Purification and Characterisation of Organic Compounds MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Test the chlorine assignment:}\) If the precipitate were silver chloride: \[ M_{\mathrm{AgCl}}=143.5\,\mathrm{g\,mol^{-1}} \] The chlorine mass would be: \[ m_{\mathrm{Cl}}=0.376\times\frac{35.5}{143.5} \] \[ m_{\mathrm{Cl}}\approx0.0930\,\mathrm{g} \] This is less than the \(0.200\,\mathrm{g}\) sample mass, so the chlorine assignment is not excluded by mass conservation. \(\textbf{Test the bromine assignment:}\) If the precipitate were silver bromide: \[ M_{\mathrm{AgBr}}=188\,\mathrm{g\,mol^{-1}} \] The bromine mass would be: \[ m_{\mathrm{Br}}=0.376\times\frac{80}{188} \] \[ m_{\mathrm{Br}}=0.160\,\mathrm{g} \] This is also less than the original sample mass, so bromine is not excluded. \(\textbf{Test the iodine assignment:}\) If the precipitate were silver iodide: \[ M_{\mathrm{AgI}}=108+127=235\,\mathrm{g\,mol^{-1}} \] The iodine mass would be: \[ m_{\mathrm I}=0.376\times\frac{127}{235} \] \[ m_{\mathrm I}=0.2032\,\mathrm{g} \] However: \[ m_{\mathrm I}\gt m_{\mathrm{sample}} \] A \(0.200\,\mathrm{g}\) sample cannot contain \(0.2032\,\mathrm{g}\) of iodine. \(\textbf{Interpretation:}\) The mass balance therefore excludes iodine under the stated assumptions. It does not uniquely distinguish chlorine from bromine because both assignments produce physically possible elemental masses. An independent observation, such as suitable precipitate colour and ammonia-solubility behaviour, would be needed for a unique chemical identification. The result illustrates how gravimetric data can reject an impossible assignment without necessarily determining the remaining identity.

Explanation: The two procedures identify or measure sulphur after converting it into different inorganic forms. During suitable sodium fusion, organically bound sulphur is converted into sodium sulphide. The resulting sulphide ions can give a characteristic violet response with sodium nitroprusside. This provides evidence for sulphur but does not by itself establish its mass percentage. The second procedure follows the gravimetric principle used for quantitative sulphur estimation. Sulphur in the organic compound is oxidised into sulphuric acid under the prescribed conditions. Barium chloride supplies barium ions, which react with sulphate ions to produce sparingly soluble barium sulphate. The ionic precipitation reaction is \(\mathrm{Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)}\). The barium sulphate is collected, dried and weighed. Since each mole of barium sulphate contains one mole of sulphur atoms, the sulphur mass can be calculated using its molar-mass fraction. The complete barium sulphate mass is not equal to the original sulphur mass because barium and oxygen also contribute to the precipitate. The correct distinction therefore involves both the different chemical forms generated and the qualitative or quantitative purpose of the observation.

Explanation: \(\textbf{Identify the measured analytical compound:}\) The recovered precipitate is barium sulphate. Its mass is: \[ m_{\mathrm{BaSO_4}}=0.699\,\mathrm{g} \] Each formula unit contains one sulphur atom. \(\textbf{Calculate the molar mass:}\) \[ M_{\mathrm{BaSO_4}}=137+32+4(16) \] \[ M_{\mathrm{BaSO_4}}=233\,\mathrm{g\,mol^{-1}} \] The mass of sulphur in one mole of barium sulphate is \(32\,\mathrm{g}\). \(\textbf{Convert precipitate mass into sulphur mass:}\) \[ m_{\mathrm S}=m_{\mathrm{BaSO_4}}\times\frac{32}{233} \] Substituting: \[ m_{\mathrm S}=0.699\times\frac{32}{233} \] \[ m_{\mathrm S}=0.0960\,\mathrm{g} \] \(\textbf{Calculate sulphur percentage:}\) The original sample mass is: \[ m_{\mathrm{sample}}=0.400\,\mathrm{g} \] Therefore: \[ \%\mathrm S=\frac{0.0960}{0.400}\times100 \] \[ \%\mathrm S=24.0\% \] \(\textbf{Check the stoichiometry:}\) The barium sulphate precipitate contains barium and oxygen in addition to sulphur. Consequently, its complete mass must not be assigned to sulphur. The correct gravimetric factor is \(32/233\), not \(233/32\). The calculation also assumes complete oxidation to sulphate and quantitative precipitation and recovery of barium sulphate. \(\textbf{Final answer:}\) \[ \boxed{24.0\%} \]

