How do you find number of atoms or molecules?

Number of particles = n × Nₐ = (m/M) × 6.022 × 10²³. For molecules of a compound, multiply n by Avogadro's number. For atoms, also multiply by the number of atoms per molecule if needed.

Samacheer Kalvi 10th Science Chapter 7 Numerical Problems — Atoms and Molecules with Solutions

All numerical problems from Samacheer Kalvi Class 10 Science Chapter 7 Atoms and Molecules with step-by-step solutions. Covers mole concept, Avogadro's number, molar mass, and percentage composition.

Updated 24 May 20264 min read

Numerical Problems — Chapter 7: Atoms and Molecules

Key Formulas:

Formula	Use
n = m/M	moles = mass / molar mass
N = n × Nₐ	number of particles = moles × Avogadro's number
m = n × M	mass = moles × molar mass
% element = (mass of element / M) × 100	percentage composition
Volume at STP = n × 22.4 L	for gases only

Atomic masses: H=1, C=12, N=14, O=16, Na=23, Cl=35.5, S=32, Ca=40, Fe=56

Problem 1. Find the number of moles in 36 g of water (H₂O).

Given

Mass, m = 36 g
Molar mass of H₂O = 2(1) + 16 = 18 g/mol

Solution

n = m/M = 36/18 = 2 moles

Answer36 g of water = 2 moles.

Problem 2. Find the mass of 0.5 moles of NaCl.

Given

n = 0.5 mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Solution

m = n × M = 0.5 × 58.5 = 29.25 g

Answer0.5 moles of NaCl has a mass of 29.25 g.

Problem 3. How many molecules are there in 18 g of water?

Given

m = 18 g; M(H₂O) = 18 g/mol

Solution

Step 1: Find moles n = m/M = 18/18 = 1 mol

Step 2: Find number of molecules N = n × Nₐ = 1 × 6.022 × 10²³ = 6.022 × 10²³ molecules

Answer18 g of water contains 6.022 × 10²³ molecules.

Problem 4. Calculate the number of atoms in 5.6 g of iron (Fe). (Atomic mass of Fe = 56 u)

Given

m = 5.6 g; M(Fe) = 56 g/mol

Solution

n = m/M = 5.6/56 = 0.1 mol

Number of atoms = 0.1 × 6.022 × 10²³ = 6.022 × 10²² atoms

Answer5.6 g of iron contains 6.022 × 10²² atoms.

Problem 5. Calculate the molar mass of H₂SO₄ (sulphuric acid).

Given

FormulaH₂SO₄

Atomic masses: H = 1, S = 32, O = 16

Solution

M(H₂SO₄) = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol

AnswerMolar mass of H₂SO₄ = 98 g/mol.

Problem 6. Calculate the percentage composition of calcium carbonate (CaCO₃). (Ca = 40, C = 12, O = 16)

Given

FormulaCaCO₃

Molar mass = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g/mol

Solution

% Ca = (40/100) × 100 = 40%

% C = (12/100) × 100 = 12%

% O = (48/100) × 100 = 48%

Check: 40 + 12 + 48 = 100% ✓

AnswerCaCO₃ contains 40% Ca, 12% C, and 48% O by mass.

Problem 7. How many moles of CO₂ are present in 44 g? How many oxygen atoms are in this amount?

Given

m = 44 g; M(CO₂) = 12 + 32 = 44 g/mol

Solution

Moles of CO₂: n = 44/44 = 1 mol

Number of CO₂ molecules: N(CO₂) = 1 × 6.022 × 10²³ = 6.022 × 10²³

Number of oxygen atoms: Each CO₂ has 2 oxygen atoms. N(O atoms) = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ atoms

Problem 8. Find the volume occupied by 2 moles of oxygen gas (O₂) at STP.

Given

n = 2 mol O₂
Molar volume at STP = 22.4 L/mol

Solution

Volume = n × 22.4 = 2 × 22.4 = 44.8 L

Answer2 moles of O₂ occupies 44.8 L at STP.

Problem 9. 3.01 × 10²³ atoms of an element weigh 12 g. What is the atomic mass of the element?

Given

Number of atoms = 3.01 × 10²³
Mass = 12 g

Solution

Number of moles = N/Nₐ = (3.01 × 10²³) / (6.022 × 10²³) = 0.5 mol

Molar mass M = m/n = 12/0.5 = 24 g/mol

Atomic mass = 24 u → This is Magnesium (Mg)

Quick Reference Table

Compound	Molar Mass
H₂O (water)	18 g/mol
CO₂	44 g/mol
NaCl	58.5 g/mol
H₂SO₄	98 g/mol
CaCO₃	100 g/mol
NH₃	17 g/mol
HCl	36.5 g/mol
Na₂CO₃	106 g/mol

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Frequently Asked Questions

n = m/M, where n = moles, m = given mass in grams, M = molar mass in g/mol. Example: 36 g of water (M = 18 g/mol) → n = 36/18 = 2 moles.