitrogen atom. Two such nitrogen atoms form one \(\mathrm{N_2}\) molecule. Therefore: \[ \frac{x+y}{2}=0.00300 \] \[ x+y=0.00600 \] \(\textbf{Convert the copper mass into moles:}\) \[ n_{\mathrm{Cu}}=\frac{0.512}{64} \] \[ n_{\mathrm{Cu}}=0.00800\,\mathrm{mol} \] \(\textbf{Apply the distinct reduction stoichiometries:}\) From: \[ \mathrm{2NO+2Cu\rightarrow N_2+2CuO} \] one mole of NO consumes one mole of Cu. From: \[ \mathrm{2NO_2+4Cu\rightarrow N_2+4CuO} \] one mole of \(\mathrm{NO_2}\) consumes two moles of Cu. Thus: \[ x+2y=0.00800 \] \(\textbf{Solve the connected constraints:}\) The nitrogen balance gives: \[ x+y=0.00600 \] The copper balance gives: \[ x+2y=0.00800 \] Subtracting: \[ y=0.00200\,\mathrm{mol} \] Therefore: \[ x=0.00600-0.00200 \] \[ x=0.00400\,\mathrm{mol} \] \(\textbf{Verify the copper consumption:}\) \[ n_{\mathrm{Cu,predicted}} = 0.00400+2(0.00200) \] \[ n_{\mathrm{Cu,predicted}} = 0.00800\,\mathrm{mol} \] The corresponding copper mass is: \[ m_{\mathrm{Cu,predicted}} = 0.00800\times64 \] \[ m_{\mathrm{Cu,predicted}} = 0.512\,\mathrm{g} \] \(\textbf{Verify nitrogen production:}\) \[ n_{\mathrm{N_2,predicted}} = \frac{0.00400+0.00200}{2} \] \[ n_{\mathrm{N_2,predicted}} = 0.00300\,\mathrm{mol} \] The expected STP volume is: \[ V_{\mathrm{N_2,predicted}} = 0.00300\times22400 \] \[ V_{\mathrm{N_2,predicted}} = 67.2\,\mathrm{mL} \] \(\textbf{Interpretation:}\) The molecular nitrogen amount determines the total number of nitrogen atoms initially present. It does not independently distinguish NO from \(\mathrm{NO_2}\). Copper consumption supplies the distinguishing information because nitrogen dioxide requires twice as much copper per molecule under the specified reactions. The two analytical quantities therefore permit the original gas composition to be reconstructed. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{NO}},n_{\mathrm{NO_2}}\right)=\left(0.00400,0.00200\right)\,\mathrm{mol}} \]

Explanation: \(\textbf{Identify the gravimetric interference:}\) Silver nitrate is intended to precipitate sample-derived chloride as silver chloride. However, the extract also contains sulphide. Silver ions can therefore form silver sulphide, which contributes to the collected precipitate mass. The total precipitate cannot be treated as pure silver chloride. The independent sulphur estimation provides the information needed to correct this interference under the stated quantitative assumptions. \(\textbf{Calculate sulphur from barium sulphate:}\) \[ M_{\mathrm{BaSO_4}}=137+32+4(16) \] \[ M_{\mathrm{BaSO_4}}=233\,\mathrm{g\,mol^{-1}} \] The recovered amount is: \[ n_{\mathrm{BaSO_4}}=\frac{0.2330}{233} \] \[ n_{\mathrm{BaSO_4}}=0.00100\,\mathrm{mol} \] Each barium sulphate formula unit contains one sulphur atom. Therefore: \[ n_{\mathrm S}=0.00100\,\mathrm{mol} \] \(\textbf{Determine silver sulphide in the mixed precipitate:}\) Each silver sulphide formula unit also contains one sulphur atom. The question