What is a mole in chemistry?

A mole is a fundamental unit in chemistry representing 6.022 × 10²³ particles (Avogadro's number). It's the amount of substance that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12.

What is the difference between atomic mass and molecular mass?

Atomic mass is the mass of a single atom (measured in amu or u), while molecular mass is the sum of atomic masses of all atoms in a molecule. For example, atomic mass of H is 1u, but molecular mass of H₂ is 2u.

How do you calculate percentage composition?

Percentage composition = (Mass of element in compound / Molar mass of compound) × 100. For example, in H₂O, percentage of H = (2/18) × 100 = 11.11%.

What is a limiting reagent?

A limiting reagent is the reactant that is completely consumed first in a chemical reaction, thus limiting the amount of product formed. The reactant in excess remains after the reaction is complete.

How is molarity different from molality?

Molarity (M) is moles of solute per liter of solution and changes with temperature. Molality (m) is moles of solute per kilogram of solvent and is temperature-independent.

Some Basic Concepts of Chemistry for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Q: What is the difference between empirical and molecular formula?

Empirical formula shows the simplest whole number ratio of atoms in a compound, while molecular formula shows the actual number of atoms. For example, glucose has empirical formula CH₂O and molecular formula C₆H₁₂O₆.

Matter and Its Nature

Matter is anything that occupies space and has mass. Understanding the basic nature of matter is fundamental to chemistry. Everything around us - air, water, rocks, living organisms - is made of matter.

1.1 Classification of Matter

Physical Classification

Solid: Definite shape & volume (e.g., iron, ice)
Liquid: Definite volume, no definite shape (e.g., water)
Gas: No definite shape or volume (e.g., air, oxygen)
Plasma: Ionized gas state (e.g., stars)

Chemical Classification

Element: Pure substance (e.g., H, O, Au)
Compound: Two or more elements (e.g., H₂O, NaCl)
Mixture: Physical combination
- Homogeneous (uniform - salt solution)
- Heterogeneous (non-uniform - sand + salt)

1.2 Properties of Matter

Property Type	Definition	Examples
Physical Properties	Observable without changing composition	Color, melting point, boiling point, density
Chemical Properties	Describe how substance reacts	Reactivity, combustibility, acidity
Intensive Properties	Independent of amount of matter	Temperature, pressure, density, refractive index
Extensive Properties	Depend on amount of matter	Mass, volume, length, heat capacity

💡 JEE Memory Trick

Intensive vs Extensive Properties:
Remember "INT" in Intensive = "INdependent of Total amount"
Extensive properties are "EXTended" when you add more substance

1.3 Laws of Chemical Combination

1. Law of Conservation of Mass (Lavoisier, 1789)

"Mass can neither be created nor destroyed in a chemical reaction"

Total mass of reactants = Total mass of products

2. Law of Definite Proportions (Proust, 1799)

"A chemical compound always contains same elements in the same proportion by mass"

Example: H₂O always has H:O = 1:8 by mass

3. Law of Multiple Proportions (Dalton, 1803)

"When two elements combine to form two or more compounds, masses of one element bear simple whole number ratio"

Example: CO (C:O = 3:4) and CO₂ (C:O = 3:8)

For fixed mass of C, O masses are in ratio 4:8 = 1:2

4. Gay-Lussac's Law of Gaseous Volumes (1808)

"Gases combine in simple whole number ratios by volume at same T and P"

H₂ + Cl₂ → 2HCl (1 vol : 1 vol : 2 vol)

5. Avogadro's Law (1811)

"Equal volumes of all gases at same T and P contain equal number of molecules"

V ∝ n (at constant T and P)

Atomic and Molecular Mass

Atomic and molecular masses are fundamental concepts for quantitative chemistry. They form the basis for mole concept and stoichiometric calculations - accounting for 15-20% of JEE Chemistry questions.

2.1 Atomic Mass Unit (amu or u)

Definition

One atomic mass unit (u) is defined as exactly 1/12th the mass of one carbon-12 atom.

