JEE Chemistry Some Basic Concepts Of Chemistry Notes Set 02

Matter and Its Classification

1. Matter is anything that has mass and occupies space. Physically it exists as solid, liquid or gas.
2. Chemically, matter is either a pure substance (element or compound) or a mixture (homogeneous or heterogeneous).
3. Elements have atoms of one kind only. Compounds contain two or more elements combined in a fixed ratio by mass.
4. Mixtures can be separated by physical methods; compounds only by chemical methods.

Laws of Chemical Combination

Atomic Mass, Molar Mass and the Mole

1. One atomic mass unit (amu) is 1/12 of the mass of one carbon-12 atom, which is \( 1.66\times10^{-24} \) g.
2. Average atomic mass = \( \sum (\text{isotopic mass}\times\text{fractional abundance}) \).
3. One mole is the amount of substance containing \( 6.022\times10^{23} \) entities (Avogadro number, \( N_A \)). It has a mass in grams equal to the formula mass.
4. \( n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{\text{number of particles}}{N_A} = \dfrac{V}{22.4\ \text{L}} \) (V at 273 K and 1 atm).
5. For any gas use \( PV = nRT \) when conditions are not standard.
6. Vapour density = molar mass of gas / molar mass of hydrogen gas, so molar mass = 2 x vapour density. Relative density is the ratio of two molar masses.

Percentage Composition and Formulae

1. Mass % of an element = (mass of the element in one mole of compound / molar mass) x 100.
2. Empirical formula: divide percentages by atomic masses, divide by the smallest, and clear fractions.
3. Molecular formula = (empirical formula)n, with \( n = \dfrac{\text{molar mass}}{\text{empirical formula mass}} \).
4. Minimum molecular mass assumes one atom of the element that is present in small amount.

Stoichiometry

1. Balance the equation, convert given data to moles, and use coefficients as mole ratios.
2. Limiting reagent: divide moles of each reactant by its coefficient. The smallest quotient identifies the limiting reagent, which decides the product amount.
3. Percentage yield = (actual yield / theoretical yield) x 100.
4. Mass is conserved in every reaction, so with a single product in a stoichiometric mixture the product mass equals the total reactant mass (at 100% yield).
5. For partial dissociation of a gas, \( \dfrac{M_{initial}}{M_{mixture}} = 1 + \alpha(n-1) \), where n is the number of moles obtained from one mole.

Worked Examples

Example 1. How many molecules are present in 4.4 g of \( CO_2 \)? Moles = 4.4/44 = 0.1, so molecules = \( 0.1\times6.022\times10^{23} = 6.022\times10^{22} \).

Example 2. A compound has 40% C, 6.7% H and 53.3% O. Mole ratio C : H : O = 3.33 : 6.7 : 3.33 = 1 : 2 : 1, so the empirical formula is \( CH_2O \).

Example 3. 5 g of \( H_2 \) reacts with 40 g of \( O_2 \) to form water. Moles are 2.5 and 1.25. Dividing by coefficients (2 and 1) gives 1.25 and 1.25, so both are used up completely and 2.5 mol (45 g) of water form.