The marks lost here are almost never lost in the chemistry. They are lost in the last line of the calculation, when a correct empirical formula is submitted as the answer to a question that asked for the molecular formula.
Percentage composition tells you mass, not counting. To get counting, divide each element's mass per cent by that element's atomic mass — this converts a mass share into a mole share. Divide every result by the smallest of them and you have the simplest whole-number ratio of atoms: the empirical formula. That is the whole of the method, and NCERT Class 11 Chemistry Chapter 8 (printed page 285) is explicit that mass per cent is the input this calculation requires, referring the procedure itself back to Unit 1, Some Basic Concepts of Chemistry.
The empirical formula is a ratio. It is not a molecule. CH₂O is the empirical formula of formaldehyde, of acetic acid and of glucose alike — three different substances that share one ratio. To name the actual molecule you need one more piece of data, and the question will always have given it to you: the molar mass. Compute the empirical formula mass, divide the molar mass by it, and you get a small whole number n. Multiply every subscript by n. That is the molecular formula.
Two habits cause most of the damage. The first is rounding the mole numbers to whole numbers before dividing by the smallest; 2.50 rounded to 3 destroys the ratio that the division was about to reveal, and a ratio of 1 : 2.5 means you scale by two, not that you round. The second is stopping at the empirical formula when a molar mass was supplied — if a question hands you a molar mass, it has told you what it wants.
Watch out for the option list. Both answers will be sitting there.
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 expresses:
Answer: C. C is correct. The empirical formula is defined as the simplest whole-number ratio of the atoms of the constituent elements, which is why several different compounds can share one empirical formula. NCERT Class 11 Chemistry Chapter 8, printed page 285, specifies mass per cent as the input to this ratio calculation.
Why A is wrong: A is wrong because the actual number of atoms per molecule is the molecular formula, not the empirical formula; the two coincide only when the scaling factor happens to be one.
Why B is wrong: B is wrong because mass per cent is the experimental input to the calculation, not its output — mass shares must be converted to mole shares before any ratio of atoms exists.
Why D is wrong: D is wrong because the empirical formula carries no information about how many moles of atoms a mole of the substance contains; that follows only once the molecular formula is known.
To convert the mass per cent of an element into the quantity needed for a formula calculation, the mass per cent is divided by:
Answer: A. A is correct. Dividing an element's mass per cent by its own atomic mass converts a mass share into a mole share, which is the only quantity from which an atom ratio can be read. NCERT Class 11 Chemistry Chapter 8, printed page 285, identifies mass per cent as the required starting data for this route.
Why B is wrong: B is wrong because dividing by the molar mass of the whole compound mixes a per-element mass with a whole-molecule mass and yields no per-element mole count.
Why C is wrong: C is wrong because the empirical formula mass is not known until after the empirical formula has been found; it is used later, to obtain the scaling factor.
Why D is wrong: D is wrong because dividing by the smallest mass per cent is a step performed on mole numbers, not on mass numbers, and doing it first skips the mass-to-mole conversion entirely.
In addition to the empirical formula, the single further datum required to determine a compound's molecular formula is its:
Answer: C. C is correct. The molecular formula is the empirical formula scaled by a whole number n, and n is obtained only by dividing the molar mass by the empirical formula mass.
Why A is wrong: A is wrong because a boiling point identifies or characterises a substance but gives no numerical route to the scaling factor n.
Why B is wrong: B is wrong because solid-state density does not yield the mass of one mole and so cannot fix how many empirical units make up a molecule.
Why D is wrong: D is wrong because the carbon percentage is already part of the composition data used to reach the empirical formula; repeating it adds nothing new about molecular size.
A compound has the empirical formula CH₂O and a molar mass of 1.80 × 10² g mol⁻¹. Its molecular formula is (atomic masses: C = 12, H = 1, O = 16):
Answer: A. A is correct. The empirical formula mass of CH₂O is 12 + 2 + 16 = 30 g mol⁻¹, so n = 180 ÷ 30 = 6 and every subscript is multiplied by six to give C₆H₁₂O₆. NCERT Class 11 Chemistry Chapter 8, printed page 285, sets out mass per cent data as the basis of this class of calculation.
Why B is wrong: B is wrong because it uses n = 2, which corresponds to a molar mass of 60 g mol⁻¹, not the 1.80 × 10² g mol⁻¹ given.