1 \text{ u} = \frac{1}{12} \times \text{mass of one } ^{12}C \text{ atom}

1 \text{ u} = 1.66054 \times 10^{-24} \text{ g} = 1.66054 \times 10^{-27} \text{ kg}

2.2 Average Atomic Mass

For elements with isotopes, average atomic mass is calculated using natural abundance:

\text{Average Atomic Mass} = \sum (\text{Isotopic mass} \times \text{Fractional abundance})

📝 Solved Example 1 (JEE Main Pattern)

Question: Chlorine has two isotopes: ³⁵Cl (75% abundance, mass 34.97 u) and ³⁷Cl (25% abundance, mass 36.97 u). Calculate average atomic mass of chlorine.

Solution:

Step 1: Convert percentages to fractions

³⁵Cl: 75% = 0.75

³⁷Cl: 25% = 0.25

Step 2: Apply formula

\text{Average atomic mass} = (34.97 \times 0.75) + (36.97 \times 0.25)

= 26.2275 + 9.2425 = 35.47 \text{ u}

\text{Answer: } 35.47 \text{ u (or 35.5 u)}

2.3 Molecular Mass and Formula Mass

Molecular Mass

Sum of atomic masses of all atoms in a molecule

Example: H₂O

= 2(1) + 1(16)

= 18 u

Used for molecular compounds (covalent)

Formula Mass

Sum of atomic masses in empirical formula

Example: NaCl

= 23 + 35.5

= 58.5 u

Used for ionic compounds

⚠️ Important Distinction

Atomic mass: Mass of ONE atom (use for elements)
Molecular mass: Mass of ONE molecule (use for covalent compounds)
Formula mass: Mass from empirical formula (use for ionic compounds)
Molar mass: Mass of ONE mole (in grams, numerically equal to atomic/molecular mass)

Mole Concept (The Heart of Chemistry)

The mole is the most important concept in chemistry. It's the bridge between the atomic world and the laboratory world. Master this, and you'll excel in JEE Chemistry!

3.1 Definition of Mole

What is a Mole?

One mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12.

1 \text{ mole} = 6.022 \times 10^{23} \text{ particles (Avogadro's number)}

Avogadro's Number (N_A) = 6.022 × 10²³ mol⁻¹

💡 Mole Analogy

Just like "dozen" means 12, "mole" means 6.022 × 10²³!
1 dozen eggs = 12 eggs
1 mole of atoms = 6.022 × 10²³ atoms

Why this number? Because 1 mole of ¹²C atoms weighs exactly 12 grams!

3.2 Molar Mass

Definition

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

\text{Molar Mass (M)} = \frac{\text{Mass in grams}}{\text{Number of moles}}

Key Point: Numerically, Molar Mass = Atomic/Molecular Mass

Example: Atomic mass of C = 12 u → Molar mass of C = 12 g/mol

3.3 Mole Concept Formulas (Must Memorize!)

The Golden Triangle of Mole Concept

Formula 1

n = \frac{m}{M}

n = moles, m = mass, M = molar mass

Formula 2

n = \frac{N}{N_A}

N = no. of particles, N_A = 6.022×10²³

Formula 3 (for gases)

n = \frac{V}{22.4}

V = volume at STP in liters

Relationship	Formula	Application
Moles from mass	n = m/M	Given mass, find moles
Moles from particles	n = N/N_A	Given number of atoms/molecules
Moles from volume (gas at STP)	n = V/22.4	1 mole gas = 22.4 L at STP
Mass from moles	m = n × M	Find mass if moles known
Particles from moles	N = n × N_A	Find number of particles

📝 Solved Example 2 (JEE Main 2023 Type)

Question: Calculate the number of molecules in 11 g of CO₂.

