1. Laws of Chemical Combination

1. Law of Conservation of Mass (Lavoisier):

Matter can neither be created nor destroyed. Total mass of reactants = Total mass of products.

2. Law of Definite Proportions (Joseph Proust):

A given compound always contains exactly the same proportion of elements by weight, irrespective of the source. (e.g., Water is always H:O = 1:8 by mass).

3. Law of Multiple Proportions (Dalton):

If two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
Example: CO (12:16) and CO₂ (12:32). Ratio of O is 16:32 = 1:2.

4. Law of Reciprocal Proportions (Richter):

The ratio of masses of two elements A and B which combine separately with a fixed mass of C is either the same or a simple multiple of the ratio of masses in which A and B combine directly.

5. Gay Lussac's Law of Gaseous Volumes:

Gases combine in a simple ratio by volume (at same T and P).
H₂(1 vol) + Cl₂(1 vol) → 2HCl(2 vol).

2. The Mole Concept

A Mole is the amount of substance containing as many entities as there are atoms in exactly 12g of C-12 isotope.

N_A = 6.022 × 10²³

The Y-Map Calculation

From Mass: n = Given Mass / Molar Mass
From Particles: n = Given Particles / N_A
From Volume (Gas at STP): n = Volume (L) / 22.4

Average Atomic Mass

For isotopes with % abundance (a, b) and atomic masses (m₁, m₂):

Avg Mass = (a×m₁ + b×m₂) / (a + b)

3. Empirical & Molecular Formula

Empirical Formula (EF): Simplest whole number ratio of atoms. (e.g., CH for Benzene).
Molecular Formula (MF): Actual number of atoms. (e.g., C₆H₆ for Benzene).

MF = n × EF
n = Molar Mass / Empirical Formula Mass

4. Stoichiometry

Calculations based on balanced chemical equations.

Limiting Reagent (LR)

The reactant which is consumed first and limits the amount of product formed.

How to find LR:

Calculate moles of all reactants.
Divide moles by their stoichiometric coefficient.
The reactant with the lowest ratio is the Limiting Reagent.

5. Concentration Terms

1. Mass Percent (% w/w):

(Mass of Solute / Mass of Solution) × 100

2. Mole Fraction (x):

x_A = n_A / (n_A + n_B)

(Independent of Temperature)

3. Molarity (M):

Moles of Solute / Vol of Solution (L)

(Changes with Temperature)

4. Molality (m):

Moles of Solute / Mass of Solvent (kg)

(Independent of Temperature)

Numericals & HOTS

Q1. Basic Mole Concept

Calculate the total number of electrons present in 1.6 g of Methane (CH₄).

Solution:
1. Moles of CH₄ = Mass / Molar Mass
n = 1.6 g / 16 g/mol = 0.1 mol.

2. Electrons in 1 molecule of CH₄:
C(6) + 4H(4) = 10 electrons.

3. Total Electrons = n × N_A × 10
= 0.1 × 6.022 × 10²³ × 10
= 1 × 6.022 × 10²³
Ans: 6.022 × 10²³ electrons

Q2. Stoichiometry & Purity

20 g of MgCO₃ sample decomposes on heating to give carbon dioxide and 8.0 g Magnesium oxide. What will be the percentage purity of Magnesium carbonate in the sample? (At. mass Mg=24).

Solution:
MgCO₃ → MgO + CO₂
Molar Mass MgCO₃ = 24+12+48 = 84 g.
Molar Mass MgO = 24+16 = 40 g.

According to stoichiometry:
84 g MgCO₃ gives 40 g MgO.
To get 8.0 g MgO, MgCO₃ needed = (84/40) × 8 = 16.8 g.

% Purity = (Pure Mass / Total Mass) × 100
% Purity = (16.8 / 20) × 100
Ans: 84%

Q3. Limiting Reagent

50.0 kg of N₂ (g) and 10.0 kg of H₂ (g) are mixed to produce NH₃ (g). Calculate the amount of NH₃ formed.

Solution:
Reaction: N₂ + 3H₂ → 2NH₃
Moles N₂ = 50,000 / 28 = 1785.7 mol.
Moles H₂ = 10,000 / 2 = 5000 mol.

