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Tamil Nadu (TNBSE)Class 10 Science★ 5-9 marks in the TN SSLC Class 10 Science public exam through MCQs, short answers and mole-concept numericals

Atoms and Molecules

Samacheer Kalvi (TN State Board) Class 10 Science — Chemistry Chapter 7, Atoms and Molecules. Atomic and molecular mass, the mole concept and Avogadro's number, molar mass and gram molecular mass, percentage composition, empirical and molecular formula, and vapour density. Notes with formulas, worked numericals, common mistakes, practice and quick revision.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Calculate atomic, molecular and molar masses
2Apply the mole concept and Avogadro's number
3Find the number of moles and particles from mass
4Use percentage composition and empirical/molecular formula
5Relate vapour density to molar mass

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Why this chapter matters

Atoms and Molecules introduces the mole concept — the counting unit of chemistry. Mole, molar mass and vapour density numericals are reliable scoring questions in the TN SSLC Science exam and are essential for all later chemistry.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 9 Science — Atomic structure

Atoms, symbols and valency.

Atoms and Molecules — Class 10 Science (Samacheer Kalvi)

TN State Board (Samacheer Kalvi) Class 10 Science, Chemistry — Chapter 7. Counting atoms and molecules using the mole concept.

1. About this chapter

This chapter builds the quantitative language of chemistry: atomic and molecular mass, the mole concept and Avogadro's number, molar mass, percentage composition, and empirical & molecular formulae, plus vapour density.

2. Mass of atoms and molecules

Atomic mass: mass of an atom relative to 1/12 the mass of a carbon-12 atom (unit u, atomic mass unit).
Molecular mass: sum of the atomic masses of all atoms in a molecule. e.g. H₂O = (2×1) + 16 = 18 u; CO₂ = 12 + (2×16) = 44 u.
Gram molecular mass (GMM): molecular mass expressed in grams.

3. The mole concept

1 mole = the amount containing Avogadro's number, Nₐ = 6.022×10²³ particles.
Molar mass = mass of 1 mole = numerically equal to atomic/molecular mass in grams (g mol⁻¹).
Key relations:
- Number of moles n = given mass / molar mass
- Number of particles = n × 6.022×10²³
Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. 1 mole of any gas at STP occupies 22.4 L.

4. Composition and formula

Percentage composition = (mass of element in 1 mole / molar mass) × 100.
Empirical formula: simplest whole-number ratio of atoms.
Molecular formula = (empirical formula) × n, where n = molecular mass / empirical formula mass.
Vapour density (V.D.): Molar mass = 2 × Vapour Density (so V.D. = molar mass / 2).

5. Worked examples

Example 1. Find the molar mass of Ca₃(PO₄)₂. (Ca=40, P=31, O=16) = 3(40) + 2[31 + 4(16)] = 120 + 2(95) = 120 + 190 = 310 g mol⁻¹.

Example 2. How many moles are in 88 g of CO₂? (M = 44) n = 88/44 = 2 moles → particles = 2 × 6.022×10²³ = 1.2044×10²⁴ molecules.

Example 3. The vapour density of a gas is 22. Find its molar mass. Molar mass = 2 × V.D. = 2 × 22 = 44 g mol⁻¹.

6. Common mistakes

Mistake: Confusing atomic mass with molar mass. Fix: They are numerically equal; molar mass carries the unit g mol⁻¹ for 1 mole.
Mistake: Forgetting the bracket multiplier in formulae. Fix: In Ca₃(PO₄)₂, multiply the whole (PO₄) group by 2.
Mistake: Using V.D. as the molar mass directly. Fix: Molar mass = 2 × V.D.

7. Practice (book-back style)

Define one mole and state Avogadro's number.
Calculate the molar mass of H₂SO₄. (H=1, S=32, O=16)
How many molecules are in 0.5 mole of water?
The vapour density of a gas is 16. Find its molecular mass.
State Avogadro's law.

8. Answer key

The amount of substance containing 6.022×10²³ particles; Nₐ = 6.022×10²³.
2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g mol⁻¹.
0.5 × 6.022×10²³ = 3.011×10²³ molecules.
Molar mass = 2 × 16 = 32 g mol⁻¹.
Equal volumes of all gases at the same T and P contain equal numbers of molecules.

