Empirical and Molecular Formulae — the ratio trap NEET exploits
The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms per molecule. The two are related by a whole-number multiplier n:
Molecular formula = n × Empirical formula, where n = Molar mass / Empirical formula mass.
NCERT Class 11 Chemistry Chapter 1, page 15 defines the empirical formula as "the simplest whole number ratio of atoms of each element present in a compound." The molecular formula may be identical to the empirical formula (e.g., H₂O) or a multiple of it (e.g., glucose C₆H₁₂O₆ has empirical formula CH₂O, with n = 6).
The standard workflow from percentage composition:
Where aspirants lose marks:
These are formula-derivation errors intrinsic to the empirical/molecular formula workflow itself. When NEET asks "find the molecular formula," the mark is lost in the ratio arithmetic, not in the chemistry.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The empirical formula of a compound represents:
Answer: D. D is correct. NCERT Class 11 Chemistry Chapter 1, page 15 defines the empirical formula as the simplest whole-number ratio of atoms of each element present in a compound.
Why A is wrong: A describes the molecular formula, not the empirical formula. The empirical formula may have fewer atoms than actually present per molecule.
Why B is wrong: B confuses atomic-mass ratios with atom-count ratios. The empirical formula expresses atom ratios, not mass ratios.
Why C is wrong: C describes the mole composition from percentage data — an intermediate calculation step, not the formula itself.
The molecular formula of glucose is C₆H₁₂O₆. Its empirical formula is:
Answer: A. A is correct. Dividing all subscripts in C₆H₁₂O₆ by the GCD (6) gives C₁H₂O₁ = CH₂O, the simplest whole-number ratio. NCERT Class 11 Chemistry Chapter 1 uses glucose as the standard example.
Why B is wrong: B drops the subscript on H — the ratio C:H:O is 1:2:1, not 1:1:1.
Why C is wrong: C is the molecular formula itself, not reduced to the simplest ratio.
Why D is wrong: D is C₂H₄O₂, which simplifies further to CH₂O. This is not the simplest ratio.
If the empirical formula of a compound is CH₂O and its molar mass is 180 g/mol, the value of the multiplier n used to obtain the molecular formula is:
Answer: B. B is correct. Empirical formula mass of CH₂O = 12 + 2(1) + 16 = 30 g/mol. n = 180/30 = 6.
Why A is wrong: A gives n = 3, implying empirical formula mass = 60, which is incorrect for CH₂O (actual = 30).
Why C is wrong: C gives n = 4, implying molar mass = 120, not the stated 180 g/mol.
Why D is wrong: D gives n = 12, implying empirical formula mass = 15, which does not match CH₂O.
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16)
Answer: A. A is correct. Assuming 100 g: C = 40.0/12 = 3.33 mol, H = 6.7/1 = 6.7 mol, O = 53.3/16 = 3.33 mol. Dividing by the smallest (3.33): C:H:O = 1:2.01:1 ≈ 1:2:1 → CH₂O.
Why B is wrong: B gives a 1:1:1 ratio, which would require H to be 3.33 mol — but 6.7 g of H gives 6.7 mol, not 3.33. This error comes from misreading percentage as moles directly.
Why C is wrong: C is C₂H₄O₂, which simplifies to CH₂O. The empirical formula must be the simplest ratio, so C₂H₄O₂ is not reduced far enough.
Why D is wrong: D gives a 1:3:1 ratio for C:H:O, which does not match the 1:2:1 ratio obtained from the calculation.
A compound has the empirical formula NO₂ and a molar mass of 92 g/mol. Its molecular formula is: (Atomic masses: N = 14, O = 16)
Answer: D. D is correct. Empirical formula mass of NO₂ = 14 + 2(16) = 46 g/mol. n = 92/46 = 2. Molecular formula = 2 × NO₂ = N₂O₄.
Why A is wrong: A assumes n = 1, which gives molar mass = 46, not 92. The multiplier was not applied.
Why B is wrong: B is an entirely different compound (nitrous oxide), not a multiple of NO₂.
Why C is wrong: C is N₂O₃ (molar mass = 76 g/mol), which is neither a multiple of NO₂ nor matches the given molar mass of 92.
A hydrocarbon contains 85.7% carbon by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1)
Answer: C. C is correct. Assuming 100 g: C = 85.7/12 = 7.14 mol, H = 14.3/1 = 14.3 mol. Ratio: 14.3/7.14 = 2.00. So C:H = 1:2 → CH₂.
Why A is wrong: A gives C:H = 1:1, which requires equal moles. The hydrogen moles (14.3) are double the carbon moles (7.14), ruling out a 1:1 ratio.
Why B is wrong: B gives C:H = 1:3, implying H% ≈ 20% for a compound with C = 80%. That does not match the stated 85.7% C / 14.3% H.
