Q.The empirical formula of benzene is
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A chemical formula can be reported at two levels of detail. The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of each atom in one molecule. This concept moves between percentage composition, the empirical formula, and the molecular formula.
1 — Percentage composition by mass. The mass percent of an element in a compound is (mass of that element in one formula unit / molar mass) × 100. In water, oxygen is 16 / 18 × 100 = 88.9%. Read it as: of every 100 g of the compound, this many grams are that element. Given the formula you can compute any element's percent; given the percent you can work backwards.
2 — Empirical formula from composition. The recipe (works from either percentages or actual combining masses):
- Step 1 — take the mass (or the percent, treated as grams per 100 g) of each element.
- Step 2 — divide each by its atomic mass to get moles (relative number of atoms).
- Step 3 — divide every mole value by the smallest of them to get a ratio.
- Step 4 — if the ratio is not already whole numbers, multiply all of them up by a small integer to clear it. A value ending in .5 multiplies by 2; .33 or .67 by 3; .25 or .75 by 4. Rounding a genuine .5 down to 1 is the classic error — clear it, do not round it.
3 — Empirical-formula mass (EFM). Add the atomic masses in the empirical formula. For CH₂O the EFM is 12 + 2(1) + 16 = 30. The EFM is the stepping stone to the molecular formula.
4 — Molecular formula. The molecular formula is a whole-number multiple of the empirical formula: molecular formula = n × (empirical formula), where
n = molar mass / empirical-formula mass.
n should come out to a whole number (round only tiny rounding error). If glucose has empirical formula CH₂O (EFM 30) and molar mass 180, then n = 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆. You can also get the molecular formula directly from percentages plus the molar mass: find the empirical formula, then scale by n.
5 — Combustion / elemental analysis. Burning a compound of C, H (and possibly O) in excess oxygen converts all the carbon to CO₂ and all the hydrogen to H₂O. Then:
- mass of C =
(12 / 44) × mass of CO₂(each CO₂ carries one C, of mass 12 out of 44); - mass of H =
(2 / 18) × mass of H₂O(each H₂O carries two H, mass 2 out of 18); - mass of O (if the compound contains oxygen) =
mass of sample − mass of C − mass of H(by difference — never from the CO₂/H₂O, whose oxygen came from the air). Convert those masses to moles and run the empirical-formula recipe; use the molar mass to getn.
6 — Water of crystallisation (hydrates). A hydrate salt·xH₂O is solved the same way: find the moles of anhydrous salt and the moles of water, take their ratio to get x. For example, from the mass lost on heating (the water driven off) you get moles of H₂O; divide by moles of salt for x in CuSO₄·5H₂O. …
The empirical formula expresses a compound's simplest whole-number ratio of atoms, which can differ from its actual molecular formula. …
The empirical formula shows the simplest whole-number ratio of atoms, while the molecular formula shows the actual number of atoms. Benzene's molecular formula is C6H6.
Step 1: Molecular formula of benzene = C6H6 (6 carbon atoms, 6 hydrogen atoms).
Step 2: Find the simplest ratio of C : H.
C : H = 6 : 6 = 1 : 1 (divide both by their HCF, 6).
Step 3: Write the empirical formula using this simplest ratio.
Empirical formula = CH.
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Showing the 12 most recent of 17 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The empirical formula of benzene is(a) C6H12(b) CH(c) C6H6(d) CH2
›Reveal solutionSolution
The empirical formula shows the simplest whole-number ratio of atoms, while the molecular formula shows the actual number of atoms. Benzene's molecular formula is C6H6.
Step 1: Molecular formula of benzene = C6H6 (6 carbon atoms, 6 hydrogen atoms).
Step 2: Find the simplest ratio of C : H.
C : H = 6 : 6 = 1 : 1 (divide both by their HCF, 6).
Step 3: Write the empirical formula using this simplest ratio.
Empirical formula = CH.
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- CBSE 2026Set ANNUAL1 markMCQQ.The empirical formula of a compound is CH2. Molar mass of compound is 42 gram. Its molecular formula is:(a) C3H6(b) C3H8(c) CH2(d) C2H2
›Reveal solutionSolution
Empirical formula mass of CH2 = 14 g/mol; 42/14=3, so molecular formula = (CH2)3=C3H6.
Step 1 — empirical formula mass: CH2 has mass =12+2(1)=14 g/mol.
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- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following is the molecular formula of sulphuric acid?(a) H2S(b) H2SO3(c) HNO3(d) H2SO4
›Reveal solutionSolution
Sulphuric acid's molecular formula is H2SO4.
Sulphuric acid is composed of 2 hydrogen atoms, 1 sulphur atom, and 4 oxygen atoms per molecule, giving the formula H2SO4. The other options are different acids/compounds: H2S …
- CBSE 2025Set ANNUAL1 markMCQQ.The empirical formula of the compound glucose is(a) CH2O(b) CHO(c) C6H12O6(d) C12H22O11
›Reveal solutionSolution
The empirical formula is the molecular formula's atoms reduced to their smallest whole-number ratio; C6H12O6 reduces to CH2O.
