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Atoms, Molecules & Mole Concept

Master Atoms, Molecules and Mole Concept for JEE Main, JEE Advanced, NEET & Board Exams. Covers Dalton's atomic theory, laws of chemical combination, molar mass, Avogadro's number, percentage composition, empirical and molecular formula, vapour density and equivalent concept with solved examples.

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The Mole Concept is the language of chemistry — it provides the bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and litres that we can measure in a laboratory. Every calculation in chemistry — whether it involves reactions, concentrations, electrochemistry, or kinetics — ultimately traces back to the mole. Established through Dalton's Atomic Theory and refined over two centuries, the mole concept is the foundation of quantitative chemistry. For JEE (Main & Advanced) and NEET, this topic appears directly in stoichiometry problems and indirectly in virtually every other chapter.

1. Dalton's Atomic Theory

John Dalton (1808) proposed the first scientific atomic theory with the following postulates:

Matter is made of tiny, indivisible particles called atoms.
Atoms of the same element are identical in mass, size, and properties.
Atoms of different elements have different masses and properties.
Atoms combine in simple whole-number ratios to form compounds.
Atoms can neither be created nor destroyed in a chemical reaction (conservation of mass).

Laws Explained by Dalton's Theory

Law	Statement	Example
Law of Conservation of Mass	Mass of reactants = Mass of products in a chemical reaction	$2 H_{2} + O_{2} \to 2 H_{2} O$ — mass balanced
Law of Definite Proportions	A compound always contains elements in a fixed mass ratio	Water is always $1 : 8$ by mass (H:O)
Law of Multiple Proportions	When two elements form more than one compound, masses of one element that combine with a fixed mass of the other are in simple ratios	CO and CO₂: O combining with fixed C is in ratio 1:2
Gay-Lussac's Law of Gaseous Volumes	Gases combine in simple whole-number ratios by volume (at same T and P)	$H_{2} : O_{2} : H_{2} O = 2 : 1 : 2$ by volume
Avogadro's Law	Equal volumes of all gases at same T and P contain equal number of molecules	1 L of H₂ and 1 L of O₂ have same number of molecules

Limitations of Dalton's Theory

Atoms are not indivisible — they contain subatomic particles (protons, neutrons, electrons).
Atoms of the same element can have different masses — isotopes (e.g., $^{12} C$ and $^{14} C$ ).
Does not explain the nature of chemical bonding.
Does not account for the existence of allotropes (e.g., diamond and graphite are both carbon).

2. Atoms and Molecules — Key Definitions

Term	Definition	Example
Atom	Smallest particle of an element that retains its chemical identity	C, Na, Fe
Molecule	Smallest particle of a substance (element or compound) that can exist independently	$O_{2}$ , $H_{2} O$ , $C O_{2}$
Element	Pure substance made of only one type of atom	Gold, Oxygen, Carbon
Compound	Pure substance made of two or more elements in fixed ratio	Water ( $H_{2} O$ ), Salt (NaCl)
Mixture	Combination of two or more substances not chemically combined	Air, seawater, alloys
Ion	Charged particle formed by gain or loss of electrons from an atom/molecule	$N a^{+}$ , $C l^{-}$ , $S O_{4}^{2 -}$

Atomic and Molecular Mass

Atomic masses are measured on the unified atomic mass unit (u) or Dalton (Da), defined as exactly $\frac{1}{12}$ of the mass of one $^{12} C$ atom:

$1 u = 1.66054 \times 10^{- 27}$ kg

Atomic mass of an element = average mass of naturally occurring isotopes, weighted by their abundance.
Molecular mass = sum of atomic masses of all atoms in a molecule.
Formula mass = sum of atomic masses of all atoms in the formula unit (used for ionic compounds, e.g., NaCl).

