Some Basic Concepts in Chemistry MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: \(\textbf{Determining the empirical-formula mass:}\) The empirical formula is \(\mathrm{CH_2O}\). Its formula mass is \[ M_{\mathrm{empirical}}=12+2(1)+16 \] \[ M_{\mathrm{empirical}}=30\,\mathrm{u} \] \(\textbf{Using the molecular-formula relationship:}\) For a molecular compound, \[ \text{Molecular formula}=(\text{Empirical formula})_k \] The multiplier is \[ k=\frac{M_{\mathrm{molecular}}}{M_{\mathrm{empirical}}} \] For a valid molecular formula, \(k\) must be a positive whole number. \(\textbf{Checking substance P:}\) \[ k_P=\frac{60}{30} \] \[ k_P=2 \] P is compatible with the molecular formula \[ \mathrm{C_2H_4O_2} \] \(\textbf{Checking substance Q:}\) \[ k_Q=\frac{90}{30} \] \[ k_Q=3 \] Q is compatible with \[ \mathrm{C_3H_6O_3} \] \(\textbf{Checking substance R:}\) \[ k_R=\frac{105}{30} \] \[ k_R=3.5 \] A multiplier of 3.5 would produce nonintegral atomic subscripts if applied to the empirical formula. Therefore, R is incompatible with the reported empirical formula and exact molar mass. \(\textbf{Checking substance S:}\) \[ k_S=\frac{150}{30} \] \[ k_S=5 \] S is compatible with \[ \mathrm{C_5H_{10}O_5} \] \(\textbf{Interpreting the consistency requirement:}\) There is no general requirement that the molecular multiplier must be two or three. Any positive whole-number multiplier can be considered arithmetically compatible, although chemical identity and stability require additional evidence. The data for R fail a necessary numerical condition for a molecular formula. The conclusion is therefore about compatibility of the reported measurements, not about proving that every compatible formula corresponds to a distinct isolable substance. P, Q and S pass the required multiplier test, whereas R does not.

Explanation: \(\textbf{Recovering the empirical atomic ratio:}\) Consider a hypothetical 100 g sample. It contains approximately 85.71 g of carbon and 14.29 g of hydrogen. The corresponding atomic amounts are \[ n_C=\frac{85.71}{12} \] \[ n_C\approx7.1425\,\mathrm{mol} \] \[ n_H=\frac{14.29}{1} \] \[ n_H=14.29\,\mathrm{mol} \] Dividing by the smaller amount, \[ C:H\approx1:2 \] The empirical formula is \[ \mathrm{CH_2} \] \(\textbf{Determining the empirical-formula mass:}\) \[ M_{\mathrm{CH_2}}=12+2(1) \] \[ M_{\mathrm{CH_2}}=14\,\mathrm{u} \] \(\textbf{Using the gas-density measurement:}\) Density equals mass divided by volume. For one mole of the gas, \[ d=\frac{M}{V_m} \] Therefore, \[ M=dV_m \] Substituting the measured density and specified molar volume, \[ M=(1.875\,\mathrm{g\,L^{-1}})(22.4\,\mathrm{L\,mol^{-1}}) \] \[ M=42.0\,\mathrm{g\,mol^{-1}} \] \(\textbf{Determining the molecular multiplier:}\) \[ k=\frac{M_{\mathrm{molecular}}}{M_{\mathrm{empirical}}} \] \[ k=\frac{42}{14} \] \[ k=3 \] Multiplying both empirical subscripts by three, \[ (\mathrm{CH_2})_3=\mathrm{C_3H_6} \] \(\textbf{Independent molar-mass verification:}\) \[ M_{\mathrm{C_3H_6}}=3(12)+6(1) \] \[ M_{\mathrm{C_3H_6}}=42\,\mathrm{g\,mol^{-1}} \] The density predicted under the stated conditions is \[ d=\frac{42}{22.4} \] \[ d=1.875\,\mathrm{g\,L^{-1}} \] The calculated density matches the measured value. \(\textbf{Checking elemental percentages:}\) The carbon percentage is \[ \%C=\frac{36}{42}\times100 \] \[ \%C\approx85.71\% \] The hydrogen percentage is \[ \%H=\frac{6}{42}\times100 \] \[ \%H\approx14.29\% \] \(\textbf{Chemical interpretation:}\) Elemental composition establishes the empirical formula. The independent gaseous density measurement establishes the molar mass under the specified gas conditions. Both are needed to determine the molecular multiplier. The density is not itself an empirical-formula measurement, and percentage composition alone cannot distinguish \(\mathrm{CH_2}\) from its possible molecular multiples. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_3H_6}} \]

Explanation: \(\textbf{Selecting a convenient molecular basis:}\) Consider exactly one mole of the gaseous mixture. Let the mole fraction of carbon dioxide be \(x\). The mixture then contains \[ n_{\mathrm{CO_2}}=x\,\mathrm{mol} \] \[ n_{\mathrm{CO}}=(1-x)\,\mathrm{mol} \] \(\textbf{Constructing the total mass:}\) The molar masses are \[ M_{\mathrm{CO}}=28\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{CO_2}}=44\,\mathrm{g\,mol^{-1}} \] Therefore, \[ m_{\text{total}}=28(1-x)+44x \] \[ m_{\text{total}}=28+16x \] \(\textbf{Constructing the oxygen mass:}\) Each mole of CO contains 16 g of oxygen. Each mole of \(\mathrm{CO_2}\) contains 32 g of oxygen. Hence, \[ m_O=16(1-x)+32x \] \[ m_O=16+16x \] \(\textbf{Applying the measured mass percentage:}\) Oxygen contributes 64% of the mixture mass. Therefore, \[ \frac{m_O}{m_{\text{total}}}=0.64 \] Substituting, \[ \frac{16+16x}{28+16x}=0.64 \] Cross-multiplying, \[ 16+16x=0.64(28+16x) \] \[ 16+16x=17.92+10.24x \] \[ 5.76x=1.92 \] \[ x=\frac13 \] Thus, \[ X_{\mathrm{CO_2}}=\frac13 \] \(\textbf{Converting molecular composition into atomic composition:}\) Each CO molecule contains one carbon atom and one oxygen atom. Each \(\mathrm{CO_2}\) molecule contains one carbon atom and two oxygen atoms. On the one-mole mixture basis, the amount of carbon atoms is \[ n_C=(1-x)+x \] \[ n_C=1\,\mathrm{mol} \] The amount of oxygen atoms is \[ n_O=(1-x)+2x \] \[ n_O=1+x \] Substituting \(x=\frac13\), \[ n_O=\frac43\,\mathrm{mol} \] The total constituent-atom amount is \[ n_{\text{atoms,total}}=1+\frac43 \] \[ n_{\text{atoms,total}}=\frac73\,\mathrm{mol} \] \(\textbf{Finding the oxygen atomic fraction:}\) \[ f_{O,\text{atoms}}=\frac{4/3}{7/3} \] \[ f_{O,\text{atoms}}=\frac47 \] \(\textbf{Independent mass-percentage check:}\) The mixture contains \(\frac23\,\mathrm{mol}\) CO and \(\frac13\,\mathrm{mol}\) \(\mathrm{CO_2}\). Its total mass is \[ m_{\text{total}}=\frac23(28)+\frac13(44) \] \[ m_{\text{total}}=\frac{100}{3}\,\text{g} \] Its oxygen mass is \[ m_O=\frac23(16)+\frac13(32) \] \[ m_O=\frac{64}{3}\,\text{g} \] Therefore, \[ \frac{m_O}{m_{\text{total}}}=\frac{64}{100}=0.64 \] The measured oxygen percentage is reproduced. \(\textbf{Chemical interpretation:}\) Oxygen mass percentage is not the same as the fraction of atoms that are oxygen. The mass percentage must first be converted into molecular composition using the different molar masses of CO and \(\mathrm{CO_2}\). The molecular composition can then be converted into constituent-atom composition using the molecular formulae. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{CO_2}}=\frac13,\quad f_{O,\text{atoms}}=\frac47} \]

