Quick Summary
Basic Concepts of Chemistry builds the quantitative foundation for all of chemistry. The mole is the central unit connecting mass, number of particles, and volume (for gases). Stoichiometry — using balanced equations to calculate amounts — is the most tested skill in NEET chemistry.
Matter and Its Classification
- Pure substances: Elements (one type of atom) and Compounds (chemically combined)
- Mixtures: Homogeneous (uniform) or Heterogeneous (non-uniform)
- Physical change: No new substance formed (e.g., melting)
- Chemical change: New substance formed; irreversible (e.g., combustion)
Laws of Chemical Combination
| Law | Statement |
|---|---|
| Conservation of mass | Mass of reactants = mass of products |
| Definite proportions | A compound always has the same ratio of elements by mass |
| Multiple proportions | Two elements forming multiple compounds — masses of one element that combine with fixed mass of other are in simple ratio |
| Reciprocal proportions (Equivalent proportions) | Equivalents of elements that react with each other |
| Gay-Lussac's law of combining volumes | Gases react in simple whole-number ratios by volume (at same T, P) |
The Mole Concept
Stoichiometry
From a balanced equation, the mole ratio of reactants and products is fixed.
Steps: 1. Write and balance the equation. 2. Convert given mass to moles. 3. Use mole ratio from the equation. 4. Convert moles to mass/volume/particles as required.
Limiting Reagent
The reagent that is completely consumed first. Determines maximum product.
Finding limiting reagent: Divide moles of each reactant by its coefficient → smallest value = limiting reagent.
% yield = (actual yield / theoretical yield) × 100
Empirical and Molecular Formula
Empirical formula: Simplest ratio of atoms. Steps from % composition: 1. Assume 100 g → % = mass in g. 2. Divide each by atomic mass → moles. 3. Divide by smallest → ratio → empirical formula.
Molecular formula: n × empirical formula, where n = molar mass / empirical formula mass.
Concentration Units
| Unit | Formula | Notes |
|---|---|---|
| Molarity (M) | mol solute / L solution | Changes with temperature |
| Molality (m) | mol solute / kg solvent | Independent of temperature |
| Mole fraction (χ) | nₐ / (nₐ + n_b) | Dimensionless; sum = 1 |
| Normality (N) | equivalents / L | N = M × n-factor |
| % w/w | (mass solute / mass solution) × 100 | — |
| % w/v | (mass solute / vol solution) × 100 | — |
Worked Examples
Example 1: Find the moles in 36 g of water (M = 18 g/mol). n = 36/18 = 2 mol. Molecules = 2 × 6.022×10²³ = 1.204 × 10²⁴ molecules.
Example 2: In the reaction N₂ + 3H₂ → 2NH₃, if 28 g N₂ and 6 g H₂ are used, find the limiting reagent. n(N₂) = 28/28 = 1 mol; n(H₂) = 6/2 = 3 mol. Ratio needed: 1:3. Ratio available: 1:3. Both are exactly stoichiometric — neither is limiting (both are used up). NH₃ formed = 2 mol = 34 g.
Example 3: A compound has 40% C, 6.67% H, 53.33% O. Find empirical formula. 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 → CH₂O (formaldehyde series). If molar mass = 180, n = 180/30 = 6 → molecular formula = C₆H₁₂O₆ (glucose).
Important Questions
One Mark
- Define Avogadro's number.
- How many moles are in 44 g of CO₂?
- State the law of conservation of mass.
- What is STP? What is the molar volume of a gas at STP?
Short Answer (3 marks)
- Calculate the percentage composition of H₂SO₄.
- Find the empirical formula of a compound containing 75% C and 25% H by mass.
- In the reaction 2H₂ + O₂ → 2H₂O, if 4g H₂ and 32g O₂ react, find the mass of water formed.
Long Answer (5 marks)
- Explain the mole concept. Calculate the number of atoms in 27 g of aluminium and the volume occupied by this many moles of any gas at STP.
- A compound is 38.7% C, 9.7% H, and 51.6% O. Its molar mass is 62 g/mol. Find its empirical and molecular formula.
Common Mistakes
- Using molecular mass in g instead of g/mol — the molar mass unit must be g/mol.
- Stoichiometry: not balancing the equation before doing calculations.
- Limiting reagent: comparing raw moles without dividing by stoichiometric coefficients.
- Empirical formula: forgetting to divide all ratios by the smallest mole value.
Board Exam Tips
- Mole concept numericals are the #1 most tested topic in Class 11 Chemistry board exams — do at least 20 practice problems.
- Empirical and molecular formula problems appear as 3-mark or 5-mark questions.
- NEET: Stoichiometry, limiting reagent, and % yield appear in 1–2 questions every year.
- Show all working clearly — partial marks are awarded for correct method even if arithmetic is wrong.