Purification and Characterisation of Organic Compounds MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: \(\textbf{Interpret the calculated mole proportions:}\) The given atomic ratio is: \[ C:H:O=1.5:3:1 \] An empirical formula must express the simplest chemically consistent whole-number ratio of atoms. A fractional subscript is therefore not retained in the completed empirical formula. \(\textbf{Determine the common multiplier:}\) The carbon term contains the simple fraction: \[ 1.5=\frac{3}{2} \] Multiplying every term by two removes this denominator without changing the relative proportions. \[ C:H:O = 2(1.5):2(3):2(1) \] \[ C:H:O=3:6:2 \] \(\textbf{Construct the formula:}\) The resulting whole-number ratio is: \[ \mathrm{C_3H_6O_2} \] The subscripts have no common factor greater than one, so the ratio cannot be reduced further. \(\textbf{Verify the elemental composition:}\) The empirical-formula mass is: \[ M=3(12)+6(1)+2(16) \] \[ M=74\,\mathrm{g\,mol^{-1}} \] The calculated carbon percentage is: \[ \%\mathrm C=\frac{36}{74}\times100 \] \[ \%\mathrm C=48.65\% \] The hydrogen percentage is: \[ \%\mathrm H=\frac{6}{74}\times100 \] \[ \%\mathrm H=8.11\% \] The oxygen percentage is: \[ \%\mathrm O=\frac{32}{74}\times100 \] \[ \%\mathrm O=43.24\% \] These reproduce the supplied percentages within rounding. \(\textbf{Interpretation:}\) Rounding the fractional carbon ratio directly would alter the experimentally determined atomic proportions. Multiplying only one term would also change the relative composition. The correct operation is to multiply every ratio term by the same smallest integer that removes the simple fraction. \(\textbf{Final answer:}\) \[ \mathrm{C_3H_6O_2} \]

Explanation: \(\textbf{Determine the empirical-formula mass:}\) The empirical formula is \(\mathrm{CH_2O}\). Therefore: \[ M_{\mathrm{emp}}=12+2(1)+16 \] \[ M_{\mathrm{emp}}=30\,\mathrm{g\,mol^{-1}} \] \(\textbf{Use the independent molecular molar mass:}\) The given molecular molar mass is: \[ M_{\mathrm{molecular}}=90\,\mathrm{g\,mol^{-1}} \] For a molecular compound: \[ M_{\mathrm{molecular}}=nM_{\mathrm{emp}} \] where \(n\) is a positive integer. \(\textbf{Calculate the multiplication factor:}\) \[ n=\frac{M_{\mathrm{molecular}}}{M_{\mathrm{emp}}} \] \[ n=\frac{90}{30} \] \[ n=3 \] \(\textbf{Construct the molecular formula:}\) Every subscript in the empirical formula must be multiplied by three. \[ (\mathrm{CH_2O})_3=\mathrm{C_3H_6O_3} \] \(\textbf{Test the student's proposal:}\) The proposed formula \(\mathrm{C_2H_4O_2}\) has molar mass: \[ M=2(12)+4(1)+2(16) \] \[ M=60\,\mathrm{g\,mol^{-1}} \] Although it has the correct empirical ratio, its molar mass does not match the independently measured value. \(\textbf{Interpretation:}\) Elemental percentage composition determines relative atomic proportions but not necessarily actual molecular size. The molar mass provides the additional constraint that distinguishes the candidate molecular formulae. The factor is three, not two, and all subscripts must be scaled consistently. \(\textbf{Final answer:}\) \[ \mathrm{C_3H_6O_3} \]

Explanation: The molecular formula of a molecular compound must be a positive integer multiple of its empirical formula. The molecular molar mass must consequently be a positive integer multiple of the empirical-formula mass. For P, the empirical formula is \(\mathrm{CH_2O}\), whose formula mass is \(30\). The ratio for P is: \[ \frac{120}{30}=4 \] Thus P has a valid integer multiplier. For Q, the empirical formula is \(\mathrm{C_2H_4O}\), whose formula mass is \(44\). For Q: \[ \frac{132}{44}=3 \] Thus Q is also consistent. For R, the empirical-formula mass is again \(30\), but the stated molecular molar mass is \(95\). For R: \[ \frac{95}{30}=\frac{19}{6} \] This ratio is not an integer. No whole-number multiplication of the empirical subscripts can therefore produce a molecule with exactly this molar mass. For S, the empirical-formula mass is \(44\), and \(88/44\) equals two. S is consequently consistent with the required relation. The inconsistency in R cannot be corrected by multiplying different elemental subscripts by different factors. Such a procedure would change the empirical atomic ratio itself. Because the data are specified as exact, the noninteger factor cannot be attributed to experimental rounding. The only inconsistent data set is R.

Explanation: The Assertion is true because an atomic ratio identifies an empirical formula rather than necessarily identifying the molecular formula. The ratio \(2:4:1\) gives the empirical formula \(\mathrm{C_2H_4O}\). A molecule with precisely that formula would be consistent with the ratio. However, a molecular formula containing two empirical-formula units would also have the same elemental ratio. The same is true of any other chemically possible positive integer multiple. Relative atomic proportions therefore do not establish how many atoms occur in one molecule. The Reason is also true because molecular formula determination requires an additional constraint on molecular size. The molar mass can supply that constraint. Dividing the molecular molar mass by the empirical-formula mass gives the multiplication factor when the data are internally consistent. That factor must be a positive integer for the stated molecular compound. Without the additional information, the student's formula is only one possible molecular formula consistent with the composition. The Reason directly explains why the original conclusion is not uniquely justified. Both statements are true, and R correctly explains A.