Explanation: \(\textbf{Interpreting the carbon dioxide measurement:}\) Every carbon dioxide molecule contains one carbon atom. The collected carbon dioxide amount is \[ n_{\mathrm{CO_2}}=\frac{8.8}{44} \] \[ n_{\mathrm{CO_2}}=0.20\,\mathrm{mol} \] This product therefore contains \[ m_{C,\mathrm{CO_2}}=0.20(12) \] \[ m_{C,\mathrm{CO_2}}=2.4\,\text{g} \] \(\textbf{Recognising incomplete carbon recovery:}\) The question states that additional carbon formed uncollected carbon monoxide. Therefore, the 2.4 g of carbon found in carbon dioxide represents only part of the original carbon inventory. The total original carbon mass must be greater than 2.4 g if a nonzero amount of carbon monoxide was formed. However, the amount of extra carbon is unknown. \(\textbf{Recovering the original hydrogen:}\) All original hydrogen is confirmed to be present in the collected water. The water amount is \[ n_{\mathrm{H_2O}}=\frac{5.4}{18} \] \[ n_{\mathrm{H_2O}}=0.30\,\mathrm{mol} \] Each water molecule contains two hydrogen atoms. Thus, \[ n_H=2(0.30) \] \[ n_H=0.60\,\mathrm{mol} \] The original hydrogen mass is therefore \[ m_H=0.60(1) \] \[ m_H=0.60\,\text{g} \] \(\textbf{Assessing the oxygen mass by difference:}\) The original sample contains only carbon, hydrogen and oxygen. Therefore, \[ m_O=9.0-m_C-m_H \] Substituting the known hydrogen mass, \[ m_O=8.4-m_C \] Because the complete original carbon mass is unknown, the oxygen mass is also unknown. \(\textbf{Why a unique empirical formula cannot be obtained:}\) An empirical formula requires the relative amounts of all three constituent elements. The hydrogen amount is known exactly. The carbon mass recovered in carbon dioxide provides only a lower bound for the original carbon mass. The oxygen mass cannot be determined independently from the stated measurements. Different possible amounts of uncollected carbon monoxide would lead to different original carbon and oxygen amounts. Consequently, a unique simplest whole-number atomic ratio cannot be established. The key scientific limitation is incomplete elemental recovery, not failure of the law of conservation of mass. The correct inference is that the original compound contained at least 2.4 g of carbon, exactly 0.6 g of hydrogen, and an undetermined amount of oxygen.

Explanation: \(\textbf{Establishing the composition of urea:}\) The urea formula contains one carbon atom, one oxygen atom, two nitrogen atoms and four hydrogen atoms. Its molar mass is \[ M_{\mathrm{urea}}=12+16+2(14)+4(1) \] \[ M_{\mathrm{urea}}=60\,\mathrm{g\,mol^{-1}} \] Nitrogen contributes 28 units of the formula mass. Therefore, \[ w_{N,\mathrm{urea}}=\frac{28}{60}=\frac7{15} \] \(\textbf{Establishing the composition of ammonium nitrate:}\) Ammonium nitrate contains two nitrogen atoms, four hydrogen atoms and three oxygen atoms per formula unit. \[ M_{\mathrm{NH_4NO_3}}=2(14)+4(1)+3(16) \] \[ M_{\mathrm{NH_4NO_3}}=80\,\mathrm{g\,mol^{-1}} \] Its nitrogen mass fraction is \[ w_{N,\mathrm{NH_4NO_3}}=\frac{28}{80} \] \[ w_{N,\mathrm{NH_4NO_3}}=\frac7{20} \] \(\textbf{Representing the mixture composition:}\) Let \(x\) grams of the original mixture be urea. The ammonium nitrate mass is then \[ m_{\mathrm{NH_4NO_3}}=80-x \] \(\textbf{Using the measured total nitrogen mass:}\) Both components contribute nitrogen. Therefore, \[ \frac7{15}x+\frac7{20}(80-x)=31.5 \] Multiplying throughout by 60, \[ 28x+21(80-x)=1890 \] \[ 28x+1680-21x=1890 \] \[ 7x=210 \] \[ x=30.0\,\text{g} \] Therefore, the mixture contains 30.0 g urea and 50.0 g ammonium nitrate. \(\textbf{Determining the urea mass percentage:}\) \[ \%\mathrm{urea}=\frac{30.0}{80.0}\times100 \] \[ \%\mathrm{urea}=37.5\% \] \(\textbf{Calculating the oxygen contribution from urea:}\) One mole of urea contains one mole of oxygen atoms. Its oxygen mass fraction is \[ w_{O,\mathrm{urea}}=\frac{16}{60} \] Thus, \[ m_{O,\mathrm{urea}}=30.0\left(\frac{16}{60}\right) \] \[ m_{O,\mathrm{urea}}=8.0\,\text{g} \] \(\textbf{Calculating oxygen from ammonium nitrate:}\) Its three oxygen atoms contribute 48 units of formula mass. \[ w_{O,\mathrm{NH_4NO_3}}=\frac{48}{80} \] Therefore, \[ m_{O,\mathrm{NH_4NO_3}}=50.0\left(\frac{48}{80}\right) \] \[ m_{O,\mathrm{NH_4NO_3}}=30.0\,\text{g} \] \(\textbf{Total oxygen percentage:}\) \[ m_{O,\mathrm{total}}=8.0+30.0 \] \[ m_{O,\mathrm{total}}=38.0\,\text{g} \] \[ \%O_{\mathrm{mixture}}=\frac{38.0}{80.0}\times100 \] \[ \%O_{\mathrm{mixture}}=47.5\% \] \(\textbf{Independent nitrogen check:}\) Urea contributes \[ m_{N,\mathrm{urea}}=30.0\left(\frac{28}{60}\right)=14.0\,\text{g} \] Ammonium nitrate contributes \[ m_{N,\mathrm{NH_4NO_3}}=50.0\left(\frac{28}{80}\right)=17.5\,\text{g} \] Their sum is \[ 14.0+17.5=31.5\,\text{g} \] This agrees with the measured nitrogen mass. \(\textbf{Chemical interpretation:}\) The overall nitrogen percentage cannot be assigned directly to either pure component. The measured nitrogen mass must first be partitioned between the two components using their different fixed nitrogen mass fractions. Only after determining the mixture composition can its oxygen percentage be calculated reliably. \(\textbf{Final answer:}\) \[ \boxed{\%\mathrm{urea}=37.5\%,\quad\%O_{\mathrm{mixture}}=47.5\%} \]

