Some Basic Concepts of Chemistry

Memory aid 'C-D-MM-RP-G': (1) Law of Conservation of Mass (Lavoisier) - matter is neither created nor destroyed. (2) Law of Definite/Constant Proportions (Proust) - a pure compound always has the same elements in the same fixed mass ratio. (3) Law of Multiple Proportions (Dalton) - when two elements form more than one compound, the masses of one combining with a fixed mass of the other are in small whole-number ratios (e.g., CO and CO2: O ratio 1:2). (4) Gay-Lussac's Law of Gaseous Volumes - gases combine in simple whole-number volume ratios at same T,P. (5) Avogadro's Law - equal volumes of gases at same T,P contain equal number of molecules. Avogadro's law resolved the atom-molecule confusion of Dalton's theory.

Avogadro number NA = 6.022 x 10^23. Number of moles can be found four ways: n = given mass/molar mass = number of particles/NA = volume at STP/22.4 L (gases) = molarity x volume(L). At STP (273.15 K, 1 bar) molar volume = 22.7 L; at 273.15 K and 1 atm it is 22.4 L. Mass of one atom = molar mass/NA. Number of molecules = n x NA; number of atoms = n x NA x (atomicity). Shortcut: 1 mole = molar mass in grams = 6.022 x 10^23 particles = 22.4 L gas at STP(1 atm). Remember total moles of atoms in a molecule = moles of molecules x number of atoms per molecule.

Carbon forms two oxides. In CO, 12 g C combines with 16 g O. In CO2, 12 g C combines with 32 g O. Fixing carbon mass at 12 g, the oxygen masses are 16 g and 32 g, giving ratio 16:32 = 1:2, a simple whole-number ratio - illustrating the Law of Multiple Proportions. Another example: H2O (2 g H : 16 g O) and H2O2 (2 g H : 32 g O) gives oxygen ratio 1:2. Tip: always fix the mass of one element, then take the ratio of the other element's masses to verify the law quickly in exams.

One atomic mass unit (amu / u) = 1/12th the mass of one C-12 atom = 1.66 x 10^-24 g. Average atomic mass accounts for isotopic abundance: Average mass = Sum(isotopic mass x fractional abundance). Example for chlorine: (35 x 0.7577) + (37 x 0.2423) = 35.45 u. This is why atomic masses on the periodic table are usually non-integers. Memory aid: 'AVERAGE = mass-weighted by abundance'. The unified mass unit 'u' has replaced 'amu' in modern usage but both are accepted. Gram atomic mass = atomic mass expressed in grams = mass of 1 mole of atoms.

Why does a chemistry textbook say "molecular mass of water = 18 u" but "formula mass of sodium chloride = 58.5 u" — never the other way round? The distinction is not pedantic. It traces back to whether the substance exists as discrete molecules or as an infinite lattice of ions, and getting it right is what separates a NEET aspirant who has understood mole concept from one who is just memorising numbers.

Definition: Molecular mass is the sum of the atomic masses of all atoms in a single molecule of a molecular (covalent) substance. Unit: atomic mass unit (u), where 1 u = 1/12 the mass of a carbon-12 atom.

Definition: Formula mass is the sum of the atomic masses of all atoms in one formula unit of an ionic compound. Unit: u, same scale.

Definition: Gram molecular mass is the molecular (or formula) mass expressed in grams; numerically it equals the mass of one mole of those molecules or formula units.

Why the word "molecule" matters

A molecule is a discrete, electrically neutral group of atoms held together by covalent bonds — H₂O, CO₂, NH₃, C₆H₁₂O₆. You can in principle pick out a single H₂O molecule.

An ionic compound like NaCl does NOT exist as a discrete NaCl molecule. In a salt crystal, every Na⁺ ion is surrounded by 6 Cl⁻ ions and vice versa, in an extended three-dimensional lattice. There is no such thing as "one molecule of NaCl"; there are only many ions arranged in a repeating pattern. The empirical formula NaCl simply tells us the ratio in which ions exist — 1 Na⁺ per 1 Cl⁻.

So when we write "NaCl = 58.5 u," we are speaking about one formula unit — the smallest collection of ions reflecting the formula's ratio. Strictly speaking, calling it "molecular mass" of NaCl is a category error; the correct term is formula mass.

