Purification and Characterisation of Organic Compounds MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Use carbon as a selective indicator of ethanoic acid:}\) Water adhering to the sample contains no carbon. Therefore, all carbon dioxide originates from ethanoic acid. The molecular formula of ethanoic acid is: \[ \mathrm{C_2H_4O_2} \] Its molecular molar mass is: \[ M_{\mathrm{acid}}=2(12)+4(1)+2(16) \] \[ M_{\mathrm{acid}}=60\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine carbon dioxide amount:}\) The collected carbon dioxide mass is \(0.220\,\mathrm{g}\). Therefore: \[ n_{\mathrm{CO_2}}=\frac{0.220}{44} \] \[ n_{\mathrm{CO_2}}=0.00500\,\mathrm{mol} \] \(\textbf{Infer the amount of pure acid:}\) Each ethanoic acid molecule contains two carbon atoms. Complete combustion therefore gives two carbon dioxide molecules per ethanoic acid molecule. Thus: \[ n_{\mathrm{acid}}=\frac{n_{\mathrm{CO_2}}}{2} \] \[ n_{\mathrm{acid}}=\frac{0.00500}{2} \] \[ n_{\mathrm{acid}}=0.00250\,\mathrm{mol} \] The pure acid mass is: \[ m_{\mathrm{acid}}=0.00250\times60 \] \[ m_{\mathrm{acid}}=0.150\,\mathrm{g} \] \(\textbf{Determine the adhering water from the wet sample mass:}\) The total wet mass is: \[ m_{\mathrm{wet}}=0.168\,\mathrm{g} \] Therefore: \[ m_{\mathrm{water,adhering}}=0.168-0.150 \] \[ m_{\mathrm{water,adhering}}=0.0180\,\mathrm{g} \] \(\textbf{Independently check the water collected:}\) Each ethanoic acid molecule contains four hydrogen atoms. Therefore, complete combustion produces two water molecules per original acid molecule. \[ n_{\mathrm{H_2O,from\ acid}}=2(0.00250) \] \[ n_{\mathrm{H_2O,from\ acid}}=0.00500\,\mathrm{mol} \] Its mass is: \[ m_{\mathrm{H_2O,from\ acid}}=0.00500\times18 \] \[ m_{\mathrm{H_2O,from\ acid}}=0.0900\,\mathrm{g} \] The water originally adhering to the crystals also reaches the collector. Therefore: \[ m_{\mathrm{H_2O,total}}=0.0900+0.0180 \] \[ m_{\mathrm{H_2O,total}}=0.108\,\mathrm{g} \] This exactly matches the experimental water measurement. \(\textbf{Interpretation:}\) The carbon dioxide result selectively measures ethanoic acid because adhering water contains no carbon. The collected water, however, has two sources: hydrogen combustion and pre-existing moisture. Treating the entire water collection as combustion-derived hydrogen would overestimate the hydrogen originally present in ethanoic acid. The two independent balances confirm both the pure-compound mass and the adhering-water mass. \(\textbf{Final answer:}\) \[ \boxed{\left(0.150\,\mathrm{g},\,0.0180\,\mathrm{g}\right)} \]

Explanation: The original fusion extract must be appropriately prepared before the silver nitrate test because cyanide and sulphide can interfere with silver-halide precipitation. The question specifies that this preparation has been completed, so the subsequent observations may be interpreted using the stated silver-halide behaviour. AgCl dissolves in dilute aqueous ammonia under the specified conditions. Therefore, the absence of detectable dissolution during the first treatment provides evidence against a detectable quantity of AgCl in the mixture. AgBr requires the concentrated-ammonia conditions specified in the problem for dissolution. The partial dissolution observed during the second treatment is consequently consistent with AgBr. When the complexing effect of ammonia is removed, silver bromide can precipitate again. The resulting pale-yellow solid provides further support for this assignment. AgI remains insoluble even in concentrated ammonia under the stated conditions. Therefore, the yellow residue surviving that treatment is consistent with AgI. The original precipitate must consequently have contained at least two distinguishable silver-halide components. The observations support bromide and iodide but not detectable chloride. The distinction depends on the sequence of treatments and the behaviour of each separated fraction, not solely on the original precipitate colour. A mixed precipitate cannot be assigned to a single halide merely because its overall appearance resembles one reference compound.

