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Some Basic Concepts of Chemistry

Ch. 1Std 11

Easy Overview

Imagine trying to count how many grains of sand are on a beach. That's the challenge chemists face working with atoms. This chapter gives you the toolkit: the mole concept, stoichiometry, and the laws of chemical combinations. The mole is the chemist's dozen — 6.022 × 10²³ particles. One mole of carbon-12 weighs exactly 12 grams, linking atomic mass (amu) to weighable samples (grams). From here you get percentage composition, empirical/molecular formulas, and stoichiometric calculations. Concentration terms (molarity, molality, normality) describe solutions precisely. The laws of chemical combination — conservation of mass, definite proportions, multiple proportions, Gay-Lussac's law — are the foundation.

Mole Concept — The Chemist's Counting Unit

A mole is 6.022 × 10²³ particles (Avogadro's constant, N_A). One mole of carbon-12 weighs exactly 12 grams. This bridges amu and grams: atomic mass in amu = mass in grams of one mole. So 1 mole O atoms = 16 g, 1 mole H₂O = 18 g.

Molar Mass and Mole Calculations

Molar mass (M) = mass of one mole (g/mol). For elements = atomic mass. For compounds = sum of atomic masses. Central formula: n = m/M. Number of particles = n × N_A. Example: 88 g CO₂ (M=44) = 2 mol = 1.204 × 10²⁴ molecules = 3.612 × 10²⁴ atoms.

Average Atomic Mass and Isotopic Abundance

Atomic masses are weighted averages of isotopes. Cl: Cl-35 (75.77%, 34.969 amu) and Cl-37 (24.23%, 36.966 amu). Average = (34.969×0.7577)+(36.966×0.2423) = 35.45 amu. Work backwards to find abundances if needed.

Percentage Composition

% element = (mass of element in 1 mol / molar mass) × 100. H₂O: %H = (2.016/18.015)×100 = 11.19%, %O = 88.81%. Sum = 100%. Common mistake: forgetting subscripts — Ca(OH)₂ has 2 O and 2 H.

Empirical and Molecular Formulas

Empirical = simplest whole-number ratio. Molecular = (empirical)_n, where n = molecular mass / empirical mass. From % composition: assume 100 g, convert to moles, divide by smallest, round. If 1.33 appears, multiply all by 3 (1.33 = 4/3). CH₂O (n=1), C₂H₄O₂ (n=2), C₆H₁₂O₆ (n=6).

Stoichiometry — The Mathematics of Reactions

Balanced equation coefficients = mole ratios. Convert given to moles, use ratio, convert back. Example: 2H₂+O₂→2H₂O: 4 mol H₂ needs 2 mol O₂, produces 4 mol H₂O. Never mass-to-mass directly — always through moles.

Limiting Reagent — The Bottleneck

The reactant that produces least product is limiting. N₂+3H₂→2NH₃: 2 mol N₂ + 3 mol H₂ — H₂ is limiting (needs 6 mol for 2 mol N₂), produces 2 mol NH₃. N₂ could produce 4 mol if enough H₂ were present.

Percentage Yield — Reality Check

% yield = (actual/theoretical) × 100. >90% excellent, 70-90% good. Multistep: overall = product of step yields. Three 80% steps = 51.2% overall.

Concentration Terms — Molarity, Molality, Normality

Molarity (M) = mol/L (temp-dependent). Molality (m) = mol/kg solvent (temp-independent). Normality (N) = n×M. For acids: n = basicity. For bases: n = acidity. For redox: n = electrons transferred. Mole fraction X = nᵢ/n_total. ppm = (mass solute/mass solution)×10⁶.

Laws of Chemical Combination

Conservation of Mass: mass of reactants = mass of products. Definite Proportions: fixed mass ratios in compounds. Multiple Proportions: mass ratios for multiple compounds are small whole numbers. Gay-Lussac's Law: gaseous volumes react in simple whole-number ratios.

Avogadro's Law and Molar Volume

Equal gas volumes at same T/P have equal molecules. At STP (0°C, 1 atm): 1 mol = 22.4 L. n = V(L)/22.4 only at STP. At RTP (25°C, 1 atm): ~24.5 L/mol. Use PV=nRT otherwise.

