Percentage Composition, Empirical and Molecular Formula

Number analogy questions reward speed, but speed comes from a checklist, not from cleverness. Once your eyes know what to scan for, two-mark questions in the RPF SI reasoning section start dropping in under twenty seconds each.

Definition: A number analogy asks you to spot the mathematical relation hidden inside the first pair of numbers (A : B) and apply the same relation to a second pair (C : ?). Your job is to find a unique, reusable operation that takes A to B and apply it to C.

The five relations that cover ~90% of analogy questions

Almost every number analogy in RPF SI, RPF Constable, SSC CHSL and SSC CGL papers belongs to one of these families. Learn them in order — the cheapest computation first.

1. Square / Cube — the second number is the second or third power of the first. 4 : 16 (4² = 16), 5 : 125 (5³ = 125). When B is much bigger than A and ends in a typical perfect-square or cube digit (0, 1, 4, 5, 6, 9 for squares; 0, 1, 7, 8, 27 endings for cubes), check this first.
2. Add / Subtract a constant — B − A is a small whole number that repeats. 7 : 11 (+4), 9 : 13 (+4). Useful when B is just slightly larger than A.
3. Multiply / Divide — B is a small integer multiple or quotient of A. 6 : 36 (×6), 8 : 4 (÷2).
4. n and n+1 / consecutive integer product — B = A × (A + 1) or similar. 5 : 30 (5 × 6), 7 : 56 (7 × 8). These look like "near squares" but are slightly bigger than A².
5. Sum / Reverse / Digit-manipulation — B is built from the digits of A. 12 : 21 (digits reversed), 23 : 5 (2 + 3), 34 : 12 (3 × 4). These are the ones that defeat people who only scan for "math operations".

The SCADM checklist — your test order

Memory aid 'SCADM' — Square, Cube, Add, Difference (and divide/multiply), Manipulate digits. Test in this order because:

• Squares and cubes either fit exactly or don't — one mental check, no ambiguity.
• Add/subtract checks are arithmetic of single digits — three seconds.
• Multiplicative relations are usually obvious from the size ratio.
• Digit manipulation is the slowest test, so leave it for last.

The discipline is: whichever operation you commit to, it must give the same answer when applied to both pairs. If 4 : 16 = 4² holds, then the second pair 5 : ? must also be 5² = 25. If 25 is not in the options but 30 is, then 4 : 16 was not really a square relation — it was 4 × 4 — but 4 × 4 and 4² give the same answer, so the test that distinguishes them lies in the second pair. Always confirm the relation against both pairs before circling.

Why it matters: RPF SI dedicates 10–15 marks to analogy and classification. These are the easiest marks on the paper, but only if you do not lose 90 seconds per question. A checklist transforms an open-ended puzzle into a closed-ended sieve, and your accuracy stays above 95% even under time pressure.

Grouped numbers — a separate trick

When the question gives a triple like (3, 9, 27) and asks you to extend it, think geometric progression: each term is the previous one multiplied by a common ratio. Here the ratio is 3, so the next term would be 81. For (2, 4, 8, ?), ratio 2 gives 16. For (1, 4, 9, 16, ?), the pattern is not GP but square of natural numbers, so the next is 25.

For arithmetic progressions like (5, 9, 13, 17, ?), the common difference is 4 and the next term is 21. Train your eye to spot the constant difference, the constant ratio, or the underlying sequence (squares, cubes, primes, Fibonacci) within two seconds.

Worked example

Question: 6 : 42 :: 8 : ? (Options: 56, 64, 72, 81)

Solution:
Step 1 (Square test): 6² = 36, not 42. Reject.
Step 2 (Cube test): 6³ = 216. Reject.
Step 3 (Add test): 42 − 6 = 36. Apply to 8: 8 + 36 = 44. Not in options. Probably not this.
Step 4 (Multiply test): 42 / 6 = 7. Apply to 8: 8 × 7 = 56. Present in options. Hold this hypothesis.
Step 5 (n(n+1) test, the elegant fit): 6 × 7 = 42. Apply to 8: 8 × 9 = 72. Also present.