Explanation: \(\textbf{Establish the extraction relation:}\) For a solute distributed between two immiscible phases: \[ K=\frac{C_{\mathrm{organic}}}{C_{\mathrm{aqueous}}} \] The fraction remaining in the aqueous phase after one equilibrium contact is: \[ f_{\mathrm{remaining}}=\frac{V_{\mathrm{aqueous}}}{V_{\mathrm{aqueous}}+KV_{\mathrm{organic}}} \] The corresponding extracted fraction is: \[ f_{\mathrm{extracted}}=1-f_{\mathrm{remaining}} \] \(\textbf{Evaluate Plan I for P:}\) Plan I uses \(100\,\mathrm{mL}\) of organic solvent in one contact. \[ f_{\mathrm{P,remaining,I}}=\frac{100}{100+3.00(100)} \] \[ f_{\mathrm{P,remaining,I}}=0.250 \] Therefore: \[ m_{\mathrm{P,extracted,I}}=1.00(1-0.250) \] \[ m_{\mathrm{P,extracted,I}}=0.750\,\mathrm{g} \] \(\textbf{Evaluate Plan I for Q:}\) \[ f_{\mathrm{Q,remaining,I}}=\frac{100}{100+0.500(100)} \] \[ f_{\mathrm{Q,remaining,I}}=\frac23 \] Thus: \[ m_{\mathrm{Q,extracted,I}}=1.00\left(1-\frac23\right) \] \[ m_{\mathrm{Q,extracted,I}}=0.3333\,\mathrm{g} \] \(\textbf{Calculate the purity under Plan I:}\) \[ \%\mathrm{purity_I}=\frac{0.750}{0.750+0.3333}\times100 \] \[ \%\mathrm{purity_I}\approx69.23\% \] \(\textbf{Evaluate two fresh-solvent contacts for P:}\) Each contact uses \(50.0\,\mathrm{mL}\). The fraction of P remaining after one such contact is: \[ f_{\mathrm{P,one\ contact}}=\frac{100}{100+3.00(50.0)} \] \[ f_{\mathrm{P,one\ contact}}=0.400 \] The second contact acts on the P remaining after the first. Therefore: \[ f_{\mathrm{P,remaining,II}}=(0.400)^2 \] \[ f_{\mathrm{P,remaining,II}}=0.160 \] Thus: \[ m_{\mathrm{P,extracted,II}}=1.00(1-0.160) \] \[ m_{\mathrm{P,extracted,II}}=0.840\,\mathrm{g} \] \(\textbf{Evaluate the corresponding impurity extraction:}\) For Q, the fraction remaining after one \(50.0\,\mathrm{mL}\) contact is: \[ f_{\mathrm{Q,one\ contact}}=\frac{100}{100+0.500(50.0)} \] \[ f_{\mathrm{Q,one\ contact}}=0.800 \] After two contacts: \[ f_{\mathrm{Q,remaining,II}}=(0.800)^2 \] \[ f_{\mathrm{Q,remaining,II}}=0.640 \] Therefore: \[ m_{\mathrm{Q,extracted,II}}=1.00(1-0.640) \] \[ m_{\mathrm{Q,extracted,II}}=0.360\,\mathrm{g} \] \(\textbf{Calculate the purity under Plan II:}\) \[ \%\mathrm{purity_{II}}=\frac{0.840}{0.840+0.360}\times100 \] \[ \%\mathrm{purity_{II}}=70.0\% \] \(\textbf{Compare the two outcomes:}\) Plan I recovers \(75.0\%\) of P with approximately \(69.23\%\) purity. Plan II recovers \(84.0\%\) of P with \(70.0\%\) purity. Although Plan II also extracts slightly more Q, its improvement in P recovery is proportionally greater. \(\textbf{Interpretation:}\) Repeated extraction with fresh solvent can improve cumulative recovery without changing the equilibrium distribution coefficients. However, greater recovery does not automatically imply greater selectivity. Both extracted component quantities must be calculated before judging the purity of the combined extracts. In this particular case, splitting the solvent improves both P recovery and the final P percentage. \(\textbf{Final answer:}\) \[ \left(\%\mathrm{recovery_{II}},\%\mathrm{purity_{II}}\right)=(84.0\%,70.0\%) \]

Explanation: \(\textbf{Distinguish proportional signal from constant background:}\) The carbon dioxide graph passes through the origin. Therefore, its measured mass gain is proportional to the organic sample mass. The water graph has a positive intercept of \(0.0180\,\mathrm{g}\). This intercept represents moisture that would enter the absorber even without an organic sample. Consequently, it must be removed before calculating hydrogen. \(\textbf{Choose a common sample basis:}\) Use the point corresponding to: \[ m_{\mathrm{sample}}=0.0600\,\mathrm{g} \] The measured absorber gains are: \[ m_{\mathrm{CO_2}}=0.0880\,\mathrm{g} \] \[ m_{\mathrm{H_2O,observed}}=0.0540\,\mathrm{g} \] \(\textbf{Correct the water measurement using the graph intercept:}\) \[ m_{\mathrm{H_2O,sample}}=m_{\mathrm{H_2O,observed}}-m_{\mathrm{blank}} \] \[ m_{\mathrm{H_2O,sample}}=0.0540-0.0180 \] \[ m_{\mathrm{H_2O,sample}}=0.0360\,\mathrm{g} \] \(\textbf{Determine carbon mass:}\) \[ m_{\mathrm C}=0.0880\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0240\,\mathrm{g} \] \(\textbf{Determine hydrogen from corrected water:}\) \[ m_{\mathrm H}=0.0360\times\frac{2}{18} \] \[ m_{\mathrm H}=0.00400\,\mathrm{g} \] \(\textbf{Determine oxygen by difference:}\) Only carbon, hydrogen and oxygen are present. Therefore: \[ m_{\mathrm O}=0.0600-0.0240-0.00400 \] \[ m_{\mathrm O}=0.0320\,\mathrm{g} \] \(\textbf{Construct the empirical atomic ratio:}\) \[ n_{\mathrm C}=\frac{0.0240}{12}=0.00200\,\mathrm{mol} \] \[ n_{\mathrm H}=\frac{0.00400}{1}=0.00400\,\mathrm{mol} \] \[ n_{\mathrm O}=\frac{0.0320}{16}=0.00200\,\mathrm{mol} \] Thus: \[ C:H:O=0.00200:0.00400:0.00200 \] \[ C:H:O=1:2:1 \] The empirical formula is \(\mathrm{CH_2O}\). \(\textbf{Apply the molecular-mass condition:}\) \[ M_{\mathrm{emp}}=12+2+16 \] \[ M_{\mathrm{emp}}=30\,\mathrm{g\,mol^{-1}} \] The molecular multiplier is: \[ n=\frac{90}{30} \] \[ n=3 \] Therefore: \[ (\mathrm{CH_2O})_3=\mathrm{C_3H_6O_3} \] \(\textbf{Check the second graph point:}\) For a \(0.120\,\mathrm{g}\) sample, corrected water is: \[ m_{\mathrm{H_2O,sample}}=0.0900-0.0180 \] \[ m_{\mathrm{H_2O,sample}}=0.0720\,\mathrm{g} \] This is exactly twice the corrected water mass obtained for the \(0.0600\,\mathrm{g}\) sample. The carbon dioxide measurement also doubles. Thus the graph slopes and the blank intercept are mutually consistent. \(\textbf{Interpretation:}\) A proportional absorber signal passes through the origin, whereas a constant additive background produces a nonzero intercept. The hydrogen calculation must use the sample-derived water rather than the entire measured absorber gain. The independently measured molecular mass then distinguishes the molecular formula from its empirical ratio. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_3H_6O_3}} \]