specifies quantitative conversion into \(\mathrm{Ag_2S}\). Therefore: \[ n_{\mathrm{Ag_2S}}=0.00100\,\mathrm{mol} \] Its molar mass is: \[ M_{\mathrm{Ag_2S}}=2(108)+32 \] \[ M_{\mathrm{Ag_2S}}=248\,\mathrm{g\,mol^{-1}} \] Thus: \[ m_{\mathrm{Ag_2S}}=0.00100\times248 \] \[ m_{\mathrm{Ag_2S}}=0.2480\,\mathrm{g} \] \(\textbf{Separate the true silver chloride contribution:}\) The mixed precipitate weighs: \[ m_{\mathrm{mixed}}=0.3915\,\mathrm{g} \] Therefore: \[ m_{\mathrm{AgCl}}=0.3915-0.2480 \] \[ m_{\mathrm{AgCl}}=0.1435\,\mathrm{g} \] \(\textbf{Determine sample-derived chlorine:}\) The molar mass of silver chloride is: \[ M_{\mathrm{AgCl}}=108+35.5 \] \[ M_{\mathrm{AgCl}}=143.5\,\mathrm{g\,mol^{-1}} \] Thus: \[ n_{\mathrm{AgCl}}=\frac{0.1435}{143.5} \] \[ n_{\mathrm{AgCl}}=0.00100\,\mathrm{mol} \] Each silver chloride formula unit contains one chlorine atom. Therefore: \[ m_{\mathrm{Cl}}=0.00100\times35.5 \] \[ m_{\mathrm{Cl}}=0.0355\,\mathrm{g} \] \(\textbf{Calculate chlorine percentage:}\) \[ \%\mathrm{Cl}=\frac{0.0355}{0.200}\times100 \] \[ \%\mathrm{Cl}=17.75\% \] \(\textbf{Interpretation:}\) The uncorrected silver precipitate is not a valid direct gravimetric measure of chlorine because two silver compounds are present. The independent sulphur result allows the sulphide-derived mass contribution to be identified. This mathematical correction is possible only because the problem establishes the complete composition and quantitative recovery of the precipitate. In ordinary qualitative halogen detection, suitable nitric-acid treatment is used to remove sulphide interference before the silver nitrate reaction. \(\textbf{Final answer:}\) \[ \boxed{17.75\%} \]

Explanation: \(\textbf{Identify the two nitrogen sources:}\) The gas collected in the sample experiment contains nitrogen derived from the organic compound and nitrogen entering through the leak. Both are present as molecular nitrogen. Therefore, gas composition alone does not distinguish their sources. The independent blank measures the atmospheric contribution under the stated reproducible conditions. \(\textbf{Correct the collected gas volume:}\) The sample experiment gives: \[ V_{\mathrm{total}}=56.0\,\mathrm{mL} \] The blank gives: \[ V_{\mathrm{blank}}=11.2\,\mathrm{mL} \] Thus: \[ V_{\mathrm{sample-derived}} = V_{\mathrm{total}}-V_{\mathrm{blank}} \] \[ V_{\mathrm{sample-derived}} = 56.0-11.2 \] \[ V_{\mathrm{sample-derived}} = 44.8\,\mathrm{mL} \] \(\textbf{Determine sample-derived molecular nitrogen:}\) \[ n_{\mathrm{N_2}} = \frac{44.8}{22400} \] \[ n_{\mathrm{N_2}} = 0.00200\,\mathrm{mol} \] \(\textbf{Convert molecular nitrogen into elemental nitrogen mass:}\) Each nitrogen molecule contains two nitrogen atoms. Therefore: \[ m_{\mathrm N} = n_{\mathrm{N_2}}\times28 \] \[ m_{\mathrm N} = 0.00200\times28 \] \[ m_{\mathrm N} = 0.0560\,\mathrm{g} \] \(\textbf{Calculate the corrected percentage:}\) \[ \%\mathrm N = \frac{0.0560}{0.280}\times100 \] \[ \%\mathrm