Explanation: \(\textbf{Determine carbon from combustion:}\) The carbon dioxide produced is: \[ m_{\mathrm{CO_2}}=0.440\,\mathrm{g} \] Each mole of carbon dioxide contains one mole of carbon atoms. Therefore: \[ m_{\mathrm C}=0.440\times\frac{12}{44} \] \[ m_{\mathrm C}=0.120\,\mathrm{g} \] \(\textbf{Determine hydrogen from water:}\) The water produced is: \[ m_{\mathrm{H_2O}}=0.180\,\mathrm{g} \] Each mole of water contains two moles of hydrogen atoms. Thus: \[ m_{\mathrm H}=0.180\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0200\,\mathrm{g} \] \(\textbf{Calculate oxygen in the original compound:}\) The compound contains only carbon, hydrogen and oxygen. Hence: \[ m_{\mathrm O}=m_{\mathrm{sample}}-m_{\mathrm C}-m_{\mathrm H} \] Substituting: \[ m_{\mathrm O}=0.300-0.120-0.0200 \] \[ m_{\mathrm O}=0.160\,\mathrm{g} \] \(\textbf{Convert elemental masses into atomic amounts:}\) For carbon: \[ n_{\mathrm C}=\frac{0.120}{12}=0.0100\,\mathrm{mol} \] For hydrogen: \[ n_{\mathrm H}=\frac{0.0200}{1}=0.0200\,\mathrm{mol} \] For oxygen: \[ n_{\mathrm O}=\frac{0.160}{16}=0.0100\,\mathrm{mol} \] \(\textbf{Establish the empirical formula:}\) \[ C:H:O=0.0100:0.0200:0.0100 \] Dividing by \(0.0100\): \[ C:H:O=1:2:1 \] The empirical formula is therefore \(\mathrm{CH_2O}\). \(\textbf{Use the independently measured molar mass:}\) The empirical-formula mass is: \[ M_{\mathrm{emp}}=12+2+16 \] \[ M_{\mathrm{emp}}=30\,\mathrm{g\,mol^{-1}} \] The given molecular molar mass is: \[ M_{\mathrm{molecular}}=90\,\mathrm{g\,mol^{-1}} \] Thus: \[ n=\frac{90}{30}=3 \] \(\textbf{Construct and verify the molecular formula:}\) \[ (\mathrm{CH_2O})_3=\mathrm{C_3H_6O_3} \] Its calculated molar mass is: \[ M=3(12)+6(1)+3(16) \] \[ M=36+6+48 \] \[ M=90\,\mathrm{g\,mol^{-1}} \] This matches the independent molecular-mass result. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_3H_6O_3}} \]

Explanation: \(\textbf{Convert percentages into masses:}\) Choose a hypothetical \(100\,\mathrm{g}\) portion. The elemental masses are: \[ m_{\mathrm C}=63.16\,\mathrm{g} \] \[ m_{\mathrm H}=12.28\,\mathrm{g} \] \[ m_{\mathrm N}=24.56\,\mathrm{g} \] \(\textbf{Determine atomic mole proportions:}\) For carbon: \[ n_{\mathrm C}=\frac{63.16}{12} \] \[ n_{\mathrm C}=5.2633\,\mathrm{mol} \] For hydrogen: \[ n_{\mathrm H}=\frac{12.28}{1} \] \[ n_{\mathrm H}=12.28\,\mathrm{mol} \] For nitrogen: \[ n_{\mathrm N}=\frac{24.56}{14} \] \[ n_{\mathrm N}=1.7543\,\mathrm{mol} \] \(\textbf{Normalise the mole quantities:}\) Divide by the smallest amount, approximately \(1.7543\): \[ C:H:N = \frac{5.2633}{1.7543}: \frac{12.28}{1.7543}: \frac{1.7543}{1.7543} \] \[ C:H:N\approx3:7:1 \] The empirical formula is: \[ \mathrm{C_3H_7N} \] \(\textbf{Calculate the empirical-formula mass:}\) \[ M_{\mathrm{emp}}=3(12)+7(1)+14 \] \[ M_{\mathrm{emp}}=57\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine the molecular multiplier:}\) The measured molecular molar mass is \(114\,\mathrm{g\,mol^{-1}}\). Thus: \[ n=\frac{114}{57} \] \[ n=2 \] \(\textbf{Construct the molecular formula:}\) \[ (\mathrm{C_3H_7N})_2=\mathrm{C_6H_{14}N_2} \] \(\textbf{Verify the molecular molar mass:}\) \[ M=6(12)+14(1)+2(14) \] \[ M=72+14+28 \] \[ M=114\,\mathrm{g\,mol^{-1}} \] \(\textbf{Interpretation:}\) The empirical formula gives the correct mass percentages but has only half the independently measured molecular molar mass. Therefore, the molecular formula must contain two empirical-formula units. The nitrogen and hydrogen subscripts must both be multiplied by two along with the carbon subscript. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_6H_{14}N_2}} \]