Two candidates survive — 56 and 72. Pick the relation that is more specific or more elegant. Multiplying by 7 is arbitrary (why 7 and not 6 or 8?). But n × (n + 1) is a rule: take the number and multiply it by its successor. Both pairs follow it: 6 × 7 = 42, 8 × 9 = 72. The rule wins.

Conclusion: The answer is 72.

The lesson is built in: when two operations both fit the first pair, the more general / more specific one usually wins on the second pair. Always test all surviving hypotheses on the second pair before circling.

Real-world example: Railway scheduling software hidden inside an RRB-managed signalling system uses simple proportional rules — if a track of 6 km handles 42 trains per day, an 8-km track of the same design handles 8 × 7 = 56 (one extra train per kilometre due to load) or 8 × 9 = 72 (square-of-successor capacity). Engineers explicitly choose between these "n × (n+1)" and "n × constant" models the same way you choose between two analogy hypotheses — by checking which one matches the second observed data point.

Common misconception: "Whatever fits the first pair must be the answer." It is not. If two operations fit the first pair, the question is asking which one also fits the implicit pattern of the second pair — exactly as the worked example showed. Never circle after checking only one pair.

Common misconception: Treating "addition" and "multiplication" as interchangeable. 4 : 8 could be +4 or ×2. 5 : ? would give 9 or 10. The first pair alone cannot decide. The exam always gives you the means to decide, but only if you compute both possibilities.

Common misconception: Ignoring the "odd one out" framing. Some questions ask which pair does not follow the rule. In that case, find the rule that fits the majority and pick the rebel. Speed-readers often mark the rebel as the "answer pair" by mistake.

Walk into any NEET Chemistry classroom and ask, "What is the formula of glucose?" Everyone says C₆H₁₂O₆ instantly. Now ask, "What is its simplest formula?" — and the room hesitates. That hesitation is exactly what the empirical-vs-molecular formula chapter exists to remove.

Definition: An empirical formula is the simplest whole-number ratio of the atoms of each element present in a compound.
Definition: A molecular formula is the actual number of atoms of each element present in one molecule of the compound.

The molecular formula is always either equal to the empirical formula or a whole-number multiple of it: Molecular formula = n × Empirical formula, where n = molecular mass ÷ empirical formula mass.

Why Two Formulas Exist

Imagine four different compounds — methanal (HCHO), acetic acid (CH₃COOH), glucose (C₆H₁₂O₆) and ribose (C₅H₁₀O₅). If you compute the ratio C : H : O in each, you get 1 : 2 : 1 every single time. So all four share the same empirical formula CH₂O. The empirical formula tells you the composition; the molecular formula tells you the identity. NEET routinely tests this distinction by giving you percentage data, asking for the empirical formula, then handing you a molar mass and asking for the molecular formula.

The Four-Step Method — Memory Hook "MMD-R-W"

Whenever you are given percentage composition of a compound, follow these four steps in order. The mnemonic MMD-R-W captures the verbs: Mass, Moles (divide by atomic mass), Divide by smallest, Round to Whole numbers.

Step 1 — Take Mass %: Assume you have a 100 g sample. Then the percentage of each element directly becomes its mass in grams. (If you are given grams instead of percent, skip this step.)

Step 2 — Divide by atomic Mass: For each element, divide the mass (g) by its atomic mass to get the number of moles. This is the bridge from grams to atoms.

Step 3 — Divide all by the smallest mole value: This gives a simple ratio of atoms. You are essentially asking, "If one atom of the smallest-mole element is present, how many atoms of the others are there?"

Step 4 — Round to Whole numbers: If the ratio comes out as 1, 2, 2.99 — round 2.99 to 3. If it comes out as 1.5, multiply every ratio by 2 to clear the fraction. For 1.33 (i.e., 4/3), multiply by 3. Never round 1.5 down to 1 or up to 2 — multiplication is the only safe move.