N=20.0\% \] \(\textbf{Compare with the uncorrected result:}\) Without the blank correction: \[ n_{\mathrm{N_2,apparent}} = \frac{56.0}{22400} \] \[ n_{\mathrm{N_2,apparent}} = 0.00250\,\mathrm{mol} \] The corresponding nitrogen mass would be: \[ m_{\mathrm{N,apparent}} = 0.00250\times28 \] \[ m_{\mathrm{N,apparent}} = 0.0700\,\mathrm{g} \] Thus: \[ \%\mathrm{N,apparent} = \frac{0.0700}{0.280}\times100 \] \[ \%\mathrm{N,apparent} = 25.0\% \] The uncorrected result is too high. \(\textbf{Interpretation:}\) An accurate gas-volume measurement does not necessarily represent nitrogen originating exclusively from the organic compound. Atmospheric contamination adds molecular nitrogen indistinguishable from sample-derived nitrogen in the collector. The reproducible apparatus blank permits that contribution to be subtracted under the stipulated conditions. \(\textbf{Final answer:}\) \[ \boxed{20.0\%} \]

Explanation: \(\textbf{Identify the need for two independent measurements:}\) The recovered solid is not pure magnesium pyrophosphate. Both components contain magnesium, but only magnesium pyrophosphate contains phosphorus. The total mass of the solid and its independently measured magnesium mass therefore provide two different composition constraints. \(\textbf{Calculate the two formula masses:}\) For magnesium pyrophosphate: \[ M_{\mathrm{Mg_2P_2O_7}} = 2(24)+2(31)+7(16) \] \[ M_{\mathrm{Mg_2P_2O_7}} = 222\,\mathrm{g\,mol^{-1}} \] For magnesium oxide: \[ M_{\mathrm{MgO}}=24+16 \] \[ M_{\mathrm{MgO}}=40\,\mathrm{g\,mol^{-1}} \] \(\textbf{Represent the component mole amounts:}\) Let: \[ x=n_{\mathrm{Mg_2P_2O_7}} \] \[ y=n_{\mathrm{MgO}} \] The total recovered mass gives: \[ 222x+40y=0.262 \] \(\textbf{Apply the independent magnesium measurement:}\) Each mole of \(\mathrm{Mg_2P_2O_7}\) contains two moles of magnesium atoms. Its magnesium contribution is \(48\,\mathrm{g}\) per mole. Each mole of MgO contains \(24\,\mathrm{g}\) of magnesium. Therefore: \[ 48x+24y=0.0720 \] \(\textbf{Solve for the two components:}\) From the magnesium equation: \[ 24y=0.0720-48x \] \[ y=0.00300-2x \] Substitute into the total-mass equation: \[ 222x+40(0.00300-2x)=0.262 \] \[ 222x+0.120-80x=0.262 \] \[ 142x=0.142 \] \[ x=0.00100\,\mathrm{mol} \] Therefore: \[ y=0.00300-2(0.00100) \] \[ y=0.00100\,\mathrm{mol} \] \(\textbf{Determine phosphorus from magnesium pyrophosphate:}\) Each pyrophosphate formula unit contains two phosphorus atoms. Therefore: \[ n_{\mathrm P}=2x \] \[ n_{\mathrm P}=2(0.00100) \] \[ n_{\mathrm P}=0.00200\,\mathrm{mol} \] The phosphorus mass is: \[ m_{\mathrm P}=0.00200\times31 \] \[ m_{\mathrm P}=0.0620\,\mathrm{g} \] \(\textbf{Calculate the original phosphorus percentage:}\) \[ \%\mathrm P = \frac{0.0620}{0.310}\times100 \] \[ \%\mathrm P=20.0\% \] \(\textbf{Verify the recovered-solid composition:}\) The magnesium pyrophosphate mass is: \[ m_{\mathrm{pyrophosphate}}=0.00100\times222 \] \[ m_{\mathrm{pyrophosphate}}=0.222\,\mathrm{g} \] The