Explanation: \(\textbf{Determine the actual sulphur mass:}\) The original sample weighs: \[ m_{\mathrm{sample}}=0.400\,\mathrm{g} \] Its actual sulphur percentage is \(20.0\%\). Therefore: \[ m_{\mathrm{S,true}}=0.400\times\frac{20.0}{100} \] \[ m_{\mathrm{S,true}}=0.0800\,\mathrm{g} \] \(\textbf{Calculate barium sulphate theoretically formed:}\) Each mole of barium sulphate contains one mole of sulphur. Therefore: \[ m_{\mathrm{BaSO_4,theoretical}} = m_{\mathrm{S,true}}\times\frac{233}{32} \] Substituting: \[ m_{\mathrm{BaSO_4,theoretical}} = 0.0800\times\frac{233}{32} \] \[ m_{\mathrm{BaSO_4,theoretical}}=0.5825\,\mathrm{g} \] \(\textbf{Apply the recovery loss:}\) Only \(80.0\%\) of the precipitate is collected. \[ m_{\mathrm{BaSO_4,recovered}} = 0.5825\times\frac{80.0}{100} \] \[ m_{\mathrm{BaSO_4,recovered}}=0.4660\,\mathrm{g} \] \(\textbf{Reconstruct the uncorrected calculation:}\) If the student assumes the recovered precipitate represents all sulphur: \[ m_{\mathrm{S,apparent}} = 0.4660\times\frac{32}{233} \] \[ m_{\mathrm{S,apparent}}=0.0640\,\mathrm{g} \] Consequently: \[ \%\mathrm{S,apparent} = \frac{0.0640}{0.400}\times100 \] \[ \%\mathrm{S,apparent}=16.0\% \] \(\textbf{Interpret the recovery factor:}\) The apparent sulphur percentage is \(80.0\%\) of the actual percentage because only \(80.0\%\) of the analytical precipitate was recovered. The missing precipitate contains the unaccounted sulphur contribution. Correcting for the known recovery loss restores the true sulphur percentage of \(20.0\%\). This illustrates why a pure precipitate can still give an inaccurate gravimetric result if its recovery is incomplete. \(\textbf{Final answer:}\) \[ \boxed{\left(16.0\%,\,20.0\%\right)} \]

Explanation: \(\textbf{Determine sulphur originally present:}\) The sample mass is: \[ m_{\mathrm{sample}}=0.310\,\mathrm{g} \] Sulphur constitutes \(20.0\%\) of the sample. Therefore: \[ m_{\mathrm S}=0.310\times\frac{20.0}{100} \] \[ m_{\mathrm S}=0.0620\,\mathrm{g} \] \(\textbf{Convert sulphur mass into amount:}\) \[ n_{\mathrm S}=\frac{0.0620}{32} \] \[ n_{\mathrm S}=0.0019375\,\mathrm{mol} \] \(\textbf{Determine barium sulphate amount:}\) Each barium sulphate formula unit contains one sulphur atom. Consequently: \[ n_{\mathrm{BaSO_4}}=0.0019375\,\mathrm{mol} \] The molar mass is: \[ M_{\mathrm{BaSO_4}}=137+32+4(16) \] \[ M_{\mathrm{BaSO_4}}=233\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculate the expected precipitate:}\) \[ m_{\mathrm{BaSO_4}}=0.0019375\times233 \] \[ m_{\mathrm{BaSO_4}}=0.4514375\,\mathrm{g} \] \[ m_{\mathrm{BaSO_4}}\approx0.451\,\mathrm{g} \] \(\textbf{Interpret the mass balance:}\) The precipitate is heavier than the original organic sample because it contains barium and oxygen in addition to sulphur. These additional atoms originate from the analytical reagents and the preceding chemical transformations. The complete mass of the precipitate therefore need not be supplied by the original sample. Conservation of mass applies to the entire reaction system, not only to the original organic compound and one isolated product. The student's objection is consequently incorrect even though the precipitate is heavier than the starting sample.

Explanation: Gravimetric sulphur estimation requires the weighed precipitate to represent the sulphur-derived barium sulphate quantitatively. The calculation assumes that the measured solid mass corresponds to a known number of moles of barium sulphate. Each mole of barium sulphate contains one mole of sulphur atoms, giving the gravimetric factor \(32/233\). In the described experiment, some barium carbonate is also present in the collected solid. Barium carbonate contains no sulphur, so its mass does not represent any additional sulphur from the organic sample. Nevertheless, treating the entire collected mass as barium sulphate assigns a positive sulphur contribution to the carbonate impurity. The inferred amount of barium sulphate therefore becomes larger than the amount actually produced from the sample's sulphur. The calculated sulphur mass and sulphur percentage are consequently overestimated. The error differs from incomplete precipitation, which would tend to produce a smaller recovered barium sulphate mass. The presence of a barium-containing precipitate does not prove that every portion of it has the required sulphate composition. Correct gravimetry therefore depends not only on collecting and drying a precipitate but also on ensuring its chemical identity and purity.