Check LR:
For N₂: 1785.7 / 1 = 1785.7
For H₂: 5000 / 3 = 1666.6
Since 1666.6 < 1785.7, H₂ is LR.

Moles NH₃ = (2/3) × Moles H₂
= (2/3) × 5000 = 3333.3 mol.
Mass NH₃ = 3333.3 × 17 = 56666 g
Ans: 56.67 kg

Q4. Mixing of Acids

If 500 mL of 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?

Solution:
Dilution Formula: M₁V₁ = M₂V₂
M₁ = 5 M, V₁ = 500 mL.
V₂ = 1500 mL.

5 × 500 = M₂ × 1500
2500 = 1500 M₂
M₂ = 2500 / 1500 = 5/3
Ans: 1.66 M

Q5. Finding Formula

An oxide of iron has 69.9% iron and 30.1% dioxygen by mass. Determine the empirical formula. (Fe=55.85, O=16.00).

Solution:
1. Moles Fe = 69.9 / 55.85 = 1.25 mol.
2. Moles O = 30.1 / 16.00 = 1.88 mol.

Simple Ratio:
Fe = 1.25 / 1.25 = 1
O = 1.88 / 1.25 = 1.5

Whole Number Ratio: Multiply by 2.
Fe = 2, O = 3.
Ans: Fe₂O₃

Q6. Minimum Molecular Mass

An enzyme contains 3.2% sulphur by mass. Calculate the minimum molecular mass of the enzyme. (Atomic mass S = 32).

Solution:
For "minimum" molar mass, the molecule must contain at least one atom of Sulphur.

Let Molar Mass = M.
(Mass of S / Total Mass) × 100 = 3.2
(32 / M) × 100 = 3.2
M = (32 × 100) / 3.2
Ans: 1000 g/mol

Q7. Concentration Conversion

Calculate the molality of 3 M solution of NaCl. The density of the solution is 1.25 g/mL. (Molar mass NaCl = 58.5).

Solution:
Consider 1 L (1000 mL) solution.
Moles Solute = 3 mol.
Mass Solute = 3 × 58.5 = 175.5 g.

Mass Solution = Volume × Density = 1000 × 1.25 = 1250 g.
Mass Solvent = Mass Solution - Mass Solute
= 1250 - 175.5 = 1074.5 g = 1.0745 kg.

Molality (m) = Moles / Mass Solvent(kg)
m = 3 / 1.0745
Ans: 2.79 m

Q8. Mole Fraction to Molality

The mole fraction of a solute in an aqueous solution is 0.2. Calculate the molality of the solution.

Solution:
Given x_solute = 0.2.
So, x_solvent (water) = 1 - 0.2 = 0.8.

Ratio of moles: n_solute / n_solvent = 0.2 / 0.8 = 1 / 4.
Let n_solute = 1 mol, n_solvent = 4 mol.

Mass Solvent (Water) = 4 mol × 18 g/mol = 72 g = 0.072 kg.

Molality = Moles Solute / Mass Solvent(kg)
m = 1 / 0.072
Ans: 13.88 m

Q9. Isotope Calculation

Boron has two stable isotopes, ¹⁰B (19%) and ¹¹B (81%). Calculate the average atomic mass of Boron in the periodic table.

Solution:
Avg Mass = (a₁m₁ + a₂m₂) / 100
= (19 × 10 + 81 × 11) / 100
= (190 + 891) / 100
= 1081 / 100
Ans: 10.81 u

Q10. V.D. Calculation

The vapor density of a mixture containing NO₂ and N₂O₄ is 27.6. Calculate the mole fraction of NO₂ in the mixture.

Solution:
Molar Mass of Mixture = 2 × V.D. = 2 × 27.6 = 55.2 g/mol.
Molar Mass NO₂ = 46 g/mol.
Molar Mass N₂O₄ = 92 g/mol.