That gives you the empirical formula. To upgrade to the molecular formula, compute:

n = (given molecular mass) ÷ (empirical formula mass), then multiply each subscript in the empirical formula by n.

A Classic Worked Example — Glucose

Question: A compound is found to contain 40.0% C, 6.7% H and 53.3% O by mass. Its molecular mass is 180 g/mol. Find the empirical and molecular formulas.

Solution:
Step 1 — Take mass: in 100 g sample → 40.0 g C, 6.7 g H, 53.3 g O.
Step 2 — Moles: 40.0 / 12 = 3.33 mol C; 6.7 / 1 = 6.7 mol H; 53.3 / 16 = 3.33 mol O.
Step 3 — Divide by smallest (3.33): C = 3.33/3.33 = 1; H = 6.7/3.33 = 2.01 ≈ 2; O = 3.33/3.33 = 1.
Step 4 — Whole-number ratio: C : H : O = 1 : 2 : 1, giving empirical formula CH₂O.
Now upgrade: empirical formula mass = 12 + 2(1) + 16 = 30 g/mol. n = 180 / 30 = 6.
Conclusion: Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ — glucose, as expected.

A Trickier Example — Handling Fractions

Question: A hydrocarbon contains 85.7% C and 14.3% H. Its molar mass is 56 g/mol. Find the molecular formula.

Solution:
Step 1: 85.7 g C, 14.3 g H in 100 g.
Step 2: 85.7 / 12 = 7.14 mol C; 14.3 / 1 = 14.3 mol H.
Step 3: Divide by smallest (7.14): C = 1, H = 14.3 / 7.14 = 2.0.
Step 4: Ratio = 1 : 2 → empirical formula CH₂. Empirical mass = 14.
n = 56 / 14 = 4.
Conclusion: Molecular formula = (CH₂)₄ = C₄H₈ (likely 1-butene or 2-butene).

When the Ratio is 1.5, 1.33, or 2.5

These fractional ratios scare students, but they are mechanical to handle.

• If a ratio comes out as 1.5 (= 3/2), multiply all ratios by 2. Example: 1 : 1.5 → 2 : 3.
• If it is 1.33 (= 4/3), multiply by 3. Example: 1 : 1.33 → 3 : 4.
• If it is 2.5, multiply by 2 to get 5.

Never round these directly — you would change the chemistry.

Why it matters

Percentage composition and formula determination is a direct one-marker in NEET almost every year, and the same logic appears in JEE and KVPY. More importantly, the conceptual idea — that composition alone does not fix a compound — underpins later topics: isomerism, polymerisation (a polymer has the same empirical formula as its monomer), and combustion analysis where you back-calculate composition from CO₂ and H₂O masses.

Real-world example

A pharmaceutical chemist analysing a new drug starts exactly the same way: burn a known mass of the compound, measure the CO₂ and H₂O produced, convert to percent composition, then apply MMD-R-W to derive an empirical formula. The molecular mass — from mass spectrometry — gives the multiplier n. This is how penicillin's formula was deduced in the 1940s.

Common misconception

Many students believe the empirical and molecular formulas are always different. They are not — in many simple compounds they are identical. Water (H₂O), methane (CH₄), carbon dioxide (CO₂), ammonia (NH₃) — for all of these, n = 1, so the empirical and molecular formulas coincide. Empirical and molecular differ only when n > 1 (e.g., glucose with n = 6, or benzene C₆H₆ with empirical CH and n = 6).

A second misconception: that you should round 1.5 to either 1 or 2. Never. Multiply by 2 instead. Rounding here would change a sulphate from SO₄²⁻ to something nonsensical.

From a few percentages and a molar mass, you can walk straight to the molecular formula of an unknown compound. This is one of the cleanest, most repeated patterns in Class 11 chemistry, and NEET loves it.

Definition: Empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. Example: the empirical formula of glucose is CH₂O.

Definition: Molecular formula gives the actual number of atoms of each element in one molecule of the compound. For glucose this is C₆H₁₂O₆ — six times the empirical unit.