Explanation: The quantitative phosphorus-estimation procedure first converts phosphorus from its original organic environment into phosphoric acid under suitable oxidising conditions. The phosphoric acid is then converted into a compound suitable for gravimetric measurement. One such product is ammonium phosphomolybdate, represented in NCERT by \(\mathrm{(NH_4)_3PO_4\cdot12MoO_3}\). The formula contains a single \(\mathrm{PO_4}\) group and therefore one phosphorus atom per formula unit. The three ammonium groups contribute nitrogen and hydrogen but do not contain phosphorus. The twelve \(\mathrm{MoO_3}\) units contribute molybdenum and oxygen rather than additional phosphorus. Consequently, the precipitate has a large formula mass relative to the mass of phosphorus that it represents. The phosphorus mass is calculated from the weighed precipitate using the ratio of one phosphorus atomic mass to the complete formula mass. Counting either ammonium groups or molybdenum oxide units as phosphorus would produce an incorrect stoichiometric factor. The whole precipitate mass cannot be treated as elemental phosphorus because the other atoms originate from analytical reagents. Correct estimation therefore requires reading the complete chemical formula before applying the gravimetric relation.

Explanation: Statement I is correct because magnesium ammonium phosphate is the product obtained before the ignition stage, whereas magnesium pyrophosphate is the compound obtained after ignition. The procedure deliberately converts the initial phosphate precipitate into a stable compound suitable for weighing. Statement II is correct because the formula \(\mathrm{Mg_2P_2O_7}\) contains two phosphorus atoms per formula unit. Consequently, one mole of magnesium pyrophosphate represents two moles of phosphorus atoms. Statement III is correct because the sulphur-free final gravimetric product contains phosphorus together with magnesium and oxygen. The phosphorus mass fraction is therefore the mass of two phosphorus atoms divided by the molar mass of magnesium pyrophosphate. Statement IV is incorrect because ignition changes the identity and composition of the original precipitate. Treating the initial magnesium ammonium phosphate and final magnesium pyrophosphate as having the same formula would ignore this chemical conversion. The correct mass relation must use the compound actually collected and weighed at the end of the procedure. This distinction is important because phosphorus estimation can employ more than one final gravimetric product. Different products have different phosphorus mass fractions even when they represent the same original phosphorus quantity. The correct statements are I, II and III.

Explanation: \(\textbf{Identify the final weighed product:}\) The analytical precipitate has been ignited to magnesium pyrophosphate. Its mass is: \[ m_{\mathrm{Mg_2P_2O_7}}=0.111\,\mathrm{g} \] The final formula contains two phosphorus atoms per formula unit. \(\textbf{Calculate its molar mass:}\) \[ M_{\mathrm{Mg_2P_2O_7}} = 2(24)+2(31)+7(16) \] \[ M_{\mathrm{Mg_2P_2O_7}} = 48+62+112 \] \[ M_{\mathrm{Mg_2P_2O_7}}=222\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine the phosphorus mass fraction:}\) Each mole of magnesium pyrophosphate contains: \[ m_{\mathrm{P\ per\ mole}}=2(31) \] \[ m_{\mathrm{P\ per\ mole}}=62\,\mathrm{g} \] Therefore: \[ m_{\mathrm P} = m_{\mathrm{Mg_2P_2O_7}}\times\frac{62}{222} \] \(\textbf{Substitute the precipitate mass:}\) \[ m_{\mathrm P}=0.111\times\frac{62}{222} \] \[ m_{\mathrm P}=0.0310\,\mathrm{g} \] \(\textbf{Calculate phosphorus percentage:}\) The original compound mass is: \[ m_{\mathrm{sample}}=0.310\,\mathrm{g} \] Hence: \[ \%\mathrm P=\frac{0.0310}{0.310}\times100 \] \[ \%\mathrm P=10.0\% \] \(\textbf{Check the formula interpretation:}\) The final weighed product contains two phosphorus atoms per formula unit. Using only one phosphorus atomic mass would underestimate the phosphorus content by a factor of two. The magnesium and oxygen contributions must also remain included in the denominator because the weighed compound is magnesium pyrophosphate, not elemental phosphorus. \(\textbf{Final answer:}\) \[ \boxed{10.0\%} \]

Explanation: \(\textbf{Determine phosphorus from Procedure I:}\) The amount of ammonium phosphomolybdate obtained is: \[ n_{\mathrm{phosphomolybdate}}=0.000500\,\mathrm{mol} \] Its formula is: \[ \mathrm{(NH_4)_3PO_4\cdot12MoO_3} \] Each formula unit contains one phosphorus atom. Therefore: \[ n_{\mathrm P}=0.000500\,\mathrm{mol} \] \(\textbf{Translate phosphorus into the alternative product:}\) Procedure II produces magnesium pyrophosphate: \[ \mathrm{Mg_2P_2O_7} \] Each formula unit contains two phosphorus atoms. Therefore: \[ n_{\mathrm{Mg_2P_2O_7}}=\frac{n_{\mathrm P}}{2} \] Substituting: \[ n_{\mathrm{Mg_2P_2O_7}}=\frac{0.000500}{2} \] \[ n_{\mathrm{Mg_2P_2O_7}}=0.000250\,\mathrm{mol} \] \(\textbf{Calculate the final product mass:}\) The molar mass is: \[ M_{\mathrm{Mg_2P_2O_7}}=2(24)+2(31)+7(16) \] \[ M_{\mathrm{Mg_2P_2O_7}}=222\,\mathrm{g\,mol^{-1}} \] Therefore: \[ m_{\mathrm{Mg_2P_2O_7}} = n_{\mathrm{Mg_2P_2O_7}}M_{\mathrm{Mg_2P_2O_7}} \] \[ m_{\mathrm{Mg_2P_2O_7}}=0.000250\times222 \] \[ m_{\mathrm{Mg_2P_2O_7}}=0.0555\,\mathrm{g} \] \(\textbf{Interpretation:}\) The two procedures use different final gravimetric compounds but must represent the same quantity of elemental phosphorus. One mole of ammonium phosphomolybdate contains one mole of phosphorus atoms, while one mole of magnesium pyrophosphate contains two moles of phosphorus atoms. Therefore, the expected magnesium pyrophosphate amount is half the phosphomolybdate amount in moles. The two precipitate masses are not interchangeable because their formula masses and phosphorus contents differ. The calculation depends on conserving phosphorus atoms across the two analytical pathways rather than treating the two product formulae as equivalent. \(\textbf{Final answer:}\) \[ \boxed{0.0555\,\mathrm{g}} \]