Solution:

Step 1: Calculate molar mass of CO₂

M_{CO_2} = 12 + 2(16) = 44 \text{ g/mol}

Step 2: Calculate number of moles

n = \frac{m}{M} = \frac{11}{44} = 0.25 \text{ mol}

Step 3: Calculate number of molecules

N = n \times N_A = 0.25 \times 6.022 \times 10^{23}

N = 1.5055 \times 10^{23} \text{ molecules}

📝 Solved Example 3 (JEE Advanced Pattern)

Question: How many atoms are present in:
(a) 52 g of He
(b) 52 u of He

Solution:

Part (a): 52 g of He

Molar mass of He = 4 g/mol

n = \frac{52}{4} = 13 \text{ mol}

N = 13 \times 6.022 \times 10^{23} = 7.83 \times 10^{24} \text{ atoms}

Part (b): 52 u of He

Atomic mass of He = 4 u

\text{Number of atoms} = \frac{52}{4} = 13 \text{ atoms}

⚠️ Critical Difference:

52 g contains 10²⁴ atoms (macroscopic)

52 u contains just 13 atoms (microscopic)

Stoichiometry and Chemical Equations

Stoichiometry deals with quantitative relationships between reactants and products in chemical reactions. This is the MOST important topic for JEE - nearly 40% of numerical problems involve stoichiometry!

4.1 Balancing Chemical Equations

Rules for Balancing

Write correct formulas for all reactants and products
Balance atoms one element at a time (except H and O)
Balance polyatomic ions as a unit if they appear on both sides
Balance H and O atoms last
Check: Number of atoms of each element must be equal on both sides
Use smallest whole number coefficients

📝 Solved Example 4

Question: Balance the equation: C₃H₈ + O₂ → CO₂ + H₂O

Solution:

Step 1: Balance carbon atoms

C₃H₈ + O₂ → 3CO₂ + H₂O

Step 2: Balance hydrogen atoms

C₃H₈ + O₂ → 3CO₂ + 4H₂O

Step 3: Balance oxygen atoms

O on right side = 3(2) + 4(1) = 10

So need 5 O₂ molecules

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Verification:

C: 3 = 3 ✓, H: 8 = 8 ✓, O: 10 = 10 ✓

4.2 Stoichiometric Calculations

General Approach

Write balanced chemical equation
Convert given quantity to moles
Use mole ratio from balanced equation
Convert moles to desired quantity

📝 Solved Example 5 (JEE Main Pattern)

Question: Calculate the mass of CO₂ formed when 5 g of CaCO₃ is heated completely.
CaCO₃ → CaO + CO₂

Solution:

Step 1: Calculate molar masses

M(CaCO₃) = 40 + 12 + 3(16) = 100 g/mol

M(CO₂) = 12 + 2(16) = 44 g/mol

Step 2: Calculate moles of CaCO₃

n_{CaCO_3} = \frac{5}{100} = 0.05 \text{ mol}

Step 3: Use mole ratio

From equation: 1 mol CaCO₃ → 1 mol CO₂

So, 0.05 mol CaCO₃ → 0.05 mol CO₂

Step 4: Calculate mass of CO₂

m_{CO_2} = n \times M = 0.05 \times 44 = 2.2 \text{ g}

\text{Answer: } 2.2 \text{ g of CO}_2

4.3 Limiting Reagent

Concept

The limiting reagent is the reactant that is completely consumed first and thus limits the amount of product formed.

How to identify Limiting Reagent:

Calculate moles of each reactant
Divide moles by stoichiometric coefficient
The reactant with SMALLEST value is limiting reagent

📝 Solved Example 6 (JEE Advanced Pattern)

Question: 10 g of H₂ reacts with 80 g of O₂ to form water. Find:
(a) Limiting reagent
(b) Mass of water formed
(c) Mass of excess reagent left
2H₂ + O₂ → 2H₂O

Solution:

Given data:

M(H₂) = 2 g/mol, M(O₂) = 32 g/mol, M(H₂O) = 18 g/mol

Step 1: Calculate moles

n_{H_2} = \frac{10}{2} = 5 \text{ mol}

n_{O_2} = \frac{80}{32} = 2.5 \text{ mol}

Step 2: Find limiting reagent

From equation: 2H₂ + O₂ → 2H₂O

\frac{n_{H_2}}{2} = \frac{5}{2} = 2.5

\frac{n_{O_2}}{1} = \frac{2.5}{1} = 2.5

Both are equal, so BOTH are completely consumed (no limiting reagent in this case!)