Explanation: \(\textbf{Determine carbon mass:}\) The measured carbon dioxide mass 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 measured water mass is: \[ m_{\mathrm{H_2O}}=0.0540\,\mathrm{g} \] Thus: \[ m_{\mathrm H}=0.0540\times\frac{2}{18} \] \[ m_{\mathrm H}=0.00600\,\mathrm{g} \] \(\textbf{Determine chlorine mass:}\) The recovered silver chloride mass is: \[ m_{\mathrm{AgCl}}=0.287\,\mathrm{g} \] Its molar mass is: \[ M_{\mathrm{AgCl}}=108+35.5 \] \[ M_{\mathrm{AgCl}}=143.5\,\mathrm{g\,mol^{-1}} \] Therefore: \[ m_{\mathrm{Cl}}=0.287\times\frac{35.5}{143.5} \] \[ m_{\mathrm{Cl}}=0.0710\,\mathrm{g} \] \(\textbf{Determine oxygen by difference:}\) Only four elements are present. \[ m_{\mathrm O} = m_{\mathrm{sample}}-m_{\mathrm C}-m_{\mathrm H}-m_{\mathrm{Cl}} \] \[ m_{\mathrm O} = 0.157-0.0480-0.00600-0.0710 \] \[ m_{\mathrm O}=0.0320\,\mathrm{g} \] \(\textbf{Convert elemental masses into atomic amounts:}\) \[ n_{\mathrm C}=\frac{0.0480}{12}=0.00400\,\mathrm{mol} \] \[ n_{\mathrm H}=\frac{0.00600}{1}=0.00600\,\mathrm{mol} \] \[ n_{\mathrm{Cl}}=\frac{0.0710}{35.5}=0.00200\,\mathrm{mol} \] \[ n_{\mathrm O}=\frac{0.0320}{16}=0.00200\,\mathrm{mol} \] \(\textbf{Establish the empirical ratio:}\) \[ C:H:Cl:O=0.00400:0.00600:0.00200:0.00200 \] Divide every value by \(0.00200\): \[ C:H:Cl:O=2:3:1:1 \] The empirical formula is: \[ \mathrm{C_2H_3ClO} \] \(\textbf{Use the molecular molar mass:}\) The empirical-formula mass is: \[ M_{\mathrm{emp}}=2(12)+3(1)+35.5+16 \] \[ M_{\mathrm{emp}}=78.5\,\mathrm{g\,mol^{-1}} \] The molecular multiplication factor is: \[ n=\frac{157}{78.5} \] \[ n=2 \] Therefore: \[ (\mathrm{C_2H_3ClO})_2=\mathrm{C_4H_6Cl_2O_2} \] \(\textbf{Check the complete molecular mass:}\) \[ M=4(12)+6(1)+2(35.5)+2(16) \] \[ M=48+6+71+32 \] \[ M=157\,\mathrm{g\,mol^{-1}} \] The composition and molecular-mass constraints are both satisfied. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_4H_6Cl_2O_2}} \]

Explanation: Combustion analysis can provide accurate information about the masses of carbon and hydrogen originally present in an organic compound. Carbon is estimated through the carbon dioxide produced, while hydrogen is estimated through the water produced. Their percentages account for \(58.0\%\) of the original sample mass. Therefore, \(42.0\%\) remains unaccounted for by these two elements. However, this remaining mass does not necessarily consist entirely of oxygen. The question explicitly states that the possible presence of sulphur and halogens has not yet been excluded. These elements, if present, would contribute to the residual mass. Assigning all remaining mass to oxygen would therefore introduce an unsupported assumption. Complete combustion does not establish that a compound contains only carbon, hydrogen and oxygen. It establishes that the specified carbon and hydrogen measurements are complete under the analytical conditions. Oxygen by difference becomes justified only when the complete list of constituent elements is known or other relevant elements have been excluded. An empirical formula constructed using an unjustified oxygen percentage could be numerically neat but chemically incorrect. The appropriate next step is to establish the remaining elemental composition before converting all mass fractions into atomic mole ratios.

Explanation: \(\textbf{Check the possible molecular multipliers:}\) A molecular formula must be a positive integer multiple of the empirical formula. For the first proposed molar mass: \[ n_1=\frac{118}{59}=2 \] Its possible molecular formula is \(\mathrm{C_4H_{10}N_2O_2}\). For the second proposed molar mass: \[ n_2=\frac{177}{59}=3 \] Its possible molecular formula is \(\mathrm{C_6H_{15}N_3O_3}\). Both multipliers are integers, so the empirical formula by itself excludes neither candidate. \(\textbf{Convert the measured Dumas volume into nitrogen amount:}\) \[ n(\mathrm{N_2})=\frac{22.4\,\mathrm{mL}}{22400\,\mathrm{mL\,mol^{-1}}} \] \[ n(\mathrm{N_2})=0.00100\,\mathrm{mol} \] The amount of nitrogen atoms represented is: \[ n(\mathrm N)=2(0.00100)=0.00200\,\mathrm{mol} \] The corresponding nitrogen mass is: \[ m(\mathrm N)=0.00200\times14=0.0280\,\mathrm{g} \] \(\textbf{Compare with the empirical composition:}\) Each empirical-formula unit has one nitrogen atom and a mass of \(59\). The mass fraction of nitrogen is therefore \(14/59\). For the original \(0.118\,\mathrm{g}\) sample, the expected nitrogen mass is: \[ m(\mathrm N)=0.118\times\frac{14}{59}=0.0280\,\mathrm{g} \] Thus the Dumas result agrees with the empirical formula regardless of whether a molecule contains two or three empirical units. The larger proposed molecule would contain more nitrogen atoms per molecule, but the same sample mass would contain proportionally fewer molecules. These effects cancel in the total nitrogen recovered from a fixed sample mass. \(\textbf{Interpret the analytical limitation:}\) Dumas estimation determines nitrogen content, not the number of empirical-formula units per molecule. The two proposed molar masses cannot both be the actual mass of one pure substance, but neither is ruled out by the supplied elemental-composition data. An independent valid molecular-mass determination is needed to select between them.