Explanation: \(\textbf{Identify the nitrogen measured by each method:}\) Dumas estimation accounts for nitrogen from both components under the stated conditions. Therefore: \[ m_{\mathrm{N,total}}=0.210\,\mathrm{g} \] Ordinary Kjeldahl estimation accounts for nitrogen from acetamide but not from the nitro group of nitrobenzene. Thus: \[ m_{\mathrm{N,acetamide}}=0.140\,\mathrm{g} \] \(\textbf{Isolate nitrogen from nitrobenzene:}\) \[ m_{\mathrm{N,nitrobenzene}} = m_{\mathrm{N,total}}-m_{\mathrm{N,acetamide}} \] \[ m_{\mathrm{N,nitrobenzene}}=0.210-0.140 \] \[ m_{\mathrm{N,nitrobenzene}}=0.0700\,\mathrm{g} \] \(\textbf{Convert nitrogen into nitrobenzene:}\) The molecular formula of nitrobenzene is: \[ \mathrm{C_6H_5NO_2} \] Its molar mass is: \[ M=6(12)+5(1)+14+2(16) \] \[ M=123\,\mathrm{g\,mol^{-1}} \] Each molecule contains one nitrogen atom. Therefore: \[ n_{\mathrm{nitrobenzene}}=\frac{0.0700}{14} \] \[ n_{\mathrm{nitrobenzene}}=0.00500\,\mathrm{mol} \] The mass of nitrobenzene is: \[ m_{\mathrm{nitrobenzene}}=0.00500\times123 \] \[ m_{\mathrm{nitrobenzene}}=0.615\,\mathrm{g} \] \(\textbf{Calculate percentage composition:}\) \[ \%\mathrm{nitrobenzene} = \frac{0.615}{1.205}\times100 \] \[ \%\mathrm{nitrobenzene}=51.04\ldots\% \] \[ \%\mathrm{nitrobenzene}\approx51.0\% \] \(\textbf{Independent consistency check:}\) The remaining mass is: \[ m_{\mathrm{acetamide}}=1.205-0.615 \] \[ m_{\mathrm{acetamide}}=0.590\,\mathrm{g} \] Acetamide has molar mass \(59\,\mathrm{g\,mol^{-1}}\). Therefore: \[ n_{\mathrm{acetamide}}=\frac{0.590}{59}=0.0100\,\mathrm{mol} \] It contains one nitrogen atom per molecule, giving: \[ m_{\mathrm{N,acetamide}}=0.0100\times14 \] \[ m_{\mathrm{N,acetamide}}=0.140\,\mathrm{g} \] This agrees with the stated Kjeldahl result. The method comparison is therefore chemically and quantitatively consistent. The difference between the two nitrogen determinations identifies the nitrogen belonging to the Kjeldahl-inapplicable component, permitting its mass to be calculated. \(\textbf{Final answer:}\) \[ \boxed{51.0\%} \]

