Basic Concepts - Class 11 Chemistry

Basic Concepts

Overview: Introduction to matter, laws of chemical combination, mole concept, and stoichiometry.

1. Laws of Chemical Combination

Law of Conservation of Mass: Matter is neither created nor destroyed (Lavoisier).
Law of Definite Proportions: Given compound always contains exactly the same proportion of elements by weight (Proust).
Law of Multiple Proportions: If two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole number ratio (Dalton).
Gay Lussac's Law: Gases combine in simple ratio by volume showing simple ratio to product volume.
Avogadro's Law: Equal volumes of all gases at same T and P contain equal number of molecules.

2. Dalton's Atomic Theory

Matter consists of indivisible atoms.
All atoms of a given element have identical properties including mass.
Compounds are formed when atoms of different elements combine in fixed ratio.

Modified later: Atoms are divisible (p, n, e), isotopes exist (different mass same element).

3. Mole Concept

Mole: Amount of substance containing as many elementary entities as there are atoms in 0.012 kg of C-12.

1 Mole = 6.022 × 10²³ particles

Moles (n) = Mass (m) / Molar Mass (M)

n = N / N_A = V_STP / 22.4 L

4. Atomic & Molecular Mass

Atomic Mass Unit (amu or u): 1/12th mass of C-12 atom. 1 amu = 1.66 × 10^-24 g.
Average Atomic Mass: Weighted average of isotopes. Σ(Abundance × Mass) / ATotal Abundance.
Molar Mass: Mass of 1 mole of substance in grams.

5. Concentration Terms

Molarity (M)

Moles of solute per liter of solution.

M = n_solute / V_solution (L)

Molality (m)

Moles of solute per kg of solvent.

m = n_solute / Mass_solvent (kg)

Mole Fraction (x)

x_A = n_A / (n_A + n_B)

Limiting Reagent

Reagent that is completely consumed in a reaction. It determines the amount of product formed.

Numericals - Basic Concepts

Numericals

Molar Mass

Q1. Calculate molecular mass of Glucose (C6H12O6).

Atomic masses: C=12, H=1, O=16

Mass = 6(12) + 12(1) + 6(16)

Mass = 72 + 12 + 96

Mass = 180 u

Mole Concept

Q2. Calculate number of atoms in 52g of Helium.

Molar mass of He = 4g.

Moles = Mass / Molar Mass = 52 / 4 = 13 moles.

No. of atoms = n × NA

N = 13 × 6.022 × 10²³ = 7.8286 × 10²⁴ atoms.

Molarity

Q3. Calculate molarity of NaOH in solution prepared by dissolving 4g in enough water to form 250ml solution.

Molar mass NaOH = 23 + 16 + 1 = 40g.

Moles = 4 g / 40 g/mol = 0.1 mol.

Volume = 250 ml = 0.25 L.

Molarity = Moles / Vol(L) = 0.1 / 0.25

M = 0.4 M

Empirical Formula

Q4. A compound contains 4.07% H, 24.27% C, and 71.65% Cl. Molar mass is 98.96g. Find Empirical and Molecular Formula.

Moles: H=4.07/1=4.07, C=24.27/12=2.02, Cl=71.65/35.5=2.02.

Simple Ratio: Divide by 2.02.

H=2, C=1, Cl=1. → Empirical Formula: CH2Cl.

Empirical Mass = 12 + 2 + 35.5 = 49.5g.

n = Molar Mass / Emp Mass = 98.96 / 49.5 = 2.

Molecular Formula = 2 × (CH2Cl) = C2H4Cl2.

Limiting Reagent

Q5. 50kg N2 and 10kg H2 form NH3. Calculate NH3 formed. Identify LR.

N2 + 3H2 → 2NH3

Moles N2 = 50000/28 = 1786 mol.

Moles H2 = 10000/2 = 5000 mol.

Required H2 for 1786 mol N2 = 1786 × 3 = 5358 mol.

But we have 5000 mol H2 only. So H2 is Limiting Reagent.

NH3 formed depends on H2. 3 mol H2 → 2 mol NH3.

NH3 = (2/3) × 5000 = 3333 moles.

Mass NH3 = 3333 × 17g = 56.6 kg.

Molality

Q6. Density of 3M solution of NaCl is 1.25 g/mL. Calculate molality.

Consider 1 L solution. Vol = 1000 mL.

Mass of solution = V × d = 1000 × 1.25 = 1250 g.

Moles of NaCl = 3 mol. Mass NaCl = 3 × 58.5 = 175.5 g.

Mass of Solvent (water) = 1250 - 175.5 = 1074.5 g = 1.0745 kg.

Molality m = Moles Solute / Mass Solvent(kg)

m = 3 / 1.0745 = 2.79 m.

Combustion

Q7. Calculate CO2 produced by combustion of 16g methane.

CH4 + 2O2 → CO2 + 2H2O

1 mol CH4 (16g) produces 1 mol CO2 (44g).

Ans: 44g CO2.

Mole Fraction

Q8. Calc mole fraction of Ethylene Glycol (C2H6O2) in solution with 20% by mass.

20g Glycol + 80g Water.