Important Atomic Masses to Memorise

Element	Symbol	Atomic Mass (u)	Element	Symbol	Atomic Mass (u)
Hydrogen	H	1	Sodium	Na	23
Carbon	C	12	Magnesium	Mg	24
Nitrogen	N	14	Aluminium	Al	27
Oxygen	O	16	Sulphur	S	32
Fluorine	F	19	Chlorine	Cl	35.5
Phosphorus	P	31	Calcium	Ca	40
Iron	Fe	56	Copper	Cu	63.5

3. The Mole — Definition and Avogadro's Number

The mole (mol) is the SI unit of amount of substance. One mole of any substance contains exactly:

$N_{A} = 6.022 \times 10^{23}$ entities (atoms, molecules, ions, electrons, etc.)

This number is called Avogadro's number ( $N_{A}$ ). The mole is defined such that one mole of $^{12} C$ has a mass of exactly 12 grams.

Key Mole Relations

Quantity	Formula	Units
Number of moles	$n = \frac{m}{M}$	$m$ = mass (g), $M$ = molar mass (g/mol)
Number of particles	$N = n \times N_{A}$	$N_{A} = 6.022 \times 10^{23}$ mol $^{- 1}$
Volume at STP (gas)	$V = n \times 22.4$ L	At STP (0°C, 1 atm) — old definition
Volume at STP (gas)	$V = n \times 22.7$ L	At new STP (0°C, 1 bar) — IUPAC 1982
Molar mass of gas	$M = d \times 22.4$ (at old STP)	$d$ = density in g/L at STP

Note for JEE/NEET: Most problems use the old STP ( $0 ° C$ , $1$ atm) with molar volume = $22.4$ L/mol. If the problem specifies $25 ° C$ and $1$ atm (SATP), use $24.5$ L/mol. Always check the conditions given.

4. Molar Mass and Percentage Composition

The molar mass (molecular weight) of a substance is the mass of one mole of that substance, numerically equal to its molecular/formula mass in grams.

Calculating Molar Mass

Example: Molar mass of $H_{2} S O_{4}$ :
$= 2 (1) + 32 + 4 (16) = 2 + 32 + 64 = 98$ g/mol.

Percentage Composition

The percentage by mass of each element in a compound:

$% of element = \frac{Mass of element in 1 mol of compound}{Molar mass of compound} \times 100$

Example: % of H in $H_{2} S O_{4}$ :
$= \frac{2 \times 1}{98} \times 100 = \frac{2}{98} \times 100 \approx 2.04 %$

5. Empirical and Molecular Formula

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule.

Molecular formula $= n \times$ Empirical formula where $n = \frac{Molecular mass}{Empirical formula mass}$

Steps to Determine Empirical Formula from % Composition

Step 1: Write down the percentage of each element (treat as grams if 100 g sample assumed).
Step 2: Divide each mass by the atomic mass of that element → gives mole ratio.
Step 3: Divide all mole values by the smallest value → gives simplest ratio.
Step 4: If ratios are not whole numbers, multiply by the appropriate integer (2, 3, etc.) to make them whole.
Step 5: Write the empirical formula using these whole-number ratios.
Step 6: Use given molecular mass to find $n$ and hence the molecular formula.

Worked Example

A compound contains C: 40%, H: 6.67%, O: 53.33%. Molecular mass = 180 g/mol. Find the molecular formula.

Mole ratios: C $= 40 / 12 = 3.33$ ; H $= 6.67 / 1 = 6.67$ ; O $= 53.33 / 16 = 3.33$ .
Divide by smallest (3.33): C:H:O $= 1 : 2 : 1$ .
Empirical formula: $C H_{2} O$ ; empirical mass $= 12 + 2 + 16 = 30$ .
$n = 180 / 30 = 6$ .
Molecular formula: $C_{6} H_{12} O_{6}$ (glucose).