Explanation: \(\textbf{Determine the original acid quantity:}\) The receiving flask contains hydrochloric acid: \[ V_{\mathrm{HCl}}=50.0\,\mathrm{mL} \] \[ M_{\mathrm{HCl}}=0.100\,\mathrm{mol\,L^{-1}} \] Therefore: \[ n_{\mathrm{HCl,initial}} = \frac{50.0\times0.100}{1000} \] \[ n_{\mathrm{HCl,initial}}=0.00500\,\mathrm{mol} \] \(\textbf{Determine the unused acid:}\) The remaining acid is neutralised by sodium hydroxide. \[ n_{\mathrm{NaOH}} = \frac{20.0\times0.100}{1000} \] \[ n_{\mathrm{NaOH}}=0.00200\,\mathrm{mol} \] Because hydrochloric acid and sodium hydroxide react in a \(1:1\) mole ratio: \[ n_{\mathrm{HCl,unused}}=0.00200\,\mathrm{mol} \] \(\textbf{Calculate ammonia from the aliquot:}\) The acid consumed by ammonia is: \[ n_{\mathrm{HCl,consumed}} = 0.00500-0.00200 \] \[ n_{\mathrm{HCl,consumed}}=0.00300\,\mathrm{mol} \] The relevant reaction is: \[ \mathrm{NH_3+HCl\rightarrow NH_4Cl} \] Thus: \[ n_{\mathrm{NH_3,aliquot}}=0.00300\,\mathrm{mol} \] \(\textbf{Restore the full digest quantity:}\) Only \(25.0\,\mathrm{mL}\) of the \(100.0\,\mathrm{mL}\) digest was analysed. The dilution factor is: \[ f=\frac{100.0}{25.0} \] \[ f=4 \] Therefore: \[ n_{\mathrm{NH_3,total}}=0.00300\times4 \] \[ n_{\mathrm{NH_3,total}}=0.0120\,\mathrm{mol} \] \(\textbf{Determine nitrogen mass:}\) Each ammonia molecule contains one nitrogen atom. \[ m_{\mathrm N}=0.0120\times14 \] \[ m_{\mathrm N}=0.168\,\mathrm{g} \] \(\textbf{Calculate nitrogen percentage:}\) \[ \%\mathrm N=\frac{0.168}{0.700}\times100 \] \[ \%\mathrm N=24.0\% \] \(\textbf{Interpretation:}\) The titration directly measures nitrogen corresponding only to the withdrawn aliquot. Failing to account for the total digest volume would underestimate nitrogen by a factor of four. The use of hydrochloric acid also means that one mole of ammonia consumes one mole of acid. The correct calculation requires acid-base stoichiometry followed by restoration of the original sample quantity. \(\textbf{Final answer:}\) \[ \boxed{24.0\%} \]

Explanation: The Dumas method requires the nitrogen in the organic compound to be represented quantitatively by molecular nitrogen collected for measurement. In Trial I, atmospheric air has been completely displaced, so there is no atmospheric nitrogen contamination. However, the heated copper gauze is absent. The question specifies that some sample nitrogen remains in the form of nitrogen dioxide. That nitrogen dioxide is subsequently absorbed by the potassium hydroxide solution and does not contribute to the measured molecular-nitrogen volume. Consequently, the collected nitrogen represents less than the total nitrogen originally present in the sample. Trial I therefore underestimates nitrogen percentage. In Trial II, the heated copper gauze effectively converts the relevant nitrogen oxides into molecular nitrogen. However, atmospheric air has not been completely displaced. Nitrogen from the trapped air survives the carbon dioxide absorption stage and reaches the graduated collection tube. The measured nitrogen volume consequently includes nitrogen that was never part of the organic compound. Assigning the full gas volume to the sample causes an overestimate. The two trials illustrate opposite errors arising at different stages of the same analytical method. Correct Dumas estimation requires both suitable chemical conversion of sample nitrogen and exclusion of nitrogen from external sources.

Explanation: \(\textbf{Calculate nitrogen obtained by Dumas estimation:}\) The dry nitrogen volume is: \[ V_{\mathrm{N_2}}=67.2\,\mathrm{mL} \] At STP: \[ n_{\mathrm{N_2}}=\frac{67.2}{22400} \] \[ n_{\mathrm{N_2}}=0.00300\,\mathrm{mol} \] \(\textbf{Determine the corresponding ammonia requirement:}\) Each nitrogen molecule contains two nitrogen atoms, whereas each ammonia molecule contains one. Therefore: \[ n_{\mathrm{NH_3}}=2n_{\mathrm{N_2}} \] \[ n_{\mathrm{NH_3}}=2(0.00300) \] \[ n_{\mathrm{NH_3}}=0.00600\,\mathrm{mol} \] The receiving solution must neutralise more than \(0.00600\,\mathrm{mol}\) of ammonia to leave excess acid. \(\textbf{Check receiving solution P:}\) \[ n_{\mathrm{H_2SO_4}}=\frac{20.0\times0.100}{1000} \] \[ n_{\mathrm{H_2SO_4}}=0.00200\,\mathrm{mol} \] Since one mole of sulphuric acid neutralises two moles of ammonia: \[ n_{\mathrm{NH_3,capacity}}=2(0.00200) \] \[ n_{\mathrm{NH_3,capacity}}=0.00400\,\mathrm{mol} \] This is insufficient. \(\textbf{Check receiving solution Q:}\) \[ n_{\mathrm{HCl}}=\frac{30.0\times0.100}{1000} \] \[ n_{\mathrm{HCl}}=0.00300\,\mathrm{mol} \] Hydrochloric acid neutralises ammonia in a \(1:1\) ratio. Therefore, Q is insufficient. \(\textbf{Check receiving solution R:}\) \[ n_{\mathrm{HCl}}=\frac{40.0\times0.100}{1000} \] \[ n_{\mathrm{HCl}}=0.00400\,\mathrm{mol} \] This is also insufficient. \(\textbf{Check receiving solution S:}\) \[ n_{\mathrm{H_2SO_4}}=\frac{35.0\times0.100}{1000} \] \[ n_{\mathrm{H_2SO_4}}=0.00350\,\mathrm{mol} \] Its ammonia-neutralisation capacity is: \[ n_{\mathrm{NH_3,capacity}}=2(0.00350) \] \[ n_{\mathrm{NH_3,capacity}}=0.00700\,\mathrm{mol} \] The excess capacity is: \[ 0.00700-0.00600=0.00100\,\mathrm{mol} \] \(\textbf{Interpretation:}\) Only S can neutralise the predicted ammonia completely while retaining some unused acid. Merely comparing the volumes of the receiving solutions would be incorrect because hydrochloric acid and sulphuric acid have different neutralisation capacities per mole. The Dumas result establishes the nitrogen quantity, which must then be translated into the corresponding Kjeldahl ammonia amount before selecting a suitable receiving solution.