Molar Mass Glycol = 24+6+32 = 62g. Moles = 20/62 = 0.322.

Moles Water = 80/18 = 4.444.

x_glycol = 0.322 / (0.322 + 4.444) = 0.322 / 4.766

x = 0.068.

Average Atomic Mass

Q9. Chlorine has two isotopes Cl-35 (75%) and Cl-37 (25%). Avg mass?

Avg Mass = (35 × 75 + 37 × 25) / 100

Avg Mass = (2625 + 925) / 100

Avg Mass = 3550 / 100 = 35.5 u.

Number of Molecules

Q10. How many molecules in 34.2g of sucrose (C12H22O11)?

MM Sucrose = 342 g/mol.

Moles = 34.2 / 342 = 0.1 mol.

Molecules = 0.1 × 6.022 × 10²³

= 6.022 × 10²².

Formulas & Facts - Basic Concepts

Equations & Formulas

Concept	Formula
Moles (mass)	n = m / M
Moles (particles)	n = N / N_A
Moles (gas STP)	n = V(L) / 22.4
Molarity (M)	M = n_solute / V_sol(L)
Molality (m)	m = n_solute / W_solvent(kg)
Mass %	(Mass solute / Mass sol) × 100
Mole Fraction	x_A = n_A / n_total
Dilution Law	M₁V₁ = M₂V₂
Avg Atomic Mass	Σ(% × Mass) / 100

50 NEET Facts

Key points for Basic Concepts.

1. Avogadro Number 6.022 × 10²³. Number of atoms in exactly 12g C-12.

2. Molality & Temp Molality is independent of temperature (mass doesn't change).

3. Molarity & Temp Molarity depends on temperature (Volume expands/contracts).

4. 1 amu 1.66056 × 10^-24 g.

5. STP Conditions New Convention: 1 bar, 273.15 K. Vm = 22.7 L. (Old: 1 atm, Vm = 22.4 L).

6. Law of Conservation of Mass Valid for chemical rxns, not nuclear rxns (E=mc^2).

7. Law of Definite Proportions Invalid for isotopes and non-stoichiometric compounds (Berthollides).

8. Limiting Reagent Reactant that gets consumed first. Decides product yield.

9. Precision Closeness of various measurements to each other.

10. Accuracy Closeness of result to true value.

11. Significant Figures (0) Zeros between non-zeros are significant. Zeros to left of first non-zero are NOT.

12. Sig Figs (Right) Zeros to right of decimal are significant.

13. Exact Numbers Have infinite significant figures (e.g., 2 balls).

14. Rounding Off If digit is 5: Increase prec if odd, leave if even.

15. Density of Water 1 g/mL at 4°C.

16. ppm Parts per million. (Mass solute / Mass sol) × 10^6.

17. Specific Gravity Density of substance / Density of water. Unitless.

18. Empirical Formula Simplest whole number ratio of atoms.

19. Molecular Formula Actual number of atoms. MF = n × EF.

20. Gay Lussac's Law Valid only for gases. Volume ratio is simple whole number.

21. Dalton's Theory Limit Could not explain Gay Lussac's law of gaseous volumes.

22. Isomers Compounds with same Molecular Formula but different structure.

23. 1 Mole of Gas 22.4 Liters at STP (1 atm). 22.7 Liters at STP (1 bar).

24. Vapor Density VD = Molar Mass / 2.

25. Mass of Electron 9.1 × 10^-31 kg. Negligible for atomic mass.

26. Proton Mass 1.672 × 10^-27 kg. ~1 amu.

27. Neutron Mass 1.674 × 10^-27 kg. Slightly heavier than proton.

28. Dulong Petit Law At mass × Sp Heat ≈ 6.4 (for solids).

29. Loschmidt Number Number of molecules in 1 mL of gas at STP.

30. Mixture Molarity M = (M1V1 + M2V2) / (V1 + V2).

31. Normality N = M × n-factor (Valency factor).

32. n-factor Acid Basicity (Number of replaceable H+).

33. n-factor Base Acidity (Number of replaceable OH-).

34. n-factor Salt Total positive or negative charge.

35. Stoichiometry Calculation of masses/volumes of reactants and products.

36. % Yield (Actual Yield / Theoretical Yield) × 100.

37. Purity % (Mass pure / Mass sample) × 100.

38. Dimensional Analysis Factor-Label method to convert units.

39. Femto 10^-15.

40. Angstrom 10^-10 m. Common for atomic radii.

41. Nanometer 10^-9 m.

42. Derived Units Units derived from 7 base SI units (e.g. Density kg/m3).

43. Base Units m, kg, s, A, K, mol, cd.

44. Temperature SI Kelvin (K). No degree sign.

45. Volume SI Cubic meter (m3). Common unit Liter (L = dm3).

46. Law of Reciprocal Proportions Related to equivalent weights.

47. Formula Mass Used for ionic compounds (NaCl) instead of Molecular Mass.

48. Mass of 1 Atom Atomic Mass (g) / N_A.

49. Mass Spectrometry Used to determine exact atomic masses.

50. Mole Fraction Sum Sum of mole fractions of all components is always 1.

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