Explanation: \(\textbf{Determine the atom amounts contributed by P:}\) The molecular formula of P is: \[ \mathrm{C_2H_4O_2} \] Each mole of P contains two moles of carbon atoms, four moles of hydrogen atoms and two moles of oxygen atoms. For \(0.120\,\mathrm{mol}\) of P: \[ n_{\mathrm{C,P}}=2(0.120)=0.240\,\mathrm{mol} \] \[ n_{\mathrm{H,P}}=4(0.120)=0.480\,\mathrm{mol} \] \[ n_{\mathrm{O,P}}=2(0.120)=0.240\,\mathrm{mol} \] \(\textbf{Determine the atom amounts contributed by Q:}\) The molecular formula of Q is: \[ \mathrm{C_3H_6O_3} \] For \(0.0800\,\mathrm{mol}\) of Q: \[ n_{\mathrm{C,Q}}=3(0.0800)=0.240\,\mathrm{mol} \] \[ n_{\mathrm{H,Q}}=6(0.0800)=0.480\,\mathrm{mol} \] \[ n_{\mathrm{O,Q}}=3(0.0800)=0.240\,\mathrm{mol} \] \(\textbf{Combine the elemental quantities:}\) Total carbon: \[ n_{\mathrm C}=0.240+0.240=0.480\,\mathrm{mol} \] Total hydrogen: \[ n_{\mathrm H}=0.480+0.480=0.960\,\mathrm{mol} \] Total oxygen: \[ n_{\mathrm O}=0.240+0.240=0.480\,\mathrm{mol} \] \(\textbf{Establish the simplest overall ratio:}\) \[ C:H:O=0.480:0.960:0.480 \] Dividing by \(0.480\): \[ C:H:O=1:2:1 \] The overall empirical atomic ratio is therefore represented by: \[ \mathrm{CH_2O} \] \(\textbf{Interpretation:}\) Both components individually have the same empirical formula, even though their molecular formulae differ. Consequently, combining them in any nonnegative proportions preserves their common elemental atomic ratio. This does not make the mixture a pure molecular compound with a unique molecular formula. The result is an overall empirical composition ratio for the mixture. The distinction between elemental-ratio information and molecular identity is essential when interpreting composition data from a sample that is not known to be a single pure substance. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{CH_2O}} \]

Explanation: \(\textbf{Express the molecular formula generally:}\) The empirical formula is: \[ \mathrm{CH_2O} \] A molecular formula consistent with this empirical composition must be a positive integer multiple: \[ (\mathrm{CH_2O})_n \] Therefore, the general molecular formula is: \[ \mathrm{C_nH_{2n}O_n} \] \(\textbf{Apply the oxygen-atom constraint:}\) The question specifies that each molecule contains exactly four oxygen atoms. In the general expression, the oxygen subscript is \(n\). Therefore: \[ n=4 \] \(\textbf{Determine the complete molecular formula:}\) Substitute \(n=4\): \[ \mathrm{C_nH_{2n}O_n} = \mathrm{C_4H_8O_4} \] \(\textbf{Check the atomic ratio:}\) The subscripts are: \[ C:H:O=4:8:4 \] Dividing by four: \[ C:H:O=1:2:1 \] This reproduces the empirical formula \(\mathrm{CH_2O}\). \(\textbf{Interpretation:}\) Molecular molar mass is one possible source of information about the empirical-formula multiplier, but it is not the only possible constraint. A reliable independent count of one constituent element per molecule can also determine the multiplier. Because the empirical formula contains one oxygen atom, the specified four oxygen atoms establish four empirical units per molecule. All subscripts must therefore be multiplied by four, not only the oxygen subscript. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_4H_8O_4}} \]

Explanation: \(\textbf{Calculate the empirical-formula mass:}\) The empirical formula is: \[ \mathrm{C_2H_4O} \] Its formula mass is: \[ M_{\mathrm{emp}}=2(12)+4(1)+16 \] \[ M_{\mathrm{emp}}=44\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determine the molecular multiplier:}\) The measured molecular molar mass is: \[ M_{\mathrm{molecular}}=88\,\mathrm{g\,mol^{-1}} \] Therefore: \[ n=\frac{88}{44} \] \[ n=2 \] \(\textbf{Apply the multiplier consistently:}\) Every empirical-formula unit contains: \[ 2C,\quad4H,\quad1O \] Two such units contain: \[ 4C,\quad8H,\quad2O \] Therefore: \[ (\mathrm{C_2H_4O})_2=\mathrm{C_4H_8O_2} \] \(\textbf{Test the student's proposed formula:}\) The proposed \(\mathrm{C_4H_8O}\) has molar mass: \[ M=4(12)+8(1)+16 \] \[ M=72\,\mathrm{g\,mol^{-1}} \] It does not match the required \(88\,\mathrm{g\,mol^{-1}}\). It also changes the oxygen-to-carbon ratio relative to the empirical formula. \(\textbf{Verify the correct formula:}\) \[ M_{\mathrm{C_4H_8O_2}}=48+8+32 \] \[ M_{\mathrm{C_4H_8O_2}}=88\,\mathrm{g\,mol^{-1}} \] \(\textbf{Interpretation:}\) A molecular formula must preserve all the atomic ratios expressed by the empirical formula. Multiplying selected subscripts while leaving others unchanged alters the elemental composition. The correct molecular formula is obtained by multiplying every empirical subscript by the same positive integer. \(\textbf{Final answer:}\) \[ \mathrm{C_4H_8O_2} \]