Why C is wrong: C is wrong because it uses n = 3, corresponding to 90 g mol⁻¹ — exactly half the stated molar mass.
Why D is wrong: D is wrong because it reports the empirical formula unchanged, ignoring the molar mass that was supplied precisely so that the scaling factor could be applied.
Combustion data give a compound as 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its empirical formula is (atomic masses: C = 12, H = 1, O = 16):
Answer: D. D is correct. The mole numbers are 40.0/12 = 3.33, 6.7/1 = 6.7 and 53.3/16 = 3.33; dividing each by the smallest, 3.33, gives 1 : 2 : 1, so the empirical formula is CH₂O.
Why A is wrong: A is wrong because CHO has H in a 1 : 1 ratio with C, while the mole numbers 3.33 : 6.7 : 3.33 give twice as much H as C.
Why B is wrong: B is wrong because C₃H₄O₃ would contain 40.9% C and 54.5% O, not the given 40.0% and 53.3%; it does not fit the data.
Why C is wrong: C is wrong because it does not reduce to a simplest ratio consistent with the mole numbers; 2 : 3 : 2 does not follow from 3.33 : 6.7 : 3.33.
Elemental analysis of a compound gives mole numbers of 0.0500 mol of element X and 0.125 mol of element Y. The simplest whole-number ratio of X to Y is:
Answer: D. D is correct. Dividing both by the smaller value gives 1 : 2.5, and a ratio ending in .5 is cleared by multiplying both terms by two, giving 2 : 5.
Why A is wrong: A is wrong because it rounds 2.5 down to 2, discarding the half-unit that the multiplication step exists to clear.
Why B is wrong: B is wrong because it rounds 2.5 up to 3, which is the same rounding error in the opposite direction and equally destroys the ratio.
Why C is wrong: C is wrong because it treats two unequal mole numbers as equal and so ignores the division by the smaller value altogether.
Methanal (formaldehyde), ethanoic acid and glucose all share the empirical formula CH₂O. This shows that the empirical formula:
Answer: B. B is correct. An empirical formula is a ratio, and the three substances named differ in the scaling factor n applied to that ratio, so the ratio alone cannot distinguish them.
Why A is wrong: A is wrong because the example given is a direct counter-case: three different substances answer to one empirical formula.
Why C is wrong: C is wrong because glucose, C₆H₁₂O₆, contains exactly those three elements and its molecular formula differs from its empirical formula by a factor of six.
Why D is wrong: D is wrong because the empirical formula follows from percentage composition alone; the molar mass is needed only for the subsequent step to the molecular formula.
A compound contains 92.3% carbon and 7.7% hydrogen by mass and has a molar mass of 78 g mol⁻¹. Its molecular formula is (atomic masses: C = 12, H = 1):
Answer: B. B is correct. Mole numbers are 92.3/12 = 7.69 and 7.7/1 = 7.7, giving a 1 : 1 ratio and the empirical formula CH of mass 13 g mol⁻¹; n = 78 ÷ 13 = 6, so the molecular formula is C₆H₆.
Why A is wrong: A is wrong because it stops at the empirical formula even though a molar mass was supplied, which is the single most common way marks are lost on this calculation.
Why C is wrong: C is wrong because it applies n = 2, corresponding to a molar mass of 26 g mol⁻¹ rather than the 78 g mol⁻¹ stated.
Why D is wrong: D is wrong because its constituent masses total 78 g mol⁻¹ only by coincidence of arithmetic and it does not preserve the 1 : 1 carbon-to-hydrogen ratio the composition data require.
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Given
• Mass per cent of carbon = 54.5%• Mass per cent of hydrogen = 9.1%• Mass per cent of oxygen = 100 − 54.5 − 9.1 = 36.4%• Molar mass = 8.8 × 10¹ g mol⁻¹• Atomic masses (exact, by definition of the scale used here): C = 12, H = 1, O = 16
Required
The molecular formula — not the empirical formula. The molar mass has been supplied, which is the signal that the answer must be scaled.
Concept
Mass per cent is a mass share. An atom ratio is a counting share. Dividing each element's mass per cent by its atomic mass performs that conversion; dividing through by the smallest result reduces the counts to a simplest ratio. The molar mass then fixes how many such ratio units make one molecule. NCERT Class 11 Chemistry Chapter 8, printed page 285, states that mass per cent is the datum this calculation requires.