Compound	Empirical Formula	Molecular Formula	$n$
Glucose	$C H_{2} O$	$C_{6} H_{12} O_{6}$	6
Benzene	$C H$	$C_{6} H_{6}$	6
Water	$H_{2} O$	$H_{2} O$	1
Hydrogen peroxide	$H O$	$H_{2} O_{2}$	2
Ethylene	$C H_{2}$	$C_{2} H_{4}$	2

6. Mole–Mass–Number Triangle

The three most common interconversions in mole concept problems:

Mass ↔ Moles: $n = \frac{m}{M}$ or $m = n \times M$
Moles ↔ Number of particles: $N = n \times N_{A}$ or $n = \frac{N}{N_{A}}$
Moles ↔ Volume (gas at STP): $V = n \times 22.4$ or $n = \frac{V}{22.4}$

Number of Atoms in a Molecule

Total atoms in $n$ moles of $X_{a} Y_{b}$ :

Atoms of X $= n \times a \times N_{A}$
Atoms of Y $= n \times b \times N_{A}$
Total atoms $= n \times (a + b) \times N_{A}$

Example: Number of H atoms in $18$ g of water ( $H_{2} O$ ):
Moles of $H_{2} O = 18 / 18 = 1$ mol.
Each $H_{2} O$ has 2 H atoms.
H atoms $= 1 \times 2 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$ .

Important Mole Values to Remember

Substance	Mass of 1 mol	Particles in 1 mol	Volume at STP
Water ( $H_{2} O$ )	18 g	$6.022 \times 10^{23}$ molecules	—
Oxygen ( $O_{2}$ )	32 g	$6.022 \times 10^{23}$ molecules	22.4 L
Carbon ( $C$ )	12 g	$6.022 \times 10^{23}$ atoms	—
NaCl	58.5 g	$6.022 \times 10^{23}$ formula units	—
Electron ( $e^{-}$ )	$\approx 0.55$ mg	$6.022 \times 10^{23}$ electrons	—

7. Vapour Density and Molar Mass

Vapour density (VD) is the ratio of the mass of a given volume of a gas to the mass of the same volume of hydrogen at the same temperature and pressure:

$Vapour Density = \frac{Mass of V litres of gas}{Mass of V litres of H_{2}}$

The relation between vapour density and molar mass:

$M = 2 \times VD$

This is because molar mass of $H_{2} = 2$ g/mol, so VD is defined relative to hydrogen.

Example: VD of $C O_{2} = 22$ . Molar mass $= 2 \times 22 = 44$ g/mol ✓

8. Equivalent Concept (Normality)

The equivalent weight (gram equivalent) is the weight of a substance that reacts with or displaces 1 g of hydrogen (or 8 g of oxygen, or 35.5 g of chlorine).

$Equivalent weight = \frac{Molar mass}{n-factor}$

where the n-factor is:

For acids: number of $H^{+}$ ions furnished per molecule (basicity).
For bases: number of $O H^{-}$ ions furnished per molecule (acidity).
For salts: total charge on cation (or anion) per formula unit.
For oxidising/reducing agents: change in oxidation state per molecule.

Substance	Molar Mass (g/mol)	n-factor	Equivalent Weight (g/eq)
HCl	36.5	1	36.5
$H_{2} S O_{4}$	98	2	49
NaOH	40	1	40
$C a (O H)_{2}$	74	2	37
$K M n O_{4}$ (acidic)	158	5	31.6
$K_{2} C r_{2} O_{7}$	294	6	49

Normality (N): $N = \frac{Number of equivalents}{Volume in litres} = Molarity \times n -factor$

Law of Equivalents: At equivalence point in a reaction: $N_{1} V_{1} = N_{2} V_{2}$

Mole Concept — Master Calculation Framework

All mole concept problems resolve into one of these pathways. Identify which quantities are given and which are asked, then trace the path:

Given	Asked	Path
Mass (g)	Moles	$n = m / M$
Mass (g)	Number of molecules/atoms	$n = m / M$ → $N = n \times N_{A}$
Volume of gas at STP	Moles	$n = V / 22.4$
Number of molecules	Mass	$n = N / N_{A}$ → $m = n \times M$
% composition	Empirical formula	Divide by atomic mass → divide by smallest → whole numbers
Empirical formula + molecular mass	Molecular formula	$n = M_{m o l e c u l a r} / M_{e m p i r i c a l}$ → multiply subscripts by $n$
Vapour density	Molar mass	$M = 2 \times VD$

