Atoms and Molecules – Numerical Problems with Stepwise Solutions

Class 9 Chemistry — Chapter 3: Atoms & Molecules | 20 Numerical Problems with Stepwise Solutions

Content Bank — Important Formulas & Constants (Chapter 3)

Molar mass (M or Mr): Sum of atomic masses of atoms in a molecule. Example: Mr(CO₂) = 12 + 16×2 = 44 g mol⁻¹

Moles (n): n = \(\dfrac{\text{mass (g)}}{\text{molar mass (g mol⁻¹)}}\)

Number of particles (N): N = n × N_A, where N_A = 6.022×10²³ mol⁻¹

Molar volume at STP (0°C, 1 atm): 1 mol gas ≈ 22.4 L

Mass from moles: mass (g) = n × M

Percent composition: % element = \dfrac{\text{mass of element in formula}}{\text{formula mass}} × 100

Empirical formula from mass%: Convert % → g → moles → divide by smallest → simplest whole-number ratio.

Atomic mass unit (u): 1 u = 1/12 × mass of ¹²C atom

Avogadro constant: N_A = 6.022×10²³ mol⁻¹

Each problem below is given with a clear, step-by-step solution. Units: masses in grams (g), volumes in litres (L), molar masses in g mol⁻¹, Avogadro's number used as 6.022×10²³.

Topics covered: Laws & conservation, Atomic & molecular mass, Mole concept, Molar volume at STP, Empirical & molecular formula, Stoichiometry & limiting reagent, Percent composition, Number of atoms/molecules.

Q1 — Topic: Atomic / Formula mass

Calculate the molar mass (formula mass) of calcium carbonate, CaCO₃.

Step 1: Atomic masses (approx): Ca = 40.0, C = 12.0, O = 16.0 (all in g mol⁻¹).

Step 2: Add contributions: Mr(CaCO₃) = 40.0 + 12.0 + 3×16.0

Step 3: Calculate: 3×16.0 = 48.0 → 40.0 + 12.0 + 48.0 = 100.0 g mol⁻¹.

Answer: 100.0 g mol⁻¹.

Q2 — Topic: Mole concept

How many moles are present in 18 g of water (H₂O)? (M_r(H₂O)=18 g mol⁻¹)

Step 1: Formula: n = mass / M.

Step 2: n = 18 g / 18 g mol⁻¹ = 1.00 mol.

Answer: 1.00 mol of H₂O.

Q3 — Topic: Number of molecules

Find the number of water molecules in 2 g of H₂O. (Use N_A = 6.022×10²³ mol⁻¹)

Step 1: Molar mass M(H₂O) = 18 g mol⁻¹.

Step 2: Moles = 2 / 18 = 0.1111111111... mol (0.111111... ≈ 1/9).

Step 3: Number of molecules = n × N_A = 0.111111... × 6.022×10²³

Step 4: Calculate: 0.111111... × 6.022×10²³ ≈ 6.69×10²² molecules.

Answer: ≈ 6.69 × 10²² molecules of H₂O.

Q4 — Topic: Mass from moles

What is the mass of 0.25 mol of NaCl? (M(NaCl) ≈ 58.5 g mol⁻¹)

Step 1: mass = n × M = 0.25 × 58.5

Step 2: 58.5 × 0.25 = 58.5/4 = 14.625 g.

Answer: 14.625 g (≈ 14.63 g).

Q5 — Topic: Molar volume at STP

What volume (in litres) is occupied by 2.00 mol of an ideal gas at STP (0 °C, 1 atm)? (Use 1 mol = 22.4 L)

Step 1: Volume = n × molar volume = 2.00 × 22.4 L

Step 2: 2.00 × 22.4 = 44.8 L

Answer: 44.8 L (at STP).

Q6 — Topic: Empirical formula from % composition

A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Find its empirical formula.

Step 1: Assume 100 g sample → masses: C = 40.0 g, H = 6.7 g, O = 53.3 g.

Step 2: Convert to moles:

C: 40.0 / 12.0 = 3.3333... mol
H: 6.7 / 1.0 = 6.7 mol
O: 53.3 / 16.0 = 3.3333... mol

Step 3: Divide by smallest (3.3333...): C = 1, H ≈ 2.01, O = 1 → nearest whole numbers → C₁H₂O₁.