Explanation: \(\textbf{Calculate carbon from carbon dioxide:}\) The combustion produces: \[ m_{\mathrm{CO_2}}=0.440\,\mathrm{g} \] Therefore: \[ m_{\mathrm C}=0.440\times\frac{12}{44} \] \[ m_{\mathrm C}=0.120\,\mathrm{g} \] \(\textbf{Calculate hydrogen from water:}\) The water produced is: \[ m_{\mathrm{H_2O}}=0.225\,\mathrm{g} \] Hence: \[ m_{\mathrm H}=0.225\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0250\,\mathrm{g} \] \(\textbf{Determine nitrogen from Dumas data:}\) The dry nitrogen volume is: \[ V_{\mathrm{N_2}}=56.0\,\mathrm{mL} \] At STP: \[ n_{\mathrm{N_2}}=\frac{56.0}{22400} \] \[ n_{\mathrm{N_2}}=0.00250\,\mathrm{mol} \] The nitrogen mass is: \[ m_{\mathrm N}=0.00250\times28 \] \[ m_{\mathrm N}=0.0700\,\mathrm{g} \] \(\textbf{Verify the sample mass:}\) The compound contains only carbon, hydrogen and nitrogen. Adding the determined masses: \[ m_{\mathrm C}+m_{\mathrm H}+m_{\mathrm N} = 0.120+0.0250+0.0700 \] \[ m_{\mathrm{total}}=0.215\,\mathrm{g} \] This exactly matches the original sample mass. \(\textbf{Convert elemental masses into atomic amounts:}\) For carbon: \[ n_{\mathrm C}=\frac{0.120}{12} \] \[ n_{\mathrm C}=0.0100\,\mathrm{mol} \] For hydrogen: \[ n_{\mathrm H}=\frac{0.0250}{1} \] \[ n_{\mathrm H}=0.0250\,\mathrm{mol} \] For nitrogen atoms: \[ n_{\mathrm N}=\frac{0.0700}{14} \] \[ n_{\mathrm N}=0.00500\,\mathrm{mol} \] \(\textbf{Determine the empirical formula:}\) \[ C:H:N=0.0100:0.0250:0.00500 \] Divide by \(0.00500\): \[ C:H:N=2:5:1 \] The empirical formula is: \[ \mathrm{C_2H_5N} \] \(\textbf{Use the independent molecular molar mass:}\) The empirical-formula mass is: \[ M_{\mathrm{emp}}=2(12)+5(1)+14 \] \[ M_{\mathrm{emp}}=43\,\mathrm{g\,mol^{-1}} \] The measured molecular molar mass is \(86\,\mathrm{g\,mol^{-1}}\). Therefore: \[ n=\frac{86}{43} \] \[ n=2 \] \(\textbf{Construct the molecular formula:}\) \[ (\mathrm{C_2H_5N})_2=\mathrm{C_4H_{10}N_2} \] \(\textbf{Verify the molecular mass:}\) \[ M=4(12)+10(1)+2(14) \] \[ M=48+10+28 \] \[ M=86\,\mathrm{g\,mol^{-1}} \] \(\textbf{Interpretation:}\) The combustion products establish the carbon and hydrogen quantities, while Dumas estimation provides nitrogen. These measurements use equal sample masses and can therefore be combined directly. The original elemental masses account for the complete sample, supporting the stipulated three-element composition. The empirical ratio is then scaled using the independently measured molecular molar mass. A molecular formula cannot be selected from elemental composition alone without this additional constraint. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_4H_{10}N_2}} \]

Explanation: \(\textbf{Establish a common analytical basis:}\) The combustion and sulphur determinations use \(0.152\,\mathrm{g}\) samples, but the nitrogen determination uses only \(0.0760\,\mathrm{g}\). The nitrogen result must therefore be scaled before combining the elemental quantities. \(\textbf{Determine carbon from combustion:}\) \[ m_{\mathrm C}=m_{\mathrm{CO_2}}\frac{12}{44} \] \[ m_{\mathrm C}=0.0880\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0240\,\mathrm{g} \] \(\textbf{Determine hydrogen from water:}\) \[ m_{\mathrm H}=m_{\mathrm{H_2O}}\frac{2}{18} \] \[ m_{\mathrm H}=0.0720\times\frac{2}{18} \] \[ m_{\mathrm H}=0.00800\,\mathrm{g} \] \(\textbf{Calculate nitrogen in the smaller sample:}\) At STP: \[ n_{\mathrm{N_2}}=\frac{22.4}{22400} \] \[ n_{\mathrm{N_2}}=0.00100\,\mathrm{mol} \] The nitrogen mass represented by this gas is: \[ m_{\mathrm N}=0.00100\times28 \] \[ m_{\mathrm N}=0.0280\,\mathrm{g} \] \(\textbf{Normalise nitrogen to the common sample mass:}\) \[ f=\frac{0.152}{0.0760}=2 \] Therefore: \[ m_{\mathrm{N,common}}=0.0280\times2 \] \[ m_{\mathrm{N,common}}=0.0560\,\mathrm{g} \] \(\textbf{Determine sulphur from barium sulphate:}\) \[ M_{\mathrm{BaSO_4}}=137+32+4(16) \] \[ M_{\mathrm{BaSO_4}}=233\,\mathrm{g\,mol^{-1}} \] Hence: \[ m_{\mathrm S}=0.466\times\frac{32}{233} \] \[ m_{\mathrm S}=0.0640\,\mathrm{g} \] \(\textbf{Check elemental mass conservation:}\) \[ m_{\mathrm{total}}=0.0240+0.00800+0.0560+0.0640 \] \[ m_{\mathrm{total}}=0.152\,\mathrm{g} \] The four determined elemental masses account for the complete sample. \(\textbf{Determine atomic mole proportions:}\) \[ n_{\mathrm C}=\frac{0.0240}{12}=0.00200\,\mathrm{mol} \] \[ n_{\mathrm H}=\frac{0.00800}{1}=0.00800\,\mathrm{mol} \] \[ n_{\mathrm N}=\frac{0.0560}{14}=0.00400\,\mathrm{mol} \] \[ n_{\mathrm S}=\frac{0.0640}{32}=0.00200\,\mathrm{mol} \] Thus: \[ C:H:N:S=0.00200:0.00800:0.00400:0.00200 \] Dividing by \(0.00200\): \[ C:H:N:S=1:4:2:1 \] The empirical formula is \(\mathrm{CH_4N_2S}\). \(\textbf{Verify the molecular formula:}\) \[ M_{\mathrm{emp}}=12+4+2(14)+32 \] \[ M_{\mathrm{emp}}=76\,\mathrm{g\,mol^{-1}} \] The molecular molar mass is also \(76\,\mathrm{g\,mol^{-1}}\). Therefore: \[ n=\frac{76}{76}=1 \] The molecular formula equals the empirical formula. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{CH_4N_2S}} \]