magnesium oxide mass is: \[ m_{\mathrm{MgO}}=0.00100\times40 \] \[ m_{\mathrm{MgO}}=0.0400\,\mathrm{g} \] Their sum is: \[ 0.222+0.0400=0.262\,\mathrm{g} \] The magnesium mass represented is: \[ 0.00100(48)+0.00100(24)=0.0720\,\mathrm{g} \] Both measurements agree. \(\textbf{Interpretation:}\) A mixed gravimetric product cannot be treated as pure magnesium pyrophosphate. However, an independent magnesium measurement permits the amounts of the two known components to be separated mathematically. Only the magnesium pyrophosphate contribution is then used to calculate phosphorus. \(\textbf{Final answer:}\) \[ \boxed{20.0\%} \]

Explanation: \(\textbf{Determine the number of molecules in the sample:}\) The sample weighs \(0.0620\,\mathrm{g}\), and the molecular molar mass is \(62\,\mathrm{g\,mol^{-1}}\). Therefore: \[ n_{\mathrm{compound}} = \frac{0.0620}{62} \] \[ n_{\mathrm{compound}} = 0.00100\,\mathrm{mol} \] \(\textbf{Determine carbon atoms per molecule:}\) \[ n_{\mathrm{CO_2}} = \frac{0.0880}{44} \] \[ n_{\mathrm{CO_2}} = 0.00200\,\mathrm{mol} \] Each carbon dioxide molecule contains one carbon atom. Therefore: \[ N_{\mathrm C} = \frac{0.00200}{0.00100} \] \[ N_{\mathrm C}=2 \] \(\textbf{Determine hydrogen atoms per molecule:}\) \[ n_{\mathrm{H_2O}} = \frac{0.0540}{18} \] \[ n_{\mathrm{H_2O}} = 0.00300\,\mathrm{mol} \] The hydrogen-atom amount is: \[ n_{\mathrm H} = 2(0.00300) \] \[ n_{\mathrm H} = 0.00600\,\mathrm{mol} \] Therefore: \[ N_{\mathrm H} = \frac{0.00600}{0.00100} \] \[ N_{\mathrm H}=6 \] \(\textbf{Identify the unresolved molecular mass:}\) The established partial molecular formula is: \[ \mathrm{C_2H_6} \] Its mass contribution is: \[ M_{\mathrm{C_2H_6}} = 2(12)+6(1) \] \[ M_{\mathrm{C_2H_6}} = 30\,\mathrm{g\,mol^{-1}} \] The residual mass contribution is: \[ M_{\mathrm{remaining}} = 62-30 \] \[ M_{\mathrm{remaining}} = 32\,\mathrm{g\,mol^{-1}} \] \(\textbf{Compare the two chemically possible assignments:}\) Two oxygen atoms contribute: \[ 2(16)=32\,\mathrm{g\,mol^{-1}} \] One sulphur atom contributes: \[ 1(32)=32\,\mathrm{g\,mol^{-1}} \] Therefore, both proposed molecular formulae have the same molar mass. Both also contain two carbon atoms and six hydrogen atoms per molecule. Consequently, they predict identical carbon dioxide and water quantities for equal molecular amounts. \(\textbf{Identify the missing evidence:}\) An appropriate independent sulphur test could distinguish the sulphur-containing candidate from the oxygen-only candidate under the stipulated choice set. Without such evidence, assigning the entire residual mass to oxygen is unjustified. \(\textbf{Interpretation:}\) An accurate molecular molar mass and complete carbon–hydrogen combustion data do not always determine a unique molecular formula. Different elemental combinations can have equal total mass contributions. The identity of the remaining element must be established rather than inferred merely from subtraction.