How to compute either quantity

The arithmetic is identical. Add up the atomic masses, weighted by the number of atoms of each kind in the chemical formula. The standard atomic masses you should have at the tip of your tongue:

• H = 1.008 (often rounded to 1)
• C = 12.01 (often 12)
• N = 14.01 (often 14)
• O = 16.00
• F = 19.00
• Na = 23.00
• Mg = 24.00
• Al = 27.00
• S = 32.07 (often 32)
• Cl = 35.45 (often 35.5)
• K = 39.10
• Ca = 40.08

For water: H₂O = 2(1.008) + 16.00 = 18.02 u. So one mole of water has a mass of 18.02 g.

For NaCl: Na + Cl = 23.0 + 35.5 = 58.5 u (formula mass). One mole of NaCl formula units = 58.5 g.

For CaCl₂: Ca + 2Cl = 40 + 2(35.5) = 111 u. One mole of CaCl₂ formula units = 111 g.

Worked example — a polyatomic compound

Question: Calculate the molecular mass of glucose (C₆H₁₂O₆) and the formula mass of magnesium nitrate Mg(NO₃)₂.

Solution:
Step 1: Glucose, C₆H₁₂O₆.

• 6 carbons: 6 × 12 = 72
• 12 hydrogens: 12 × 1 = 12
• 6 oxygens: 6 × 16 = 96
• Total: 72 + 12 + 96 = 180 u.

Step 2: Magnesium nitrate, Mg(NO₃)₂.

• 1 magnesium: 1 × 24 = 24
• 2 nitrogens: 2 × 14 = 28
• 6 oxygens (3 oxygens per nitrate × 2 nitrate groups): 6 × 16 = 96
• Total: 24 + 28 + 96 = 148 u.

Conclusion: Molecular mass of glucose = 180 u (one mole = 180 g). Formula mass of Mg(NO₃)₂ = 148 u (one mole = 148 g). Glucose is molecular (covalent); magnesium nitrate is ionic — same maths, different terminology.

Why this distinction matters in chemistry

Why it matters: when you melt or dissolve NaCl, you get free Na⁺ and Cl⁻ ions. When you melt or dissolve water (well, dissolve sugar in water), the molecules stay intact. This affects colligative properties — boiling-point elevation and freezing-point depression — through the van't Hoff factor (i), which counts the number of particles each formula unit produces. For NaCl in water, i ≈ 2; for glucose, i = 1. NEET regularly tests this distinction.

It also matters for electrical conduction. Molten NaCl conducts electricity because the ions are free to move; molten sucrose does not, because the molecules carry no net charge. Realising that "molecular" and "ionic" describe different physical realities — not just different naming conventions — is what makes the rest of solid-state chemistry click.

Real-world example

Real-world example: India's iodised salt is essentially NaCl with a trace of potassium iodate (KIO₃) added — about 30 ppm. Both NaCl and KIO₃ are ionic. The reason iodine stays in salt over months of storage is that it is locked into the IO₃⁻ ion of a lattice, not roaming free as I₂ molecule which would sublime away. Thinking in terms of formula units rather than molecules is exactly what predicts this stability.

By contrast, sugar (C₁₂H₂₂O₁₁, molecular mass 342 u) is a molecular solid; that is why it dissolves cleanly without dissociating and gives sweetness rather than electrolyte behaviour.

Common misconception

Common misconception: "NaCl molecule has mass 58.5 u." Strictly false. There is no NaCl molecule. The formula unit NaCl has formula mass 58.5 u. While most textbooks use "molecular mass" loosely for ionic compounds, NEET prefers the precise term "formula mass."

A second misconception: "Average atomic masses are integers." They are not. Chlorine's atomic mass is 35.45 because natural chlorine is a mixture of ³⁵Cl (about 76 percent) and ³⁷Cl (about 24 percent). Use 35.5 (not 35 or 37) in calculations involving chlorine.

A third misconception: "Molecular mass and molar mass are different." They are numerically the same but with different units — molecular mass is in u (per molecule), molar mass is in g/mol (per mole). Avogadro's number (6.022 × 10²³) is the conversion factor that makes them numerically equal in the chosen unit system.