Let mole fraction of NO₂ = x.
x(46) + (1-x)(92) = 55.2
46x + 92 - 92x = 55.2
92 - 55.2 = 46x
36.8 = 46x
x = 36.8 / 46
Ans: 0.8

Important Formulae

1. The Mole Calculation (n)

From Mass:

n = Given Mass (w) / Molar Mass (M)

From Number of Particles (N):

n = N / N_A

(N_A = 6.022 × 10²³)

From Volume of Gas (at STP):

n = Vol (L) / 22.4 or Vol (mL) / 22400

2. Concentration Terms

Molarity (M):

M = n_solute / V_sol(L)

Molality (m):

m = n_solute / Mass_solvent(kg)

Mass Percent (% w/w):

% = (Mass_solute / Mass_solution) × 100

3. Dilution & Mixing Laws

Dilution Formula:

M₁V₁ = M₂V₂

Mixing of Solutions (Same solute):

M_mix = (M₁V₁ + M₂V₂) / (V₁ + V₂)

4. Average Atomic Mass

For isotopes with abundance % a and b:

Avg. Mass = (aA₁ + bA₂) / (a + b)

5. Relation: M and m

m = (1000 × M) / (1000d - M × M_solute)

d = Density of solution (g/mL)
M_solute = Molar Mass of solute

20 Golden Facts (NEET)

1. STP Conditions: Standard Temperature and Pressure (STP) implies T = 273.15 K (0°C) and P = 1 bar. Molar volume of ideal gas = 22.7 L (Modern) or 22.4 L (Old convention, often used in exams).
2. Limiting Reagent: The reactant which is completely consumed determines the amount of product. Always convert given amounts to moles and divide by stoichiometric coefficient to find LR.
3. Temperature Dependence: Molarity (M), Normality (N), and % w/V change with temperature (Volume expands). Molality (m), Mole Fraction (x), and % w/w are independent of Temperature.
4. 1 amu (u): Defined as 1/12th of the mass of one C-12 atom. 1 amu = 1.66 × 10^-24 g. It is the reciprocal of Avogadro's Number.
5. Dulong and Petit's Law: For solid elements, Atomic Mass × Specific Heat ≈ 6.4. Used to find approximate atomic mass.
6. Water Density: For water, 1 g = 1 mL (approx). Therefore, Molarity ≈ Molality for very dilute aqueous solutions.
7. Vapor Density: V.D. = Molar Mass / 2. This relation is valid for gases. It helps to find molecular mass from vapor density.
8. Law of Multiple Proportions: Explains formation of oxides like N₂O, NO, NO₂, N₂O₅. Fixed mass of N combines with variable masses of O in simple ratio.
9. Mass of 1 atom: To find mass of a single atom, divide Atomic Mass by N_A. (e.g., Mass of 1 atom of C = 12 / 6.022×10²³ g).
10. Stoichiometric Coefficients: They represent the MOLE ratio, not the mass ratio. (e.g., 2H₂ + O₂ means 2 moles H₂ react with 1 mole O₂).
11. Parts Per Million (ppm): Used for very dilute concentrations (e.g., pollutants). ppm = (Mass of solute / Mass of solution) × 10⁶.
12. Empirical Formula: It represents the simplest whole number ratio of atoms. Molecular Formula = n × Empirical Formula, where n = (Molar Mass / Empirical Mass).
13. Isotopes: Cl exists as Cl-35 (75%) and Cl-37 (25%). This explains why atomic mass of Chlorine is 35.5 (Fractional).
14. Number of Atoms: Max number of atoms is present in the sample which has the least atomic mass for the same given weight (since n = w/M, smaller M gives larger n).
15. Law of Reciprocal Proportions: Example: H₂S, SO₂, and H₂O. (Ratio of S and O reacting with fixed H).
16. Molarity of Pure Water: 1 L water = 1000 g. Moles = 1000/18 = 55.55 mol. So, Molarity = 55.55 M.
17. Loschmidt Number: The number of molecules present in 1 mL of a gas at STP. (2.68 × 10¹⁹).
18. Specific Gravity: Another name for relative density. For liquids/solids, Sp. Gr = Density (g/mL).
19. Concentration of Ions: Molarity of ions = Molarity of Salt × Number of ions per formula unit. (e.g., in 1M AlCl₃, [Cl^-] = 3 M).
20. Significant Figures: In addition/subtraction, result has same decimal places as the term with least decimal places. In multiplication/division, result has same sig figs as term with least sig figs.

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