(b) Mass of water formed:

From equation: 2 mol H₂ → 2 mol H₂O

So, 5 mol H₂ → 5 mol H₂O

m_{H_2O} = 5 \times 18 = 90 \text{ g}

(c) Excess reagent:

None (both completely consumed)

Answers:

(a) No limiting reagent (both completely consumed)

(b) 90 g of H₂O formed

🎯 JEE Shortcut for Limiting Reagent

For reaction: aA + bB → products

Calculate for each reactant: moles/coefficient

Limiting reagent = minimum value

Example: If n(A)/a = 3 and n(B)/b = 2, then B is limiting reagent

Percentage Composition

Percentage composition tells us what percentage of a compound's mass comes from each element. This is crucial for determining empirical formulas and purity of samples.

5.1 Formula for Percentage Composition

\text{Percentage of element} = \frac{\text{Mass of element in compound}}{\text{Molar mass of compound}} \times 100

Or, more specifically:

\%\text{ of element} = \frac{n \times \text{Atomic mass of element}}{\text{Molecular mass of compound}} \times 100

where n = number of atoms of that element in formula

📝 Solved Example 7

Question: Calculate percentage composition of elements in H₂SO₄.
(Atomic masses: H = 1, S = 32, O = 16)

Solution:

Step 1: Calculate molar mass of H₂SO₄

M = 2(1) + 1(32) + 4(16) = 2 + 32 + 64 = 98 \text{ g/mol}

Step 2: Calculate % of each element

% of Hydrogen:

\%H = \frac{2 \times 1}{98} \times 100 = 2.04\%

% of Sulfur:

\%S = \frac{1 \times 32}{98} \times 100 = 32.65\%

% of Oxygen:

\%O = \frac{4 \times 16}{98} \times 100 = 65.31\%

Verification: 2.04 + 32.65 + 65.31 = 100% ✓

5.2 Reverse Calculation: Finding Mass from Percentage

📝 Solved Example 8 (JEE Main Type)

Question: A sample of pure CaCO₃ weighs 5 g. What is the mass of calcium in it?
(Ca = 40, C = 12, O = 16)

Solution:

Step 1: Calculate molar mass of CaCO₃

M_{CaCO_3} = 40 + 12 + 3(16) = 100 \text{ g/mol}

Step 2: Calculate % of Ca

\%Ca = \frac{40}{100} \times 100 = 40\%

Step 3: Calculate mass of Ca in 5 g sample

m_{Ca} = \frac{40}{100} \times 5 = 2 \text{ g}

\text{Answer: } 2 \text{ g of calcium}

💡 Quick Tip

In percentage composition problems:
• Sum of all percentages MUST equal 100%
• If only one element is unknown, subtract others from 100%
• Always verify your answer by adding all percentages

Empirical and Molecular Formula

Understanding the difference between empirical and molecular formulas is crucial. This topic appears in nearly every JEE paper!

6.1 Definitions

Empirical Formula

Shows the simplest whole number ratio of atoms in a compound

Examples:

• Glucose: CH₂O

• Benzene: CH

• Hydrogen peroxide: HO

Molecular Formula

Shows the actual number of atoms in a molecule

Examples:

• Glucose: C₆H₁₂O₆

• Benzene: C₆H₆

• Hydrogen peroxide: H₂O₂

Key Relationship

\text{Molecular Formula} = n \times \text{Empirical Formula}

n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}

where n is a whole number (1, 2, 3, ...)

