Some Basic Concepts in Chemistry

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.

Empirical formula (EF) gives the simplest whole-number ratio of atoms in a compound; molecular formula (MF) gives the actual number of atoms per molecule. Relationship: MF = n × EF, where n = Molar mass of compound / Empirical formula mass, and n is always a positive integer (1, 2, 3...). Example: glucose has EF = CH2O (formula mass 30) but MF = C6H12O6 (molar mass 180), so n = 180/30 = 6. Note that ionic compounds (NaCl) and many solids are written only as empirical formulas. Memory aid: 'Empirical = Easiest ratio, Molecular = Multiple of it.' For elements like benzene (C6H6) and acetylene (C2H2), both share EF = CH but differ in n. Always reduce subscripts by their GCD to get EF from MF.

Given mass % composition: (1) Assume 100 g sample, so % becomes grams. (2) Divide each element's mass by its atomic mass to get moles. (3) Divide all mole values by the smallest to get a ratio. (4) If ratios aren't whole numbers, multiply all by a small integer (e.g. 1.5 → ×2, 1.33 → ×3, 1.25 → ×4). This gives the empirical formula. (5) For MF: n = M(molar) / M(empirical), then MF = (EF)n. Multiplier cheatsheet: decimal .5 → ×2, .33/.67 → ×3, .25/.75 → ×4, .2 → ×5. Common trap: round only at the final ratio step, never round intermediate mole values prematurely (e.g. don't call 2.49 → 2; it's likely 2.5 → ×2 = 5).

A compound contains C = 40.0%, H = 6.7%, O = 53.3%; molar mass = 180 g/mol. Find MF.
Step 1 (moles): C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.
Step 2 (divide by smallest 3.33): C = 1, H = 2.01 ≈ 2, O = 1.
Empirical formula = CH2O, EF mass = 12 + 2 + 16 = 30.
Step 3: n = 180/30 = 6.
Molecular formula = (CH2O)6 = C6H12O6 (glucose).
Verification: 6×12 + 12×1 + 6×16 = 72 + 12 + 96 = 180. Correct. JEE tip: if combustion data is given, all C goes to CO2 (mass C = 12/44 × mass CO2) and all H goes to H2O (mass H = 2/18 × mass H2O); O is found by difference from total sample mass.

Solids have definite shape, volume, and rigidity due to strong intermolecular forces. Two broad types:

• Crystalline: ordered, repeating arrangement (e.g., NaCl, quartz, ice).
• Amorphous: no long-range order (e.g., glass, rubber, plastic). Sometimes called "super-cooled liquids."

CRYSTAL CLASSIFICATION (by particle and forces):

UNIT CELL: smallest repeating unit that builds the entire crystal by translation.

7 crystal systems (based on axes and angles): cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, rhombohedral.

14 Bravais lattices total (subdividing the 7 systems with body-centered, face-centered variations).

CUBIC UNIT CELLS:

Simple cubic (SC): atoms only at corners.

• Atoms per unit cell: 8 × (1/8) = 1.
• Coordination number: 6.
• Packing efficiency: 52.4%.

Body-centered cubic (BCC): atoms at corners + 1 at body center.

• Atoms per unit cell: 8 × (1/8) + 1 = 2.
• Coordination number: 8.
• Packing efficiency: 68%.

Face-centered cubic (FCC): atoms at corners + 1 at each face center.

• Atoms per unit cell: 8 × (1/8) + 6 × (1/2) = 4.
• Coordination number: 12.
• Packing efficiency: 74% (the densest possible for spheres).

RELATIONSHIPS (for cubic cells with edge length a):

SC: a = 2r (where r = atomic radius).
BCC: a√3 = 4r → r = (a√3)/4.
FCC: a√2 = 4r → r = (a√2)/4.

DENSITY of unit cell:

ρ = (Z × M) / (a³ × N_A)

where Z = atoms per unit cell, M = molar mass, N_A = Avogadro's number.

Worked example. Copper crystallizes as FCC with a = 361 pm. Find density.
Z = 4, M = 63.5 g/mol, a³ = (3.61 × 10⁻⁸)³ = 4.7 × 10⁻²³ cm³.
ρ = (4 × 63.5) / (4.7 × 10⁻²³ × 6.022 × 10²³) = 254 / 28.3 = 8.97 g/cm³ ✓ (matches copper's known density).