Explanation: \(\textbf{Recovering oxygen mass from reduction:}\) The original substance is a pure oxide containing only iron and oxygen. The reduction removes all oxygen and leaves elemental iron. Therefore, the difference between original oxide mass and recovered iron mass equals the original oxygen mass. \[ m_O=m_{\text{oxide}}-m_{\mathrm{Fe}} \] Substituting, \[ m_O=23.2-16.8 \] \[ m_O=6.4\,\text{g} \] \(\textbf{Converting iron mass into atomic amount:}\) \[ n_{\mathrm{Fe}}=\frac{16.8\,\text{g}}{56\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{Fe}}=0.300\,\mathrm{mol} \] \(\textbf{Converting oxygen mass into atomic amount:}\) \[ n_O=\frac{6.4\,\text{g}}{16\,\mathrm{g\,mol^{-1}}} \] \[ n_O=0.400\,\mathrm{mol} \] \(\textbf{Determining the simplest whole-number ratio:}\) \[ n_{\mathrm{Fe}}:n_O=0.300:0.400 \] Dividing by 0.100, \[ n_{\mathrm{Fe}}:n_O=3:4 \] Therefore, the empirical formula is \[ \mathrm{Fe_3O_4} \] \(\textbf{Calculating the measured oxygen percentage:}\) \[ \%O=\frac{m_O}{m_{\text{oxide}}}\times100 \] \[ \%O=\frac{6.4}{23.2}\times100 \] \[ \%O\approx27.59\% \] \(\textbf{Formula-mass verification:}\) The formula mass of \(\mathrm{Fe_3O_4}\) is \[ M_{\mathrm{Fe_3O_4}}=3(56)+4(16) \] \[ M_{\mathrm{Fe_3O_4}}=168+64 \] \[ M_{\mathrm{Fe_3O_4}}=232\,\mathrm{u} \] The oxygen mass percentage predicted by this formula is \[ \%O_{\mathrm{formula}}=\frac{64}{232}\times100 \] \[ \%O_{\mathrm{formula}}\approx27.59\% \] \(\textbf{Checking the sample amount:}\) The oxide sample contains \[ n_{\mathrm{Fe_3O_4}}=\frac{23.2}{232} \] \[ n_{\mathrm{Fe_3O_4}}=0.100\,\mathrm{mol} \] The corresponding iron amount is \[ n_{\mathrm{Fe}}=3(0.100)=0.300\,\mathrm{mol} \] This gives an iron mass of \[ m_{\mathrm{Fe}}=0.300(56) \] \[ m_{\mathrm{Fe}}=16.8\,\text{g} \] The calculated iron mass reproduces the experimental result. \(\textbf{Chemical interpretation:}\) The oxide formula is inferred from the masses of its constituent elements after complete separation. The ratio of iron mass to oxygen mass is not itself the empirical atomic ratio. Conversion to atomic amounts is essential before determining the formula subscripts. The independently calculated oxygen percentage verifies the resulting empirical formula. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{Fe_3O_4},\quad\%O=27.59\%} \]

Explanation: \(\textbf{Determining the molecular molar mass:}\) The sample weighs 7.40 g and contains \(0.100\,\mathrm{mol}\) of molecules. Therefore, \[ M_X=\frac{m_X}{n_X} \] Substituting, \[ M_X=\frac{7.40\,\text{g}}{0.100\,\mathrm{mol}} \] \[ M_X=74\,\mathrm{g\,mol^{-1}} \] \(\textbf{Recovering the carbon subscript:}\) The sample contains \(0.300\,\mathrm{mol}\) of carbon atoms and \(0.100\,\mathrm{mol}\) of molecules. Therefore, the number of carbon atoms per molecule is \[ a=\frac{n_{\mathrm{C\ atoms}}}{n_{\mathrm{molecules}}} \] \[ a=\frac{0.300}{0.100} \] \[ a=3 \] \(\textbf{Recovering the hydrogen subscript:}\) Similarly, \[ b=\frac{n_{\mathrm{H\ atoms}}}{n_{\mathrm{molecules}}} \] \[ b=\frac{0.600}{0.100} \] \[ b=6 \] Thus, the molecular formula has the preliminary form \[ \mathrm{C_3H_6O_c} \] where \(c\) is the unknown oxygen subscript. \(\textbf{Using molecular mass to determine oxygen:}\) The carbon contribution to the molecular molar mass is \[ M_C=3(12)=36\,\mathrm{g\,mol^{-1}} \] The hydrogen contribution is \[ M_H=6(1)=6\,\mathrm{g\,mol^{-1}} \] The remaining molar-mass contribution must come from oxygen. \[ M_O=74-(36+6) \] \[ M_O=32\,\mathrm{g\,mol^{-1}} \] Each oxygen atom contributes \(16\,\mathrm{g\,mol^{-1}}\). Therefore, \[ c=\frac{32}{16} \] \[ c=2 \] The molecular formula is \[ \mathrm{C_3H_6O_2} \] \(\textbf{Testing empirical-formula reducibility:}\) The three molecular subscripts are 3, 6 and 2. Their greatest common integer divisor is \[ \gcd(3,6,2)=1 \] Therefore, they cannot all be divided by a common integer greater than one. The molecular formula is already in its simplest atomic ratio. Hence, the empirical formula is also \[ \mathrm{C_3H_6O_2} \] \(\textbf{Independent mass verification:}\) The calculated molecular molar mass is \[ 3(12)+6(1)+2(16)=74\,\mathrm{g\,mol^{-1}} \] A \(0.100\,\mathrm{mol}\) sample consequently weighs \[ m=0.100(74) \] \[ m=7.40\,\text{g} \] The predicted carbon-atom and hydrogen-atom amounts are \[ n_C=3(0.100)=0.300\,\mathrm{mol} \] \[ n_H=6(0.100)=0.600\,\mathrm{mol} \] All three measurements are reproduced. \(\textbf{Chemical interpretation:}\) The carbon and hydrogen atomic amounts independently determine two molecular subscripts. The sample mass and molecular amount determine molar mass, from which the remaining oxygen subscript is inferred. The final empirical-formula decision requires a separate whole-number divisibility check. A molecular formula does not have to be a multiple greater than one of its empirical formula. \(\textbf{Final answer:}\) \[ \boxed{\text{Molecular formula}=\text{Empirical formula}=\mathrm{C_3H_6O_2}} \]

