MPSC 2023 Prelims Paper-I · Question 64 of 100

Correct — option (3). The atomic masses supplied on the second line of the stem, H = 1 and C = 12, are the instruction: this item is to be computed, not recalled. The percentage of carbon by mass in a hydrocarbon CₙHₘ is the mass contributed by carbon divided by the total molar mass, that is 12n ÷ (12n + m), multiplied by a hundred. Applying it to ethane, C₂H₆: the two carbons contribute 2 × 12 = 24, the six hydrogens contribute 6 × 1 = 6, the molar mass is 30, and the carbon percentage is 24 ÷ 30 = 0.80, which is 80 per cent exactly. That is the keyed option, and the other three fail the same test — benzene C₆H₆ gives 72 ÷ 78 = 92.3 per cent, methane CH₄ gives 12 ÷ 16 = 75 per cent, and cyclohexane C₆H₁₂ gives 72 ÷ 84 = 85.7 per cent — so ethane is the only hydrocarbon among them with 80 per cent carbon. There is a faster route worth learning, because it turns the question round and answers it in one step. Percentage composition depends only on the ratio of the atoms, that is on the empirical formula, and not on how large the molecule is. If carbon is 80 per cent by mass then hydrogen is the remaining 20 per cent, and converting those mass shares to a mole ratio by dividing each by its atomic mass gives 80 ÷ 12 = 6.67 for carbon against 20 ÷ 1 = 20 for hydrogen; dividing both by the smaller gives 1 : 3, so the empirical formula must be CH₃. The hydrocarbon whose empirical formula is CH₃ is ethane, C₂H₆, since CH₃ on its own is a methyl radical and not a stable molecule. The same insight explains why the other options can be dismissed almost by inspection once their empirical formulae are seen: benzene reduces to CH and cyclohexane to CH₂, and any hydrocarbon sharing one of those empirical formulae will share its percentage composition exactly — ethyne is also 92.3 per cent carbon because it too is CH, and ethene is also 85.7 per cent carbon because it too is CH₂. Option (3) is the answer.

• (1)Benzene — Benzene is C₆H₆, so its molar mass is 72 + 6 = 78 and the carbon fraction is 72 ÷ 78, which is 92.3 per cent — the highest of the four and well above the 80 per cent the question specifies. Its empirical formula is CH, one carbon to one hydrogen, and because percentage composition depends only on that ratio, every hydrocarbon with the empirical formula CH has the same figure: ethyne, C₂H₂, is also 92.3 per cent carbon. Benzene is offered first because it is the hydrocarbon candidates recognise fastest and because its high carbon content is a familiar fact, aromatic compounds being notably carbon-rich; but a question that states a precise percentage is asking for the one formula that produces it, and recognition is no substitute for the division.
• (2)Methane — Methane is CH₄, with a molar mass of 12 + 4 = 16 and a carbon fraction of 12 ÷ 16, which is 75 per cent. That is the lowest carbon percentage any hydrocarbon can have, and it is worth understanding why: methane has the largest possible number of hydrogens per carbon, four, and every longer alkane in the series CₙH₂ₙ₊₂ carries proportionally fewer, so the carbon percentage rises steadily along the series — 75 per cent for methane, 80 for ethane, 81.8 for propane, 82.8 for butane — approaching but never reaching 85.7 per cent. Methane attracts the candidate who reasons that the simplest hydrocarbon should give the roundest answer, but 75 and 80 are different numbers and the stem asks for one of them.
• (4)Cyclohexane — Cyclohexane is C₆H₁₂, so its molar mass is 72 + 12 = 84 and the carbon fraction is 72 ÷ 84, which is 85.7 per cent. Its empirical formula is CH₂, and every hydrocarbon reducing to CH₂ — the whole family of alkenes and cycloalkanes, from ethene upwards — has that identical composition, which is a useful check in itself. This option is the one that most often draws a candidate who has not done the arithmetic, because 'cyclohexane' looks like the technical answer among four options of which two are the most elementary hydrocarbons in the syllabus, and difficulty of name is mistaken for correctness. The stem's parenthetical line of atomic masses is the signal that the item is decided by a division and not by which name feels most advanced.

Percentage composition by mass converts a molecular formula into the proportions in which the elements are present, and it is calculated by dividing the total mass contributed by each element by the molar mass of the compound. For a hydrocarbon CₙHₘ the carbon percentage is 12n ÷ (12n + m) × 100. The crucial property of the result is that it depends only on the ratio n : m and not on their absolute values, so percentage composition identifies the empirical formula and never the molecular formula on its own. Benzene C₆H₆ and ethyne C₂H₂ both reduce to CH and are indistinguishable by composition alone; ethene C₂H₄, propene C₃H₆ and cyclohexane C₆H₁₂ all reduce to CH₂ and share the figure 85.7 per cent. Determining the molecular formula therefore requires a second measurement, the molar mass, obtained from vapour density or a colligative property, after which the molecular formula is the empirical formula multiplied by the whole number that reconciles the two. The calculation runs in either direction and the reverse direction is the one examinations use most. Given the mass percentages of the elements, divide each by the element's atomic mass to obtain relative numbers of moles, divide those through by the smallest, and clear any remaining fraction by multiplying to whole numbers — the result is the empirical formula. This is the classical procedure of combustion analysis, in which a weighed sample of an organic compound is burnt completely, the carbon dioxide and water produced are trapped and weighed, and the carbon and hydrogen content of the original sample is worked back out from them.

