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.

Diamond is the hardest natural substance; graphite writes on paper and lubricates machines — yet both are pure carbon. The difference is entirely in how the atoms are arranged. This lesson covers the solid state from first principles: crystal classification, packing geometry, density calculation, defects, and the electrical and magnetic properties that flow from structure — a 4–8 mark guaranteed topic in JEE and NEET.

Definition — Crystalline solid: particles arranged in a long-range, repeating, ordered three-dimensional pattern (e.g., NaCl, quartz, diamond, ice).
Definition — Amorphous solid: no long-range order; structure resembles a frozen liquid (e.g., glass, rubber, plastic). Properties are isotropic (same in all directions).
Definition — Unit cell: the smallest repeating unit of a crystal lattice that, when translated in three dimensions, builds the entire crystal.

Classifying crystals by bonding

Why brittleness in ionic crystals? Striking NaCl shifts one layer so that like-charge ions face each other — repulsion shatters the crystal. Metals deform instead because the electron sea re-accommodates the shift.

Crystal systems and Bravais lattices

There are 7 crystal systems (classified by edge lengths a, b, c and angles α, β, γ): cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, and rhombohedral (trigonal). These 7 systems give 14 Bravais lattices when body-centred and face-centred variations are included.

For exam purposes, the three cubic cells dominate.

Cubic unit cells — atoms, geometry, and packing

Simple cubic (SC):

• Atoms: only at the 8 corners. Each corner atom is shared among 8 cells → 8 × (1/8) = 1 atom per unit cell.
• Coordination number: 6 (each atom touches 4 in its plane, 1 above, 1 below).
• Edge length relation: a = 2r (atoms touch along the edge).
• Packing efficiency: 52.4% — the least efficient, lots of wasted space.
• Example: Polonium (the only element with simple cubic structure at room temperature).

Body-centred cubic (BCC):

• Atoms: 8 corners + 1 body centre → 8 × (1/8) + 1 = 2 atoms.
• Coordination number: 8 (the body-centre atom touches all 8 corner atoms).
• Edge relation: √3 · a = 4r (atoms touch along the body diagonal).
• Packing efficiency: 68%.
• Examples: Fe (at room temperature), Na, K, Cr, W.

Face-centred cubic (FCC) / cubic close packing (CCP):

• Atoms: 8 corners + 6 face centres → 8 × (1/8) + 6 × (1/2) = 4 atoms.
• Coordination number: 12 — the highest for equal spheres.
• Edge relation: √2 · a = 4r (atoms touch along the face diagonal).
• Packing efficiency: 74% — the densest possible packing for identical spheres (proved by Kepler conjecture, confirmed mathematically in 1998).
• Examples: Cu, Al, Au, Ag, Ni.

Hexagonal close packing (HCP):

• Also achieves 74% packing with coordination number 12.
• Layer sequence ABAB… (FCC has ABCABC…).
• Examples: Zn, Mg, Ti, Be.

:::compare Cubic cells at a glance

Density of a unit cell

ρ = (Z × M) / (a³ × N_A)

where Z = atoms per unit cell, M = molar mass (g/mol), a = edge length (cm), N_A = 6.022 × 10²³ mol⁻¹.

This formula connects the macroscopic density you can measure with the microscopic structure — use it both ways: given structure → calculate density; given density → find Z (and hence identify the cell type).

Question: Copper crystallises FCC with a = 361 pm. Find its density. (M_Cu = 63.5 g/mol)

Solution:
Step 1: FCC → Z = 4.
Step 2: Convert a to cm: 361 pm = 3.61 × 10⁻⁸ cm; a³ = (3.61)³ × 10⁻²⁴ = 47.1 × 10⁻²⁴ cm³.
Step 3: ρ = (4 × 63.5) / (47.1 × 10⁻²⁴ × 6.022 × 10²³) = 254 / 28.36.
Conclusion: ρ ≈ 8.96 g/cm³, matching copper's known experimental density — confirming the FCC assignment.