Explanation: \(\textbf{Recovering carbon from carbon dioxide:}\) Complete recovery establishes that every original carbon atom is represented in the collected carbon dioxide. \[ n_{\mathrm{CO_2}}=\frac{8.80\,\mathrm{g}}{44\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{CO_2}}=0.200\,\mathrm{mol} \] Each carbon dioxide molecule contains one carbon atom. Therefore, \[ n_C=0.200\,\mathrm{mol} \] \(\textbf{Recovering hydrogen from water:}\) The collected water contains all the hydrogen originally present. \[ n_{\mathrm{H_2O}}=\frac{6.30\,\mathrm{g}}{18\,\mathrm{g\,mol^{-1}}} \] \[ n_{\mathrm{H_2O}}=0.350\,\mathrm{mol} \] Since each water molecule contains two hydrogen atoms, \[ n_H=2(0.350) \] \[ n_H=0.700\,\mathrm{mol} \] \(\textbf{Recovering nitrogen from dinitrogen:}\) The molar mass of \(\mathrm{N_2}\) is \(28\,\mathrm{g\,mol^{-1}}\). \[ n_{\mathrm{N_2}}=\frac{1.40}{28} \] \[ n_{\mathrm{N_2}}=0.050\,\mathrm{mol} \] Each nitrogen molecule contains two nitrogen atoms. \[ n_N=2(0.050) \] \[ n_N=0.100\,\mathrm{mol} \] \(\textbf{Constructing the empirical atomic ratio:}\) \[ C:H:N=0.200:0.700:0.100 \] Dividing throughout by 0.100, \[ C:H:N=2:7:1 \] The empirical formula is \[ \mathrm{C_2H_7N} \] \(\textbf{Checking the molecular multiplier:}\) The empirical-formula mass is \[ M_{\mathrm{empirical}}=2(12)+7(1)+14 \] \[ M_{\mathrm{empirical}}=45\,\mathrm{u} \] The measured molecular molar mass is \(45\,\mathrm{g\,mol^{-1}}\). Therefore, \[ k=\frac{45}{45}=1 \] The molecular formula is identical to the empirical formula. \(\textbf{Independent sample-mass check:}\) The recovered elemental masses are \[ m_C=0.200(12)=2.40\,\mathrm{g} \] \[ m_H=0.700(1)=0.70\,\mathrm{g} \] \[ m_N=0.100(14)=1.40\,\mathrm{g} \] Their sum is \[ 2.40+0.70+1.40=4.50\,\mathrm{g} \] This reproduces the original compound mass. The oxygen present in the collected carbon dioxide and water need not have originated in the compound because oxygen was supplied during analysis. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_7N}} \]

Explanation: \(\textbf{Accounting for both carbon-containing products:}\) The carbon inventory must include carbon monoxide and carbon dioxide. Although carbon dioxide is not the only carbon-containing product, the problem specifies quantitative collection of both products. Thus, complete carbon recovery is established. \(\textbf{Carbon contributed by carbon dioxide:}\) \[ n_{\mathrm{CO_2}}=\frac{13.20}{44} \] \[ n_{\mathrm{CO_2}}=0.300\,\mathrm{mol} \] This represents \(0.300\,\mathrm{mol}\) of carbon atoms. \(\textbf{Carbon contributed by carbon monoxide:}\) \[ n_{\mathrm{CO}}=\frac{2.80}{28} \] \[ n_{\mathrm{CO}}=0.100\,\mathrm{mol} \] This represents another \(0.100\,\mathrm{mol}\) of carbon atoms. Therefore, \[ n_C=0.300+0.100 \] \[ n_C=0.400\,\mathrm{mol} \] The original carbon mass is \[ m_C=0.400(12)=4.80\,\mathrm{g} \] \(\textbf{Recovering hydrogen from water:}\) The collected water contains all the original hydrogen. \[ n_{\mathrm{H_2O}}=\frac{7.20}{18} \] \[ n_{\mathrm{H_2O}}=0.400\,\mathrm{mol} \] Therefore, \[ n_H=2(0.400) \] \[ n_H=0.800\,\mathrm{mol} \] The corresponding hydrogen mass is \[ m_H=0.800(1)=0.800\,\mathrm{g} \] \(\textbf{Determining oxygen originally present:}\) The original compound contains only carbon, hydrogen and oxygen. Its oxygen mass is obtained by difference. \[ m_O=8.80-(4.80+0.80) \] \[ m_O=3.20\,\mathrm{g} \] Therefore, \[ n_O=\frac{3.20}{16} \] \[ n_O=0.200\,\mathrm{mol} \] \(\textbf{Constructing the empirical formula:}\) \[ C:H:O=0.400:0.800:0.200 \] Dividing by 0.200, \[ C:H:O=2:4:1 \] Thus, the empirical formula is \[ \mathrm{C_2H_4O} \] \(\textbf{Mass-balance verification:}\) The total collected product mass is \[ 13.20+2.80+7.20=23.20\,\mathrm{g} \] This exceeds the original 8.80 g because oxygen is supplied during combustion. The additional oxygen mass is \[ 23.20-8.80=14.40\,\mathrm{g} \] The result is compatible with oxygen incorporation from the surroundings. It would be incorrect to assign all oxygen in the collected products to the original compound. It would also be incorrect to ignore carbon recovered as CO merely because \(\mathrm{CO_2}\) was collected. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_4O}} \]

