Some Basic Concepts of Chemistry

Matter is anything that occupies space and has mass. Physical classification: solid, liquid, gas. Chemical classification: pure substances (elements, compounds) and mixtures (homogeneous, heterogeneous).

Laws of Chemical Combination:

Atomic Mass: Mass of one atom relative to $1/12$ mass of $^{12}$C atom. Unit: u (or amu). $1$ u $= 1.66 \times 10^{-24}$ g.

Average atomic mass: weighted mean of isotopic masses. For Cl: $0.75(35) + 0.25(37) = 35.5$ u.

Molecular Mass: Sum of atomic masses of atoms in the molecule. Example: $H_2O = 2(1) + 16 = 18$ u.

Formula Mass: For ionic compounds (no discrete molecules), e.g., NaCl = $23 + 35.5 = 58.5$ u.

Mole: The amount of substance containing as many entities as the number of atoms in exactly $12$ g of $^{12}$C.

Avogadro Number: $N_A = 6.022 \times 10^{23}$ /mol.

Key Relations:

Note: STP (modern IUPAC) = $0°$C, $10^5$ Pa → molar volume $22.7$ L. NCERT uses old STP $= 22.4$ L. In JEE, both appear; check context.

Molar Mass: Mass of $1$ mole, numerically equal to atomic/molecular mass in grams.

Percentage Composition: $$\%\text{ of element} = \frac{\text{Mass of element in 1 mol}}{\text{Molar mass}} \times 100$$

Empirical Formula: Simplest whole-number ratio of atoms. Steps:

1. Convert % to grams (assume $100$ g)
2. Divide by atomic mass → moles
3. Divide by smallest → ratio
4. Multiply if needed to get whole numbers

Molecular Formula: $\text{Molecular formula} = n \times \text{Empirical formula}$, where $n = \dfrac{\text{Molar mass}}{\text{Empirical formula mass}}$.

Stoichiometry: Quantitative relationship between reactants and products. Based on balanced chemical equation; coefficients give mole ratios.

Steps to solve stoichiometry problems:

1. Write balanced equation
2. Convert given quantities to moles
3. Use mole ratio to find moles of desired species
4. Convert to required unit (mass, volume, molecules)

Limiting Reagent: Reactant that is completely consumed first; determines maximum amount of product. Other reactant(s) are in excess.

Identification: Compute moles of each reactant ÷ its stoichiometric coefficient. The smallest quotient is the limiting reagent.

Percentage Yield: $$\%\text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$

Theoretical yield: Maximum product assuming complete conversion. Actual yield: What is obtained experimentally.

Reactions in Solution: When dealing with solutions, use $n = MV$ (where $M$ = molarity, $V$ = volume in L).

Reactions in Gas Phase: At constant $T, P$, mole ratio = volume ratio (Avogadro).

Concentration Terms:

Molarity vs Temperature: Molarity depends on $T$ (volume changes); molality does not (mass-based).

Relation: $M = \dfrac{m \times \rho \times 1000}{1000 + m \times M_{\text{solute}}}$, where $\rho$ = density of solution.

Normality and Equivalent Weight:

• $E = M/n$, where $n$ = valency for acid/base, or change in oxidation number for redox.
• $N = M \times n$.

Examples:

• H₂SO₄: $n = 2$, $E = 98/2 = 49$
• NaOH: $n = 1$, $E = 40$
• KMnO₄ in acidic medium: $n = 5$, $E = 158/5 = 31.6$

Dilution formula: $M_1V_1 = M_2V_2$ (when only diluted).

Mixing different concentrations: $M_{\text{final}} = \dfrac{M_1V_1 + M_2V_2}{V_1+V_2}$.

Oxidation Number (ON): Hypothetical charge on an atom if all bonds were ionic.

Rules to find ON:

1. ON of free element = $0$
2. ON of monoatomic ion = its charge
3. ON of H = $+1$ (except $-1$ in metal hydrides)
4. ON of O = $-2$ (except: $-1$ in peroxides, $+2$ in OF₂, $-1/2$ in superoxides)
5. ON of F = $-1$ always
6. Sum of ON in neutral molecule = $0$; in ion = charge of ion

Oxidation: Increase in ON (loss of e⁻). Reduction: Decrease in ON (gain of e⁻). Oxidizing agent: Gets reduced (oxidizes others). Reducing agent: Gets oxidized.