Voids in close-packed structures

In FCC or HCP packing of N atoms, two types of holes appear:

• Tetrahedral void: surrounded by 4 atoms (tetrahedral arrangement); radius ratio r/R ≈ 0.225. There are 2N tetrahedral voids per N atoms.
• Octahedral void: surrounded by 6 atoms (octahedral arrangement); radius ratio r/R ≈ 0.414. There are N octahedral voids per N atoms.

Small cations occupy these voids in many ionic solids. In NaCl, Na⁺ occupies octahedral voids in a Cl⁻ FCC lattice. In ZnS (zinc blende), Zn²⁺ occupies half the tetrahedral voids in an S²⁻ FCC lattice.

Point defects

Stoichiometric (ratio preserved):

• Schottky defect: equal numbers of cations and anions are missing from their lattice sites (maintaining electrical neutrality). The crystal loses density — there are now fewer particles in the same volume. Common in NaCl, KCl, CsCl (where both ions are similar in size and can leave together).
• Frenkel defect: a cation (usually the smaller ion) leaves its lattice site and squeezes into a nearby interstitial void. No ions are missing, so density is unchanged. Common in AgCl, ZnS (large anions, small cations → cations fit into interstitials).

Non-stoichiometric (ratio shifts):

• Metal excess — F-centres: an anion vacancy is occupied by an electron to maintain neutrality. These electrons absorb visible light → crystal appears coloured. Classic example: NaCl heated in sodium vapour turns yellow (the excess Na ionises; the extra electrons occupy Cl⁻ vacancies).
• Metal deficiency: more anions than stoichiometry predicts; compensated by some cations having higher charge. Example: Fe₍₁₋ₓ₎O — some Fe²⁺ sites replaced by Fe³⁺ to balance charge.

Electrical properties

• Conductors (metals): overlapping valence and conduction bands; electrons flow freely.
• Insulators: large band gap (> ~5 eV); electrons cannot jump to the conduction band at room temperature.
• Semiconductors (Si, Ge): small band gap (~1 eV); some electrons thermally excited across.
  • n-type doping: add a pentavalent element (P, As, Sb) → one extra electron per dopant atom → majority carriers are electrons.
  • p-type doping: add a trivalent element (B, Al, Ga) → one missing electron (a "hole") per dopant → majority carriers are holes.

Magnetic properties

The Curie temperature: heating a ferromagnet above this point randomises the domains → it becomes paramagnetic. For iron, T_Curie ≈ 770 °C.

Why it matters: structure dictates every property you measure — density, hardness, melting point, conductivity, colour, and magnetism all follow from packing and bonding. This is a recurring 4–8 marks in JEE/NEET, with numerical density questions and defect identification the most common formats.

Real-world example: table salt shatters cleanly when struck because shifting ionic layers brings like-charges face to face; they repel catastrophically. Metals bend instead — the electron sea flows around the new configuration. The contrast between these two materials at a dinner table is solid-state chemistry in action.

Common misconception: "glass is a slowly flowing liquid — old window panes are thicker at the bottom." Glass is an amorphous solid, not a flowing liquid. The thickness variation in old panes came from historical manufacturing methods (glass was poured and one side is thicker by design), not from centuries of flow. Calling glass a "supercooled liquid" describes its disordered structure, not literal viscous flow.

:::keypoints Key points

• Four solid types — ionic, covalent-network, molecular, metallic — explain hardness, MP, and conductivity.
• Cubic cells: SC (1 atom, 52.4%), BCC (2 atoms, 68%), FCC (4 atoms, 74%); FCC = HCP = densest packing.
• Density ρ = ZM/(a³N_A) — use it to find density from structure, or to identify Z from a known density.
• Close packing: 2N tetrahedral voids, N octahedral voids per N atoms.
• Schottky → ions missing → density decreases; Frenkel → cation displaced to interstitial → density unchanged.
• F-centres (metal excess) → colour; NaCl in Na vapour → yellow.
• n-type = pentavalent dopant (extra electron); p-type = trivalent dopant (hole).
• Ferromagnetism disappears above the Curie temperature → becomes paramagnetic.
  :::