Explanation: \(\textbf{Separating the known component from the elemental analysis:}\) The measured elemental masses belong to the complete mixture. They must not be assigned directly to compound X. The carbon and oxygen contributions of the known carbon dioxide must first be removed. \(\textbf{Amount of known carbon dioxide:}\) \[ n_{\mathrm{CO_2}}=\frac{4.4}{44} \] \[ n_{\mathrm{CO_2}}=0.100\,\mathrm{mol} \] Its carbon mass is \[ m_{C,\mathrm{CO_2}}=0.100(12) \] \[ m_{C,\mathrm{CO_2}}=1.20\,\mathrm{g} \] Its oxygen mass is \[ m_{O,\mathrm{CO_2}}=0.100(32) \] \[ m_{O,\mathrm{CO_2}}=3.20\,\mathrm{g} \] Carbon dioxide contributes no hydrogen. \(\textbf{Recovering the composition of X:}\) Subtracting the known contributions from the complete elemental inventory, \[ m_{C,X}=6.0-1.2 \] \[ m_{C,X}=4.8\,\mathrm{g} \] \[ m_{H,X}=0.8\,\mathrm{g} \] \[ m_{O,X}=6.4-3.2 \] \[ m_{O,X}=3.2\,\mathrm{g} \] The recovered masses sum to \[ 4.8+0.8+3.2=8.8\,\mathrm{g} \] This matches the independently stated mass of X. \(\textbf{Converting the recovered masses to atomic amounts:}\) \[ n_C=\frac{4.8}{12}=0.400\,\mathrm{mol} \] \[ n_H=\frac{0.8}{1}=0.800\,\mathrm{mol} \] \[ n_O=\frac{3.2}{16}=0.200\,\mathrm{mol} \] \(\textbf{Determining the empirical formula:}\) \[ C:H:O=0.400:0.800:0.200 \] Dividing by 0.200, \[ C:H:O=2:4:1 \] The empirical formula is \[ \mathrm{C_2H_4O} \] \(\textbf{Using the independently measured molar mass:}\) The empirical-formula mass is \[ M_{\mathrm{empirical}}=2(12)+4(1)+16 \] \[ M_{\mathrm{empirical}}=44\,\mathrm{u} \] Therefore, \[ k=\frac{88}{44} \] \[ k=2 \] Multiplying all empirical subscripts by two, \[ (\mathrm{C_2H_4O})_2=\mathrm{C_4H_8O_2} \] \(\textbf{Independent molecular verification:}\) The molecular molar mass is \[ M=4(12)+8(1)+2(16) \] \[ M=88\,\mathrm{g\,mol^{-1}} \] Thus, the resulting molecular formula satisfies both the recovered elemental composition and the independently measured molar mass. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_4H_8O_2}} \]

Explanation: \(\textbf{Converting phosphorus mass into atomic amount:}\) \[ n_P=\frac{6.2\,\mathrm{g}}{31\,\mathrm{g\,mol^{-1}}} \] \[ n_P=0.200\,\mathrm{mol} \] \(\textbf{Converting oxygen mass into atomic amount:}\) \[ n_O=\frac{8.0\,\mathrm{g}}{16\,\mathrm{g\,mol^{-1}}} \] \[ n_O=0.500\,\mathrm{mol} \] \(\textbf{Recovering the simplest atomic ratio:}\) \[ P:O=0.200:0.500 \] Dividing by 0.100, \[ P:O=2:5 \] Therefore, the empirical formula is \[ \mathrm{P_2O_5} \] \(\textbf{Determining its empirical-formula mass:}\) \[ M_{\mathrm{P_2O_5}}=2(31)+5(16) \] \[ M_{\mathrm{P_2O_5}}=62+80 \] \[ M_{\mathrm{P_2O_5}}=142\,\mathrm{u} \] \(\textbf{Recovering the molecular multiplier:}\) The measured molecular molar mass is \(284\,\mathrm{g\,mol^{-1}}\). Therefore, \[ k=\frac{284}{142} \] \[ k=2 \] The molecular formula must contain twice the atoms represented by the empirical formula. \[ (\mathrm{P_2O_5})_2=\mathrm{P_4O_{10}} \] \(\textbf{Verifying the molecular mass:}\) \[ M_{\mathrm{P_4O_{10}}}=4(31)+10(16) \] \[ M_{\mathrm{P_4O_{10}}}=124+160 \] \[ M_{\mathrm{P_4O_{10}}}=284\,\mathrm{u} \] \(\textbf{Checking the elemental composition:}\) The phosphorus mass percentage predicted by the molecular formula is \[ \%P=\frac{124}{284}\times100 \] \[ \%P\approx43.66\% \] The original sample gives \[ \%P=\frac{6.2}{14.2}\times100 \] \[ \%P\approx43.66\% \] The two values agree. \(\textbf{Chemical interpretation:}\) The empirical formula expresses the simplest phosphorus-to-oxygen atomic ratio. The molecular formula describes the actual numbers of atoms in one molecule. Both formulae represent the same elemental proportions, but their formula masses differ. The independent molecular molar-mass measurement is essential for selecting the molecular multiplier. \(\textbf{Final answer:}\) \[ \boxed{\text{Empirical formula}=\mathrm{P_2O_5},\quad\text{Molecular formula}=\mathrm{P_4O_{10}}} \]

Explanation: \(\textbf{Recovering the molecular amount of the original compound:}\) The original sample weighs 4.60 g. Its measured molar mass is \(46\,\mathrm{g\,mol^{-1}}\). Therefore, \[ n_X=\frac{4.60}{46} \] \[ n_X=0.100\,\mathrm{mol} \] \(\textbf{Recovering carbon from carbon dioxide:}\) \[ n_{\mathrm{CO_2}}=\frac{8.80}{44} \] \[ n_{\mathrm{CO_2}}=0.200\,\mathrm{mol} \] Each carbon dioxide molecule contains one carbon atom. Thus, \[ n_C=0.200\,\mathrm{mol} \] \(\textbf{Recovering hydrogen from water:}\) \[ n_{\mathrm{H_2O}}=\frac{5.40}{18} \] \[ n_{\mathrm{H_2O}}=0.300\,\mathrm{mol} \] Therefore, \[ n_H=2(0.300) \] \[ n_H=0.600\,\mathrm{mol} \] \(\textbf{Accounting for oxygen in both products:}\) The carbon dioxide contains \[ n_{O,\mathrm{CO_2}}=2(0.200)=0.400\,\mathrm{mol} \] The water contains \[ n_{O,\mathrm{H_2O}}=0.300\,\mathrm{mol} \] Hence, the complete products contain \[ n_{O,\text{products}}=0.400+0.300 \] \[ n_{O,\text{products}}=0.700\,\mathrm{mol} \] \(\textbf{Separating externally supplied oxygen:}\) The oxygen gas consumed weighs 9.60 g. \[ n_{\mathrm{O_2,consumed}}=\frac{9.60}{32} \] \[ n_{\mathrm{O_2,consumed}}=0.300\,\mathrm{mol} \] It contributes \[ n_{O,\text{external}}=2(0.300) \] \[ n_{O,\text{external}}=0.600\,\mathrm{mol} \] \(\textbf{Recovering original oxygen by atomic conservation:}\) The oxygen atoms in the products must originate either from the original compound or the external oxygen gas. Therefore, \[ n_{O,\text{original}}=n_{O,\text{products}}-n_{O,\text{external}} \] \[ n_{O,\text{original}}=0.700-0.600 \] \[ n_{O,\text{original}}=0.100\,\mathrm{mol} \] \(\textbf{Constructing the molecular formula:}\) The original sample contains \(0.100\,\mathrm{mol}\) of molecules. The constituent atomic amounts are \[ n_C:n_H:n_O=0.200:0.600:0.100 \] Thus, each molecule contains \[ C:H:O=2:6:1 \] The molecular formula is \[ \mathrm{C_2H_6O} \] \(\textbf{Complete mass-balance check:}\) The initial compound and external oxygen together weigh \[ 4.60+9.60=14.20\,\mathrm{g} \] The products together weigh \[ 8.80+5.40=14.20\,\mathrm{g} \] The mass balance agrees exactly. The original oxygen contribution must not be confused with the entire oxygen inventory of the combustion products. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_6O},\quad n_{O,\text{original}}=0.100\,\mathrm{mol}} \]