Explanation: Phosphorus detection in the prescribed organic-analysis procedure depends on conversion of organically bound phosphorus into phosphate. The ammonium molybdate reaction is then performed under suitable acidic conditions. Test I begins with an authentic phosphate solution, so it checks the final phosphate-detection stage directly. The characteristic yellow precipitate shows that the ammonium molybdate detection system can respond to phosphate under the stated conditions. Test II checks a different and longer analytical pathway. Its starting material is an organic reference known to contain phosphorus. Before the final molybdate reaction can occur, that phosphorus must be converted into the relevant phosphate species. Test II fails even though the reference contains detectable phosphorus when properly converted. Therefore, at least one stage of the complete oxidation-dependent procedure is not functioning as required under the conditions used. Test III passes through that same unverified preparation procedure. Its negative result consequently cannot distinguish true absence of phosphorus from failure to produce detectable phosphate. The positive result in Test I does not establish that oxidation in Tests II and III was successful. Conversely, the failed oxidation-dependent reference test does not imply that all organic phosphorus compounds are fundamentally unsuitable for phosphate detection. The defensible conclusion is that the full analytical sequence requires correction or verification before the unknown sample is classified as phosphorus-free. This is a distinction between the functioning of a final detecting reagent and the reliability of the chemical conversion needed to make the analyte detectable.

Explanation: Quantitative sulphur estimation relies on converting the sample-derived sulphur into a suitable sulphate-containing analytical species. The resulting sulphate is precipitated as barium sulphate under conditions that permit quantitative recovery. The first student's solution contains both sulphate and phosphate. The question explicitly establishes that direct addition of barium chloride produces a precipitate containing more than one barium compound. Consequently, the entire precipitate mass cannot be interpreted as pure barium sulphate. Barium phosphate contributes mass but does not contain the sulphur being estimated. Applying the barium sulphate gravimetric factor to that mixed solid would therefore overestimate the sulphur content. The second student's procedure changes the precipitation conditions before barium chloride is added. Under the stipulated acidic conditions, phosphate no longer persists as an insoluble barium phosphate product. Sulphate nevertheless forms barium sulphate quantitatively. The resulting solid can therefore represent sulphur more selectively. The original amounts of sulphur may be identical in the two portions, but identical analyte amounts do not guarantee identical analytical validity when one precipitate is contaminated. Hydrochloric acid does not provide a basis for claiming that sulphate is reduced to sulphide. The decisive distinction is whether the chemical conditions allow only the intended gravimetric product to contribute significantly to the measured solid mass. This example illustrates why precipitation selectivity is as important as complete analyte conversion.