Quick computation shortcut

Always group atoms within polyatomic ions before multiplying. For Mg(NO₃)₂, treat NO₃ as a single unit of (14 + 48) = 62, then 62 × 2 = 124, add Mg = 24 → 148. Same answer, fewer slips. For Al₂(SO₄)₃, SO₄ = 32 + 64 = 96; Al₂(SO₄)₃ = 2(27) + 3(96) = 54 + 288 = 342 u. Practice this method until grouping is automatic.

:::compare

:::keypoints

• Molecular mass and formula mass are computed identically — sum atomic masses, weighted by atom counts.
• Use "molecular mass" for covalent substances (real molecules); use "formula mass" for ionic compounds (no discrete molecule).
• The unit is u (atomic mass unit), defined as 1/12 the mass of a ¹²C atom.
• Gram molecular/formula mass = molecular/formula mass in grams = mass of one mole.
• Atomic masses to memorise: H=1, C=12, N=14, O=16, Na=23, Mg=24, Al=27, S=32, Cl=35.5, K=39, Ca=40.
• Group polyatomic ions before multiplying — fewer arithmetic errors.
• The molecular-versus-ionic distinction governs van't Hoff factor, conductivity in melt, and colligative properties.
• One mole = 6.022 × 10²³ particles (Avogadro's number) — the bridge between u and grams.
  :::

:::memory
"Molecules for COVALENT, Formula for IONIC." If atoms are sharing electrons → molecular mass. If atoms are giving/taking electrons → formula mass. The arithmetic is the same; only the language changes.
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:::recap

• Molecular mass applies to covalent substances; formula mass applies to ionic lattices.
• Computation is identical: add atomic masses weighted by atom counts.
• Gram molecular (or formula) mass equals the mass of one mole — your gateway to mole-concept problems.
• NaCl is 58.5 u as a formula unit, not a molecule — this language matters in NEET.
  :::

Why does the periodic table list boron's atomic mass as 10.81 and not as a clean whole number? Because boron is a mixture — a weighted average of two isotopes. Let us do the calculation step by step the way NEET expects, then extract a method that works for any element.

Definition: Average atomic mass (also called relative atomic mass) is the weighted mean of the masses of all naturally occurring isotopes of an element, weighted by their fractional abundances.
Definition: Isotopes are atoms of the same element having the same atomic number (Z) but different mass numbers (A), because they differ in the number of neutrons.

The Data for Boron

Boron has two naturally occurring stable isotopes:

:::compare

Notice the abundances add to exactly 100% (19.9 + 80.1 = 100). If a NEET problem gives you two values that don't add to 100, recheck the question or look for a third isotope.

The Weighted-Average Formula

For an element with isotopes of masses m1, m2, ... and fractional abundances f1, f2, ... (where the fractions sum to 1):

Average atomic mass = m1·f1 + m2·f2 + m3·f3 + …

Or, if abundances are in percentages p1, p2, ... (summing to 100):

Average atomic mass = (m1·p1 + m2·p2 + …) / 100

Both forms are identical — choose whichever feels safer under exam pressure.

Worked Solution

Question: Boron has two isotopes: B-10 (mass 10.013 u, abundance 19.9%) and B-11 (mass 11.009 u, abundance 80.1%). Calculate the average atomic mass of boron.

Solution:
Step 1: Convert percentages to fractions. f(B-10) = 19.9/100 = 0.199 and f(B-11) = 80.1/100 = 0.801. Quick check: 0.199 + 0.801 = 1.000. Good.
Step 2: Multiply each isotopic mass by its fraction.

• Contribution of B-10 = 10.013 × 0.199 = 1.99259 ≈ 1.992 u
• Contribution of B-11 = 11.009 × 0.801 = 8.81821 ≈ 8.818 u
  Step 3: Add the contributions. Average atomic mass = 1.992 + 8.818 = 10.810 u.
  Conclusion: The average atomic mass of boron is 10.81 u, exactly matching the periodic-table value.

Sanity Check: Does the Answer Make Sense?