Explanation: Statement I is correct because different molecular compounds can share the same molecular formula while having different arrangements of constituent atoms. Such compounds necessarily have the same numbers and types of atoms per molecule. They therefore have identical calculated molecular masses and elemental mass percentages when the same atomic masses are used. For example, ethanol and dimethyl ether have the same molecular formula, \(\mathrm{C_2H_6O}\), but differ in atomic connectivity. Statement II is correct because an empirical formula gives the simplest atomic ratio and an empirical-formula mass. Dividing the experimentally measured molecular molar mass by the corresponding empirical-formula molar mass gives the molecular multiplier. When that multiplier is a valid positive whole number, multiplying all empirical subscripts by it gives the molecular formula. Statement III is incorrect because a molecular formula specifies the numbers of each kind of atom, not necessarily how those atoms are connected. Different structural arrangements can share one molecular formula. Additional structural evidence is therefore required to establish a unique chemical identity in such cases. Statement IV is incorrect because compounds having the same molecular formula must also have the same empirical formula. Reducing the same molecular subscripts produces the same simplest whole-number ratio. Differences in chemical identity do not alter this mathematical relationship. The key distinction is between determining elemental composition, determining molecular atomic counts and identifying molecular structure. Consequently, only Statements I and II are correct.

Explanation: \(\textbf{Interpreting the original mixture percentage:}\) P contains 30% oxygen, while Q contains 60%. The original mixture contains 45% oxygen. Because 45% lies exactly halfway between 30% and 60%, equal masses of P and Q would give the observed original percentage. This can also be established algebraically without assuming the component masses. \(\textbf{Representing the original mixture:}\) Let the original total mass be \(M\) grams. Let the mass of P be \(x\) grams. The mass of Q is then \[ m_Q=M-x \] The initial oxygen mass is \[ m_{O,\text{initial}}=0.30x+0.60(M-x) \] Using the reported original oxygen fraction, \[ 0.30x+0.60(M-x)=0.45M \] Expanding, \[ 0.30x+0.60M-0.60x=0.45M \] \[ 0.15M=0.30x \] Therefore, \[ x=0.50M \] The original mixture contains equal masses of P and Q. \(\textbf{Applying the addition of Q:}\) An additional 20 g of pure Q is introduced. Its oxygen contribution is \[ m_{O,\text{added}}=0.60(20) \] \[ m_{O,\text{added}}=12\,\mathrm{g} \] The new mixture mass is \[ M_{\text{final}}=M+20 \] The final oxygen mass is \[ m_{O,\text{final}}=0.45M+12 \] \(\textbf{Using the new oxygen percentage:}\) The final mixture contains 50% oxygen by mass. Therefore, \[ \frac{0.45M+12}{M+20}=0.50 \] Cross-multiplying, \[ 0.45M+12=0.50M+10 \] \[ 0.05M=2 \] \[ M=40\,\mathrm{g} \] \(\textbf{Recovering the original amount of P:}\) The original P mass is half the initial mixture mass. \[ m_{P,\text{initial}}=0.50(40) \] \[ m_{P,\text{initial}}=20\,\mathrm{g} \] \(\textbf{Independent final-composition check:}\) Originally, 20 g of P contributes 6 g oxygen. The original 20 g of Q contributes 12 g oxygen. The added 20 g of Q contributes another 12 g oxygen. Therefore, \[ m_{O,\text{final}}=6+12+12=30\,\mathrm{g} \] The final total mass is 60 g. \[ \%O_{\text{final}}=\frac{30}{60}\times100 \] \[ \%O_{\text{final}}=50\% \] Both experimental observations are satisfied. The increase in oxygen percentage arises from adding a component richer in oxygen than the initial mixture. \(\textbf{Final answer:}\) \[ \boxed{M_{\text{initial}}=40\,\mathrm{g},\quad m_{P,\text{initial}}=20\,\mathrm{g}} \]

Explanation: \(\textbf{Representing the unknown hydrocarbon:}\) Let its molecular formula be \(\mathrm{C}_{x}\mathrm{H}_{y}\). Complete combustion converts its carbon into carbon dioxide and hydrogen into water. At the same temperature and pressure, ideal-gas volume ratios equal the corresponding molar ratios. \(\textbf{Determining the carbon subscript:}\) One hydrocarbon molecule produces \(x\) carbon dioxide molecules. Therefore, \[ x=\frac{V_{\mathrm{CO_2}}}{V_{\mathrm{hydrocarbon}}} \] \[ x=\frac{5}{1} \] \[ x=5 \] \(\textbf{Determining the hydrogen subscript:}\) One hydrocarbon molecule produces \(y/2\) water molecules on complete combustion. Hence, \[ \frac{y}{2}=\frac{V_{\mathrm{H_2O(g)}}}{V_{\mathrm{hydrocarbon}}} \] \[ \frac{y}{2}=\frac{6}{1} \] \[ y=12 \] The molecular formula indicated by the gaseous products is \(\mathrm{C_5H_{12}}\). \(\textbf{Verifying the independent density measurement:}\) At a fixed temperature and pressure, ideal-gas density is directly proportional to molar mass. Thus, \[ \frac{d_{\mathrm{hydrocarbon}}}{d_{\mathrm{O_2}}} =\frac{M_{\mathrm{hydrocarbon}}}{M_{\mathrm{O_2}}} \] Given that the density ratio is \(\frac{9}{4}\), \[ M_{\mathrm{hydrocarbon}}=\frac{9}{4}(32) \] \[ M_{\mathrm{hydrocarbon}}=72\,\mathrm{g\,mol^{-1}} \] For the proposed formula, \[ M_{\mathrm{C_5H_{12}}}=5(12)+12(1) \] \[ M_{\mathrm{C_5H_{12}}}=72\,\mathrm{g\,mol^{-1}} \] The measured density independently supports the formula. \(\textbf{Balancing the combustion reaction:}\) Five carbon atoms form five carbon dioxide molecules. Twelve hydrogen atoms form six water molecules. The balanced equation is \[ \mathrm{C_5H_{12}+8O_2\rightarrow5CO_2+6H_2O} \] \(\textbf{Calculating oxygen consumption:}\) Eight moles of oxygen are required per mole of hydrocarbon. At the common reference temperature and pressure, \[ \frac{V_{\mathrm{O_2}}}{V_{\mathrm{hydrocarbon}}}=\frac{8}{1} \] Therefore, \[ V_{\mathrm{O_2}}=8(1) \] \[ V_{\mathrm{O_2}}=8\,\text{L} \] \(\textbf{Checking oxygen-atom conservation:}\) The products contain \(5(2)+6(1)=16\) oxygen atoms per hydrocarbon molecule. Eight \(\mathrm{O_2}\) molecules supply the required sixteen oxygen atoms. The volume, density and atom-balance conditions are mutually consistent. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_5H_{12}},\quad V_{\mathrm{O_2}}=8\,\text{L}} \]