Explanation: \(\textbf{Identify the different copper requirements:}\) The two nitrogen oxides contain different numbers of oxygen atoms. Consequently, their reduction reactions consume different amounts of copper per mole of nitrogen oxide. The supplied balanced equations must be applied separately. \(\textbf{Determine copper required for nitric oxide:}\) The first reaction is: \[ \mathrm{2NO+2Cu\rightarrow N_2+2CuO} \] The mole ratio is: \[ n_{\mathrm{Cu}}:n_{\mathrm{NO}}=1:1 \] Therefore: \[ n_{\mathrm{Cu,NO}}=0.00400\,\mathrm{mol} \] \(\textbf{Determine copper required for nitrogen dioxide:}\) The second reaction is: \[ \mathrm{2NO_2+4Cu\rightarrow N_2+4CuO} \] The mole ratio is: \[ n_{\mathrm{Cu}}:n_{\mathrm{NO_2}}=2:1 \] Therefore: \[ n_{\mathrm{Cu,NO_2}}=2(0.00200) \] \[ n_{\mathrm{Cu,NO_2}}=0.00400\,\mathrm{mol} \] \(\textbf{Calculate total copper consumed:}\) \[ n_{\mathrm{Cu,total}}=0.00400+0.00400 \] \[ n_{\mathrm{Cu,total}}=0.00800\,\mathrm{mol} \] The corresponding copper mass is: \[ m_{\mathrm{Cu}}=0.00800\times64 \] \[ m_{\mathrm{Cu}}=0.512\,\mathrm{g} \] \(\textbf{Determine nitrogen gas formed from NO:}\) Two nitric oxide molecules form one molecular nitrogen molecule. Therefore: \[ n_{\mathrm{N_2,from\ NO}}=\frac{0.00400}{2} \] \[ n_{\mathrm{N_2,from\ NO}}=0.00200\,\mathrm{mol} \] \(\textbf{Determine nitrogen gas formed from nitrogen dioxide:}\) Similarly: \[ n_{\mathrm{N_2,from\ NO_2}}=\frac{0.00200}{2} \] \[ n_{\mathrm{N_2,from\ NO_2}}=0.00100\,\mathrm{mol} \] \(\textbf{Calculate the total collected nitrogen:}\) \[ n_{\mathrm{N_2,total}}=0.00200+0.00100 \] \[ n_{\mathrm{N_2,total}}=0.00300\,\mathrm{mol} \] At STP: \[ V_{\mathrm{N_2}}=0.00300\times22400 \] \[ V_{\mathrm{N_2}}=67.2\,\mathrm{mL} \] \(\textbf{Check nitrogen-atom conservation:}\) The nitrogen oxides originally contain: \[ n_{\mathrm{N\ atoms}}=0.00400+0.00200 \] \[ n_{\mathrm{N\ atoms}}=0.00600\,\mathrm{mol} \] The final nitrogen gas contains: \[ 2(0.00300)=0.00600\,\mathrm{mol} \] Thus nitrogen is conserved. \(\textbf{Interpretation:}\) Copper consumption depends on oxygen removal and therefore differs between NO and \(\mathrm{NO_2}\). In contrast, molecular nitrogen formation depends on the original nitrogen-atom inventory. The two quantities must be calculated through different stoichiometric relationships before the final experimental predictions are combined. \(\textbf{Final answer:}\) \[ \boxed{\left(m_{\mathrm{Cu}},V_{\mathrm{N_2}}\right)=\left(0.512\,\mathrm{g},67.2\,\mathrm{mL}\right)} \]

Explanation: Statement I is correct because ammonium chloride contains nitrogen in an inorganic ammonium ion. When excess sodium hydroxide is added under suitable conditions, ammonium ions can liberate ammonia. The liberated gas is chemically the same ammonia that is obtained from nitrogen successfully converted during Kjeldahl digestion of P. Consequently, the receiving acid does not distinguish the source of individual ammonia molecules. Statement II is correct because the total ammonia collected represents nitrogen contributed by both P and ammonium chloride. Assigning all of it to P would include impurity-derived nitrogen in the organic-compound result. This would overestimate the nitrogen quantity attributable specifically to P. Statement III is incorrect because quantitative recovery addresses whether the available ammonia has been collected completely. It does not establish that every collected ammonia molecule originated in the desired organic compound. Accurate gas collection and correct analytical selectivity are different requirements. Statement IV is correct because the ammonium impurity must be accounted for if the purpose is to determine the nitrogen contribution of P alone. Suitable separation or an independently validated correction could establish the required source-specific quantity. The result must also be related to the appropriate mass of P rather than automatically to the total crude mass. A precise titration can therefore produce an analytically misleading answer when the sample contains another substance capable of supplying the measured product. Statements I, II and IV are correct.

Explanation: Thin-layer chromatography separates substances according to their relative migration in a chosen chromatographic system. The \(R_f\) value compares the distance travelled by a component with the distance travelled by the solvent front. Under the stated conditions, P and Q occupy different positions and are therefore chromatographically resolved. The two spots support the presence of at least two distinguishable components in the mixture. However, the \(R_f\) values describe migration behaviour rather than the amounts of substances present. A larger \(R_f\) does not imply a larger recovered mass or a greater concentration in the starting mixture. Similarly, the appearance of a spot cannot automatically be treated as a quantitative mass measurement. The question explicitly states that the detecting responses of P and Q may differ. Therefore, equal masses of the two compounds might not produce equally dark or equally large spots. Without a suitable quantitative calibration and controlled detection conditions, the observed appearances cannot establish their mass ratio. The fact that the two \(R_f\) values sum to one is incidental and has no quantitative composition significance. The defensible conclusion is that the chromatogram provides separation and detection information but not the requested relative mass information. This distinction is important when interpreting chromatography as an analytical method rather than assuming every visual feature provides a calibrated quantitative measurement.

