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).
- Mass, m = 36 g
- Molar mass of H₂O = 2(1) + 16 = 18 g/mol
n = m/M = 36/18 = 2 moles
Problem 2. Find the mass of 0.5 moles of NaCl.
- n = 0.5 mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
m = n × M = 0.5 × 58.5 = 29.25 g
Problem 3. How many molecules are there in 18 g of water?
- m = 18 g; M(H₂O) = 18 g/mol
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
Problem 4. Calculate the number of atoms in 5.6 g of iron (Fe). (Atomic mass of Fe = 56 u)
- m = 5.6 g; M(Fe) = 56 g/mol
n = m/M = 5.6/56 = 0.1 mol
Number of atoms = 0.1 × 6.022 × 10²³ = 6.022 × 10²² atoms
Problem 5. Calculate the molar mass of H₂SO₄ (sulphuric acid).
- Atomic masses: H = 1, S = 32, O = 16
M(H₂SO₄) = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol
Problem 6. Calculate the percentage composition of calcium carbonate (CaCO₃). (Ca = 40, C = 12, O = 16)
- Molar mass = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g/mol
% Ca = (40/100) × 100 = 40%
% C = (12/100) × 100 = 12%
% O = (48/100) × 100 = 48%
Check: 40 + 12 + 48 = 100% ✓
Problem 7. How many moles of CO₂ are present in 44 g? How many oxygen atoms are in this amount?
- m = 44 g; M(CO₂) = 12 + 32 = 44 g/mol
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.
- n = 2 mol O₂
- Molar volume at STP = 22.4 L/mol
Volume = n × 22.4 = 2 × 22.4 = 44.8 L
Problem 9. 3.01 × 10²³ atoms of an element weigh 12 g. What is the atomic mass of the element?
- Number of atoms = 3.01 × 10²³
- Mass = 12 g
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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