:::memory
"SC–BCC–FCC go 1–2–4" atoms per cell; packing climbs 52→68→74%.
For defects: "Schottky Shrinks density, Frenkel Freezes it."
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:::recap

• A solid's properties follow directly from how its particles pack and bond.
• Packing efficiency rises SC < BCC < FCC = HCP (52.4% → 68% → 74%).
• Defects shift density (Schottky) or colour (F-centres) and can create conductivity and magnetism.
• Glass is amorphous; calling it a flowing liquid is a myth.
• Master the density formula — it appears in JEE Mains almost every year.
  :::

"How much stuff is dissolved in how much liquid?" sounds simple, but chemistry needs several precise answers depending on which amounts and which base you choose — and picking the wrong unit is a classic exam trap. This lesson sorts out molarity, molality, mole fraction, normality and ppm, explains when each is the right tool, and shows you the worked maths.

Definition: Concentration expresses the amount of solute (the dissolved substance) in a given quantity of solution (solute + solvent) or solvent.
Definition: Solute is the substance being dissolved; solvent is the substance doing the dissolving.

The fundamental choice: per litre of solution vs per kg of solvent

This is the central fork:

• Molarity (M) = moles of solute per litre of solution → M = n ÷ V(L)
• Molality (m) = moles of solute per kilogram of solvent → m = n ÷ mass(kg)

The critical difference is temperature sensitivity:

Liquids expand when heated and contract when cooled. A litre of solution at 25 °C is not the same as a litre at 50 °C. So molarity changes with temperature — a 1 M solution at 25 °C becomes slightly less than 1 M at 50 °C because the volume grew.

Mass, on the other hand, never changes with temperature. So molality is temperature-independent — this is exactly why it is the required unit for colligative property calculations (boiling-point elevation, freezing-point depression, osmotic pressure).

Rule of thumb: Is temperature changing or does precision across temperatures matter? Use molality. Are you simply dispensing a volume at room conditions? Use molarity.

Mole fraction

Mole fraction (x_A) = moles of component A ÷ total moles in the mixture.

x_A = n_A / (n_A + n_B + …)

The mole fractions of all components always sum to 1 (Σx = 1). Mole fraction is dimensionless and is used in Raoult's law for vapour-pressure calculations in ideal solutions.

Worked example:
Question: Find the mole fraction of glucose in a solution of 18 g glucose (M = 180 g/mol) in 90 g water.
Solution:
Step 1: n(glucose) = 18 ÷ 180 = 0.1 mol. n(water) = 90 ÷ 18 = 5 mol.
Step 2: x(glucose) = 0.1 ÷ (0.1 + 5) = 0.1 ÷ 5.1.
Conclusion: x(glucose) ≈ 0.0196 (about 1.96%).

Percentage units

Three percentage expressions are used in different contexts:

• % w/w (mass/mass): (mass of solute ÷ mass of solution) × 100. Used in concentrated industrial acids (e.g. "98% H₂SO₄").
• % v/v (volume/volume): used for liquid-in-liquid mixtures. Alcoholic drinks: "40% v/v ethanol" means 40 mL ethanol per 100 mL solution.
• % w/v (mass/volume): (mass of solute in g ÷ volume of solution in mL) × 100. Dominant in medicine and pharmacy. Normal saline = 0.9% w/v = 0.9 g NaCl in every 100 mL — nurses rely on this because it lets them calculate a dose directly from the bag's volume.

Parts per million (ppm) and parts per billion (ppb)

For trace contaminants (pollutants in water, pesticides in food), percentage is too coarse a unit.

ppm = (mass of solute ÷ mass of solution) × 10⁶

For dilute aqueous solutions, 1 ppm ≈ 1 mg/L (because 1 L of dilute water ≈ 1 kg).

The WHO drinking-water limit for arsenic is 10 ppb (0.01 mg/L); the BIS standard in India is the same. Environmental monitoring and BIS/FSSAI food-safety rules cite ppm and ppb constantly — know these for exams.