Explanation: \(\textbf{Identify the mixed absorber signal:}\) Potassium hydroxide absorbs both carbon dioxide and sulphur dioxide under the stated conditions. Therefore, the observed mass gain cannot be assigned entirely to carbon dioxide. The independent sulphur determination provides the information required to separate the two contributions. \(\textbf{Determine sulphur from barium sulphate:}\) The molar mass of barium sulphate is: \[ M_{\mathrm{BaSO_4}}=137+32+4(16) \] \[ M_{\mathrm{BaSO_4}}=233\,\mathrm{g\,mol^{-1}} \] The recovered amount is: \[ n_{\mathrm{BaSO_4}}=\frac{0.466}{233} \] \[ n_{\mathrm{BaSO_4}}=0.00200\,\mathrm{mol} \] Each formula unit contains one sulphur atom. Thus: \[ n_{\mathrm S}=0.00200\,\mathrm{mol} \] \(\textbf{Determine sulphur dioxide formed during combustion:}\) Every sulphur atom is converted into one sulphur dioxide molecule under the stipulated conditions. Therefore: \[ n_{\mathrm{SO_2}}=0.00200\,\mathrm{mol} \] The molar mass of sulphur dioxide is: \[ M_{\mathrm{SO_2}}=32+2(16) \] \[ M_{\mathrm{SO_2}}=64\,\mathrm{g\,mol^{-1}} \] Thus: \[ m_{\mathrm{SO_2}}=0.00200\times64 \] \[ m_{\mathrm{SO_2}}=0.128\,\mathrm{g} \] \(\textbf{Isolate the carbon dioxide absorber contribution:}\) The total potassium hydroxide mass gain is: \[ m_{\mathrm{absorbed,total}}=0.304\,\mathrm{g} \] Therefore: \[ m_{\mathrm{CO_2}}=0.304-0.128 \] \[ m_{\mathrm{CO_2}}=0.176\,\mathrm{g} \] \(\textbf{Determine the carbon mass:}\) Each carbon dioxide molecule contains one carbon atom. \[ m_{\mathrm C}=0.176\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0480\,\mathrm{g} \] \(\textbf{Calculate carbon percentage:}\) \[ \%\mathrm C=\frac{0.0480}{0.124}\times100 \] \[ \%\mathrm C=38.7097\ldots\% \] \[ \%\mathrm C\approx38.7\% \] \(\textbf{Check elemental mass consistency:}\) The sulphur mass is: \[ m_{\mathrm S}=0.00200\times32 \] \[ m_{\mathrm S}=0.0640\,\mathrm{g} \] The remaining original sample mass is: \[ m_{\mathrm H}=0.124-0.0480-0.0640 \] \[ m_{\mathrm H}=0.0120\,\mathrm{g} \] This is a physically possible positive hydrogen mass, consistent with the stipulated three-element composition. \(\textbf{Interpretation:}\) A potassium hydroxide absorber mass gain represents carbon dioxide alone only when other absorbed substances are absent or have been removed. Here sulphur dioxide provides an additional mass contribution. Sulphur gravimetry makes it possible to determine that contribution independently and obtain the corrected carbon result. \(\textbf{Final answer:}\) \[ \boxed{38.7\%} \]

Explanation: Simple distillation separates suitable volatile liquids from nonvolatile substances through differences in their tendencies to enter the vapour phase. P is the volatile component of the mixture. Q is explicitly identified as a nonvolatile solid with negligible vapour pressure at the operating temperature. Under appropriately controlled heating, P vaporises and passes to the condenser while Q remains in the original vessel. This explains the uncontaminated distillate collected in Trial I. Trial II differs because heating is excessively vigorous. Rapid boiling can mechanically project droplets of the liquid mixture or fine suspended particles into the vapour path. Such physical carrying-over can transport nonvolatile material without requiring it to form its own vapour. The question also establishes that the collected Q remains chemically unchanged. Therefore, its appearance in the distillate does not establish a change in molecular formula or boiling point. Nor does it prove that the two substances have identical volatility. The appropriate correction concerns control of boiling and prevention of physical entrainment. The experiment demonstrates that a purification method can be chemically suitable in principle yet produce a contaminated product if the apparatus is operated improperly. Successful distillation requires both an appropriate volatility difference and conditions that prevent unintended mechanical transfer.