Formula
• moles of element ∝ (mass per cent) ÷ (atomic mass)• n = (molar mass) ÷ (empirical formula mass)• molecular formula = (empirical formula) × n
Substitution
• C: 54.5 ÷ 12• H: 9.1 ÷ 1• O: 36.4 ÷ 16
Calculation
Mole numbers:• C: 54.5 ÷ 12 = 4.54• H: 9.1 ÷ 1 = 9.1• O: 36.4 ÷ 16 = 2.28
Divide by the smallest, 2.28 — and carry the decimals through this division. Rounding 4.54 to 5 or 9.1 to 9 before dividing is the error the dossier names, and it changes the answer:• C: 4.54 ÷ 2.28 = 1.99 ≈ 2• H: 9.1 ÷ 2.28 = 3.99 ≈ 4• O: 2.28 ÷ 2.28 = 1.00 = 1
Empirical formula: C₂H₄O
Empirical formula mass = (2 × 12) + (4 × 1) + (1 × 16) = 44 g mol⁻¹
n = 88 ÷ 44 = 2
The atomic masses and the subscripts 2, 4 and 1 are exact counting values and carry no measurement uncertainty, so they do not contribute to the significant-figure count; the data limiting the result are the three mass percentages and the molar mass.
Final answer
Molecular formula = C₂H₄O scaled by 2 = C₄H₈O₂
(Check: 4 × 12 + 8 × 1 + 2 × 16 = 48 + 8 + 32 = 88 g mol⁻¹ ✓)
Common trap
Two traps, both named in the dossier's negative-marking notes, and both will be sitting in the option list.
The first is submitting C₂H₄O. It is a correct empirical formula and a wrong answer, because a molar mass was given. Treat the presence of a molar mass in the stem as an instruction: the question is asking for the molecular formula.
The second is rounding early. Round 4.54 → 5, 9.1 → 9 and 2.28 → 2 at the mole-number stage and the ratio becomes 5 : 9 : 2, which reduces to nothing sensible and produces a formula that fails the molar-mass check. Rounding is the last operation, applied to the quotients after division by the smallest, and only when the quotient is already within about 0.02 of a whole number. A quotient near .5, .33 or .67 is not a rounding candidate — it is a signal to multiply the whole set by 2 or 3.
Similar NEET-style question
A compound contains 26.7% carbon, 2.2% hydrogen and 71.1% oxygen by mass, and its molar mass is 9.0 × 10¹ g mol⁻¹. Determine its molecular formula. (Atomic masses: C = 12, H = 1, O = 16. Answer: mole numbers 2.22, 2.2, 4.44 → ratio 1 : 1 : 2 → empirical formula CHO₂, empirical formula mass 45 → n = 2 → C₂H₂O₄.)
Quantitative analysis of compounds is very important in organic chemistry. It helps chemists in the determination of mass per cent of elements present in a compound. You have learnt in Unit-1 that mass per cent of elements is required for the determination of emperical and molecular formula. The percentage composition of elements present in an organic compound is determined by the following methods:
-- NCERT Class 11 Chemistry, Ch. 8, p. 285An 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. 19Problem 1.2 A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas? Solution Step 1. Conversion of mass per cent to grams Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present. Step 2. Convert into number moles of each element Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound Moles of hydrogen = 4.07 g / 1.008 g = 4.04 Moles of carbon = 24.27 g / 12.01 g = 2.021 Moles of chlorine = 71.65 g / 35.453 g = 2.021 Step 3. Divide each of the mole values obtained above by the smallest number amongst them Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl. In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient. Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements CH2Cl is, thus, the empirical formula of the above compound. Step 5. Writing molecular formula (a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula. For CH2Cl, empirical formula mass is 12.01 + (2 ×1.008) + 35.453 = 49.48 g (b) Divide Molar mass by empirical formula mass = 2 = (n) (c) Multiply empirical formula by n obtained above to get the molecular formula Empirical formula = CH2Cl, n = 2. Hence molecular formula is C2H4Cl2.
-- NCERT Class 11 Chemistry, Ch. 1, p. 191 question from NEET 2021. Answers verified against NTA official keys.
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