Mole concept and stoichiometry

A mole is the amount of substance containing Avogadro's number NA = 6.022 x 10^23 elementary entities (atoms, molecules, ions, electrons). Key relations: Number of moles n = given mass / molar mass = (mass in g)/M. Also n = N/NA (number of particles over Avogadro's number) and, for gases at STP, n = V/22.4 (volume in litres). Molar mass M (g/mol) is numerically equal to atomic/molecular mass in u (amu). STP (NTP) now defined as 273.15 K and 1 bar gives molar volume 22.7 L, but JEE traditionally uses 22.4 L at 1 atm, 273 K — read the question's convention. Memory triangle: moles sits at the centre, connected to mass (divide by M), particles (divide by NA), and gas volume (divide by 22.4 L). Always convert the given quantity to MOLES first — moles is the universal currency of all stoichiometry.

Empirical formula = simplest whole-number ratio of atoms; molecular formula = (empirical formula) x n, where n = molar mass / empirical formula mass. Steps: convert each element's % to moles (divide % by atomic mass), divide all by the smallest, round to whole numbers. Percentage composition of an element = (atoms x atomic mass / molar mass) x 100. Concentration terms: Molarity M = moles of solute / volume of solution (L) — temperature dependent. Molality m = moles of solute / mass of solvent (kg) — temperature independent (preferred for colligative properties). Mole fraction x = moles of component / total moles (sum of all mole fractions = 1). ppm = (mass of solute/mass of solution) x 10^6. Useful link: Molarity = (10 x density x % by mass) / molar mass, with density in g/mL. Remember: molality uses solvent MASS, molarity uses solution VOLUME.

Limiting reagent is the reactant that runs out first and decides the maximum product. Method: convert each reactant to moles, divide by its coefficient, the smallest ratio is limiting. Example: Burn 4 g H2 with 32 g O2: 2H2 + O2 -> 2H2O. Moles H2 = 4/2 = 2; moles O2 = 32/32 = 1. Ratio: H2 = 2/2 = 1, O2 = 1/1 = 1 — exactly stoichiometric, none left over. Water formed = 2 mol = 36 g. Example 2: 28 g N2 + 6 g H2 in N2 + 3H2 -> 2NH3. Moles N2 = 1, H2 = 3; required H2 for 1 mol N2 = 3 mol = 6 g — perfectly balanced, NH3 = 2 mol = 34 g. Tip: if you change the limiting reagent, product changes; excess reagent is wasted. Always (1) balance the equation, (2) convert to moles, (3) find limiting reagent, (4) use mole ratio for product. This four-step routine solves nearly every stoichiometry question.

Cover the answer, recall, then check. 12 cards on the mole concept and stoichiometry for JEE Main.

Q1. What is Avogadro's number and what does one mole contain?
A1. 6.022 × 10²³ (N_A). One mole contains exactly N_A entities (atoms, molecules, ions or electrons).

Q2. Molar volume of an ideal gas at STP (273.15 K, 1 bar)?
A2. 22.7 L mol⁻¹ (current NCERT/JEE). At 1 atm (old STP) it is 22.4 L mol⁻¹ — know which the paper uses.

Q3. Define one atomic mass unit (u) in terms of carbon-12.
A3. 1 u = 1/12 the mass of one ¹²C atom = 1.66 × 10⁻²⁴ g. Atomic/molar masses are relative to ¹²C = 12.

Q4. Moles = ? (three standard routes)
A4. n = mass/molar mass = number of particles/N_A = volume of gas(STP)/22.7 L. For gases also n = PV/RT.

Q5. How many atoms of oxygen are in 1 mole of H₂SO₄?
A5. 4 × N_A = 2.409 × 10²⁴ oxygen atoms (4 O atoms per formula unit).

Q6. Define limiting reagent.
A6. The reactant that is completely consumed first; it fixes the maximum product. Find it by dividing moles of each reactant by its stoichiometric coefficient — smallest ratio limits.

Q7. In N₂ + 3H₂ → 2NH₃, 1 mol N₂ reacts with 2 mol H₂. Which is limiting?
A7. H₂. It needs 3 mol but only 2 are available (ratio 2/3 < 1/1), so H₂ limits; N₂ is in excess.

Q8. Define percentage yield.
A8. % yield = (actual yield / theoretical yield) × 100. Theoretical yield comes from the limiting reagent.

Q9. How many molecules are in 4.4 g of CO₂?
A9. n = 4.4/44 = 0.1 mol → 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules.

Q10. Gram-equivalent: how is equivalent mass of an acid found?
A10. Equivalent mass = molar mass / basicity (number of replaceable H⁺). H₂SO₄: 98/2 = 49 g eq⁻¹.

Q11. State the law of definite proportions.
A11. A given compound always contains the same elements in the same fixed mass ratio, regardless of source or method of preparation.

Q12. 11.2 L of any ideal gas at old STP (1 atm) equals how many moles?
A12. 0.5 mol (11.2/22.4). At 1 bar STP, 11.2 L ≈ 0.493 mol (÷22.7).

Every quantitative chemistry question — stoichiometry, concentration, gas laws, electrochemistry, equilibrium — bottoms out in the mole. In JEE Main this is guaranteed marks: 1–2 direct numericals plus the mole is a hidden step in dozens more. Master it and the whole of Physical Chemistry gets easier.

The central bridge

The mole links the microscopic (atoms/molecules) to the macroscopic (grams, litres) through Avogadro's number, N_A = 6.022 × 10²³.

To find moles from…	Formula
Mass	n = mass / molar mass
Number of particles	n = N / N_A
Gas volume at STP	n = V / 22.7 L (1 bar)
Gas at any P, T	n = PV / RT

Key constants: N_A = 6.022 × 10²³; molar volume = 22.7 L at 273.15 K & 1 bar (22.4 L at 1 atm); 1 u = 1.66 × 10⁻²⁴ g; R = 0.083 L bar K⁻¹ mol⁻¹.

Stoichiometry workflow

1. Balance the equation. 2. Convert given data to moles. 3. Identify the limiting reagent (smallest mole ÷ coefficient). 4. Use mole ratios to get product moles. 5. Convert back to the required unit. 6. Apply % yield if asked.

Exam Tricks & Tips

• 🎯 Always find the limiting reagent the moment two reactant amounts are given — the excess one is a decoy.
• 🎯 Check which STP the paper assumes: 22.7 L (1 bar) is current NCERT; older keys use 22.4 L. A wrong choice shifts every answer ~1.3%.
• 🎯 Atoms of an element in a compound = (moles of compound) × (subscript) × N_A. Don't forget the subscript.
• 🎯 Mass is conserved, moles are not — total moles of gas can change in a reaction, so never equate reactant and product moles unless coefficients say so.
• 🎯 % yield uses the limiting reagent's theoretical yield, never the excess reagent.
• ❌ Common mistake: dividing by molecular mass when the species is atomic (or vice-versa) — e.g. treating 32 g O as 1 mol O₂ when it is really 2 mol O atoms.

Expected exam pattern

Direct single-numerical: "number of atoms/molecules in X g", "limiting reagent and mass of product", "volume of gas at STP". Often a percentage-composition or % yield twist. Integer-type answers common.

Quick recap

Moles connect grams, particles and litres via N_A and molar volume. Balance → moles → limiting reagent → ratio → answer. Watch the STP convention and never confuse atoms with molecules.