Explanation: Chromatographic separation and visual detection are different stages of an analytical experiment. During development, P and Q migrate according to their interactions with the stationary and mobile phases. The stated \(R_f\) values show that the two compounds occupy distinct positions on the developed plate. However, both compounds are colourless, so their spots may not be directly visible. A visualisation method must produce a detectable response under the conditions used. The question establishes that ultraviolet examination reveals P but not Q. Therefore, one ultraviolet-visible spot does not establish that only one compound was separated. The subsequent iodine-vapour treatment provides a different detecting response. Because both compounds respond to that treatment, both chromatographic positions become visible. The appearance of the second spot does not require Q to be generated from P. Nor does it require further movement of the solvent front. The observation reflects a difference in detectability rather than a difference in the number of compounds originally present. A valid conclusion about mixture composition must therefore consider the detection properties of the individual substances as well as their chromatographic separation.

Explanation: \(\textbf{Determine the amount of original compound:}\) The molecular molar mass is: \[ M=6(12)+6(1)+2(14)+2(16) \] \[ M=72+6+28+32 \] \[ M=138\,\mathrm{g\,mol^{-1}} \] Each sample weighs: \[ m_{\mathrm{sample}}=0.138\,\mathrm{g} \] Therefore: \[ n_{\mathrm{compound}} = \frac{0.138}{138} \] \[ n_{\mathrm{compound}} = 0.00100\,\mathrm{mol} \] \(\textbf{Determine nitrogen represented by Dumas estimation:}\) The molecular nitrogen volume is: \[ V_{\mathrm{N_2}}=22.4\,\mathrm{mL} \] Thus: \[ n_{\mathrm{N_2}} = \frac{22.4}{22400} \] \[ n_{\mathrm{N_2}} = 0.00100\,\mathrm{mol} \] The nitrogen-atom amount is: \[ n_{\mathrm{N,total}} = 2(0.00100) \] \[ n_{\mathrm{N,total}} = 0.00200\,\mathrm{mol} \] Therefore, each original molecule contributes: \[ N_{\mathrm{N,total}} = \frac{0.00200}{0.00100} \] \[ N_{\mathrm{N,total}}=2 \] This agrees with the molecular formula. \(\textbf{Determine the nitrogen represented by Kjeldahl estimation:}\) The recovered ammonia amount is: \[ n_{\mathrm{NH_3}}=0.00100\,\mathrm{mol} \] Each ammonia molecule contains one nitrogen atom. Therefore: \[ n_{\mathrm{N,Kjeldahl}} = 0.00100\,\mathrm{mol} \] The number of nitrogen atoms recovered per original compound molecule is: \[ N_{\mathrm{N,Kjeldahl}} = \frac{0.00100}{0.00100} \] \[ N_{\mathrm{N,Kjeldahl}}=1 \] \(\textbf{Compare the represented nitrogen quantities:}\) The complete Dumas method represents two nitrogen atoms per molecule. The stipulated Kjeldahl method represents one nitrogen atom per molecule. Therefore: \[ f_{\mathrm{Kjeldahl}} = \frac12 \] \[ f_{\mathrm{Kjeldahl}}=50.0\% \] \(\textbf{Identify the structural limitation:}\) The data support a distinction between nitrogen recovered and not recovered by the stipulated Kjeldahl procedure. However, they do not uniquely identify the bonding environment of each nitrogen atom. Several nitrogen-containing structural arrangements could require consideration. The analytical comparison cannot by itself establish one particular pair of functional groups. \(\textbf{Interpretation:}\) An elemental method can provide information about analytical suitability without supplying a complete molecular structure. A complete nitrogen determination establishes the total nitrogen inventory. The selective Kjeldahl response establishes how much of that inventory is represented under the specified conditions. The remaining structural interpretation requires independent chemical evidence.