Definition: Percentage composition is the mass percent of each element in the compound, found by elemental analysis (in the lab) or given to you (in the exam).

The five-step recipe

Every problem of this type follows the same five steps:

1. Assume 100 g of compound. Each percentage then converts directly into grams.
2. Divide each element's mass by its atomic mass to get moles.
3. Divide all the mole values by the smallest of them. You now have a mole ratio normalised so the smallest atom is 1.
4. If the ratios are close to whole numbers, you have the empirical formula straight away. If they end in clean fractions (0.5, 0.33, 0.67, 0.25, 0.75), multiply every ratio by the matching small integer (2, 3, 3, 4, 4) to clear them.
5. To go from empirical to molecular, compute the empirical formula mass, divide the given molar mass by it to get n = (molar mass) ÷ (empirical mass), and multiply every subscript in the empirical formula by n.

That is the whole technique. The rest is arithmetic.

The worked problem

A compound contains 40% C, 6.7% H and 53.3% O. Its molar mass is 60 g/mol. Find its molecular formula.

Question: Determine the molecular formula of this compound.

Solution:

Step 1: Assume 100 g of compound. Then the elements are present as 40 g of C, 6.7 g of H, and 53.3 g of O.

Step 2: Convert grams to moles using atomic masses (C = 12, H = 1, O = 16).

• Moles of C = 40 ÷ 12 = 3.33
• Moles of H = 6.7 ÷ 1 = 6.70
• Moles of O = 53.3 ÷ 16 = 3.33

Step 3: Divide each by the smallest mole value, 3.33.

• C : 3.33 / 3.33 = 1
• H : 6.70 / 3.33 ≈ 2
• O : 3.33 / 3.33 = 1

Step 4: The ratio C : H : O is 1 : 2 : 1, all whole numbers. Hence the empirical formula is CH₂O. Empirical formula mass = 12 + (2 × 1) + 16 = 30 g/mol.

Step 5: n = (molar mass) ÷ (empirical mass) = 60 ÷ 30 = 2.

Multiply each subscript in CH₂O by 2 → C₂H₄O₂.

Conclusion: The molecular formula is C₂H₄O₂ — which is acetic acid (CH₃COOH), the active ingredient in vinegar.

Always verify with a back-calculation

A topper's habit is to verify by computing the molar mass of the final formula and comparing it with the given value:

Molar mass of C₂H₄O₂ = (2 × 12) + (4 × 1) + (2 × 16) = 24 + 4 + 32 = 60 g/mol. Matches. The answer is internally consistent.

This 30-second check catches arithmetic slips and saves marks in NEET, where one wrong subscript spoils the whole question.

Reading the decimal — when ratios aren't whole numbers

In Step 4 the ratios came out as whole numbers, but that is not always the case. Train your eye to recognise common decimal endings:

• .50 → multiply all ratios by 2.
• .33 or .67 → multiply by 3.
• .25 or .75 → multiply by 4.
• .20, .40, .60, .80 → multiply by 5.

For example, if your ratios came out as C : H : O = 1 : 1.33 : 1, multiply by 3 to get 3 : 4 : 3 — empirical formula C₃H₄O₃. Never round 1.33 to 1; that would lose a whole carbon atom and break the molecule.

Why it matters: questions of exactly this shape appear in NEET, JEE Main, and CBSE Class 11 board papers every year, often as a 4-mark numerical or a 1-mark MCQ where the molecular formula is one of four options. The arithmetic is light, so the question is really about method discipline — assume 100 g, moles, divide by smallest, clear fractions, scale to molar mass.

Real-world example: acetic acid (C₂H₄O₂) is the very acid that gives vinegar its sour taste and is used in pickling. Its empirical and molecular formulas are the textbook pair to remember. Glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) all share the same empirical formula CH₂O — that is the very point of distinguishing empirical from molecular: percentage composition alone cannot identify the compound; you also need the molar mass.

