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).