Glucose's molecular formula is C6H12O6. To get the empirical formula, divide every subscript by their greatest common factor (here, 6):
C: 6/6 = 1
H: 12/6 = 2
O: 6/6 = 1
So the empirical formula is CH2O (empirical formula mass = 12 + 2 + 16 = 30; the molecular formula C6H12O6 is exactly 6 times this, 180 g/mol, confirming the ratio is correct).
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- CBSE 2025Set hz1 markMCQQ.Select the correct one: A compound with empirical formula C3H4N has molecular mass 108 amu, the molecular formula is:(a) C3H4N(b) C6H8N2(c) C9H12N3(d) C12H16N4
›Reveal solutionSolution
Divide the given molecular mass by the empirical formula mass to get the whole-number multiplier n, then multiply every atom count in the empirical formula by n.
Step 1: Find the empirical formula mass of C3H4N.
C3H4N = 3(12) + 4(1) + 1(14) = 36 + 4 + 14 = 54 amu.
Step 2: Find n = Molecular mass / Empirical formula mass = 108 / 54 = 2. …
- CBSE 2025Set ANNUAL1 markMCQQ.Empirical formula of glucose is:(a) C6H12O6(b) CHO(c) CH2O(d) C2H4O2
›Reveal solutionSolution
Glucose's molecular formula C6H12O6 reduces to the empirical formula CH2O.
The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound, while the molecular formula shows the actual number of atoms.
Molecular formula of glucose = C6H12O6
Dividing all subscripts (6, 12, 6) by their highest common factor, 6: …
- CBSE 2024Set ANNUAL1 markMCQQ.The empirical formula of the compound CH3COOH is(a) C2H4O2(b) C4H8O4(c) C2H2O2(d) CH2O
›Reveal solutionSolution
The empirical formula is the molecular formula reduced to its lowest whole-number ratio; C2H4O2 divides down to CH2O.
The molecular formula of acetic acid, CH3COOH, written out atom-by-atom is C2H4O2 (2 C, 4 H, 2 O).
To get the empirical formula, divide each subscript by the highest common factor of 2, 4, 2, which is 2:
C: 2/2 = 1 …
- CBSE 2024Set sz1 markMCQQ.Select the correct one: The empirical formula of Benzene is:(a) C6H6(b) CH(c) CH3(d) C2H6
›Reveal solutionSolution
Benzene's molecular formula C6H6 reduces to the simplest whole-number atom ratio CH, which is its empirical formula.
The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound, while the molecular formula shows the actual number of atoms of each element in one molecule.
For benzene, the molecular formula is C6H6. Dividing both subscripts by their common factor (6) gives C1H1, i.e. CH. This is the simplest ratio in which carbon and hydrogen combine in benzene.
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- CBSE 2024Set ANNUAL1 markQ.Position isomer of propan-1-ol (C3H8O) is ___________.
›Reveal solutionSolution
[!TLDR]
propan-2-ol
Method
Propan-2-ol has the same molecular formula (C3H8O) but the -OH group at a different (2nd) carbon position, makin …
- CBSE 2023Set ANNUAL1 markMCQQ.An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be(a) CHO(b) CH2O(c) CH3O(d) CH4O
›Reveal solutionSolution
Converting the mass percentages to a mole ratio gives C:H:O = 1:3:1, so the empirical formula is CH3O.
Step 1 — find %O by difference: 100 − 38.71 − 9.67 = 51.62% O.
Step 2 — convert each mass percentage (per 100 g of compound) to moles, using atomic masses C=12, H=1, O=16:
- mol C = 38.71/12 = 3.226
- mol H = 9.67/1 = 9.670
- mol O = 51.62/16 = 3.226
Step 3 — divide by the smallest value (3.226) to get the simplest whole-number ratio:
- C: 3.226/3.226 = 1
- H: 9.670/3.226 ≈ 3 …
- CBSE 2023Set ANNUAL1 markMCQQ.The empirical formula and molecular mass of a compound are CH2O and 180g respectively. What will be the molecular formula of the compound?(a) C9H18O9(b) CH2O(c) C6H12O6(d) C2H4O2
›Reveal solutionSolution
The molecular formula is C6H12O6, six times the empirical formula CH2O.
Empirical formula mass of CH2O = 12 (C) + 2x1 (H) + 16 (O) = 30 g/mol.
n = molecular mass / empirical formula mass = 180/30 = 6.
Molecular formula = (CH2O) x 6 = C6H12O6 (this is, in fact, the molecular formula of glucose).
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- CBSE 2023Set ANNUAL1 markMCQQ.The empirical formula of a compound is CH2. Its gram molecular mass is 42 g. Its molecular formula will be:(a) CH4(b) C2H2(c) C3H6(d) C3H8
›Reveal solutionSolution
Molecular formula = (Empirical formula)n, where n = Molecular mass / Empirical formula mass; here n = 3, giving C3H6.
Step 1: Empirical formula mass of CH2 = 12 (C) + 2(1) (H) = 14 g/mol.
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