Explanation: \(\textbf{Identify the required blank correction:}\) The calcium chloride mass gain contains water from two sources. One contribution arises from combustion of the organic sample. The other is established independently by the blank experiment. Therefore: \[ m_{\mathrm{H_2O,sample}}=m_{\mathrm{observed}}-m_{\mathrm{blank}} \] \[ m_{\mathrm{H_2O,sample}}=0.126-0.0360 \] \[ m_{\mathrm{H_2O,sample}}=0.0900\,\mathrm{g} \] \(\textbf{Calculate carbon from carbon dioxide:}\) \[ m_{\mathrm C}=0.220\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0600\,\mathrm{g} \] \(\textbf{Calculate hydrogen from corrected water:}\) \[ m_{\mathrm H}=0.0900\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0100\,\mathrm{g} \] \(\textbf{Determine oxygen by difference:}\) Because the compound contains only carbon, hydrogen and oxygen: \[ m_{\mathrm O}=m_{\mathrm{sample}}-m_{\mathrm C}-m_{\mathrm H} \] \[ m_{\mathrm O}=0.150-0.0600-0.0100 \] \[ m_{\mathrm O}=0.0800\,\mathrm{g} \] \(\textbf{Convert the elemental masses into atom amounts:}\) \[ n_{\mathrm C}=\frac{0.0600}{12}=0.00500\,\mathrm{mol} \] \[ n_{\mathrm H}=\frac{0.0100}{1}=0.0100\,\mathrm{mol} \] \[ n_{\mathrm O}=\frac{0.0800}{16}=0.00500\,\mathrm{mol} \] Therefore: \[ C:H:O=0.00500:0.0100:0.00500 \] \[ C:H:O=1:2:1 \] The empirical formula is \(\mathrm{CH_2O}\). \(\textbf{Use the independent molecular mass:}\) \[ M_{\mathrm{emp}}=12+2+16 \] \[ M_{\mathrm{emp}}=30\,\mathrm{g\,mol^{-1}} \] The multiplication factor is: \[ n=\frac{60}{30} \] \[ n=2 \] Therefore: \[ (\mathrm{CH_2O})_2=\mathrm{C_2H_4O_2} \] \(\textbf{Verify the corrected analytical result:}\) The calculated carbon, hydrogen and oxygen masses total: \[ 0.0600+0.0100+0.0800=0.150\,\mathrm{g} \] The molecular-formula mass is: \[ 2(12)+4(1)+2(16)=60\,\mathrm{g\,mol^{-1}} \] Both constraints are satisfied. \(\textbf{Interpretation:}\) The blank moisture must not be assigned to hydrogen originally present in the sample. Failing to remove it would inflate the apparent hydrogen amount and reduce the oxygen determined by difference. The blank correction is therefore essential before constructing the empirical formula. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_4O_2}} \]

Explanation: \(\textbf{Calculate carbon from combustion:}\) \[ m_{\mathrm C}=m_{\mathrm{CO_2}}\frac{12}{44} \] \[ m_{\mathrm C}=0.792\times\frac{12}{44} \] \[ m_{\mathrm C}=0.216\,\mathrm{g} \] \(\textbf{Calculate hydrogen from water:}\) \[ m_{\mathrm H}=m_{\mathrm{H_2O}}\frac{2}{18} \] \[ m_{\mathrm H}=0.135\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0150\,\mathrm{g} \] \(\textbf{Identify phosphorus in the gravimetric product:}\) The final phosphorus-estimation product is magnesium pyrophosphate. Its formula contains two phosphorus atoms: \[ \mathrm{Mg_2P_2O_7} \] 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: \[ m_{\mathrm P}=0.111\times\frac{62}{222} \] \[ m_{\mathrm P}=0.0310\,\mathrm{g} \] \(\textbf{Check the stipulated three-element composition:}\) \[ m_{\mathrm C}+m_{\mathrm H}+m_{\mathrm P} = 0.216+0.0150+0.0310 \] \[ m_{\mathrm{total}}=0.262\,\mathrm{g} \] The elemental masses account exactly for the original sample. \(\textbf{Determine atomic mole amounts:}\) \[ n_{\mathrm C}=\frac{0.216}{12}=0.0180\,\mathrm{mol} \] \[ n_{\mathrm H}=\frac{0.0150}{1}=0.0150\,\mathrm{mol} \] \[ n_{\mathrm P}=\frac{0.0310}{31}=0.00100\,\mathrm{mol} \] \(\textbf{Construct the empirical formula:}\) \[ C:H:P=0.0180:0.0150:0.00100 \] Dividing by \(0.00100\): \[ C:H:P=18:15:1 \] The empirical formula is: \[ \mathrm{C_{18}H_{15}P} \] \(\textbf{Use the molecular-mass constraint:}\) \[ M_{\mathrm{emp}}=18(12)+15(1)+31 \] \[ M_{\mathrm{emp}}=262\,\mathrm{g\,mol^{-1}} \] Thus: \[ n=\frac{262}{262}=1 \] The empirical formula is also the molecular formula. \(\textbf{Interpretation:}\) The phosphorus mass must be calculated from two phosphorus atoms per formula unit of magnesium pyrophosphate. The complete pyrophosphate mass cannot be assigned to phosphorus. The independent combustion and phosphorus measurements account for the entire sample, making an additional oxygen atom in the molecular formula inconsistent with the stated composition. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_{18}H_{15}P}} \]

