The Mole Concept

• The mole as the SI amount-of-substance unit linking microscopic particles to macroscopic mass.
• Avogadro's number N_A ≈ 6.022 × 10²³ mol⁻¹ and its role in counting atoms and molecules.
• To convert between mass, moles, and number of particles using molar mass.
• How molar mass (g/mol) equals numerically the atomic/molecular mass in u.

Level 1 — Definition and conversions

Verbal: One mole contains exactly as many elementary entities as there are atoms in 12 g of carbon-12 — Avogadro's constant N_A. It bridges the atomic scale and the laboratory balance.

Symbolic:

• n = m / M (moles = mass / molar mass)
• N = n × N_A (number of particles)
• m = n × M

Examples:

• Molar mass of H₂O = 18.0 g/mol → 1 mol H₂O = 18.0 g = 6.022×10²³ molecules
• 0.5 mol CO₂ = 0.5 × 44 g = 22 g

Level 2 — Atoms, molecules, ions, and gas volume

Molar volume of ideal gas (STP, NCERT convention): 1 mol ≈ 22.4 L at 273 K, 1 bar (check your textbook for exact STP definition).

Percent composition: Mass % of element = (mass of element in formula / molar mass) × 100.

Empirical vs molecular formula: Empirical gives simplest ratio; molecular = (empirical formula) × n where n = molecular mass / empirical mass.

NCERT spotlight — Avogadro applications

At STP (273 K, 1 bar in updated NCERT), one mole of ideal gas occupies about 22.7 L; older 22.4 L at 273 K, 1 atm still appears in legacy problems — check question data.

Molecular vs atomic mass: O2 molar mass is 32 g/mol; atomic oxygen is 16 u. Always specify entity counted.

Empirical formula from percent composition: Convert mass percent to moles, divide by smallest mole value, obtain ratio, then multiply to integers if needed.

How many molecules are in 9.0 g of water? How many hydrogen atoms?

Step 1 — M(H₂O) = 18.0 g/mol.
Step 2 — n = m/M = 9.0/18.0 = 0.50 mol H₂O.
Step 3 — Molecules N = 0.50 × 6.022×10²³ = 3.011×10²³ molecules.
Step 4 — Each H₂O has 2 H atoms → H atoms = 2 × 3.011×10²³ = 6.022×10²³.
Step 5 — Check: 0.50 mol H₂O contains 1.0 mol H atoms (2×0.5) ✓

Applications — laboratory calculations

Prepare 250 mL of 0.1 M NaCl: moles needed = 0.1 x 0.250 = 0.025 mol; mass = 0.025 x 58.5 g/mol = 1.46 g dissolved and made up to mark in volumetric flask. Analytical chemistry reports concentration in mol/L — mole concept is daily laboratory arithmetic for CBSE practicals and NEET numericals.

Deep dive — concentration units and gas laws link

Molarity M = n/V(L); molality m = n/mass solvent(kg) — temperature independent molality preferred in colligative property chapter preview. Mole fraction X_A = n_A / (n_A + n_B) — sum equals 1 for all components. ppm ppb trace concentrations environmental chemistry. Ideal gas PV = nRT connects moles to P V T — R = 0.0821 L atm mol⁻¹ K⁻¹ or 8.314 J mol⁻¹ K⁻¹. Dalton law partial pressure P_i = X_i P_total — mole fraction in gas mixture. Empirical formula from combustion: 0.24 g C, 0.04 g H in sample → moles C 0.02 H 0.04 ratio 1:2 → CH2 empirical; if molecular mass 42 u, n = 42/14 = 3 → molecular C3H6. Avogadro hypothesis equal volumes gases same T P contain equal moles — foundation molar volume concept. Distinguish atomic mass unit u (1/12 C-12) from molar mass g/mol numerically equal for same entity but different physical meaning and units in calculations.

Review and practice drill

Review checklist: (1) n = m/M. (2) N = n N_A. (3) Molar mass numerically equals formula mass in g/mol. (4) Specify particle type. Practice: Moles in 11 g CO2 — M=44, n=0.25 mol.

• Calculate moles in 5.6 g of N₂ (M = 28 g/mol).
• How many oxygen atoms in 0.25 mol CO₂?
• Define Avogadro's number.

Open the Practice tab for graded questions on Mole.