Determining Molecular Mass — Vapor Density

Vapor density = density of gas / density of H₂ = M(gas)/2. So M(gas) = 2 × vapor density. Vapor density = 14 → M = 28 g/mol (N₂, CO, or C₂H₄).

Significant Figures and Scientific Notation

Sig figs convey precision. Non-zero digits significant. Zeros between non-zero digits significant. Leading zeros not significant. Trailing zeros after decimal significant. 0.00450 has 3 sig figs. Multiplication/division: same sig figs as least precise. Addition/subtraction: same decimal places as fewest.

Empirical Formula from Combustion Data

Burn sample in O₂. Mass C = (12/44)×mass CO₂. Mass H = (2/18)×mass H₂O. If O present: mass O = total - mass C - mass H. Find mole ratio C:H:O, simplify.

Stoichiometry in Solution — Titrations

M₁V₁/n₁ = M₂V₂/n₂. For H₂SO₄ + 2NaOH: 1:2 ratio, so 2M₁V₁ = M₂V₂. Or use N₁V₁ = N₂V₂ (simpler since N accounts for stoichiometry).

Law of Equivalents and Equivalent Mass

Equivalent mass: elements = atomic mass/valency. Acids = molar mass/basicity. Bases = molar mass/acidity. Redox = molar mass/n-factor. N₁V₁ = N₂V₂ at equivalence. N = n × M.

Concentration Conversions and Dilution

M₁V₁ = M₂V₂ (moles constant). To prepare 500 mL 0.1 M HCl from 2 M: V₁ = (0.1×500)/2 = 25 mL. For M→m conversion, need density. Dilute aqueous: M ≈ m.

Numericals on Mole Concept — A Worked Approach

Step 1: Identify given and required. Step 2: Convert to moles. Step 3: Apply relation. Step 4: Calculate with correct units. Step 5: Check reasonableness. Common errors: °C not K, mL not L, wrong molar mass, ignoring coefficients.

Key Points

  • •One mole = 6.022 × 10²³ particles (Avogadro's number)
  • •Molar mass (g/mol) = atomic/molecular mass in amu
  • •n = m/M is the central formula; Particles = n × N_A
  • •Average atomic mass = Σ (isotopic mass × fractional abundance)
  • •% composition = (mass of element in 1 mol / molar mass) × 100
  • •Empirical = simplest ratio; Molecular = n × empirical
  • •Combustion: mass C = (12/44)×mass COâ‚‚; mass H = (2/18)×mass Hâ‚‚O
  • •Limiting reagent → theoretical yield; % yield = (actual/theoretical)×100
  • •Molarity (M) = mol/L; Molality (m) = mol/kg; Normality (N) = n×M
  • •M₁V₁ = Mâ‚‚Vâ‚‚ for dilution; N₁V₁ = Nâ‚‚Vâ‚‚ for titrations
  • •STP: 1 mol gas = 22.4 L; Molecular mass = 2 × vapor density
  • •Four laws: Conservation, Definite Proportions, Multiple Proportions, Gay-Lussac
  • •Equivalent mass = molar mass / n-factor
  • •Significant figures: result precision ≤ least precise input
  • •Always convert mL to L, °C to K

Practice Questions

  • Calculate moles, molecules, and atoms in 36 g water.
  • A compound has C=54.54%, H=9.09%, O=36.37%. Molecular mass=88 g/mol. Find empirical and molecular formulas.
  • In 2Al+3Cl₂→2AlCl₃, 5.4 g Al reacts with 21.3 g Clâ‚‚. Find limiting reagent and mass of AlCl₃ formed.
  • Calculate average atomic mass of Cu: Cu-63 (62.93 amu, 69.09%) and Cu-65 (64.93 amu, 30.91%).
  • Calculate molarity and normality of 4.9 g Hâ‚‚SOâ‚„ in 250 mL solution.
  • 0.246 g organic compound gives 0.352 g COâ‚‚ and 0.180 g Hâ‚‚O. Determine empirical formula.
  • What volume of 0.1 M HCl neutralizes 25 mL of 0.05 M Ca(OH)â‚‚?
  • How many grams Oâ‚‚ react with 24 g CHâ‚„? Volume of COâ‚‚ at STP?