JEE/NEET Power Tips — Avoid These Common Mistakes

Avogadro's number applies to any entity — atoms, molecules, ions, electrons, formula units. One mole of $N a C l$ contains $6.022 \times 10^{23}$ formula units, which means $6.022 \times 10^{23}$ $N a^{+}$ ions AND $6.022 \times 10^{23}$ $C l^{-}$ ions — total $1.204 \times 10^{24}$ ions.
Molar mass of a diatomic element = 2 × atomic mass. One mole of $O_{2}$ = 32 g (not 16 g). One mole of $C l_{2}$ = 71 g. This is a very common source of error.
At STP, 22.4 L/mol applies only to ideal gases. For liquids and solids, there is no standard molar volume — use mass and molar mass only.
The empirical formula can be the same as the molecular formula — e.g., $H_{2} O$ , $N a C l$ , $C O_{2}$ . Not every compound has a different molecular and empirical formula.
% composition adds up to 100%. If C = 40%, H = 6.67%, the remaining 53.33% must be another element (usually O). Always check if all elements are accounted for.
n-factor for $K M n O_{4}$ depends on the medium — acidic medium: n = 5 ( $M n^{7 +} \to M n^{2 +}$ ); neutral/basic medium: n = 3 ( $M n^{7 +} \to M n^{4 +}$ ); strongly basic: n = 1 ( $M n^{7 +} \to M n^{6 +}$ ). This is a very commonly tested point in JEE titration problems.
Molecular formula and structural formula are different. $C_{2} H_{5} O H$ and $C H_{3} O C H_{3}$ have the same molecular formula ( $C_{2} H_{6} O$ ) but are different compounds (isomers).

Practice Questions (JEE / NEET Level)

Q1: How many molecules are present in 4.4 g of CO₂? ( $N_{A} = 6.022 \times 10^{23}$ )

A) $6.022 \times 10^{22}$
B) $1.204 \times 10^{23}$
C) $6.022 \times 10^{23}$
D) $3.011 \times 10^{23}$

Answer: A) $6.022 \times 10^{22}$ .

Explanation:

First, calculate the molar mass of CO₂:
Molar mass = $12 + 2 (16) = 44 g/mol$ .

Next, find the number of moles:
Moles of CO₂ = $\frac{4.4}{44} = 0.1 mol$ .

Finally, multiply by Avogadro's number to find the number of molecules:
Molecules = $0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ .

Q2: A compound has an empirical formula of CH₂O and a molecular mass of 60 g/mol. Its molecular formula is:

A) CH₂O
B) C₂H₄O₂
C) C₃H₆O₃
D) C₂H₆O

Answer: B) C₂H₄O₂.

Explanation:

First, find the empirical formula mass of CH₂O:
Mass = $12 + 2 (1) + 16 = 30 g/mol$ .

Next, find the multiplier ( $n$ ) by dividing the molecular mass by the empirical mass:
$n = \frac{60}{30} = 2$ .

Multiply the empirical formula by $n$ to get the molecular formula:
Molecular formula = $2 \times ({CH}_{2} O) = C_{2} H_{4} O_{2}$ (which is acetic acid).

Q3: Calculate the number of oxygen atoms in 0.5 mol of H₂SO₄.

A) $2 \times 6.022 \times 10^{23}$
B) $6.022 \times 10^{23}$
C) $4 \times 6.022 \times 10^{23}$
D) $0.5 \times 6.022 \times 10^{23}$

Answer: A) $2 \times 6.022 \times 10^{23}$ .

Explanation:

Each individual molecule of H₂SO₄ contains exactly 4 oxygen atoms.

Therefore, the total number of oxygen atoms is:
O atoms = $0.5 mol \times 4 \times N_{A}$
$= 2 \times 6.022 \times 10^{23}$
$= 1.204 \times 10^{24}$ oxygen atoms.