Explanation: \(\textbf{Determine carbon from combustion:}\) The carbon dioxide produced is: \[ m_{\mathrm{CO_2}}=0.440\,\mathrm{g} \] Therefore: \[ m_{\mathrm C}=0.440\times\frac{12}{44} \] \[ m_{\mathrm C}=0.120\,\mathrm{g} \] \(\textbf{Determine hydrogen from combustion:}\) The water produced is: \[ m_{\mathrm{H_2O}}=0.180\,\mathrm{g} \] Thus: \[ m_{\mathrm H}=0.180\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0200\,\mathrm{g} \] \(\textbf{Determine nitrogen from Dumas estimation:}\) The collected nitrogen volume is: \[ V_{\mathrm{N_2}}=44.8\,\mathrm{mL} \] At STP: \[ n_{\mathrm{N_2}}=\frac{44.8}{22400} \] \[ n_{\mathrm{N_2}}=0.00200\,\mathrm{mol} \] The mass of molecular nitrogen is: \[ m_{\mathrm N}=0.00200\times28 \] \[ m_{\mathrm N}=0.0560\,\mathrm{g} \] \(\textbf{Apply the four-element mass balance:}\) The compound is specified to contain only carbon, hydrogen, nitrogen and oxygen. Therefore: \[ m_{\mathrm O} = m_{\mathrm{sample}}-m_{\mathrm C}-m_{\mathrm H}-m_{\mathrm N} \] Substituting: \[ m_{\mathrm O} = 0.300-0.120-0.0200-0.0560 \] \[ m_{\mathrm O}=0.104\,\mathrm{g} \] \(\textbf{Calculate oxygen percentage:}\) \[ \%\mathrm O=\frac{0.104}{0.300}\times100 \] \[ \%\mathrm O=34.666\ldots\% \] \[ \%\mathrm O\approx34.7\% \] \(\textbf{Consistency check:}\) The independently estimated elemental percentages are approximately: \[ \%\mathrm C=40.0\% \] \[ \%\mathrm H=6.67\% \] \[ \%\mathrm N=18.67\% \] The remaining \(34.67\%\) belongs to oxygen. This deduction is valid because the question excludes sulphur, halogens, phosphorus and other possible constituent elements. The oxygen atoms present in combustion products cannot be used directly to determine the original oxygen content because external oxygen is introduced during combustion. The method therefore combines independent carbon, hydrogen and nitrogen estimations with a justified elemental mass balance. \(\textbf{Final answer:}\) \[ \boxed{34.7\%} \]

Explanation: \(\textbf{Determine the complete nitrogen quantity:}\) The Dumas experiment is stated to be quantitative. The dry nitrogen volume is: \[ V_{\mathrm{N_2}}=44.8\,\mathrm{mL} \] At STP: \[ n_{\mathrm{N_2}}=\frac{44.8}{22400} \] \[ n_{\mathrm{N_2}}=0.00200\,\mathrm{mol} \] \(\textbf{Convert the nitrogen quantity into ammonia:}\) Every \(\mathrm{N_2}\) molecule contains two nitrogen atoms. Every \(\mathrm{NH_3}\) molecule contains one nitrogen atom. Therefore, complete Kjeldahl conversion should produce: \[ n_{\mathrm{NH_3,theoretical}}=2n_{\mathrm{N_2}} \] \[ n_{\mathrm{NH_3,theoretical}}=2(0.00200) \] \[ n_{\mathrm{NH_3,theoretical}}=0.00400\,\mathrm{mol} \] \(\textbf{Identify the absorbed ammonia:}\) The experimentally determined acid consumption corresponds to: \[ n_{\mathrm{NH_3,absorbed}}=0.00320\,\mathrm{mol} \] This is smaller than the theoretical ammonia amount. \(\textbf{Calculate absorption recovery:}\) \[ \%\mathrm{recovery} = \frac{n_{\mathrm{NH_3,absorbed}}} {n_{\mathrm{NH_3,theoretical}}}\times100 \] Substituting: \[ \%\mathrm{recovery} = \frac{0.00320}{0.00400}\times100 \] \[ \%\mathrm{recovery}=80.0\% \] \(\textbf{Interpretation:}\) The Dumas result establishes the complete amount of nitrogen in the portion. Converting that nitrogen-atom quantity into the expected ammonia amount gives a reference against which the Kjeldahl collection can be assessed. The smaller measured ammonia amount is attributed to transfer loss because complete ammonia formation and all other experimental conditions are specified. The percentage recovered is therefore the absorbed ammonia amount divided by the expected ammonia amount. A valid quantitative comparison requires conservation of nitrogen atoms, not equality of the mole amounts of \(\mathrm{N_2}\) and \(\mathrm{NH_3}\). \(\textbf{Final answer:}\) \[ \boxed{80.0\%} \]