A weighted average must always lie between the two isotopic masses, leaning toward the more abundant one. Here, 10.81 lies between 10.013 and 11.009, and it leans toward 11.009 (because B-11 is 80.1% abundant). This is the most powerful one-second check you can do in the exam hall — if your answer is outside [10.013, 11.009], you made an arithmetic slip.

A second check: since B-11 is roughly four times as abundant as B-10, the average should sit roughly four-fifths of the way from B-10 to B-11. Geometrically, 10.013 + (4/5)(11.009 − 10.013) = 10.013 + 0.797 ≈ 10.810. Matches perfectly.

A Faster Shortcut for the Exam

If both abundances are given as percentages, you can skip the fraction conversion and write directly:

Average mass = (10.013 × 19.9 + 11.009 × 80.1) / 100 = (199.26 + 881.82) / 100 = 1081.08 / 100 ≈ 10.81 u.

This is one multiplication, one addition and a divide-by-100 — fast enough for a one-mark NEET item.

Why it matters: every mole calculation, every stoichiometry problem, every empirical formula derivation in chemistry depends on the atomic mass you pull from the periodic table. That number is already a weighted average of isotopes — so you must understand how it was built, otherwise mass-balance problems with isotopic mixtures (chlorine, copper, silver, magnesium, lead) will trip you.

Real-world example: chlorine has two isotopes Cl-35 (about 75.77%) and Cl-37 (about 24.23%), giving an average atomic mass of about 35.45 u. That is why every chlorine compound you encounter — common salt, hydrochloric acid in your stomach — is really a statistical mixture, and why mass-spectrometry peaks in chlorine-containing molecules come in 3:1 doublets.

Common misconception: "Average atomic mass means the simple average (sum/2) of isotopic masses." Wrong. A simple average for boron would give (10.013 + 11.009)/2 = 10.511, which is far from the true 10.81. You must weight by abundance.

Common misconception: "Isotopes differ in atomic number." No — they have the same Z but different A (i.e., different number of neutrons). Different Z would make them different elements.

Common misconception: "Mass number A and atomic mass are the same thing." The mass number A is an integer count of nucleons. The atomic mass is measured in u, includes the small mass defect from nuclear binding, and is close to A but never exactly A.

:::keypoints

• Average atomic mass = sum of (isotopic mass × fractional abundance).
• Convert percentages to fractions by dividing by 100, OR multiply masses by percentages and divide the final sum by 100.
• The answer must lie between the two isotopic masses, biased toward the more abundant isotope.
• Abundance fractions must add to 1 (or percentages to 100). If not, recheck.
• Isotopes share Z but differ in A; atomic mass is a weighted mean, not a simple mean.
• The same idea explains the non-integer atomic masses of Cl (35.45), Cu (63.55), Mg (24.31) etc.
  :::

:::memory
"Mass × Fraction, then Sum" — three steps, no shortcuts. Or remember: "the heavier isotope pulls the average toward itself in proportion to how common it is".
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:::recap

• Apply the formula m1·f1 + m2·f2 to get 10.81 u for boron.
• The answer lies between 10.013 and 11.009, closer to 11.009 because B-11 is more abundant.
• The same technique works for chlorine, copper, silver and every NEET stoichiometry problem.
• Always verify your weighted mean sits inside the isotope-mass range before circling an option.
  :::

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.

:::compare

:::keypoints

• Analogy = find the operation that maps A to B, then apply it to C.
• SCADM order: Square → Cube → Add → Difference/Divide/Multiply → Manipulate digits.
• A square's ones-digit must be 0, 1, 4, 5, 6 or 9 — a quick filter before squaring.
• A cube's ones-digit follows a fixed map (1→1, 2→8, 3→7, …) — equally useful.
• If two operations fit the first pair, apply BOTH to the second pair before circling.
• Grouped numbers usually hide an AP, GP, or known integer sequence (squares, cubes, primes).
• 'Odd one out' framing flips the rule — find the majority pattern and pick the outlier.
• Eliminate options first; many analogies are solvable by elimination alone in 10 seconds.
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:::memory
SCADM — Square, Cube, Add, Difference (and Divide/Multiply), Manipulate digits. Test in this order; the first hit that survives both pairs is your answer.
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:::recap

• Number analogy is a checklist game; the checklist is SCADM plus digit tricks.
• The relation must hold for BOTH pairs — that is the test that breaks ties.
• Grouped numbers signal AP / GP / known sequences; identify the family first.
• Speed comes from filter order, not from raw arithmetic ability.
  :::

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.