6.2 How to Find Empirical Formula

Step-by-Step Method

Write the percentage (or mass) of each element
Divide each by its atomic mass to get moles
Divide all moles by the smallest value
If ratios are not whole numbers, multiply by appropriate factor
Write the empirical formula using whole number ratios

📝 Solved Example 9 (JEE Main 2022 Type)

Question: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.
(C = 12, H = 1, O = 16)

Solution:

Element	% by mass	÷ Atomic mass	= Moles	÷ Smallest	Simple ratio
C	40	÷ 12	= 3.33	÷ 3.33	= 1
H	6.7	÷ 1	= 6.7	÷ 3.33	≈ 2
O	53.3	÷ 16	= 3.33	÷ 3.33	= 1

\text{Empirical Formula: } CH_2O

📝 Solved Example 10 (Complete Problem)

Question: A compound has empirical formula CH₂O and molar mass 180 g/mol. Find its molecular formula.

Solution:

Step 1: Calculate empirical formula mass

\text{EFM} = 12 + 2(1) + 16 = 30 \text{ g/mol}

Step 2: Calculate n

n = \frac{\text{Molar mass}}{\text{EFM}} = \frac{180}{30} = 6

Step 3: Find molecular formula

\text{Molecular Formula} = n \times (CH_2O) = 6 \times (CH_2O)

\text{Molecular Formula: } C_6H_{12}O_6 \text{ (Glucose)}

⚠️ Common Mistakes

Not dividing by smallest number of moles
Rounding off too early (0.49 ≠ 0.5, wait for final ratio)
Forgetting to multiply all elements when converting decimal ratios
Confusing empirical formula with molecular formula

💡 Multiplying Factor Guide

When you get decimal ratios, multiply by:

If ratio ≈ 0.5 → multiply by 2
If ratio ≈ 0.33 or 0.67 → multiply by 3
If ratio ≈ 0.25 or 0.75 → multiply by 4
If ratio ≈ 0.2, 0.4, 0.6, 0.8 → multiply by 5

Concentration Terms

Concentration describes how much solute is present in a solution. JEE frequently tests interconversions between different concentration units.

7.1 Important Concentration Terms

Term	Definition	Formula	Unit
Molarity (M)	Moles of solute per liter of solution	M = n/V	mol/L or M
Molality (m)	Moles of solute per kg of solvent	m = n/W	mol/kg or m
Mole Fraction (χ)	Ratio of moles of component to total moles	χ_A = n_A/(n_A + n_B)	Unitless
Mass Percentage	Mass of solute per 100g solution	(m_solute/m_solution) × 100	% (w/w)
Volume Percentage	Volume of solute per 100mL solution	(V_solute/V_solution) × 100	% (v/v)
ppm	Parts per million	(m_solute/m_solution) × 10⁶	ppm
Normality (N)	Gram equivalents per liter	N = (n × n-factor)/V	eq/L or N

7.2 Detailed Formulas

1. Molarity (M)

M = \frac{\text{Moles of solute}}{\text{Volume of solution in L}}

M = \frac{w \times 1000}{M \times V_{(mL)}}

• Changes with temperature (volume changes)

• Most common concentration unit in chemistry

2. Molality (m)

m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

m = \frac{w \times 1000}{M \times W_{solvent(g)}}

• Does NOT change with temperature

• Used in colligative property calculations

3. Mole Fraction (χ)

\chi_A = \frac{n_A}{n_A + n_B + n_C + ...}

• Sum of all mole fractions = 1

• Unitless, temperature independent

• For binary solution: χ_A + χ_B = 1

📝 Solved Example 11

Question: Calculate molarity of solution containing 4g NaOH in 250 mL solution.
(Na = 23, O = 16, H = 1)

Solution:

Step 1: Calculate molar mass of NaOH

M_{NaOH} = 23 + 16 + 1 = 40 \text{ g/mol}

Step 2: Calculate moles of NaOH

n = \frac{4}{40} = 0.1 \text{ mol}

Step 3: Convert volume to liters

V = 250 \text{ mL} = 0.25 \text{ L}

Step 4: Calculate molarity

M = \frac{n}{V} = \frac{0.1}{0.25} = 0.4 \text{ M}

\text{Answer: } 0.4 \text{ M or } 0.4 \text{ mol/L}

7.3 Interconversion Formulas (Very Important for JEE)

Key Conversions

Molarity to Molality:

m = \frac{M \times 1000}{1000d - M \times M_B}

where d = density of solution, M_B = molar mass of solute

Molality to Mole Fraction:

