Basic Concepts of Chemistry

In this lecture, we covered the fundamental concepts from the 'Some Basic Concepts of Chemistry' chapter according to the NCERT.

Laws of Chemical Combination

Law of Conservation of Mass

• Mass can neither be created nor destroyed in a chemical or physical reaction.
• If the reactants are completely converted into products, the mass of reactants equals the mass of products.
• Example: Heating 100 grams of mercury oxide results in mercury and oxygen, totaling 100 grams.

Law of Definite Proportions (Constant Composition)

• A chemical compound always contains the same elements in a fixed proportion by mass.
• Example: Water (H2O) always has a mass ratio of hydrogen to oxygen as 1:8.
• Example: In CO2, the mass ratio of carbon to oxygen is always 3:8.

Law of Multiple Proportions

• Two elements can combine to form more than one compound.
• Example: H2 and O2 can form H2O and H2O2 where the ratio of oxygen in the two compounds is 1:2.
• Elements combine in ratios of small whole numbers.

Gay-Lussac's Law of Gaseous Volumes

• Gases combine or are produced in chemical reactions in simple ratios by volume, provided all gases are at the same temperature and pressure.
• Example: 100 mL hydrogen + 100 mL chlorine = 200 mL HCl (1:1:2 ratio).

Avogadro's Law

• Equal volumes of gases at the same temperature and pressure contain an equal number of molecules.
• Example: 1 liter of hydrogen and 1 liter of chlorine will have the same number of molecules.

Atomic Mass

• Relative atomic mass is obtained by comparing the mass of an atom with one-twelfth of the mass of one carbon-12 isotope atom.
• Example: 1 atomic mass unit (amu) = 1/12 the mass of one carbon-12 atom.
• Some important elemental atomic masses: Hydrogen (1 u), Carbon (12 u), Oxygen (16 u).

Average Atomic Mass

• Average atomic mass is calculated considering the isotopic masses and their relative abundance.
• Example: Carbon has isotopes C-12, C-13, and C-14; the average atomic mass of carbon is approximately 12.011 u.

Molecular Mass

• Sum of the atomic masses of the elements present in a molecule.
• Examples:
  • Methane (CH4): 16 u
  • Water (H2O): 18 u
  • Glucose (C6H12O6): 180 u

Formula Mass

• Used for ionic compounds that do not form molecules but 3D structures.
• Example: NaCl formula unit mass is 23 (Na) + 35.5 (Cl) = 58.5 u.

Mole Concept

• 1 mole contains Avogadro's number (6.022 × 10^23) of particles.
• Molar mass = mass of 1 mole of a substance in grams.
• Example: 1 mole of Na = 23 g, 1 mole of CO2 = 44 g.
• Conversion: number of moles = given mass / molar mass.
• The molar mass can also be used to convert between volume and number of particles.
• Example: 1 mole of gas = 22.4 liters at STP.

Percentage Composition

• % Mass of an element = (mass of the element in the compound / molar mass of the compound) × 100
• Example: Ethanol (C2H5OH):
  • % C: (24/46) × 100 = 52.17%
  • % H: (6/46) × 100 = 13.04%
  • % O: (16/46) × 100 = 34.78%

Empirical & Molecular Formula

• Molecular Formula: Exact number of different atoms present in a molecule.
• Empirical Formula: Simplest whole-number ratio of atoms in a compound.
• Example: Glucose molecular formula = C6H12O6; empirical formula = CH2O.

Stoichiometry & Calculations

• Involves mass and sometimes volume calculations of reactants and products in a chemical reaction.
• Requires balanced chemical equations.
• Example: Combustion of methane (CH4): CH4 + 2O2 → CO2 + 2H2O.

Limiting Reagent

• The reactant that gets completely consumed first limits the amount of product formed.
• Example: Reacting 50 kg nitrogen and 10 kg hydrogen yields ammonia (NH3), where hydrogen is the limiting reagent.

Concentration Expressions in Solutions

• Mass Percent: (mass of solute / mass of solution) × 100.
• Mole Fraction: Ratio of the number of moles of a component to the total moles of all components.
• Molarity (M): Number of moles of solute per liter of solution (temperature dependent).
• Molality (m): Number of moles of solute per kg of solvent (temperature independent).
• Example Calculation: Given 44 grams CO2 and 48 grams O2, find the mole fraction.

Dilution

• When solvent is added to solution, moles of solute remain constant.
• Formula: M1V1 = M2V2

Mixing Solutions

• When two solutions are mixed, total moles of solute = (M1V1 + M2V2) / (V1+V2)

Question Examples

• Calculation of moles and mass based on given reactants and reactions.

Hope you understood the basic concepts of chemistry. Similar resources for other chapters can be found in the provided links.