Explanation: \(\textbf{Identify the separated silver halides:}\) The total silver-halide precipitate weighs: \[ m_{\mathrm{total}}=0.475\,\mathrm{g} \] Under the specified ammonia-treatment conditions, silver chloride dissolves completely, while silver bromide remains undissolved. Therefore: \[ m_{\mathrm{AgBr}}=0.188\,\mathrm{g} \] \(\textbf{Calculate silver chloride mass:}\) \[ m_{\mathrm{AgCl}}=m_{\mathrm{total}}-m_{\mathrm{AgBr}} \] \[ m_{\mathrm{AgCl}}=0.475-0.188 \] \[ m_{\mathrm{AgCl}}=0.287\,\mathrm{g} \] \(\textbf{Determine chlorine mass:}\) The molar mass of silver chloride is: \[ M_{\mathrm{AgCl}}=108+35.5 \] \[ M_{\mathrm{AgCl}}=143.5\,\mathrm{g\,mol^{-1}} \] Each mole of silver chloride contains one mole of chlorine atoms. \[ m_{\mathrm{Cl}}=0.287\times\frac{35.5}{143.5} \] \[ m_{\mathrm{Cl}}=0.0710\,\mathrm{g} \] \(\textbf{Determine bromine mass:}\) The molar mass of silver bromide is: \[ M_{\mathrm{AgBr}}=108+80 \] \[ M_{\mathrm{AgBr}}=188\,\mathrm{g\,mol^{-1}} \] Therefore: \[ m_{\mathrm{Br}}=0.188\times\frac{80}{188} \] \[ m_{\mathrm{Br}}=0.0800\,\mathrm{g} \] \(\textbf{Calculate the individual percentages:}\) The original sample weighs \(0.300\,\mathrm{g}\). For chlorine: \[ \%\mathrm{Cl}=\frac{0.0710}{0.300}\times100 \] \[ \%\mathrm{Cl}=23.666\ldots\% \] For bromine: \[ \%\mathrm{Br}=\frac{0.0800}{0.300}\times100 \] \[ \%\mathrm{Br}=26.666\ldots\% \] \(\textbf{Interpretation:}\) The initial precipitate mass cannot be converted using a single silver-halide gravimetric factor because it contains two different compounds. The selective ammonia treatment provides the additional information needed to determine the two masses separately. The chlorine and bromine contributions must then be calculated using their respective silver-halide molar masses. The sample mass is the denominator for both elemental percentages. \(\textbf{Final answer:}\) \[ \boxed{\left(\%\mathrm{Cl},\%\mathrm{Br}\right)=(23.7\%,\,26.7\%)} \]

Explanation: \(\textbf{Determine carbon mass:}\) The carbon dioxide produced is: \[ m_{\mathrm{CO_2}}=0.176\,\mathrm{g} \] Therefore: \[ m_{\mathrm C}=0.176\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0480\,\mathrm{g} \] \(\textbf{Determine hydrogen mass:}\) The water produced is: \[ m_{\mathrm{H_2O}}=0.0540\,\mathrm{g} \] Hence: \[ m_{\mathrm H}=0.0540\times\frac{2}{18} \] \[ m_{\mathrm H}=0.00600\,\mathrm{g} \] \(\textbf{Determine bromine mass:}\) The recovered silver bromide weighs: \[ m_{\mathrm{AgBr}}=0.376\,\mathrm{g} \] Its molar mass is: \[ M_{\mathrm{AgBr}}=108+80 \] \[ M_{\mathrm{AgBr}}=188\,\mathrm{g\,mol^{-1}} \] The corresponding bromine mass is: \[ m_{\mathrm{Br}}=0.376\times\frac{80}{188} \] \[ m_{\mathrm{Br}}=0.160\,\mathrm{g} \] \(\textbf{Calculate oxygen by difference:}\) The sample contains only carbon, hydrogen, bromine and oxygen. Therefore: \[ m_{\mathrm O} = m_{\mathrm{sample}}-m_{\mathrm C}-m_{\mathrm H}-m_{\mathrm{Br}} \] Substituting: \[ m_{\mathrm O} = 0.278-0.0480-0.00600-0.160 \] \[ m_{\mathrm O}=0.0640\,\mathrm{g} \] \(\textbf{Calculate oxygen percentage:}\) \[ \%\mathrm O=\frac{0.0640}{0.278}\times100 \] \[ \%\mathrm O=23.0215\ldots\% \] \[ \%\mathrm O\approx23.0\% \] \(\textbf{Interpretation:}\) The calculation combines combustion measurements with Carius gravimetry. Carbon and hydrogen are determined from their oxidation products, whereas bromine is determined from the mass of silver bromide. The masses of the analytical products cannot be added directly because they contain atoms supplied by reagents. Each product must first be converted into the mass of the relevant original element. Oxygen can then be calculated by difference because the elemental composition is explicitly restricted to four elements. \(\textbf{Final answer:}\) \[ \boxed{23.0\%} \]