Explanation: \(\textbf{Identify the two acid-consuming species:}\) Hydrochloric acid in the receiving flask reacts with ammonia. However, sodium hydroxide carried mechanically from the distillation flask also neutralises hydrochloric acid. The total acid consumed therefore cannot be assigned entirely to ammonia. \(\textbf{Distinguish the two neutralisation reactions:}\) Ammonia reacts according to: \[ \mathrm{NH_3+HCl\rightarrow NH_4Cl} \] Sodium hydroxide reacts according to: \[ \mathrm{NaOH+HCl\rightarrow NaCl+H_2O} \] Both reactions consume one mole of hydrochloric acid per mole of the respective base. \(\textbf{Determine acid consumed specifically by ammonia:}\) The measured total acid consumption is: \[ n_{\mathrm{HCl,total}}=0.00500\,\mathrm{mol} \] The sodium hydroxide contribution is: \[ n_{\mathrm{HCl,NaOH}}=0.00100\,\mathrm{mol} \] Therefore: \[ n_{\mathrm{HCl,NH_3}} = 0.00500-0.00100 \] \[ n_{\mathrm{HCl,NH_3}} = 0.00400\,\mathrm{mol} \] \(\textbf{Determine the ammonia amount:}\) Because ammonia reacts with hydrochloric acid in a \(1:1\) mole ratio: \[ n_{\mathrm{NH_3}} = 0.00400\,\mathrm{mol} \] \(\textbf{Convert ammonia into elemental nitrogen:}\) Each ammonia molecule contains one nitrogen atom. Thus: \[ n_{\mathrm N} = 0.00400\,\mathrm{mol} \] The nitrogen mass is: \[ m_{\mathrm N} = 0.00400\times14 \] \[ m_{\mathrm N} = 0.0560\,\mathrm{g} \] \(\textbf{Calculate corrected nitrogen percentage:}\) \[ \%\mathrm N = \frac{0.0560}{0.280}\times100 \] \[ \%\mathrm N=20.0\% \] \(\textbf{Check the uncorrected interpretation:}\) If all acid consumption were assigned to ammonia: \[ m_{\mathrm{N,apparent}} = 0.00500\times14 \] \[ m_{\mathrm{N,apparent}} = 0.0700\,\mathrm{g} \] The apparent nitrogen percentage would be: \[ \%\mathrm{N,apparent} = \frac{0.0700}{0.280}\times100 \] \[ \%\mathrm{N,apparent} = 25.0\% \] Thus, mechanical alkali carry-over produces an overestimate if it is ignored. \(\textbf{Interpretation:}\) Back-titration measures the total acid remaining after all acid-consuming processes in the receiver. It does not identify the source of each consumed acid equivalent. Under the stated conditions, sodium hydroxide contributes to the measurement without contributing nitrogen. Correcting that contribution restores the quantity of acid consumed specifically by ammonia. \(\textbf{Final answer:}\) \[ \boxed{20.0\%} \]

Explanation: Pairing I is appropriate because the desired compound in P sublimes readily while the impurity does not. Controlled heating permits preferential vapour formation, and a suitably cooled surface allows recovery of the deposited solid. The question explicitly requires recovery without chemical change. Pairing II is appropriate because Q contains two volatile liquids with close boiling points. Their small boiling-point difference calls for fractional separation rather than relying on simple distillation alone. The desired component is also thermally sensitive near its normal boiling temperature. Suitable pressure reduction lowers the operating temperature, while the fractionating column addresses the difficulty of separating the close-boiling liquids. Pairing III is appropriate because differential extraction exploits the different distributions of the two solutes between immiscible phases. A suitable solvent can preferentially transfer one solute and enrich the recovered fraction. This mechanism does not require the solutes to vaporise. Pairing IV is appropriate because adsorption chromatography exploits different interactions of the compounds with silica gel. The components may be separated even though they are colourless, provided an appropriate visualisation or detecting method is used. Simple distillation does not automatically reproduce the selectivity of sublimation, extraction or chromatography. It also does not remove the thermal-stability and close-boiling difficulties in Q. The student is therefore correct that method choice must follow the relevant distinguishing properties of the substances. All four pairings are defensible under the stated conditions.