Common misconception: students often confuse the empirical and the molecular formula, treating CH₂O as the answer when the question asked for the molecular formula. The empirical formula is only the simplest ratio. Without the molar mass, you cannot reach the molecular formula. The other frequent slip is rounding too aggressively — for instance, treating 2.5 as 2 or 3. The rule is: never round mid-way; clear the fraction by multiplying.

Cover the answer, recall, then check. 11 cards on composition and formula determination.

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

Q2. Define molecular formula.
A2. The formula giving the actual number of atoms of each element in one molecule (e.g. C₆H₁₂O₆ for glucose).

Q3. How are molecular and empirical formulas related?
A3. Molecular formula = n × empirical formula, where n = molecular mass ÷ empirical formula mass (n is a whole number ≥ 1).

Q4. Give the steps to find an empirical formula from percentage composition.
A4. (1) Take % as grams, (2) divide each by its atomic mass to get moles, (3) divide all by the smallest, (4) round to nearest whole numbers (multiply up if needed).

Q5. How do you calculate mass % of an element in a compound?
A5. (mass of that element in 1 mol ÷ molar mass of compound) × 100.

Q6. Find % of O in water (H₂O).
A6. (16 ÷ 18) × 100 = 88.9%.

Q7. A compound is 40% C, 6.7% H, 53.3% O. Find its empirical formula.
A7. Moles: C 40/12 = 3.33, H 6.7/1 = 6.7, O 53.3/16 = 3.33. Divide by 3.33 → C₁H₂O₁ = CH₂O.

Q8. If the compound in Q7 has molar mass 180, find its molecular formula.
A8. Empirical mass CH₂O = 30. n = 180/30 = 6 → C₆H₁₂O₆.

Q9. When can the empirical and molecular formulas be identical?
A9. When n = 1, i.e. the molecular mass equals the empirical formula mass (e.g. H₂O, CO₂, NH₃).

Q10. What ratio do you round in step 3, and when do you multiply instead of round?
A10. The mole ratio; if a value ends near .5 (e.g. 2.5), multiply all by 2 rather than rounding, to keep whole numbers.

Q11. In combustion analysis, how are C and H masses found from CO₂ and H₂O?
A11. mass C = mass CO₂ × (12/44); mass H = mass H₂O × (2/18); O is found by difference from the sample mass.

This topic turns experimental data (mass %, combustion products) into a chemical formula. The procedure is mechanical once memorised, and it appears every year — often as a two-step "empirical then molecular" problem.

The two formulas

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

Percentage composition

Mass % of an element = (mass of element in 1 mol of compound ÷ molar mass) × 100. Reverse the process to go from % to formula.

Empirical formula in four steps

Step	Action
1	Treat each % as grams (assume 100 g sample)
2	Divide each mass by the element's atomic mass → moles
3	Divide all mole values by the smallest
4	Round to whole numbers; if a ratio is ~x.5, multiply all by 2

Then compute empirical mass, find n = M/empirical mass, and scale up for the molecular formula.

Exam Tricks & Tips

• 🎯 Assume a 100 g sample so percentages become grams directly — the cleanest setup.
• 🎯 Divide by the smallest mole value in step 3; never by an atomic mass again.
• 🎯 Don't force-round 2.5 → 3. Multiply the whole ratio by 2 (→ 5) to preserve the true stoichiometry.
• 🎯 Combustion analysis: C from CO₂ (×12/44), H from H₂O (×2/18), O by difference — a standard organic question.
• 🎯 n must be a whole number ≥ 1. If you get 2.9, round to 3; a badly non-integer n signals an arithmetic slip.
• ❌ Common mistake: reporting the empirical formula when the molecular mass is given — the question wants the molecular formula (multiply by n).

Expected exam pattern

One structured numerical: given mass % (or combustion data) and molar mass, find the molecular formula. Occasionally a direct mass-% calculation. Fully scorable with the four-step drill.

Quick recap

% → grams → moles (÷ atomic mass) → ÷ smallest → whole-number ratio = empirical formula. n = molar mass ÷ empirical mass; molecular formula = n × empirical. Watch for x.5 ratios (multiply, don't round) and combustion C/H/O bookkeeping.