Answer: Empirical formula = CH₂O.

Q7 — Topic: Empirical → Molecular formula

A compound has empirical formula CH₂O and its molar mass is 180 g mol⁻¹. What is its molecular formula?

Step 1: Empirical mass = C (12) + H₂ (2×1) + O (16) = 30 g mol⁻¹.

Step 2: n = molecular mass / empirical mass = 180 / 30 = 6.

Step 3: Multiply empirical formula by 6: C₆H₁₂O₆.

Answer: C₆H₁₂O₆ (glucose).

Q8 — Topic: Stoichiometry (combustion)

5.00 g of carbon is burned to form CO₂. Find the mass of CO₂ produced. (M(C)=12.0, M(CO₂)=44.0)

Step 1: Moles of C = 5.00 / 12.0 = 0.416666... mol.

Step 2: Reaction: C + O₂ → CO₂ ⇒ 1 mol C → 1 mol CO₂. So moles CO₂ = 0.416666... mol.

Step 3: Mass CO₂ = moles × M = 0.416666... × 44.0 = 18.3333... g.

Answer: ≈ 18.33 g CO₂.

Q9 — Topic: Reacting masses

What mass of O₂ is required to react completely with 4.00 g of H₂ to form water? (2H₂ + O₂ → 2H₂O)

Step 1: Moles H₂ = 4.00 / 2.00 = 2.00 mol.

Step 2: Stoichiometry: 2 mol H₂ require 1 mol O₂. Therefore O₂ needed = 2.00 × (1/2) = 1.00 mol.

Step 3: Mass O₂ = moles × M(O₂) = 1.00 × 32.0 = 32.0 g.

Answer: 32.0 g of O₂.

Q10 — Topic: Number of atoms in a compound

How many atoms are present in 0.500 mol of Al₂O₃?

Step 1: One formula unit Al₂O₃ contains 2 Al atoms + 3 O atoms = 5 atoms.

Step 2: Number of formula units in 0.500 mol = 0.500 × N_A = 0.500 × 6.022×10²³ = 3.011×10²³ units.

Step 3: Total atoms = 3.011×10²³ × 5 = 1.5055×10²4 atoms. (Because 3.011×5 = 15.055 → ×10²³ = 1.5055×10²4)

Answer: ≈ 1.5055 × 10²⁴ atoms.

Q11 — Topic: Percent composition

Calculate the percentage by mass of each element in CaCO₃ (use Mr = 100 g mol⁻¹ from Q1).

Step 1: Mass contributions: Ca = 40.0, C = 12.0, O₃ = 48.0 → total = 100.0 g.

Step 2: %Ca = (40.0 / 100.0) × 100 = 40.0%.

Step 3: %C = (12.0 / 100.0) × 100 = 12.0%.

Step 4: %O = (48.0 / 100.0) × 100 = 48.0%.

Answer: Ca = 40.0%, C = 12.0%, O = 48.0%.

Q12 — Topic: Formula mass

Calculate the formula mass (molar mass) of magnesium hydroxide, Mg(OH)₂. (Mg ≈ 24.3)

Step 1: Mg = 24.3; OH group: O = 16.0, H = 1.0 → OH = 17.0. Two OH groups → 2 × 17.0 = 34.0.

Step 2: Mr = 24.3 + 34.0 = 58.3 g mol⁻¹.

Answer: 58.3 g mol⁻¹.

Q13 — Topic: Moles from mass

How many moles are there in 44.0 g of CO₂? (M(CO₂) = 44.0 g mol⁻¹)

Step 1: n = mass / M = 44.0 / 44.0 = 1.00 mol.

Answer: 1.00 mol.

Q14 — Topic: Volume ↔ moles at STP

What number of moles of gas is present in 11.2 L at STP? (1 mol = 22.4 L)

Step 1: n = volume / molar volume = 11.2 / 22.4 = 0.5 mol.

Answer: 0.50 mol.

Q15 — Topic: Number of atoms

Find the number of sodium atoms in 2.20 g of Na. (Atomic mass Na ≈ 23.0 g mol⁻¹)

Step 1: Moles of Na = 2.20 / 23.0 = 0.0956521739 mol.

