JEE Chemistry Some Basic Concepts Of Chemistry Notes Set 01

Matter and Its Classification

1. Matter is anything that has mass and occupies space. It exists as solid, liquid or gas.
2. Pure substances are elements (one kind of atom) or compounds (fixed ratio of different atoms). Mixtures may be homogeneous (uniform, like air or brine) or heterogeneous (like sand in water).
3. Mixtures are separated by physical methods such as filtration, distillation and chromatography. Compounds can be split only by chemical methods.

Units and Measurement

1. The SI system has seven base units, including metre (length), kilogram (mass), second (time), kelvin (temperature), mole (amount of substance), ampere and candela.
2. Common prefixes: kilo = 10³, centi = 10^-2, milli = 10^-3, micro = 10^-6, nano = 10^-9, pico = 10^-12.
3. Volume: 1 L = 1 dm³ = 1000 cm³. Temperature: K = °C + 273.15.
4. Use unit conversion factors, and keep the correct number of significant figures: leading zeros do not count, zeros between digits count, and the result of multiplication or division keeps the least number of significant figures.

The Mole Concept

1. One mole is the amount of substance that contains 6.022 × 10²³ elementary entities (Avogadro number, N_A).
2. Moles = mass/molar mass = number of particles/N_A = volume of gas at STP/22.4 L.
3. One gram atom is one mole of atoms, and one gram molecule is one mole of molecules.
4. Mass of one atom or molecule = molar mass/N_A. 1 amu = 1.66 × 10^-24 g.
5. Atoms in a sample = moles × atomicity × N_A.

Atomic, Molecular and Formula Mass

1. Relative atomic mass is the mass of an atom compared with 1/12th of the mass of a carbon-12 atom. Average atomic mass is the abundance-weighted mean of the isotopic masses.
2. Molecular mass is the sum of the atomic masses of all atoms in a molecule. Vapour density = molar mass/2, and relative density compares the molar masses of two gases.
3. Example: a sample with isotopes of mass 35 (75%) and 37 (25%) has average mass 0.75 × 35 + 0.25 × 37 = 35.5.

Percentage Composition, Empirical and Molecular Formula

1. Percentage of an element = (mass of the element in one mole/molar mass) × 100.
2. To find the empirical formula: convert percentages to masses (take 100 g), divide each by its atomic mass, divide all by the smallest result and, if needed, multiply by a small integer to get whole numbers.
3. Molecular formula = n × empirical formula, where n = molecular mass/empirical formula mass.
4. Example: 80% C and 20% H gives C : H = 6.67 : 20 = 1 : 3, so the empirical formula is CH₃ and, with molar mass 30, the molecular formula is C₂H₆.

Stoichiometry and Limiting Reagent

1. The coefficients of a balanced equation show the mole ratio of reactants and products (also the volume ratio for gases at the same temperature and pressure), not the mass ratio.
2. Steps: write and balance the equation, convert given quantities to moles, use the mole ratio, then convert to the required unit.
3. The limiting reagent is the reactant that is used up first. Divide each reactant's moles by its coefficient; the smallest value shows the limiting reagent, which decides the amount of product.
4. Percentage yield = (actual yield/theoretical yield) × 100. Percentage purity = (mass of pure substance/mass of sample) × 100.
5. Example: 4 g H₂ and 4 g O₂ form 2H₂O. O₂ = 0.125 mol is limiting, giving 0.25 mol = 4.5 g of water.

Equivalent Weight

1. Equivalent weight of an element = atomic mass/valency. For an acid, E = molar mass/basicity. For a base, E = molar mass/acidity. For an oxidising or reducing agent, E = molar mass/number of electrons transferred.
2. Equivalent weight of hydrogen = 1, oxygen = 8 and chlorine = 35.5. One equivalent of hydrogen occupies 11.2 L at STP.
3. Substances react in equal numbers of equivalents. Number of equivalents = mass/E = normality × volume (L).
4. Methods to find E: hydrogen displacement, oxide formation, metal displacement and chloride formation. Dulong and Petit's law (atomic mass × specific heat is about 6.4) gives an approximate atomic mass, from which valency = approximate atomic mass/E and exact atomic mass = valency × E.

Determination of Molecular Mass of Volatile Substances

1. Victor Meyer's method: the vapour of a known mass of the substance displaces an equal volume of air. Molar mass = (mass × 22400)/(volume of vapour in mL at STP).
2. The density method: molar mass = density in g/L at STP × 22.4. Vapour density = molar mass/2.

Laws of Chemical Combination

1. Law of conservation of mass (Lavoisier): mass is neither created nor destroyed in a chemical reaction.
2. Law of definite proportions (Proust): a pure compound always has the same elements in the same mass ratio.
3. Law of multiple proportions (Dalton): when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios, as in the oxides of nitrogen or the chlorides SnCl₂ and SnCl₄.
4. Law of reciprocal proportions (Richter): elements combine in the ratio of their equivalent weights, or in simple multiples of them.
5. Gay-Lussac's law of gaseous volumes: gases react in simple whole-number volume ratios at the same temperature and pressure.
6. Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.