Explanation: \(\textbf{Determine total compound molecules:}\) Both P and Q have molar mass: \[ M_{\mathrm P}=M_{\mathrm Q}=60\,\mathrm{g\,mol^{-1}} \] Therefore, the total amount of molecules in the mixture is: \[ n_{\mathrm{total}}=\frac{0.600}{60} \] \[ n_{\mathrm{total}}=0.0100\,\mathrm{mol} \] \(\textbf{Determine the carbon dioxide amount:}\) The measured carbon dioxide mass is: \[ m_{\mathrm{CO_2}}=1.056\,\mathrm{g} \] Therefore: \[ n_{\mathrm{CO_2}}=\frac{1.056}{44} \] \[ n_{\mathrm{CO_2}}=0.0240\,\mathrm{mol} \] \(\textbf{Use the distinct carbon counts:}\) Each molecule of P contains two carbon atoms. Each molecule of Q contains three carbon atoms. Let their mole amounts be \(x\) and \(y\), respectively. The total molecular amount gives: \[ x+y=0.0100 \] The carbon dioxide amount gives: \[ 2x+3y=0.0240 \] \(\textbf{Solve the simultaneous composition constraints:}\) From the first relation: \[ x=0.0100-y \] Substitute into the second: \[ 2(0.0100-y)+3y=0.0240 \] \[ 0.0200-2y+3y=0.0240 \] \[ y=0.00400\,\mathrm{mol} \] Therefore: \[ x=0.0100-0.00400 \] \[ x=0.00600\,\mathrm{mol} \] \(\textbf{Determine the mass of Q:}\) \[ m_{\mathrm Q}=0.00400\times60 \] \[ m_{\mathrm Q}=0.240\,\mathrm{g} \] The corresponding P mass is: \[ m_{\mathrm P}=0.00600\times60 \] \[ m_{\mathrm P}=0.360\,\mathrm{g} \] Their sum is: \[ 0.240+0.360=0.600\,\mathrm{g} \] \(\textbf{Independently predict water from the component amounts:}\) P contains four hydrogen atoms per molecule. Therefore, one mole of P produces two moles of water on complete combustion. Q contains eight hydrogen atoms per molecule and consequently produces four moles of water. Thus: \[ n_{\mathrm{H_2O}}=2x+4y \] Substituting: \[ n_{\mathrm{H_2O}}=2(0.00600)+4(0.00400) \] \[ n_{\mathrm{H_2O}}=0.0120+0.0160 \] \[ n_{\mathrm{H_2O}}=0.0280\,\mathrm{mol} \] The predicted water mass is: \[ m_{\mathrm{H_2O}}=0.0280\times18 \] \[ m_{\mathrm{H_2O}}=0.504\,\mathrm{g} \] This agrees with the independent experimental result. \(\textbf{Interpretation:}\) P and Q have identical molar masses but different numbers of carbon and hydrogen atoms per molecule. The total mixture mass therefore determines the combined mole amount, while the carbon dioxide measurement supplies a separate composition constraint. The water result can then test the inferred mixture ratio independently rather than being used merely as another copy of the same calculation. \(\textbf{Final answer:}\) \[ \boxed{\left(m_{\mathrm Q},m_{\mathrm{H_2O,predicted}}\right)=\left(0.240\,\mathrm{g},0.504\,\mathrm{g}\right)} \]

Explanation: Thin-layer chromatography separates compounds according to their relative interactions with the stationary and mobile phases. An \(R_f\) value describes migration under a specified chromatographic system. It is not a universal identifying constant independent of the solvent conditions. In Solvent I, authentic P and authentic Q have identical \(R_f\) values. Therefore, a mixture of the two can produce one unresolved spot even though two molecular species are present. The single spot cannot establish chemical purity under those conditions. In Solvent II, the authentic compounds have substantially different migration values. The unknown sample produces two resolved spots corresponding to those reference positions. Under the stated detection and stability assumptions, this supports the presence of both P and Q. The result does not require chemical decomposition during chromatography. Nor does the greater migration of Q imply that P is absent. The important additional evidence is the separation obtained after changing the mobile-phase conditions. A chromatographic system that fails to resolve two components cannot independently demonstrate their absence from one another. The comparison illustrates why confirmation with a second suitable solvent system can reveal a mixture that appears homogeneous in the first.

Explanation: \(\textbf{Identify the purpose of washing:}\) The desired compound is already present in the organic phase. The objective is therefore different from extracting P out of an aqueous solution. Water washing should preferentially remove Q while retaining most of P in the organic solvent. \(\textbf{Establish the retained-fraction relation:}\) For a solute initially present in the organic phase: \[ K=\frac{m_{\mathrm{organic}}/V_{\mathrm{organic}}}{m_{\mathrm{aqueous}}/V_{\mathrm{aqueous}}} \] Using the total solute mass balance: \[ f_{\mathrm{organic}}=\frac{KV_{\mathrm{organic}}}{KV_{\mathrm{organic}}+V_{\mathrm{aqueous}}} \] \(\textbf{Calculate retention of P:}\) \[ f_P=\frac{4.00(100)}{4.00(100)+50.0} \] \[ f_P=\frac{400}{450} \] \[ f_P=0.88889 \] The mass of P retained is: \[ m_{\mathrm{P,retained}}=2.00(0.88889) \] \[ m_{\mathrm{P,retained}}=1.7778\,\mathrm{g} \] Therefore: \[ \%\mathrm{P\ retained}=0.88889\times100 \] \[ \%\mathrm{P\ retained}=88.9\% \] \(\textbf{Calculate retention of Q:}\) \[ f_Q=\frac{0.250(100)}{0.250(100)+50.0} \] \[ f_Q=\frac{25.0}{75.0} \] \[ f_Q=\frac13 \] Thus: \[ m_{\mathrm{Q,retained}}=1.00\times\frac13 \] \[ m_{\mathrm{Q,retained}}=0.3333\,\mathrm{g} \] \(\textbf{Calculate the purity after washing:}\) The recovered solute mass is: \[ m_{\mathrm{total}}=1.7778+0.3333 \] \[ m_{\mathrm{total}}=2.1111\,\mathrm{g} \] Therefore: \[ \%\mathrm{purity_P}=\frac{1.7778}{2.1111}\times100 \] \[ \%\mathrm{purity_P}=84.21\% \] \(\textbf{Compare with the unwashed extract:}\) Initially: \[ \%\mathrm{purity_{initial}}=\frac{2.00}{2.00+1.00}\times100 \] \[ \%\mathrm{purity_{initial}}=66.67\% \] The washing step increases the P purity from \(66.67\%\) to approximately \(84.2\%\). However, some P also transfers into the aqueous wash. \(\textbf{Interpretation:}\) A useful washing solvent need not leave the desired compound completely untouched. It must provide a favourable relative distribution between the desired compound and the impurity. Here, Q is removed more extensively than P, so purity increases despite incomplete P recovery. \(\textbf{Final answer:}\) \[ \boxed{\left(88.9\%,\,84.2\%\right)} \]