Explanation: \(\textbf{Determine nitrogen molecules from Dumas data:}\) The collected molecular nitrogen volume is: \[ V_{\mathrm{N_2}}=22.4\,\mathrm{mL} \] At STP: \[ n_{\mathrm{N_2}}=\frac{22.4}{22400} \] \[ n_{\mathrm{N_2}}=0.00100\,\mathrm{mol} \] \(\textbf{Convert nitrogen molecules into nitrogen atoms:}\) Each \(\mathrm{N_2}\) molecule contains two nitrogen atoms. Therefore: \[ n_{\mathrm{N\ atoms}}=2n_{\mathrm{N_2}} \] \[ n_{\mathrm{N\ atoms}}=2(0.00100) \] \[ n_{\mathrm{N\ atoms}}=0.00200\,\mathrm{mol} \] \(\textbf{Determine bromine from Carius data:}\) The silver bromide mass is: \[ m_{\mathrm{AgBr}}=0.188\,\mathrm{g} \] Its molar mass is: \[ M_{\mathrm{AgBr}}=108+80 \] \[ M_{\mathrm{AgBr}}=188\,\mathrm{g\,mol^{-1}} \] Therefore: \[ n_{\mathrm{AgBr}}=\frac{0.188}{188} \] \[ n_{\mathrm{AgBr}}=0.00100\,\mathrm{mol} \] Each silver bromide formula unit contains one bromine atom. Thus: \[ n_{\mathrm{Br\ atoms}}=0.00100\,\mathrm{mol} \] \(\textbf{Compare the elemental atom quantities:}\) \[ n_{\mathrm N}:n_{\mathrm{Br}} = 0.00200:0.00100 \] Dividing both terms by \(0.00100\): \[ n_{\mathrm N}:n_{\mathrm{Br}}=2:1 \] \(\textbf{Interpretation:}\) Dumas provides the amount of molecular nitrogen, whereas Carius estimation provides the amount of silver bromide. Their formulae contain different numbers of the elements being estimated. Two nitrogen atoms occur in each \(\mathrm{N_2}\) molecule, but only one bromine atom occurs in each \(\mathrm{AgBr}\) formula unit. The measured molar quantities of the two analytical products happen to be equal, but the atomic quantities are not. The result therefore requires converting both products into amounts of their constituent elements before forming the ratio. \(\textbf{Final answer:}\) \[ \boxed{N:Br=2:1} \]

Explanation: \(\textbf{Determine sulphur amount:}\) 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 measured precipitate is: \[ m_{\mathrm{BaSO_4}}=0.233\,\mathrm{g} \] Therefore: \[ n_{\mathrm{BaSO_4}}=\frac{0.233}{233} \] \[ n_{\mathrm{BaSO_4}}=0.00100\,\mathrm{mol} \] Each barium sulphate formula unit contains one sulphur atom. Hence: \[ n_{\mathrm S}=0.00100\,\mathrm{mol} \] \(\textbf{Determine bromine amount:}\) The molar mass of silver bromide is: \[ M_{\mathrm{AgBr}}=108+80 \] \[ M_{\mathrm{AgBr}}=188\,\mathrm{g\,mol^{-1}} \] The precipitate mass is: \[ m_{\mathrm{AgBr}}=0.188\,\mathrm{g} \] Therefore: \[ n_{\mathrm{AgBr}}=\frac{0.188}{188} \] \[ n_{\mathrm{AgBr}}=0.00100\,\mathrm{mol} \] Each silver bromide formula unit contains one bromine atom. Thus: \[ n_{\mathrm{Br}}=0.00100\,\mathrm{mol} \] \(\textbf{Compare elemental masses:}\) The sulphur mass is: \[ m_{\mathrm S}=0.00100\times32 \] \[ m_{\mathrm S}=0.0320\,\mathrm{g} \] The bromine mass is: \[ m_{\mathrm{Br}}=0.00100\times80 \] \[ m_{\mathrm{Br}}=0.0800\,\mathrm{g} \] \(\textbf{Interpretation:}\) The equal molar quantities of barium sulphate and silver bromide represent equal numbers of sulphur and bromine atoms because both precipitates contain one relevant atom per formula unit. However, bromine atoms are heavier than sulphur atoms. Therefore, the bromine mass is greater even though the atomic amounts are equal. The larger molar mass of barium sulphate does not imply a larger mass of sulphur, because much of the precipitate mass is contributed by barium and oxygen.