Explanation: Statement I is incorrect because a single chromatographic spot does not always establish chemical purity. Two different compounds can have the same or nearly the same migration under a particular set of chromatographic conditions. Such compounds may appear as one unresolved spot even when both are present in the sample. Statement II is correct because changing the mobile solvent changes the relative interactions of P and Q with the stationary and mobile phases. Their migration values can therefore become sufficiently different to permit resolution. The second chromatogram directly demonstrates two distinguishable components under the stated conditions. Statement III is correct because the sample also has a broad melting interval beginning below the melting point of authentic P. Impurities commonly alter the melting behaviour of crystalline organic compounds. Although melting behaviour alone is not an infallible identification method, it provides independent evidence consistent with the chromatographic observation. The agreement between these two different types of evidence strengthens the conclusion that the sample is impure. Statement IV is incorrect because the question explicitly states that neither substance decomposes during chromatography. The second spot is explained by improved separation rather than chemical formation of Q. The appropriate conclusion is that the sample contains P and Q, while Solvent X fails to resolve them. Only Statements II and III are correct.

Explanation: \(\textbf{Identify the sources of oxygen in the combustion products:}\) The carbon dioxide and water contain oxygen derived from both the original ethanol and copper(II) oxide. Therefore, the total oxygen mass in the collected products cannot be attributed entirely to the copper(II) oxide. The oxygen originally present in ethanol must first be accounted for. \(\textbf{Determine the amount of ethanol:}\) The molar mass of ethanol is: \[ M_{\mathrm{C_2H_6O}}=2(12)+6+16 \] \[ M_{\mathrm{C_2H_6O}}=46\,\mathrm{g\,mol^{-1}} \] Thus: \[ n_{\mathrm{ethanol}}=\frac{0.0460}{46} \] \[ n_{\mathrm{ethanol}}=0.00100\,\mathrm{mol} \] \(\textbf{Determine oxygen contained in the carbon dioxide:}\) Each carbon dioxide molecule contains two oxygen atoms. Therefore: \[ m_{\mathrm{O,in\ CO_2}}=0.0880\times\frac{32}{44} \] \[ m_{\mathrm{O,in\ CO_2}}=0.0640\,\mathrm{g} \] \(\textbf{Determine oxygen contained in the water:}\) Each water molecule contains one oxygen atom. Therefore: \[ m_{\mathrm{O,in\ H_2O}}=0.0540\times\frac{16}{18} \] \[ m_{\mathrm{O,in\ H_2O}}=0.0480\,\mathrm{g} \] \(\textbf{Calculate total oxygen in the collected products:}\) \[ m_{\mathrm{O,products}}=0.0640+0.0480 \] \[ m_{\mathrm{O,products}}=0.1120\,\mathrm{g} \] \(\textbf{Account for oxygen already present in ethanol:}\) Each ethanol molecule contains one oxygen atom. Therefore: \[ m_{\mathrm{O,ethanol}}=0.00100\times16 \] \[ m_{\mathrm{O,ethanol}}=0.0160\,\mathrm{g} \] The oxygen supplied by copper(II) oxide is: \[ m_{\mathrm{O,from\ CuO}}=0.1120-0.0160 \] \[ m_{\mathrm{O,from\ CuO}}=0.0960\,\mathrm{g} \] \(\textbf{Determine the amount of copper(II) oxide reduced:}\) Each CuO formula unit supplies one oxygen atom. Therefore: \[ n_{\mathrm{CuO}}=\frac{0.0960}{16} \] \[ n_{\mathrm{CuO}}=0.00600\,\mathrm{mol} \] The molar mass of CuO is: \[ M_{\mathrm{CuO}}=64+16 \] \[ M_{\mathrm{CuO}}=80\,\mathrm{g\,mol^{-1}} \] Thus: \[ m_{\mathrm{CuO}}=0.00600\times80 \] \[ m_{\mathrm{CuO}}=0.480\,\mathrm{g} \] \(\textbf{Verify the complete reaction:}\) The balanced reaction under the stipulated conditions is: \[ \mathrm{C_2H_6O+6CuO\rightarrow2CO_2+3H_2O+6Cu} \] One mole of ethanol requires six moles of CuO. Therefore, \(0.00100\,\mathrm{mol}\) of ethanol requires \(0.00600\,\mathrm{mol}\) of CuO, agreeing with the oxygen balance. \(\textbf{Interpretation:}\) The combustion products contain oxygen supplied by the oxidising reagent as well as oxygen originally present in the organic compound. An oxygen-atom balance distinguishes these contributions. The copper(II) oxide requirement cannot be obtained by treating the combined combustion-product oxygen mass as entirely reagent-derived. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{CuO}}=0.480\,\mathrm{g}} \]

Explanation: \(\textbf{Infer the distribution coefficient from the first extraction:}\) The distribution coefficient is: \[ K=\frac{C_{\mathrm{organic}}}{C_{\mathrm{aqueous}}} \] After the first equilibrium contact: \[ m_{\mathrm{organic}}=1.00\,\mathrm{g} \] \[ m_{\mathrm{aqueous}}=1.00\,\mathrm{g} \] The respective phase volumes are \(50.0\,\mathrm{mL}\) and \(100\,\mathrm{mL}\). Therefore: \[ K=\frac{1.00/50.0}{1.00/100} \] \[ K=2.00 \] \(\textbf{Evaluate the fresh-solvent contact in Procedure I:}\) After removal of the first organic extract, the aqueous phase contains \(1.00\,\mathrm{g}\) of P. The fraction transferred into fresh organic solvent is: \[ f_{\mathrm{extracted}}= \frac{KV_{\mathrm{organic}}} {V_{\mathrm{aqueous}}+KV_{\mathrm{organic}}} \] Substituting: \[ f_{\mathrm{extracted}}= \frac{2.00(50.0)}{100+2.00(50.0)} \] \[ f_{\mathrm{extracted}}=0.500 \] Therefore: \[ m_{\mathrm{P,second\ extract}}=1.00(0.500) \] \[ m_{\mathrm{P,second\ extract}}=0.500\,\mathrm{g} \] The combined organic extracts contain: \[ m_{\mathrm{P,I}}=1.00+0.500 \] \[ m_{\mathrm{P,I}}=1.50\,\mathrm{g} \] \(\textbf{Examine the returned solvent in Procedure II:}\) The first organic extract contains \(1.00\,\mathrm{g}\) of P in \(50.0\,\mathrm{mL}\). The remaining aqueous phase contains \(1.00\,\mathrm{g}\) of P in \(100\,\mathrm{mL}\). These are precisely the concentrations that were present at the first equilibrium. Therefore: \[ \frac{C_{\mathrm{organic}}}{C_{\mathrm{aqueous}}} = \frac{1.00/50.0}{1.00/100} \] \[ \frac{C_{\mathrm{organic}}}{C_{\mathrm{aqueous}}}=2.00 \] The returned phases already satisfy the required equilibrium distribution. With no change in composition, phase volume or conditions, there is no net driving force for further transfer. Consequently: \[ m_{\mathrm{P,II}}=1.00\,\mathrm{g} \] \(\textbf{Interpretation:}\) Fresh solvent provides an organic phase initially containing no P, permitting additional transfer from the aqueous solution. Returning the unchanged equilibrium extract merely restores the original equilibrium conditions. Repeated shaking does not automatically increase recovery when the concentration distribution already satisfies equilibrium. The distinction is between renewing the extracting phase and mechanically repeating contact with the same equilibrium-loaded phase. \(\textbf{Final answer:}\) \[ \boxed{\left(m_{\mathrm{P,I}},m_{\mathrm{P,II}}\right)=\left(1.50\,\mathrm{g},1.00\,\mathrm{g}\right)} \]