\chi_{\text{solute}} = \frac{m}{m + \frac{1000}{M_A}}

where M_A = molar mass of solvent

Mass % to Molarity:

M = \frac{\text{Mass \%} \times d \times 10}{M_B}

where d = density in g/mL

💡 Remember the Difference

Molarity (M)

✓ Per liter of SOLUTION
✗ Changes with temperature
✓ Easy to prepare in lab

Molality (m)

✓ Per kg of SOLVENT
✓ Temperature independent
✓ Used in colligative properties

Types of Chemical Reactions

Understanding different types of chemical reactions helps in predicting products and balancing equations - essential for JEE problem solving.

8.1 Classification of Reactions

1. Combination Reaction

Two or more substances combine to form a single product

A + B → AB

Example: 2H₂ + O₂ → 2H₂O

2. Decomposition Reaction

Single compound breaks down into two or more products

AB → A + B

Example: 2H₂O → 2H₂ + O₂

3. Displacement Reaction

More reactive element displaces less reactive

A + BC → AC + B

Example: Zn + CuSO₄ → ZnSO₄ + Cu

4. Double Displacement

Exchange of ions between reactants

AB + CD → AD + CB

Example: NaCl + AgNO₃ → NaNO₃ + AgCl

8.2 Redox Reactions

Key Concepts

Oxidation

• Loss of electrons
• Increase in oxidation number
• Addition of oxygen
• Removal of hydrogen

Reduction

• Gain of electrons
• Decrease in oxidation number
• Removal of oxygen
• Addition of hydrogen

💡 Memory Aid for Redox

OIL RIG
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)

LEO says GER
Loss of Electrons = Oxidation
Gain of Electrons = Reduction

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Mole Concept & Stoichiometry: 45%
✓ Percentage Composition: 20%
✓ Empirical/Molecular Formula: 15%
✓ Concentration Terms: 12%
✓ Limiting Reagent: 8%

JEE Advanced (Last 5 Years)

✓ Complex Stoichiometry: 40%
✓ Mixture & Purity Problems: 25%
✓ Concentration Interconversions: 20%
✓ Multi-concept Problems: 15%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 10-15 marks (3-4 questions)

Difficulty Level: Easy to Medium

Time Required: 2.5-3 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Calculate number of atoms in 23 g of sodium (Na = 23)
Find molar mass of H₂SO₄ and calculate moles in 49 g
Balance: Fe + O₂ → Fe₂O₃
Calculate % of oxygen in CaCO₃
A solution contains 5.85 g NaCl in 500 mL. Find molarity
How many molecules are in 11.2 L of O₂ at STP?
Calculate empirical formula: C=40%, H=6.7%, O=53.3%
Find molality if 10 g NaOH dissolved in 200 g water

Level 2: Intermediate (JEE Main/Advanced)

10 g H₂ + 64 g O₂ → H₂O. Find limiting reagent and mass of water
A compound's empirical formula is CH₂. Molecular mass is 84. Find molecular formula
25% H₂SO₄ solution has density 1.2 g/mL. Calculate molarity
Calculate mass of CO₂ from 20 g CaCO₃: CaCO₃ → CaO + CO₂
Find mole fraction of ethanol in solution with 2 mol ethanol and 3 mol water
Interconvert 2M NaOH to molality (density = 1.08 g/mL)
If 5 g impure CaCO₃ gives 1.1 g CO₂, find % purity
Calculate average atomic mass of Cl if ³⁵Cl (75%) and ³⁷Cl (25%)

Level 3: Advanced (JEE Advanced/Olympiad)