Explanation: The empirical formula is obtained by dividing all molecular subscripts by their greatest common integer factor. The molecular multiplier is the number by which every empirical subscript must be multiplied to recover the molecular formula. For P, \(\mathrm{P_4O_{10}}\), the subscripts 4 and 10 have a greatest common factor of 2. Dividing both by 2 gives \(\mathrm{P_2O_5}\). Its molecular multiplier is 2, so P matches entry 3. For Q, \(\mathrm{C_6H_6}\), the subscripts have a greatest common factor of 6. The empirical formula is therefore \(\mathrm{CH}\), and the molecular multiplier is 6. Thus, Q matches entry 4. For R, \(\mathrm{C_2H_4O_2}\), all subscripts can be divided by 2. Its empirical formula is \(\mathrm{CH_2O}\), and the molecular multiplier is 2. Therefore, R matches entry 1. For S, \(\mathrm{N_2O_4}\), both subscripts can be divided by 2. Its empirical formula is \(\mathrm{NO_2}\), and its molecular multiplier is 2. Consequently, S matches entry 2. The empirical formula retains the same elemental atomic ratios as the molecular formula. However, it does not necessarily retain the actual numbers of atoms in one molecule. The multiplier applies uniformly to every subscript. A match in empirical formula alone would not establish the molecular multiplier without the actual molecular formula or corresponding molar-mass information. The correct matching is P–3, Q–4, R–1 and S–2.

Explanation: \(\textbf{Interpreting atomic percentages correctly:}\) The percentages refer to numbers of atoms. They therefore specify relative atomic populations directly. They must not initially be divided by atomic molar masses as though they represented elemental masses. \(\textbf{Constructing the empirical atomic ratio:}\) The reported atomic percentages give \[ C:H:O=20:60:20 \] Dividing by 20, \[ C:H:O=1:3:1 \] Therefore, the empirical formula is \[ \mathrm{CH_3O} \] \(\textbf{Calculating the empirical-formula mass:}\) \[ M_{\mathrm{CH_3O}}=12+3(1)+16 \] \[ M_{\mathrm{CH_3O}}=31\,\mathrm{u} \] \(\textbf{Determining the molecular multiplier:}\) The measured molecular molar mass is \(62\,\mathrm{g\,mol^{-1}}\). Therefore, \[ k=\frac{62}{31} \] \[ k=2 \] Multiplying every empirical subscript by two gives \[ (\mathrm{CH_3O})_2=\mathrm{C_2H_6O_2} \] \(\textbf{Checking the molecular molar mass:}\) \[ M_{\mathrm{C_2H_6O_2}}=2(12)+6(1)+2(16) \] \[ M_{\mathrm{C_2H_6O_2}}=24+6+32 \] \[ M_{\mathrm{C_2H_6O_2}}=62\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculating oxygen mass percentage:}\) Two oxygen atoms contribute 32 units of molecular mass. Therefore, \[ \%O_{\mathrm{mass}}=\frac{32}{62}\times100 \] \[ \%O_{\mathrm{mass}}\approx51.61\% \] \(\textbf{Verifying the atomic percentages:}\) The molecular formula contains \[ 2+6+2=10 \] constituent atoms per molecule. Carbon contributes two atoms, hydrogen contributes six and oxygen contributes two. Therefore, \[ \%C_{\mathrm{atoms}}=\frac2{10}\times100=20\% \] \[ \%H_{\mathrm{atoms}}=\frac6{10}\times100=60\% \] \[ \%O_{\mathrm{atoms}}=\frac2{10}\times100=20\% \] The calculated atomic percentages reproduce the measurements. \(\textbf{Chemical interpretation:}\) Oxygen accounts for only 20% of the constituent atoms but approximately 51.61% of the molecular mass. The difference arises because oxygen atoms are substantially heavier than hydrogen atoms. Atomic percentages and mass percentages must not be used interchangeably. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_6O_2},\quad\%O_{\mathrm{mass}}=51.61\%} \]

Explanation: \(\textbf{Identifying the two oxide compositions:}\) The mixture contains FeO and \(\mathrm{Fe_2O_3}\), each with its own fixed elemental mass fraction. Their formula molar masses are \[ M_{\mathrm{FeO}}=56+16=72\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{Fe_2O_3}}=2(56)+3(16) \] \[ M_{\mathrm{Fe_2O_3}}=160\,\mathrm{g\,mol^{-1}} \] \(\textbf{Calculating oxygen mass fractions:}\) For FeO, \[ w_{O,\mathrm{FeO}}=\frac{16}{72} \] \[ w_{O,\mathrm{FeO}}=\frac29 \] For \(\mathrm{Fe_2O_3}\), \[ w_{O,\mathrm{Fe_2O_3}}=\frac{48}{160} \] \[ w_{O,\mathrm{Fe_2O_3}}=\frac3{10} \] \(\textbf{Representing the unknown component masses:}\) Let \(x\) grams of the mixture be FeO. The mass of \(\mathrm{Fe_2O_3}\) is \[ m_{\mathrm{Fe_2O_3}}=38-x \] \(\textbf{Applying the measured oxygen inventory:}\) Both oxides contribute oxygen to the elemental analysis. Therefore, \[ \frac29x+\frac3{10}(38-x)=10 \] Multiplying throughout by 90, \[ 20x+27(38-x)=900 \] \[ 20x+1026-27x=900 \] \[ 7x=126 \] \[ x=18\,\mathrm{g} \] Therefore, \[ m_{\mathrm{FeO}}=18\,\mathrm{g} \] \[ m_{\mathrm{Fe_2O_3}}=20\,\mathrm{g} \] \(\textbf{Converting component masses into formula-unit amounts:}\) For FeO, \[ n_{\mathrm{FeO}}=\frac{18}{72} \] \[ n_{\mathrm{FeO}}=0.250\,\mathrm{mol} \] For \(\mathrm{Fe_2O_3}\), \[ n_{\mathrm{Fe_2O_3}}=\frac{20}{160} \] \[ n_{\mathrm{Fe_2O_3}}=0.125\,\mathrm{mol} \] Thus, \[ n_{\mathrm{FeO}}:n_{\mathrm{Fe_2O_3}}=0.250:0.125 \] \[ n_{\mathrm{FeO}}:n_{\mathrm{Fe_2O_3}}=2:1 \] \(\textbf{Finding the FeO mass percentage:}\) \[ \%\mathrm{FeO}=\frac{18}{38}\times100 \] \[ \%\mathrm{FeO}\approx47.37\% \] \(\textbf{Independent oxygen-mass check:}\) FeO contributes \[ m_{O,\mathrm{FeO}}=18\left(\frac29\right)=4.0\,\mathrm{g} \] Iron(III) oxide contributes \[ m_{O,\mathrm{Fe_2O_3}}=20\left(\frac3{10}\right)=6.0\,\mathrm{g} \] Their combined oxygen mass is \[ 4.0+6.0=10.0\,\mathrm{g} \] This reproduces the experimental result. \(\textbf{Chemical interpretation:}\) The formula-unit amount ratio is not equal to the component mass ratio because the oxides have different formula molar masses. The measured oxygen mass must first be partitioned between the two different fixed compositions. Only then can the formula-unit amounts be compared correctly. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{FeO}}:n_{\mathrm{Fe_2O_3}}=2:1,\quad\%\mathrm{FeO}=47.37\%} \]