HEXAGONAL CLOSE PACKING (HCP) — alternative dense packing, 74% efficient like FCC. Coord number 12.
Examples: Zn, Mg.

TYPES OF VOIDS:

In close-packed structures (FCC, HCP), there are:

• Tetrahedral voids: surrounded by 4 atoms. 2 per atom in close packing.
• Octahedral voids: surrounded by 6 atoms. 1 per atom in close packing.

Smaller cations occupy these voids in many ionic compounds.

For 1 mole of FCC packing of N atoms: N tetrahedral voids + N octahedral voids... wait that's not right. Let me redo: 2N tetrahedral, N octahedral voids per N atoms.

DEFECTS IN CRYSTALS:

Point defects (in stoichiometric compounds):

1. Schottky defect: equal numbers of cations and anions missing. Density decreases. Common in NaCl, KCl.

2. Frenkel defect: cation displaced from lattice site to interstitial position. Density unchanged. Common in AgCl, ZnS.

Non-stoichiometric defects:

• Metal excess: Anion vacancy filled by electron (F-center, gives color). Example: NaCl with extra Na in vapor — yellow color.
• Metal deficiency: more anions than cations expected. Example: FeO is actually Fe_(1−x)O.

ELECTRICAL PROPERTIES:

• Conductors: low resistance (metals); free electrons.
• Insulators: high resistance (NaCl, glass).
• Semiconductors: intermediate. Si, Ge.
  • n-type: doped with pentavalent (P, As) → extra electron.
  • p-type: doped with trivalent (B, Al) → hole.

MAGNETIC PROPERTIES:

• Diamagnetic: repelled by magnet. No unpaired electrons. Example: NaCl, H₂O.
• Paramagnetic: weakly attracted. Unpaired electrons. Example: O₂, Fe³⁺.
• Ferromagnetic: strongly attracted, permanent magnets. Fe, Co, Ni, Gd.
• Antiferromagnetic: opposing spins cancel out. MnO.
• Ferrimagnetic: opposing spins but unequal magnitudes → net moment. Fe₃O₄ (magnetite).

Curie point: temperature above which ferromagnetic material becomes paramagnetic.

Molarity (M): moles of solute per litre of solution.

M = moles of solute / V(L) of solution

Temperature-dependent (volumes change). SI unit: mol/L = mol/dm³.

Molality (m): moles of solute per kilogram of solvent.

m = moles of solute / mass of solvent (kg)

Temperature-independent (mass doesn't change). Preferred for colligative property problems.

Mole fraction (x):

x_A = moles A / total moles

Σ x_i = 1. Used for Raoult's law.

Mass percent (% w/w): mass of solute / mass of solution × 100.

Volume percent (% v/v): common for alcohol concentrations.

Mass / volume percent (% w/v): mass of solute / volume of solution × 100. Common for IV drugs (0.9% saline = 0.9 g NaCl / 100 mL).

ppm (parts per million): mass of solute / mass of solution × 10⁶. Used for trace contaminants. 1 ppm ≈ 1 mg/L for dilute aqueous solutions.

Normality (N): outdated but appears in older problems and titration.

N = gram-equivalents of solute / V(L) of solution

Equivalent weight depends on reaction:

• For acids: molar mass / number of H⁺ released. (H₂SO₄: 98/2 = 49)
• For bases: molar mass / number of OH⁻ released.
• For redox: molar mass / electrons transferred.

N₁V₁ = N₂V₂ is the classic titration equation.

Worked examples:

1. Molarity of 5.85 g NaCl in 500 mL solution.
n = 5.85 / 58.5 = 0.1 mol.
M = 0.1 / 0.5 = 0.2 M.

2. Molality of the same solution (assume density ~1 g/mL, so 500 mL of 0.2 M ≈ 500 g):
Mass of solvent ≈ 500 − 5.85 ≈ 494 g = 0.494 kg.
m = 0.1 / 0.494 = 0.202 m (≈ M for dilute aqueous solutions).

3. Mole fraction. 18 g glucose (M = 180) in 100 g water.
n_glucose = 18/180 = 0.1. n_water = 100/18 = 5.56.
x_glucose = 0.1 / 5.66 = 0.0177.

Conversion shortcut. For dilute aqueous solutions (density ≈ 1 g/mL):

Molarity ≈ Molality ≈ Mole fraction × 55.5 (because 1 L water = 55.5 mol).

This shortcut breaks down for non-aqueous solvents or concentrated solutions.