Why D is wrong: D has an empirical formula mass of 27, but the simplest ratio of 1:2 gives CH₂. C₂H₃ has ratio 2:3, which does not match the 1:2 calculation.
An organic compound contains 40.0% C, 6.7% H, and 53.3% O by mass. If its molar mass is 180 g/mol, what is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16)
Answer: B. B is correct. Step 1: Empirical formula from % composition — C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. Ratio 1:2:1 → CH₂O (EF mass = 30). Step 2: n = 180/30 = 6. Molecular formula = C₆H₁₂O₆.
Why A is wrong: A uses n = 3 (molar mass would be 90, not 180). This halves the correct multiplier — a common arithmetic slip.
Why C is wrong: C uses n = 4 (molar mass would be 120, not 180). This does not divide 180 evenly by 30.
Why D is wrong: D uses n = 2 (molar mass would be 60, not 180). Often results from forgetting to complete the division 180/30.
When calculating the empirical formula from percentage composition, dividing each element's mole value by the smallest mole value gives ratios of 1 : 1.5 : 1. The correct next step is:
Answer: C. C is correct. When a ratio is X.5, multiply all ratios by 2 to obtain whole numbers: 1×2 : 1.5×2 : 1×2 = 2:3:2. Rounding a non-integer ratio loses the actual atom count.
Why A is wrong: A rounds 1.5 to 1, which changes the compound's identity. Empirical formula ratios must be exact whole numbers, not approximations from rounding non-integer values.
Why B is wrong: B rounds 1.5 to 2 instead of multiplying through, giving a ratio of 1:2:1 — a different compound entirely. Rounding introduces error when the fractional part is exactly 0.5.
Why D is wrong: D multiplies by 3, giving 3:4.5:3 — still not whole numbers. The correct multiplier for a 0.5 fraction is 2, not 3.
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Given
A compound contains 26.7% carbon, 2.2% hydrogen, and 71.1% oxygen by mass. Its molar mass is 90 g/mol.
Atomic masses: C = 12.0 g/mol, H = 1.0 g/mol, O = 16.0 g/mol.
Required
Determine the molecular formula of the compound.
Concept
Convert mass percentages to moles, find the simplest ratio (empirical formula), compute empirical formula mass, then use n = molar mass / empirical formula mass to get the molecular formula.
Formula
• n(element) = mass / atomic mass• Empirical formula from simplest whole-number ratio• Multiplier: n = M(compound) / M(empirical formula)
Substitution
Assume 100 g of compound:• C: 26.7 g / 12.0 g/mol = 2.225 mol• H: 2.2 g / 1.0 g/mol = 2.2 mol• O: 71.1 g / 16.0 g/mol = 4.444 mol
Divide by smallest (2.2):• C: 2.225 / 2.2 = 1.011 ≈ 1• H: 2.2 / 2.2 = 1.000• O: 4.444 / 2.2 = 2.020 ≈ 2
Empirical formula: CHO₂
Calculation
Empirical formula mass of CHO₂ = 12 + 1 + 2(16) = 45 g/mol
n = 90 / 45 = 2
Note on exact values: The atomic masses (C = 12.0, H = 1.0, O = 16.0) are given as exact problem-defined values and do not limit significant figures in the calculation. The multiplier n = 2 is an exact integer.
Final answer
Molecular formula = 2 × CHO₂ = C₂H₂O₄ (oxalic acid)
Common trap
Stopping at CHO₂ (the empirical formula) and writing it as the final answer. The question asks for the molecular formula — you must always compute the multiplier n when molar mass is given. Another common error: rounding 2.02 to 2 is appropriate here (within rounding tolerance), but rounding 1.5 to 2 would be wrong — that signals a ×2 multiplier is needed on all ratios.
Similar NEET-style question
"A compound has 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen by mass. If the molar mass is 46 g/mol, find the molecular formula."
(Answer: C: 52.2/12 = 4.35, H: 13.0/1 = 13.0, O: 34.8/16 = 2.175. Divide through by the smallest, 2.175: C = 2.0, H = 5.98 ≈ 6, O = 1.0. Empirical formula C₂H₆O, EF mass = 46, so n = 46/46 = 1 and the molecular formula is C₂H₆O — ethanol. Note the check that matters: if you round the mole ratios before dividing you get CH₃O, whose EF mass of 31 gives n = 46/31 = 1.48, not a whole number. A non-integer n is the signal that the rounding was done too early, not that the data is faulty.)
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An empirical formula represents the simplest whole number ratio of various atoms present in a compound, whereas, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. If the mass per cent of various elements present in a compound is known, its empirical formula can be determined. Molecular formula can further be obtained if the molar mass is known.
-- NCERT Class 11 Chemistry, Ch. 1, p. 19More in Some Basic Concepts of Chemistry: 7 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
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