Step 2: Number of atoms = 0.0956521739 × 6.022×10²³ ≈ 5.763×10²² atoms.

Answer: ≈ 5.76 × 10²² sodium atoms.

Q16 — Topic: Stoichiometry (combustion)

If 3.00 g of methane (CH₄) is burned completely, what mass of CO₂ is formed? (M(CH₄)=16.0, M(CO₂)=44.0)

Step 1: Moles CH₄ = 3.00 / 16.0 = 0.1875 mol.

Step 2: Reaction: CH₄ + 2O₂ → CO₂ + 2H₂O. 1 mol CH₄ gives 1 mol CO₂ → moles CO₂ = 0.1875.

Step 3: Mass CO₂ = 0.1875 × 44.0 = 8.25 g.

Answer: 8.25 g of CO₂.

Q17 — Topic: Empirical formula (another example)

A compound contains 70.0% Fe and 30.0% O by mass. Determine its empirical formula. (Atomic masses: Fe = 56.0, O = 16.0)

Step 1: Assume 100 g → Fe = 70.0 g, O = 30.0 g.

Step 2: Convert to moles:

Fe: 70.0 / 56.0 = 1.25 mol
O: 30.0 / 16.0 = 1.875 mol

Step 3: Divide both by smallest (1.25): Fe = 1.25/1.25 = 1; O = 1.875/1.25 = 1.5.

Step 4: Multiply to get whole numbers: ×2 → Fe₂O₃.

Answer: Empirical formula = Fe₂O₃.

Q18 — Topic: Limiting reagent & mass of product

If 4.00 g of H₂ reacts with 32.0 g of O₂, determine the limiting reagent, and calculate mass of H₂O produced. (2H₂ + O₂ → 2H₂O)

Step 1: Moles H₂ = 4.00 / 2.00 = 2.00 mol. Moles O₂ = 32.0 / 32.0 = 1.00 mol.

Step 2: Stoichiometry requires 2 mol H₂ per 1 mol O₂. Here H₂ : O₂ = 2.00 : 1.00 = exactly 2:1 → both are present in stoichiometric proportion; neither is in excess.

Step 3: Moles H₂O formed: From stoichiometry, 2 mol H₂ → 2 mol H₂O: with 2.00 mol H₂ → 2.00 mol H₂O.

Step 4: Mass H₂O = 2.00 × 18.0 = 36.0 g.

Answer: Neither reagent is limiting (both consumed). Mass of H₂O formed = 36.0 g.

Q19 — Topic: Mass of single atom

Estimate the mass of one carbon atom in kilograms. (At. mass C ≈ 12.0 g mol⁻¹)

Step 1: Molar mass C = 12.0 g mol⁻¹ → mass per mole = 12.0 g.

Step 2: Mass of one atom = 12.0 g / N_A = 12.0 / 6.022×10²³ g.

Step 3: Calculate in grams: 12.0 / 6.022×10²³ ≈ 1.993×10⁻²³ g.

Step 4: Convert to kilograms: 1 g = 10⁻³ kg → 1.993×10⁻²³ g = 1.993×10⁻²6 kg.

Answer: ≈ 1.99 × 10⁻²⁶ kg per carbon atom.

Q20 — Topic: Atoms in given moles of compound

How many hydrogen atoms are present in 0.200 mol of H₂SO₄?

Step 1: H₂SO₄ has 2 H atoms per formula unit.

Step 2: Moles of H atoms = 0.200 mol × 2 = 0.400 mol.

Step 3: Number of hydrogen atoms = 0.400 × 6.022×10²³ = 2.409×10²³ atoms.

Answer: ≈ 2.41 × 10²³ hydrogen atoms.

Tip: For stepwise practice, try redoing each calculation without looking at the steps. These problems strictly follow NCERT concepts for Chapter 3 and are ideal for CBSE Class 9 revision.

If you want these problems in printable PDF or as a worksheet with answers hidden (for quizzes), tell me and I’ll prepare a downloadable file for you.

Atoms and Molecules – Numerical Problems with Stepwise Solutions

Content Bank — Important Formulas & Constants (Chapter 3)

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