Explanation: \(\textbf{Interpret relative vapour density:}\) Equal volumes of ideal gases at the same temperature and pressure contain equal numbers of molecules. Consequently, their mass ratio equals their molar-mass ratio. For the compound and hydrogen gas: \[ D_{\mathrm{H_2}}=\frac{M_{\mathrm{compound}}}{M_{\mathrm{H_2}}} \] \(\textbf{Determine the molecular molar mass:}\) The molar mass of hydrogen gas is: \[ M_{\mathrm{H_2}}=2\,\mathrm{g\,mol^{-1}} \] The given relative vapour density is: \[ D_{\mathrm{H_2}}=45.0 \] Therefore: \[ M_{\mathrm{compound}}=D_{\mathrm{H_2}}\times M_{\mathrm{H_2}} \] \[ M_{\mathrm{compound}}=45.0\times2 \] \[ M_{\mathrm{compound}}=90.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculate the empirical-formula mass:}\) The empirical formula is \(\mathrm{C_2H_5O}\). Therefore: \[ M_{\mathrm{emp}}=2(12)+5(1)+16 \] \[ M_{\mathrm{emp}}=45\,\mathrm{g\,mol^{-1}} \] \(\textbf{Find the molecular multiplier:}\) \[ n=\frac{M_{\mathrm{compound}}}{M_{\mathrm{emp}}} \] \[ n=\frac{90}{45} \] \[ n=2 \] \(\textbf{Construct the molecular formula:}\) All empirical subscripts must be doubled: \[ (\mathrm{C_2H_5O})_2=\mathrm{C_4H_{10}O_2} \] \(\textbf{Verify the result:}\) \[ M_{\mathrm{C_4H_{10}O_2}}=4(12)+10(1)+2(16) \] \[ M_{\mathrm{C_4H_{10}O_2}}=90\,\mathrm{g\,mol^{-1}} \] The molecular mass agrees with that obtained from relative vapour density. \(\textbf{Interpretation:}\) Relative vapour density supplies information about molecular mass rather than the simplest elemental ratio. The empirical formula independently supplies that ratio. Combining the two measurements gives the number of empirical-formula units in the molecule. The multiplier must be applied to carbon, hydrogen and oxygen equally. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_4H_{10}O_2}} \]

Explanation: \(\textbf{Distinguish elemental proportions from absolute atom counts:}\) P and Q share the empirical formula \(\mathrm{CH_2O}\). Therefore, they have identical carbon, hydrogen and oxygen mass percentages. However, each molecule of P contains two carbon atoms, whereas each molecule of Q contains three. \(\textbf{Use the known molecular amount:}\) The specimen contains: \[ n_{\mathrm{compound}}=0.0100\,\mathrm{mol} \] Complete combustion produces: \[ n_{\mathrm{CO_2}}=0.0300\,\mathrm{mol} \] Each carbon atom in the organic compound forms one carbon dioxide molecule. \(\textbf{Determine carbon atoms per molecule:}\) \[ N_{\mathrm{C\ per\ molecule}} = \frac{n_{\mathrm{CO_2}}}{n_{\mathrm{compound}}} \] \[ N_{\mathrm{C\ per\ molecule}} = \frac{0.0300}{0.0100} \] \[ N_{\mathrm{C\ per\ molecule}}=3 \] Therefore, the specimen must contain three carbon atoms per molecule. \(\textbf{Check the hydrogen information:}\) The measured water amount is: \[ n_{\mathrm{H_2O}}=0.0300\,\mathrm{mol} \] Each water molecule contains two hydrogen atoms. The amount of hydrogen atoms is: \[ n_{\mathrm H}=2n_{\mathrm{H_2O}} \] \[ n_{\mathrm H}=2(0.0300) \] \[ n_{\mathrm H}=0.0600\,\mathrm{mol} \] The hydrogen atoms per original molecule are: \[ N_{\mathrm{H\ per\ molecule}} = \frac{0.0600}{0.0100} \] \[ N_{\mathrm{H\ per\ molecule}}=6 \] \(\textbf{Compare the candidate formulae:}\) The results establish three carbon and six hydrogen atoms per molecule. P, \(\mathrm{C_2H_4O_2}\), does not satisfy these absolute atom counts. Q, \(\mathrm{C_3H_6O_3}\), does. \(\textbf{Interpretation:}\) Equal elemental percentages do not imply equal numbers of combustion-product molecules per mole of the original compound. The independently measured amount of original molecules provides the additional information needed to distinguish P from Q. The result does not depend on measuring their molecular masses directly.