Explanation: Statement I is correct because the usual steam-distillation relation for immiscible liquids applies while separate liquid phases of water and the organic compound coexist. Under those conditions, each component can contribute its saturated vapour pressure at the common temperature. Boiling occurs when the combined vapour pressure equals the external pressure. Statement II is incorrect because the existence of a separate liquid organic phase is an important condition for using the original saturated-vapour-pressure relation in this form. Once P is exhausted as a separate liquid phase, its previous vapour contribution cannot automatically be retained in a calculation. The remaining phase conditions must be considered. Statement III is correct because the organic contribution to the vapour may diminish substantially after depletion of the organic liquid. The condensed material can consequently become predominantly water. Statement IV is correct because water itself can continue to vaporise and condense. Collection of additional liquid in the receiver does not establish that additional desired organic compound is being recovered. A student must therefore distinguish continued distillation from continued recovery of P. The original immiscible-liquid vapour-pressure relation should not be extended beyond its stated physical conditions without justification. The experiment illustrates how depletion of a component can change the appropriate description of a separation process.

Explanation: \(\textbf{Determine the first-crop purity:}\) The first crop contains: \[ m_{\mathrm{P,1}}=7.50\,\mathrm{g} \] \[ m_{\mathrm{Q,1}}=0.150\,\mathrm{g} \] Its total mass is: \[ m_1=7.50+0.150 \] \[ m_1=7.65\,\mathrm{g} \] Therefore: \[ \%\mathrm{purity_1}=\frac{7.50}{7.65}\times100 \] \[ \%\mathrm{purity_1}=98.04\% \] The first crop satisfies the required \(95.0\%\) purity. \(\textbf{Determine the second-crop purity:}\) The second crop contains: \[ m_{\mathrm{P,2}}=2.00\,\mathrm{g} \] \[ m_{\mathrm{Q,2}}=0.600\,\mathrm{g} \] Therefore: \[ \%\mathrm{purity_2}=\frac{2.00}{2.00+0.600}\times100 \] \[ \%\mathrm{purity_2}=76.92\% \] This crop is substantially less pure. \(\textbf{Calculate the consequence of combining both crops:}\) The combined amount of P is: \[ m_{\mathrm{P,combined}}=7.50+2.00 \] \[ m_{\mathrm{P,combined}}=9.50\,\mathrm{g} \] The combined impurity mass is: \[ m_{\mathrm{Q,combined}}=0.150+0.600 \] \[ m_{\mathrm{Q,combined}}=0.750\,\mathrm{g} \] The total combined solid weighs: \[ m_{\mathrm{combined}}=9.50+0.750 \] \[ m_{\mathrm{combined}}=10.25\,\mathrm{g} \] Thus: \[ \%\mathrm{purity_{combined}}=\frac{9.50}{10.25}\times100 \] \[ \%\mathrm{purity_{combined}}=92.68\% \] \(\textbf{Check recovery against the initial P amount:}\) The combined recovery would be: \[ \%\mathrm{recovery_{combined}}=\frac{9.50}{10.0}\times100 \] \[ \%\mathrm{recovery_{combined}}=95.0\% \] However, this higher recovery is accompanied by a combined purity below the specified requirement. \(\textbf{Apply the product specification:}\) The first crop already meets the \(95.0\%\) purity requirement. Combining it immediately with the second crop would produce a nonconforming product. The second crop may instead be purified separately before a pooling decision is made. \(\textbf{Interpretation:}\) Further cooling can recover additional desired compound from a mother liquor, but impurities may also crystallise. A second crop therefore need not have the same purity as the first. Pooling decisions should be based on the composition of the combined material rather than on recovery alone. The calculation supports retaining the first crop separately while evaluating further purification of the second.

Explanation: Fractional distillation exploits differences in the volatility of components in a liquid mixture. The vapour and distillate may become enriched in the more volatile component relative to the original liquid. However, enrichment is a comparison between compositions, not a requirement that the enriched component become the majority constituent. The original mixture contains only \(10.0\%\) P. The early collected fraction contains \(35.0\%\) P. Therefore, the fraction has a substantially greater proportion of P than the starting mixture. Student I correctly recognises this as enrichment. Nevertheless, Q still constitutes \(65.0\%\) of the collected fraction. Student II incorrectly assumes that preferential vaporisation always produces a distillate containing more than half of the more volatile component. The actual composition depends on the original mixture and its vapour–liquid behaviour under the operating conditions. A component can be preferentially transferred while remaining the minority component of the transferred material. The observation is consequently compatible with effective fractional enrichment without implying pure P recovery. Further fractionation may be needed if a highly purified P fraction is required. The distinction between relative enrichment and absolute majority composition is essential when evaluating distillation data.

