Mole concept and stoichiometry for NEET

The mole concept is the first chapter of Class 11 chemistry and the quiet foundation of the whole physical chemistry syllabus. Solutions, equilibrium, electrochemistry, thermodynamics and redox titrations all begin with "how many moles?". A candidate who is slow here is slow everywhere else.

NEET questions from this chapter are usually direct: count the atoms, find the limiting reagent, work out an empirical formula, convert between concentration terms. They are quick marks if the conversions are automatic, and easy losses if you mix up molecules with atoms or molarity with molality.

The mole and its three bridges

One mole is the amount of substance that contains 6.022 × 10²³ elementary entities (atoms, molecules, ions or formula units). This number is the Avogadro constant, N_A.

Three conversions cover almost every question:

The molar mass in g mol⁻¹ has the same number as the atomic or molecular mass in u. H₂O is 18 u, so 18 g of water is 1 mol.

Molar volume: an ideal gas occupies about 22.4 L per mole at 273 K and 1 atm, the value most questions use. NCERT also gives 22.7 L at 273.15 K and 1 bar, the current standard. Use whichever the question implies; if the options only fit one, that settles it.

Laws of chemical combination

Empirical and molecular formula

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual numbers. Molecular formula = n × empirical formula, where n = molar mass/empirical formula mass.

Method:

1. Take 100 g of the compound, so percentages become grams.
2. Divide each mass by the atomic mass to get moles.
3. Divide every result by the smallest.
4. If a ratio ends in about .5, .33 or .25, multiply all of them by 2, 3 or 4 to reach whole numbers.

Stoichiometry and the limiting reagent

A balanced equation gives the mole ratio in which substances react. The limiting reagent is the one used up first, and it decides how much product forms. The other reactant is left in excess.

To find it, divide the moles of each reactant by its coefficient in the balanced equation. The smallest value marks the limiting reagent.

Concentration terms

Two relations appear often:

• Dilution: M₁V₁ = M₂V₂, because the moles of solute do not change when you add solvent.
• Molarity to molality: find the mass of 1 L of solution from its density, subtract the mass of solute to get the mass of solvent, then divide.

Worked NEET-style questions

Example 1: counting atoms

How many atoms are present in 4.4 g of CO₂?

• Molar mass of CO₂ = 12 + 32 = 44 g mol⁻¹, so n = 4.4/44 = 0.1 mol of molecules.
• Each molecule has 3 atoms: 0.3 mol of atoms.
• Atoms = 0.3 × 6.022 × 10²³ ≈ 1.81 × 10²³.

Example 2: limiting reagent

28 g of N₂ reacts with 10 g of H₂: N₂ + 3H₂ → 2NH₃. Find the mass of NH₃ formed and the reactant left over.

• N₂: 28/28 = 1 mol; divided by its coefficient 1, this gives 1.
• H₂: 10/2 = 5 mol; divided by its coefficient 3, this gives about 1.67.
• N₂ has the smaller value, so it is the limiting reagent.
• NH₃ formed = 2 × 1 = 2 mol = 2 × 17 = 34 g.
• H₂ used = 3 mol, so H₂ left = 5 − 3 = 2 mol = 4 g.
• Check the mass balance: 28 + 10 = 38 g in; 34 + 4 = 38 g out.

Example 3: empirical and molecular formula

A compound contains 40.0% C, 6.67% H and 53.33% O. Its molar mass is 180 g mol⁻¹. Find its molecular formula.

• Moles in 100 g: C = 40/12 = 3.33; H = 6.67/1 = 6.67; O = 53.33/16 = 3.33.
• Divide by 3.33: C : H : O = 1 : 2 : 1. Empirical formula CH₂O (mass 30).
• n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.

Example 4: molarity to molality

A 3.0 M aqueous NaCl solution has a density of 1.25 g mL⁻¹. Find its molality. (Molar mass of NaCl = 58.5 g mol⁻¹.)

• Take 1 L of solution: mass = 1000 × 1.25 = 1250 g.
• NaCl in it = 3 × 58.5 = 175.5 g.
• Water = 1250 − 175.5 = 1074.5 g = 1.0745 kg.
• Molality = 3/1.0745 ≈ 2.79 m.

Practice MCQs

1. The number of moles in 9 g of water is: (a) 2 (b) 1 (c) 0.5 (d) 0.25
2. 1 mol of glucose is dissolved in 500 g of water. The molality is: (a) 0.5 m (b) 1 m (c) 2 m (d) 5 m
3. The number of molecules in 3.4 g of NH₃ is about: (a) 6.0 × 10²² (b) 1.2 × 10²³ (c) 3.4 × 10²³ (d) 6.0 × 10²³
4. Which contains the largest number of atoms? (a) 4 g He (b) 16 g O₂ (c) 14 g N₂ (d) 18 g H₂O
5. A solution has 46 g of ethanol (C₂H₅OH) and 36 g of water. The mole fraction of ethanol is: (a) 0.25 (b) 0.33 (c) 0.50 (d) 0.67
6. 100 mL of 0.5 M NaOH is diluted to 250 mL. The new molarity is: (a) 0.1 M (b) 0.2 M (c) 0.25 M (d) 1.25 M
7. 50 g of CaCO₃ is heated completely: CaCO₃ → CaO + CO₂. The mass of CO₂ formed is: (a) 11 g (b) 44 g (c) 28 g (d) 22 g
8. A hydrocarbon contains 80% C and 20% H, and its molar mass is 30 g mol⁻¹. Its molecular formula is: (a) C₂H₆ (b) C₂H₄ (c) CH₃ (d) C₃H₈

Answers

1. (c) 9/18 = 0.5 mol.
2. (c) 1 mol/0.5 kg = 2 m.
3. (b) 3.4/17 = 0.2 mol; 0.2 × 6.022 × 10²³ ≈ 1.2 × 10²³.
4. (d) 1 mol of H₂O has 3 mol of atoms; each of the others has 1 mol of atoms.
5. (b) Ethanol 46/46 = 1 mol; water 36/18 = 2 mol; x = 1/3.
6. (b) 0.5 × 100 = M₂ × 250, so M₂ = 0.2 M.
7. (d) 50/100 = 0.5 mol CaCO₃ gives 0.5 mol CO₂ = 22 g.
8. (a) C = 80/12 = 6.67, H = 20/1 = 20; ratio 1 : 3 gives CH₃ (mass 15); n = 2, so C₂H₆.

What to do next

• Solve 20 mass–mole–particle conversions without a calculator until each takes under 30 seconds.
• For every limiting-reagent question, write the moles ÷ coefficient line before anything else.
• Make a card with the five concentration terms and mark which depend on temperature.
• Carry these skills into solutions and redox reactions, and next read structure of the atom.