Empirical and molecular formula

Empirical formula (EF) gives the simplest whole-number ratio of atoms in a compound; molecular formula (MF) gives the actual number of atoms per molecule. Relationship: MF = n × EF, where n = Molar mass of compound / Empirical formula mass, and n is always a positive integer (1, 2, 3...). Example: glucose has EF = CH2O (formula mass 30) but MF = C6H12O6 (molar mass 180), so n = 180/30 = 6. Note that ionic compounds (NaCl) and many solids are written only as empirical formulas. Memory aid: 'Empirical = Easiest ratio, Molecular = Multiple of it.' For elements like benzene (C6H6) and acetylene (C2H2), both share EF = CH but differ in n. Always reduce subscripts by their GCD to get EF from MF.

Given mass % composition: (1) Assume 100 g sample, so % becomes grams. (2) Divide each element's mass by its atomic mass to get moles. (3) Divide all mole values by the smallest to get a ratio. (4) If ratios aren't whole numbers, multiply all by a small integer (e.g. 1.5 → ×2, 1.33 → ×3, 1.25 → ×4). This gives the empirical formula. (5) For MF: n = M(molar) / M(empirical), then MF = (EF)n. Multiplier cheatsheet: decimal .5 → ×2, .33/.67 → ×3, .25/.75 → ×4, .2 → ×5. Common trap: round only at the final ratio step, never round intermediate mole values prematurely (e.g. don't call 2.49 → 2; it's likely 2.5 → ×2 = 5).

A compound contains C = 40.0%, H = 6.7%, O = 53.3%; molar mass = 180 g/mol. Find MF.
Step 1 (moles): C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.
Step 2 (divide by smallest 3.33): C = 1, H = 2.01 ≈ 2, O = 1.
Empirical formula = CH2O, EF mass = 12 + 2 + 16 = 30.
Step 3: n = 180/30 = 6.
Molecular formula = (CH2O)6 = C6H12O6 (glucose).
Verification: 6×12 + 12×1 + 6×16 = 72 + 12 + 96 = 180. Correct. JEE tip: if combustion data is given, all C goes to CO2 (mass C = 12/44 × mass CO2) and all H goes to H2O (mass H = 2/18 × mass H2O); O is found by difference from total sample mass.

Cover the answer, recall, then check. 11 cards on empirical and molecular formulae for JEE Main.

Q1. Define empirical formula.
A1. The simplest whole-number ratio of atoms of each element in a compound (e.g. CH for benzene, CH₂O for glucose).

Q2. Define molecular formula in terms of the empirical formula.
A2. Molecular formula = n × (empirical formula), where n = molar mass / empirical formula mass (n is a positive integer).

Q3. Steps to get the empirical formula from percentage composition.
A3. (1) % → grams (assume 100 g). (2) grams → moles (÷ atomic mass). (3) divide all moles by the smallest. (4) round to whole numbers (×2, ×3 if needed).

Q4. Glucose is 40% C, 6.7% H, 53.3% O. Find its empirical formula.
A4. C: 40/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33 → ratio 1:2:1 → CH₂O.

Q5. Glucose's molar mass is 180. Find its molecular formula.
A5. Empirical mass CH₂O = 30; n = 180/30 = 6 → C₆H₁₂O₆.

Q6. Why can two different compounds share an empirical formula?
A6. Empirical formula gives only the ratio, not the actual count. Ethyne C₂H₂ and benzene C₆H₆ both have empirical formula CH.

Q7. In combustion analysis, how is carbon mass found from CO₂?
A7. mass C = (12/44) × mass of CO₂ produced (each CO₂ carries one C).

Q8. In combustion analysis, how is hydrogen mass found from H₂O?
A8. mass H = (2/18) × mass of H₂O produced (each H₂O carries two H).

Q9. A hydrocarbon gives 0.44 g CO₂ and 0.18 g H₂O. Find mole ratio C:H.
A9. C = (12/44)(0.44) = 0.12 g → 0.01 mol; H = (2/18)(0.18) = 0.02 g → 0.02 mol → C:H = 1:2 → CH₂.

Q10. When do empirical and molecular formulae coincide?
A10. When n = 1, i.e. molar mass equals empirical formula mass (e.g. H₂O, CO₂, NH₃).

Q11. After dividing by the smallest, you get a ratio like 1 : 1.5. What do you do?
A11. Multiply all by 2 to clear the decimal → 2 : 3. Never round 1.5 to 2.

Empirical and molecular formulae turn raw experimental data (percentage composition, combustion products) into a chemical formula. JEE Main almost always has one such numerical, and combustion analysis is a favourite because it packages stoichiometry, mole concept and formula-finding into one problem.

Two formulae, one relationship

• Empirical formula: simplest whole-number atom ratio (CH₂O).
• Molecular formula: actual atom count (C₆H₁₂O₆).
• Link: Molecular formula = n × empirical, where n = molar mass ÷ empirical formula mass.

The universal method

Step	Action
1	Assume 100 g → % becomes grams
2	Divide each mass by its atomic mass → moles
3	Divide all mole values by the smallest
4	Multiply to clear decimals → whole-number ratio
5	Compute n = M / (empirical mass) for the molecular formula

Combustion analysis: mass C = (12/44) × m(CO₂); mass H = (2/18) × m(H₂O); mass O = total − (C + H).

Exam Tricks & Tips

• 🎯 Never round 1.33, 1.5, 1.25, 1.66 — multiply by 3, 2, 4, 3 respectively to reach whole numbers.
• 🎯 In combustion problems, oxygen mass is found by difference, not measured, because combustion adds O₂.
• 🎯 The 12/44 and 2/18 factors are worth memorising — they appear in every combustion question.
• 🎯 If asked only for empirical formula, you don't need the molar mass; if asked for molecular formula, you must have it.
• 🎯 Same empirical formula ≠ same compound — always check whether the question wants the ratio or the real molecule.
• ❌ Common mistake: dividing percentages by molecular mass instead of atomic mass, or forgetting that one H₂O carries two hydrogen atoms.

Expected exam pattern

"A compound contains X% … , molar mass Y; find its molecular formula," or a combustion problem giving masses of CO₂ and H₂O. Answers are often integer subscripts, suiting numerical-type questions.

Quick recap

% → g → moles → divide by smallest → clear decimals → empirical formula. Then n = M / empirical mass gives the molecular formula. In combustion, use 12/44 (C) and 2/18 (H), and get O by difference.