9. Quick revision

Chemistry Ch 7 · atomic/molecular mass, mole, molar mass.
1 mole = 6.022×10²³ particles = molar mass in grams; 22.4 L at STP.
n = mass / molar mass; particles = n × Nₐ.
Molar mass = 2 × vapour density.
Molecular formula = empirical formula × (mol. mass / emp. mass).

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Avogadro's number

Nₐ = 6.022×10²³ per mole

Particles in 1 mole.

Number of moles

n = given mass / molar mass

Molar mass in g mol⁻¹.

Number of particles

N = n × 6.022×10²³

Atoms, molecules or ions.

Vapour density

Molar mass = 2 × V.D.

For gases.

Molar volume

1 mole gas = 22.4 L at STP

Avogadro's law.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing atomic mass with molar mass

✓ They are numerically equal; molar mass carries the unit g mol⁻¹ for one mole.

WATCH OUT

✗ Forgetting the bracket multiplier in formulae

✓ In Ca₃(PO₄)₂ multiply the whole (PO₄) group by 2.

WATCH OUT

✗ Using vapour density as the molar mass

✓ Molar mass = 2 × vapour density.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Multiple choice questions

MCQs on mole, molar mass and Avogadro's number.

Questions

Short answer questions

Definitions, Avogadro's law and vapour density.

Questions

Numerical problems

Molar mass, moles, particles and vapour-density sums.

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Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Concept

Define one mole and state Avogadro's number.

▶ Show solution

One mole is the amount of substance containing 6.022×10²³ particles (Avogadro's number).

Q2MEDIUM· Numerical

Calculate the molar mass of H₂SO₄ (H=1, S=32, O=16).

▶ Show solution

2(1) + 32 + 4(16) = 98 g mol⁻¹.

Q3HARD· Numerical

Find the molar mass of Ca₃(PO₄)₂ (Ca=40, P=31, O=16).

▶ Show solution

3(40) + 2[31 + 4(16)] = 120 + 190 = 310 g mol⁻¹.

Q4MEDIUM· Numerical

How many molecules are in 0.5 mole of water?

▶ Show solution

0.5 × 6.022×10²³ = 3.011×10²³ molecules.

Q5EASY· Numerical

The vapour density of a gas is 16. Find its molecular mass.

▶ Show solution

Molar mass = 2 × V.D. = 2 × 16 = 32 g mol⁻¹.

Q6MEDIUM· Numerical

How many moles are in 88 g of CO₂ (M = 44)?

▶ Show solution

n = 88/44 = 2 moles.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Chemistry Chapter 7 of Samacheer Kalvi Class 10 Science.
•Molecular mass = sum of atomic masses (H₂O = 18, CO₂ = 44).
•1 mole = 6.022×10²³ particles = molar mass in grams = 22.4 L gas at STP.
•n = mass / molar mass; particles = n × Nₐ.
•Molar mass = 2 × vapour density.
•Molecular formula = empirical formula × (mol. mass / emp. mass).

Tamil Nadu (TNBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-9 marks across MCQ, short answer and numerical questions

Question type	Marks each	Typical count	What it tests
MCQ	1	1-2	Mole, Avogadro number, units
Short Answer	2-3	1-2	Avogadro's law, vapour density
Numerical	2-3	1-2	Molar mass, moles, particles

Prep strategy

Memorise Nₐ = 6.022×10²³ and 22.4 L at STP
Practise molar mass for compounds with brackets
Use n = mass/molar mass for mole sums
Remember molar mass = 2 × vapour density

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Chemical calculations

The mole lets chemists measure exact amounts for reactions and medicines.

Industry

Manufacturing fertilisers, fuels and drugs relies on mole ratios.

Laboratory work

Preparing solutions of known concentration uses molar mass.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the formula and atomic masses before calculating
Handle brackets in molar-mass sums carefully
Keep units g mol⁻¹ and show steps
Use molar mass = 2 × V.D. for gas problems

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Find the empirical formula of a compound from its percentage composition.
Calculate the number of atoms in a given mass of a compound.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

TN SSLC Class 10 Public ExamHigh

Foundation / NTSE ChemistryMedium

School unit testsHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Atoms and molecules are far too small and numerous to count individually, so the mole lets us measure them by mass — one mole always contains 6.022×10²³ particles.

For a gas, molar mass = 2 × vapour density, because vapour density is the mass of a gas compared with that of hydrogen, whose molar mass is 2.

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