MPSC's science section carries a small number of genuine calculations among its recall questions, and this is one of them. The parenthetical line of atomic masses printed beneath the stem is the Commission's standard signal that arithmetic is expected, and it also removes the excuse of not remembering them. The item is worth about thirty seconds to a prepared candidate: write the four formulae, form four fractions, and compare. What makes it fail in practice is the temptation to answer from association rather than from the numbers — benzene feels carbon-rich, methane feels like the standard first example, cyclohexane sounds advanced — and any of those instincts produces a confident wrong answer at a cost of a quarter mark under this paper's negative marking. The deeper reason to master the reverse calculation is that it converts this question from four divisions into one inference: 80 per cent carbon means the empirical formula CH₃, and only one of the four options has it. A candidate who can move in that direction also handles the empirical-formula and combustion-analysis questions that appear in the same block of the syllabus. Note that the stem here prints the singular 'hydrocarbon' after 'Which of the following', which is one of several grammatical slips in this paper's English column and carries no bearing on the meaning.

• The carbon percentage of a hydrocarbon CₙHₘ is 12n ÷ (12n + m) × 100; for ethane C₂H₆ this is 24 ÷ 30, exactly 80 per cent, which is why ethane is the keyed answer.
• Percentage composition depends only on the empirical formula, so benzene C₆H₆ and ethyne C₂H₂ share the figure 92.3 per cent carbon, while ethene, propene and cyclohexane all reduce to CH₂ and share 85.7 per cent.
• Methane at 75 per cent carbon has the lowest carbon percentage of any hydrocarbon; along the alkane series CₙH₂ₙ₊₂ the figure rises to 80 for ethane, 81.8 for propane and 82.8 for butane, approaching 85.7 per cent as the chain lengthens.
• Working backwards from percentages, dividing each element's mass percentage by its atomic mass and reducing the result to the smallest whole-number ratio gives the empirical formula: 80 per cent carbon and 20 per cent hydrogen yield a C : H ratio of 1 : 3, that is CH₃, whose hydrocarbon is ethane.
• The empirical formula alone cannot fix the molecular formula; a separate determination of the molar mass is required, and the molecular formula is then a whole-number multiple of the empirical formula.

• Answering from association rather than arithmetic, choosing benzene because aromatic compounds feel carbon-rich or cyclohexane because the name sounds more advanced than the alternatives
• Overlooking the line of atomic masses printed under the stem, which is the Commission's signal that the item is a calculation and not a recall question
• Assuming a percentage composition identifies a unique compound, when it fixes only the empirical formula and is shared by every compound reducing to it
• Miscounting hydrogens in a cyclic formula, most often writing cyclohexane as C₆H₆ instead of C₆H₁₂ and so confusing it with benzene
• Reading the percentage as a percentage by number of atoms rather than by mass, which for a hydrocarbon gives an entirely different figure

Mole-concept and composition questions appear in MPSC science sections in a handful of standard shapes: compute a percentage by mass from a formula, deduce an empirical formula from given percentages, find the number of moles or molecules in a stated mass, or compare two compounds by the content of one element. The Commission signals every such item by printing the relevant atomic masses beneath the stem, so a candidate can identify the computational questions at a glance while scanning the paper and decide when to attempt them. The compounds chosen are almost always from the elementary organic list — methane, ethane, benzene, ethanol, glucose, urea — whose formulae are expected to be known without prompting, since the atomic masses are supplied but the formulae are not. That is the real preparation this family demands: the arithmetic is trivial, and a candidate loses the mark only by not knowing that ethane is C₂H₆ or that cyclohexane is C₆H₁₂.

No directly related past PYQ was found.

• practice — not a real PYQ

  An organic compound is found on analysis to contain 85.7% carbon and 14.3% hydrogen by mass. What is its empirical formula ? (H = 1, C = 12)

  • (a)CH
  • (b)CH₂
  • (c)CH₃
  • (d)CH₄

  Answer(b) CH₂ — dividing each mass percentage by the atomic mass of the element gives 85.7 ÷ 12 = 7.14 for carbon and 14.3 ÷ 1 = 14.3 for hydrogen, and dividing both by the smaller gives a ratio of 1 : 2. Every alkene and every cycloalkane reduces to this empirical formula, so the data are consistent with ethene, propene or cyclohexane alike, and the molecular formula cannot be settled without a separate determination of the molar mass.

• practice — not a real PYQ

  Which of the following pairs of hydrocarbons have exactly the same percentage of carbon by mass ?

  • (a)Methane and ethane
  • (b)Benzene and ethyne
  • (c)Ethane and propane
  • (d)Benzene and cyclohexane

  Answer(b) Benzene and ethyne — benzene is C₆H₆ and ethyne is C₂H₂, and both reduce to the empirical formula CH, so both are 92.3 per cent carbon by mass. Percentage composition is fixed by the ratio of the atoms alone and is blind to the size of the molecule, which is exactly why it can never establish a molecular formula on its own. Benzene and cyclohexane differ, at 92.3 and 85.7 per cent, because their empirical formulae are CH and CH₂ respectively.