:::compare

:::keypoints

• Empirical = simplest whole-number ratio; molecular = actual atoms per molecule.
• Molecular formula = n × empirical, where n = molar mass / empirical formula mass.
• Use the MMD-R-W sequence: Mass → Moles → Divide by smallest → Whole numbers.
• For fractional ratios like 1.5 or 1.33, multiply all by 2 or 3 — never round directly.
• Empirical and molecular formulas can be identical when n = 1.
• Compounds with the same empirical formula can be very different (HCHO vs glucose).
• Combustion analysis is just MMD-R-W applied to indirectly derived percentages.
  :::

:::memory
"MMD-R-W" — Mass percent, Mole conversion (divide by atomic mass), Divide by smallest, Ratio, round to Whole numbers.
"Same empirical, different molecules" — methanal, acetic acid, ribose and glucose all share CH₂O.
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:::recap

• Empirical formula gives composition; molecular formula gives identity.
• Convert percentages to moles, divide by smallest, polish to whole numbers.
• Use molar mass to find n, then multiply the empirical subscripts by n.
• Fractional ratios are cleared by multiplication, never by rounding.
  :::

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.

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:::keypoints

• Assume 100 g of compound — percentages become grams in one step.
• Convert grams to moles using atomic masses, then divide every mole by the smallest mole.
• Clear fractional ratios by multiplying through by 2, 3, 4, or 5 as the decimal demands.
• The result, with whole-number subscripts, is the empirical formula.
• n = (given molar mass) ÷ (empirical formula mass).
• Multiply each subscript of the empirical formula by n to get the molecular formula.
• Always back-check: the molar mass of your final formula must equal the given molar mass.
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:::memory
"Mass → Moles → Min-divide → Multiply → Match."
The five M's mirror the five steps. The last step, Match, is the verification — never skip it.
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:::recap

• Percentage composition + molar mass → unique molecular formula in five steps.
• CH₂O (empirical) × 2 = C₂H₄O₂ (acetic acid) in the worked example.
• Empirical formula alone does not identify a compound — three different compounds can share CH₂O.
• Decimal endings .33, .5, .25 are signals to scale, not to round.
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Molarity (M) = moles of solute / volume of solution in litres (temperature dependent). Molality (m) = moles of solute / mass of solvent in kg (temperature independent - preferred for accuracy). Mole fraction (x) = moles of component / total moles (xA + xB = 1, dimensionless). Mass percent = (mass of solute / mass of solution) x 100. ppm = (mass of solute / mass of solution) x 10^6. Memory aid: 'Molarity = Litres of Solution; molality = kg of Solvent'. Useful relation: Molarity = (10 x density x mass%)/molar mass. Dilution: M1V1 = M2V2.

The limiting reagent is the reactant that is completely consumed first and thus determines (limits) the amount of product formed; the other reactant is in excess. STEPS: (1) Write the balanced equation. (2) Convert all given masses/volumes to moles. (3) Divide moles of each reactant by its stoichiometric coefficient. (4) The smallest value identifies the limiting reagent. (5) Calculate product using the limiting reagent's moles. Memory aid: 'Least ratio LIMITS'. Always base product calculations on the limiting reagent, never on the excess reagent. Excess reagent leftover = initial moles - moles reacted.

Reaction: N2 + 3H2 -> 2NH3. Given 28 g N2 and 6 g H2. Moles N2 = 28/28 = 1; moles H2 = 6/2 = 3. Divide by coefficients: N2 = 1/1 = 1; H2 = 3/3 = 1. Both ratios are equal, so neither is in excess - they react completely. NH3 formed = 2 x 1 = 2 mol = 2 x 17 = 34 g. If instead H2 were 4 g (2 mol), then H2/3 = 0.67 < N2/1 = 1, so H2 is limiting and NH3 = 2 x (2/3) = 1.33 mol. Always recompute the ratio when quantities change.