Q4: The vapour density of a gas is 14. Its molar mass is:

A) 7 g/mol
B) 14 g/mol
C) 28 g/mol
D) 56 g/mol

Answer: C) 28 g/mol.

Explanation:

The relationship between molar mass ( $M$ ) and vapour density (VD) is:
$M = 2 \times VD$

Substitute the given value:
$M = 2 \times 14 = 28 g/mol$ .

Note: This molar mass corresponds to gases like nitrogen (N₂) or carbon monoxide (CO), both of which have a molar mass of 28 g/mol.

Q5: A compound contains 75% carbon and 25% hydrogen by mass. Its empirical formula is:

A) CH₄
B) CH₃
C) CH
D) C₂H₆

Answer: A) CH₄.

Explanation:

Assume a 100 g sample, which gives 75 g of Carbon and 25 g of Hydrogen. Convert these masses to moles:
Moles of C = $\frac{75}{12} = 6.25$
Moles of H = $\frac{25}{1} = 25$

Find the simplest whole-number ratio by dividing by the smallest value (6.25):
C : H ratio = $\frac{6.25}{6.25} : \frac{25}{6.25}$
C : H ratio = $1 : 4$ .

The empirical formula is CH₄ (methane).

Q6: 10 mL of 2 N H₂SO₄ solution is neutralised by 20 mL of NaOH solution. The normality of NaOH is:

A) 0.5 N
B) 1 N
C) 2 N
D) 4 N

Answer: B) 1 N.

Explanation:

According to the law of equivalents for neutralization:
$N_{1} V_{1} = N_{2} V_{2}$

Substitute the known values for the acid and the base:
$2 \times 10 = N_{2} \times 20$
$20 = 20 N_{2}$

Solve for the normality of the base ( $N_{2}$ ):
$N_{2} = \frac{20}{20} = 1 N$ .

Q7 (JEE Advanced type): 1 g of a metal carbonate MCO₃ on heating gives 0.56 g of metal oxide MO and CO₂. The atomic mass of the metal M is:

A) 24
B) 40
C) 56
D) 28

Answer: B) 40.

Explanation:

Write the balanced chemical equation for the thermal decomposition:
${MCO}_{3} \to MO + {CO}_{2}$

Let the atomic mass of the unknown metal M be $m$ .
Molar mass of MCO₃ = $m + 12 + 48 = m + 60$
Molar mass of MO = $m + 16$

From the stoichiometry, 1 mole of MCO₃ yields exactly 1 mole of MO. Therefore, the ratio of their masses equals the ratio of their molar masses:
$\frac{Mass of MO}{{Mass of MCO}_{3}} = \frac{Molar mass of MO}{{Molar mass of MCO}_{3}}$

$\frac{0.56}{1} = \frac{m + 16}{m + 60}$

Cross-multiply and solve for $m$ :
$m + 16 = 0.56 (m + 60)$
$m + 16 = 0.56 m + 33.6$
$0.44 m = 17.6$
$m = \frac{17.6}{0.44} = 40 g/mol$

An atomic mass of 40 corresponds to Calcium (Ca).

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Atoms, Molecules & Mole Concept

1. Dalton's Atomic Theory

Laws Explained by Dalton's Theory

Limitations of Dalton's Theory

2. Atoms and Molecules — Key Definitions

Atomic and Molecular Mass

Important Atomic Masses to Memorise

3. The Mole — Definition and Avogadro's Number

Key Mole Relations

4. Molar Mass and Percentage Composition

Calculating Molar Mass

Percentage Composition

5. Empirical and Molecular Formula

Steps to Determine Empirical Formula from % Composition

Worked Example

6. Mole–Mass–Number Triangle

Number of Atoms in a Molecule

Important Mole Values to Remember

7. Vapour Density and Molar Mass

8. Equivalent Concept (Normality)

Mole Concept — Master Calculation Framework

Practice Questions (JEE / NEET Level)

Related Study Material

Practice Questions

Topic Information

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