Explanation: The two experiments examine chlorine under different chemical conditions. Sodium chloride, if present in the crude sample, supplies chloride ions directly to water. Those ions can react with silver nitrate to form white silver chloride. The characteristic ammonia solubility of the precipitate supports its identification as silver chloride, but it does not establish the original source of the chloride. Therefore, the first experiment may detect inorganic contamination rather than covalently bonded chlorine in P. The second experiment begins with suitably purified organic material. Sodium fusion converts suitable organically bound chlorine into an inorganic chloride species that can be detected through the prescribed silver nitrate procedure. The question also specifies appropriate pretreatment before the final silver nitrate test. Thus the second positive result supports chlorine associated with the purified organic compound. The two observations are compatible because the crude sample may contain chlorine in more than one chemical form. A detecting reaction establishes the presence of an analytical species after sample treatment; it does not automatically identify every original bonding environment. Sample preparation and the origin of the detected ion must therefore be considered alongside precipitate colour and solubility. The distinction prevents a contaminating inorganic salt from being mistaken for conclusive evidence about the covalent structure of the desired organic compound.

Explanation: The two trials begin with equal amounts of the same pure reference compound. Consequently, their original phosphorus quantities are identical. The prescribed detection procedure requires suitable conversion of organically bound phosphorus into phosphate before the final ammonium molybdate reaction. Trial I uses the intended oxidation treatment and produces a yellow precipitate. Trial II deliberately changes an earlier stage of the analytical sequence. A shortened oxidation treatment may leave a portion of the phosphorus insufficiently converted into the required detectable form. The amount of yellow product subsequently formed can therefore be smaller even though the original phosphorus inventory is unchanged. A smaller precipitate need not establish a smaller elemental content. It may instead reflect less complete chemical conversion or recovery. The comparison does not establish the exact fraction of phosphorus converted because no quantitative conversion efficiency is supplied. It also cannot be used to determine a new molecular formula for the same reference compound. The final detection reagent responds to the appropriately prepared phosphate species rather than determining how much phosphorus was originally present independently of preparation. A valid quantitative comparison requires equivalent, verified conversion conditions and suitable recovery of the analytical product. The operator has confused the measured analytical response with the original elemental quantity.

Explanation: \(\textbf{Determine the initial silver chloride amount:}\) The molar mass of silver chloride is: \[ M_{\mathrm{AgCl}}=108+35.5 \] \[ M_{\mathrm{AgCl}}=143.5\,\mathrm{g\,mol^{-1}} \] The initial pure precipitate weighs \(0.287\,\mathrm{g}\). Therefore: \[ n_{\mathrm{AgCl,initial}}=\frac{0.287}{143.5} \] \[ n_{\mathrm{AgCl,initial}}=0.00200\,\mathrm{mol} \] \(\textbf{Determine the actual original chlorine content:}\) Each silver chloride formula unit contains one chlorine atom. Thus: \[ n_{\mathrm{Cl,true}}=0.00200\,\mathrm{mol} \] The chlorine mass is: \[ m_{\mathrm{Cl,true}}=0.00200\times35.5 \] \[ m_{\mathrm{Cl,true}}=0.0710\,\mathrm{g} \] Therefore: \[ \%\mathrm{Cl,true}=\frac{0.0710}{0.200}\times100 \] \[ \%\mathrm{Cl,true}=35.5\% \] \(\textbf{Account for photochemical decomposition:}\) Exactly half the original AgCl decomposes. The amount remaining unchanged is: \[ n_{\mathrm{AgCl,remaining}}=0.00100\,\mathrm{mol} \] The amount converted into metallic silver is also: \[ n_{\mathrm{Ag,formed}}=0.00100\,\mathrm{mol} \] \(\textbf{Calculate the final weighed solid mass:}\) The mass of unchanged silver chloride is: \[ m_{\mathrm{AgCl,remaining}}=0.00100\times143.5 \] \[ m_{\mathrm{AgCl,remaining}}=0.1435\,\mathrm{g} \] The metallic silver mass is: \[ m_{\mathrm{Ag}}=0.00100\times108 \] \[ m_{\mathrm{Ag}}=0.1080\,\mathrm{g} \] Thus: \[ m_{\mathrm{final}}=0.1435+0.1080 \] \[ m_{\mathrm{final}}=0.2515\,\mathrm{g} \] \(\textbf{Determine the analyst's apparent chlorine result:}\) The analyst mistakenly applies the AgCl gravimetric factor to the entire final solid. \[ m_{\mathrm{Cl,apparent}}=0.2515\times\frac{35.5}{143.5} \] \[ m_{\mathrm{Cl,apparent}}\approx0.06222\,\mathrm{g} \] Therefore: \[ \%\mathrm{Cl,apparent}=\frac{0.06222}{0.200}\times100 \] \[ \%\mathrm{Cl,apparent}\approx31.1\% \] \(\textbf{Interpretation:}\) The initial gravimetric determination represents the correct original chlorine quantity. Photochemical decomposition changes the chemical composition and mass of the solid subsequently weighed. Some chlorine is lost while metallic silver remains. Consequently, the final solid is no longer pure silver chloride and cannot legitimately be treated with the original gravimetric factor. The incorrect calculation underestimates the chlorine percentage. The true chlorine percentage is \(35.5\%\); treating the light-exposed residue as pure silver chloride gives an apparent value of \(31.1\%\). \(\textbf{Final answer:}\) \[ \boxed{\left(\%\mathrm{Cl,true},\%\mathrm{Cl,apparent}\right)=\left(35.5\%,31.1\%\right)} \]