A mixture of CaCO₃ and MgCO₃ weighs 10 g. On heating, 2.24 L CO₂ (STP) evolved. Find % of each
20 g mixture of NaOH and Na₂CO₃ requires 0.2 mol HCl for neutralization. Find composition
60% H₂SO₄ (d=1.5) is diluted to 20%. What volume of water per 100mL acid?
Fe₂O₃ + CO → Fe + CO₂. If 1.6 kg Fe₂O₃ with 80% pure CO gives 70% yield. Find Fe obtained
Calculate molarity, molality, mole fraction for 20% NaOH (d=1.2 g/mL)
A hydrocarbon has C:H = 6:1 by mass. Vapor density is 39. Find molecular formula
Two solutions: 100mL 2M HCl and 200mL 1.5M HCl are mixed. Find final molarity
3.42 g mixture of NaCl and KCl in AgNO₃ gives 7.2 g AgCl. Find % of each salt

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Topic Focus
2024	3 Questions (12 marks)	4 Questions (14 marks)	Limiting reagent, Concentration interconversion
2023	2 Questions (8 marks)	3 Questions (12 marks)	Mole concept, Mixture problems
2022	3 Questions (12 marks)	3 Questions (10 marks)	Stoichiometry, Percentage yield

📑 Quick Navigation - Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry JEE Main & Advanced 2025-26

Matter and Its Nature

1.1 Classification of Matter

Physical Classification

Chemical Classification

1.2 Properties of Matter

💡 JEE Memory Trick

1.3 Laws of Chemical Combination

1. Law of Conservation of Mass (Lavoisier, 1789)

2. Law of Definite Proportions (Proust, 1799)

3. Law of Multiple Proportions (Dalton, 1803)

4. Gay-Lussac's Law of Gaseous Volumes (1808)

5. Avogadro's Law (1811)

Atomic and Molecular Mass

2.1 Atomic Mass Unit (amu or u)

Definition

2.2 Average Atomic Mass

📝 Solved Example 1 (JEE Main Pattern)

2.3 Molecular Mass and Formula Mass

Molecular Mass

Formula Mass

⚠️ Important Distinction

Mole Concept (The Heart of Chemistry)

3.1 Definition of Mole

What is a Mole?

💡 Mole Analogy

3.2 Molar Mass

Definition

3.3 Mole Concept Formulas (Must Memorize!)

The Golden Triangle of Mole Concept

Formula 1

Formula 2

Formula 3 (for gases)

📝 Solved Example 2 (JEE Main 2023 Type)

📝 Solved Example 3 (JEE Advanced Pattern)

Stoichiometry and Chemical Equations

4.1 Balancing Chemical Equations

Rules for Balancing

📝 Solved Example 4

4.2 Stoichiometric Calculations

General Approach

📝 Solved Example 5 (JEE Main Pattern)

4.3 Limiting Reagent

Concept

How to identify Limiting Reagent:

📝 Solved Example 6 (JEE Advanced Pattern)

🎯 JEE Shortcut for Limiting Reagent

Percentage Composition

5.1 Formula for Percentage Composition

📝 Solved Example 7

5.2 Reverse Calculation: Finding Mass from Percentage

📝 Solved Example 8 (JEE Main Type)

💡 Quick Tip

Empirical and Molecular Formula

6.1 Definitions

Empirical Formula

Molecular Formula

Key Relationship

6.2 How to Find Empirical Formula

Step-by-Step Method

📝 Solved Example 9 (JEE Main 2022 Type)

📝 Solved Example 10 (Complete Problem)

⚠️ Common Mistakes

💡 Multiplying Factor Guide

Concentration Terms

7.1 Important Concentration Terms

7.2 Detailed Formulas

1. Molarity (M)

2. Molality (m)

3. Mole Fraction (χ)

📝 Solved Example 11

7.3 Interconversion Formulas (Very Important for JEE)

Key Conversions

Molarity to Molality:

Molality to Mole Fraction:

Mass % to Molarity:

💡 Remember the Difference

Types of Chemical Reactions

8.1 Classification of Reactions