Explanation: \(\textbf{Determine the individual molar masses:}\) For P: \[ M_{\mathrm P}=2(12)+4(1)+2(16) \] \[ M_{\mathrm P}=60\,\mathrm{g\,mol^{-1}} \] For Q: \[ M_{\mathrm Q}=3(12)+6(1)+3(16) \] \[ M_{\mathrm Q}=90\,\mathrm{g\,mol^{-1}} \] \(\textbf{Use the independently known total amount:}\) Let the mole amounts of P and Q be \(x\) and \(y\), respectively. The total mole amount gives: \[ x+y=0.00200 \] Therefore: \[ y=0.00200-x \] \(\textbf{Construct the mass balance:}\) The total mass is the sum of the masses of the two components. \[ 60x+90y=0.150 \] Substituting for \(y\): \[ 60x+90(0.00200-x)=0.150 \] \[ 60x+0.180-90x=0.150 \] \[ 30x=0.0300 \] \[ x=0.00100\,\mathrm{mol} \] \(\textbf{Determine the amount of Q:}\) \[ y=0.00200-0.00100 \] \[ y=0.00100\,\mathrm{mol} \] \(\textbf{Calculate the mole ratio:}\) \[ P:Q=x:y \] \[ P:Q=0.00100:0.00100 \] \[ P:Q=1:1 \] \(\textbf{Check the masses:}\) The mass of P is: \[ m_{\mathrm P}=0.00100\times60 \] \[ m_{\mathrm P}=0.0600\,\mathrm{g} \] The mass of Q is: \[ m_{\mathrm Q}=0.00100\times90 \] \[ m_{\mathrm Q}=0.0900\,\mathrm{g} \] Their sum is: \[ 0.0600+0.0900=0.150\,\mathrm{g} \] \(\textbf{Interpretation:}\) The two compounds have identical empirical compositions. Consequently, elemental percentage analysis alone cannot determine their mixing ratio. The additional information about the total number of moles provides an independent constraint. Combined with total mass, it permits the individual mole amounts to be determined. Equal mole amounts do not mean equal component masses because the compounds have different molar masses. \(\textbf{Final answer:}\) \[ \boxed{P:Q=1:1} \]

Explanation: \(\textbf{Determine carbon in the combustion portion:}\) The combustion sample weighs \(0.150\,\mathrm{g}\). Its carbon dioxide product is: \[ m_{\mathrm{CO_2}}=0.176\,\mathrm{g} \] Therefore: \[ m_{\mathrm C}=0.176\times\frac{12}{44} \] \[ m_{\mathrm C}=0.0480\,\mathrm{g} \] \(\textbf{Determine hydrogen in the same portion:}\) \[ m_{\mathrm H}=0.0900\times\frac{2}{18} \] \[ m_{\mathrm H}=0.0100\,\mathrm{g} \] \(\textbf{Determine ammonia from the separate Kjeldahl portion:}\) The initial hydrochloric acid amount is: \[ n_{\mathrm{HCl,initial}} = \frac{60.0\times0.100}{1000} \] \[ n_{\mathrm{HCl,initial}}=0.00600\,\mathrm{mol} \] The unused acid amount is obtained from the back-titration: \[ n_{\mathrm{HCl,unused}} = \frac{20.0\times0.100}{1000} \] \[ n_{\mathrm{HCl,unused}}=0.00200\,\mathrm{mol} \] Thus: \[ n_{\mathrm{HCl,consumed}}=0.00600-0.00200 \] \[ n_{\mathrm{HCl,consumed}}=0.00400\,\mathrm{mol} \] The reaction is: \[ \mathrm{NH_3+HCl\rightarrow NH_4Cl} \] Therefore: \[ n_{\mathrm{NH_3}}=0.00400\,\mathrm{mol} \] \(\textbf{Convert ammonia into nitrogen mass:}\) Each ammonia molecule contains one nitrogen atom. \[ m_{\mathrm{N,larger\ portion}}=0.00400\times14 \] \[ m_{\mathrm{N,larger\ portion}}=0.0560\,\mathrm{g} \] \(\textbf{Normalise nitrogen to the combustion sample mass:}\) The Kjeldahl sample is \(0.300\,\mathrm{g}\), twice the combustion sample mass. Therefore: \[ m_{\mathrm{N,common}}=0.0560\times\frac{0.150}{0.300} \] \[ m_{\mathrm{N,common}}=0.0280\,\mathrm{g} \] \(\textbf{Determine oxygen by difference:}\) The compound contains only carbon, hydrogen, nitrogen and oxygen. Thus: \[ m_{\mathrm O}=0.150-0.0480-0.0100-0.0280 \] \[ m_{\mathrm O}=0.0640\,\mathrm{g} \] \(\textbf{Convert the elemental masses into atomic proportions:}\) \[ n_{\mathrm C}=\frac{0.0480}{12}=0.00400\,\mathrm{mol} \] \[ n_{\mathrm H}=\frac{0.0100}{1}=0.0100\,\mathrm{mol} \] \[ n_{\mathrm N}=\frac{0.0280}{14}=0.00200\,\mathrm{mol} \] \[ n_{\mathrm O}=\frac{0.0640}{16}=0.00400\,\mathrm{mol} \] Therefore: \[ C:H:N:O=0.00400:0.0100:0.00200:0.00400 \] Dividing by \(0.00200\): \[ C:H:N:O=2:5:1:2 \] The empirical formula is \(\mathrm{C_2H_5NO_2}\). \(\textbf{Apply the molecular-mass condition:}\) \[ M_{\mathrm{emp}}=2(12)+5+14+2(16) \] \[ M_{\mathrm{emp}}=75\,\mathrm{g\,mol^{-1}} \] Thus: \[ n=\frac{75}{75}=1 \] The molecular formula is the same as the empirical formula. \(\textbf{Interpretation:}\) The combustion and Kjeldahl measurements use different sample masses. Their elemental results cannot be combined until they refer to a common mass of the original compound. The nitrogen value must be obtained from the acid consumed by ammonia, not from all the acid originally added. Only after these steps can oxygen be determined by difference and the atomic ratio constructed. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_5NO_2}} \]