Normality — the older acid-base unit

Normality (N) = gram-equivalents of solute per litre of solution.

N = gram-equivalents ÷ V(L)

Gram-equivalent weight depends on the reaction:

• Acid: equivalent weight = molar mass ÷ number of H⁺ ions released per molecule.
  • H₂SO₄: 98 ÷ 2 = 49 g/equiv; so 1 M H₂SO₄ = 2 N H₂SO₄.
  • HCl: 36.5 ÷ 1 = 36.5 g/equiv; so 1 M HCl = 1 N HCl.
• Base: equivalent weight = molar mass ÷ number of OH⁻ released.
• Redox: equivalent weight = molar mass ÷ electrons transferred.

The key titration shortcut: N₁V₁ = N₂V₂ (normalities × volumes of acid and base at equivalence point).

Why normality matters: older lab manuals, pharmacopoeia entries and some exam questions still use it. A "10 N H₂SO₄" solution is 5 M — confusing if you haven't converted.

Worked examples

Worked example 1:
Question: Find the molarity of 5.85 g of NaCl dissolved in 500 mL of solution. (Molar mass NaCl = 58.5 g/mol)
Solution:
Step 1: n(NaCl) = 5.85 ÷ 58.5 = 0.1 mol.
Step 2: V = 500 mL = 0.5 L. M = 0.1 ÷ 0.5.
Conclusion: Molarity = 0.2 M.

Worked example 2 (molality of same solution):
Question: Using the same solution (density ≈ 1 g/mL, so total mass ≈ 500 g), find the molality.
Solution:
Step 1: Mass of solvent = 500 − 5.85 = 494.15 g = 0.494 kg.
Step 2: m = 0.1 ÷ 0.494.
Conclusion: Molality ≈ 0.202 m — almost equal to molarity because this is a dilute aqueous solution. They diverge in concentrated solutions.

Shortcut for dilute aqueous solutions

For any dilute aqueous solution (density ≈ 1 g/mL):

Molarity ≈ Molality ≈ mole fraction × 55.5

(55.5 mol is the approximate molar concentration of pure water.) This breaks down for concentrated solutions or non-aqueous solvents — never apply it there.

:::compare Units at a glance

Common misconception: Students treat molarity and molality as interchangeable. They are nearly equal only for dilute aqueous solutions (density ≈ 1 g/mL, solvent mass ≈ solution mass). For concentrated sulphuric acid or any organic solvent, the two diverge sharply — and only molality stays valid when temperature changes, since it is built on mass, not volume.

Real-world example: A saline drip labelled "0.9% w/v Normal Saline" contains 0.9 g NaCl per 100 mL. The ICU nurse uses % w/v, not molarity, to calculate how much salt a patient receives per hour — straightforward from the drip rate without any mole arithmetic. Meanwhile the laboratory chemist titrating that same salt solution uses normality and N₁V₁ = N₂V₂ to find its exact strength.

:::keypoints Key points

• Molarity = mol/L of solution; changes with temperature.
• Molality = mol/kg of solvent; temperature-independent.
• Use molality for any problem involving temperature-dependent properties.
• Mole fractions sum to 1; they underpin Raoult's law.
• % w/v is the medical-dose unit; ppm/ppb measures trace contaminants.
• Normality uses equivalents; N₁V₁ = N₂V₂ solves acid-base titrations.
• H₂SO₄: equivalent weight = 49, so 1 M = 2 N.
  :::

:::memory
"My Mole Naps Peacefully" → Molarity (per Litre solution), Molality (per kg solvent), Normality (equivalents), Percentage (w/w, w/v, v/v), ppm (trace).
:::

:::recap

• Molarity changes with temperature; molality does not — that single fact directs most unit-choice problems.
• Mole fractions are dimensionless and always sum to 1.
• % w/v dominates in medicine; ppm/ppb in environmental and food safety standards.
• Normality extends molarity to reactions involving proton or electron transfer.
• For dilute water solutions, M ≈ m ≈ x × 55.5; for anything else, calculate carefully.
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