Explanation: \(\textbf{Separating hydrogen from the anhydrous component:}\) The formula \(\mathrm{MgSO_4\cdot xH_2O}\) contains no hydrogen in the anhydrous magnesium sulphate portion. All hydrogen atoms therefore belong to the water of crystallisation. This makes the measured hydrogen percentage a direct constraint on the hydration number. \(\textbf{Molar mass of anhydrous magnesium sulphate:}\) \[ M_{\mathrm{MgSO_4}}=24+32+4(16) \] \[ M_{\mathrm{MgSO_4}}=120\,\mathrm{g\,mol^{-1}} \] \(\textbf{Molar mass of the unknown hydrate:}\) Each water molecule contributes \(18\,\mathrm{g\,mol^{-1}}\). Thus, \[ M_{\text{hydrate}}=120+18x \] \(\textbf{Hydrogen contribution from water:}\) Each water molecule contains two hydrogen atoms. Therefore, the hydrogen mass per mole of hydrated formula units is \[ m_{H,\text{per mole}}=2x\,\mathrm{g} \] \(\textbf{Constructing the hydrogen percentage equation:}\) \[ \%H=\frac{2x}{120+18x}\times100 \] The reported value is approximately 4.76%. Therefore, \[ 4.76\approx\frac{2x}{120+18x}\times100 \] Using the decimal fraction, \[ 0.0476\approx\frac{2x}{120+18x} \] Cross-multiplying, \[ 0.0476(120+18x)\approx2x \] \[ 5.712+0.8568x\approx2x \] \[ 1.1432x\approx5.712 \] \[ x\approx4.9965 \] Since \(x\) must be a positive whole number and the percentage was rounded, \[ x=5 \] \(\textbf{Calculating the water mass percentage:}\) The hydrate is \(\mathrm{MgSO_4\cdot5H_2O}\). Its molar mass is \[ M_{\text{hydrate}}=120+5(18) \] \[ M_{\text{hydrate}}=210\,\mathrm{g\,mol^{-1}} \] The water of crystallisation contributes \[ m_{\text{water,per mole}}=5(18) \] \[ m_{\text{water,per mole}}=90\,\mathrm{g} \] Therefore, \[ \%\text{water}=\frac{90}{210}\times100 \] \[ \%\text{water}\approx42.86\% \] \(\textbf{Checking the original hydrogen percentage:}\) Five water molecules contain ten hydrogen atoms. \[ \%H=\frac{10}{210}\times100 \] \[ \%H\approx4.7619\% \] This agrees with the reported approximately 4.76%. \(\textbf{Chemical interpretation:}\) The percentage of hydrogen is not the percentage of water of crystallisation. Water contributes oxygen as well as hydrogen to the hydrate mass. The hydration coefficient is determined from the fixed hydrogen contribution of each associated water molecule. \(\textbf{Final answer:}\) \[ \boxed{x=5,\quad\%\text{water}=42.86\%} \]

Explanation: \(\textbf{Comparing the two molecular molar masses:}\) Ethene has molecular formula \(\mathrm{C_2H_4}\). \[ M_{\mathrm{C_2H_4}}=2(12)+4(1) \] \[ M_{\mathrm{C_2H_4}}=28\,\mathrm{g\,mol^{-1}} \] Butene has molecular formula \(\mathrm{C_4H_8}\). \[ M_{\mathrm{C_4H_8}}=4(12)+8(1) \] \[ M_{\mathrm{C_4H_8}}=56\,\mathrm{g\,mol^{-1}} \] \(\textbf{Recovering the molecular mixture composition:}\) Let \(x\) be the mole fraction of butene. The ethene mole fraction is \(1-x\). For this non-reacting mixture, \[ \overline{M}=(1-x)M_{\mathrm{C_2H_4}}+xM_{\mathrm{C_4H_8}} \] Substituting the measured average molar mass, \[ 35=28(1-x)+56x \] \[ 35=28-28x+56x \] \[ 35=28+28x \] \[ 28x=7 \] \[ x=0.25 \] Thus, \[ X_{\mathrm{C_4H_8}}=0.25 \] \(\textbf{Choosing a one-mole mixture basis:}\) One mole of the mixture contains \[ n_{\mathrm{C_2H_4}}=0.75\,\mathrm{mol} \] \[ n_{\mathrm{C_4H_8}}=0.25\,\mathrm{mol} \] The total mixture mass is \[ m_{\text{total}}=0.75(28)+0.25(56) \] \[ m_{\text{total}}=21+14 \] \[ m_{\text{total}}=35\,\mathrm{g} \] \(\textbf{Calculating the carbon contribution:}\) Each mole of ethene contains 24 g of carbon. Each mole of butene contains 48 g of carbon. Therefore, \[ m_C=0.75(24)+0.25(48) \] \[ m_C=18+12 \] \[ m_C=30\,\mathrm{g} \] \(\textbf{Determining the mixture's carbon percentage:}\) \[ \%C=\frac{30}{35}\times100 \] \[ \%C\approx85.71\% \] \(\textbf{Recognising the composition invariant:}\) Ethene has empirical formula \(\mathrm{CH_2}\). Butene also has empirical formula \(\mathrm{CH_2}\). Consequently, both pure compounds have the same elemental mass percentages. Their carbon mass fraction is \[ w_C=\frac{12}{12+2} \] \[ w_C=\frac67 \] Therefore, any non-reacting mixture containing only these two compounds also has this carbon mass fraction, regardless of the mole fractions. The average molar mass changes with molecular composition, but the carbon mass percentage does not. This distinguishes molecular population from elemental composition. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{C_4H_8}}=0.25,\quad\%C=85.71\%} \]