Atoms and Molecules — Practice

Question 1

5.3 g of sodium carbonate reacts with 6 g of acetic acid. The products are 8.2 g of sodium acetate, 2.2 g of CO₂ and 0.9 g of water. Verify the law of conservation of mass.

Accepted Answer

Step 1 — Sum the reactant masses.
  m(reactants) = 5.3 + 6 = 11.3 g.

Step 2 — Sum the product masses.
  m(products) = 8.2 + 2.2 + 0.9 = 11.3 g.

Step 3 — Compare.
  Mass of reactants = mass of products = 11.3 g. ✓

✦ Answer: The law of conservation of mass is verified — total mass is unchanged at 11.3 g.

Question 2

Compute the molecular mass of CO₂.

Accepted Answer

Step 1 — List atomic masses.
  C = 12 u, O = 16 u.

Step 2 — Add per atom count.
  M(CO₂) = 1 × 12 + 2 × 16 = 12 + 32 = 44 u.

✦ Answer: 44 u.

Question 3

Write the chemical formula of calcium chloride.

Accepted Answer

Step 1 — Identify valencies.
  Ca: +2. Cl: −1.

Step 2 — Cross-multiply.
  Ca₁Cl₂. Simplify (already minimal) → CaCl₂.

✦ Answer: CaCl₂.

Question 4

Give the atomicity of: (a) O₃, (b) P₄, (c) H₂SO₄, (d) Argon.

Accepted Answer

Step 1 — Atomicity = total atoms in one molecule.

Step 2 — Count for each.
  (a) O₃ = 3 atoms → triatomic.
  (b) P₄ = 4 → tetra-atomic.
  (c) H₂SO₄ = 2 + 1 + 4 = 7 atoms → 7-atomic.
  (d) Ar = 1 atom → monatomic (noble gas).

✦ Answer: (a) 3, (b) 4, (c) 7, (d) 1.

Question 5

How many moles are in 36 g of water? (M(H₂O) = 18 g/mol)

Accepted Answer

Step 1 — Use n = m/M.

Step 2 — Substitute.
  n = 36 / 18 = 2 mol.

✦ Answer: 2 moles.

Question 6

How many oxygen atoms are present in 9 g of water? (M(H₂O) = 18 g/mol, Nₐ = 6.022 × 10²³)

Accepted Answer

Step 1 — Find moles of water.
  n = m/M = 9/18 = 0.5 mol.

Step 2 — Each H₂O molecule has 1 oxygen atom.
  Moles of O atoms = 0.5 mol.

Step 3 — Number of O atoms.
  N = 0.5 × 6.022 × 10²³ = 3.011 × 10²³.

✦ Answer: 3.011 × 10²³ oxygen atoms.

Watch out: hydrogen atoms would be 6.022 × 10²³ (double the O atoms, since each H₂O has 2 H atoms).

Question 7

Calculate the mass of 0.2 mole of glucose (C₆H₁₂O₆). (Atomic masses: C = 12, H = 1, O = 16)

Accepted Answer

Step 1 — Compute molar mass.
  M = 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g/mol.

Step 2 — Use m = n × M.
  m = 0.2 × 180 = 36 g.

✦ Answer: 36 g.

Question 8

Calculate the percentage by mass of nitrogen in urea, CO(NH₂)₂. (Atomic masses: C = 12, O = 16, N = 14, H = 1)

Accepted Answer

Step 1 — Compute molar mass of urea.
  CO(NH₂)₂: 1 C, 1 O, 2 N, 4 H.
  M = 12 + 16 + 2 × 14 + 4 × 1 = 12 + 16 + 28 + 4 = 60 g/mol.

Step 2 — Mass of nitrogen in 1 mole.
  Mass(N) = 2 × 14 = 28 g.

Step 3 — Percentage.
  %N = (28 / 60) × 100 = 46.67 %.

✦ Answer: ≈ 46.67 %.

Fertiliser context: urea's high N% (~47%) makes it the most popular nitrogenous fertiliser in Indian agriculture.

Question 9

Write the chemical formulas for: (a) aluminium nitrate, (b) ammonium sulphate, (c) magnesium hydroxide.

Accepted Answer

Step 1 — Identify ions and valencies.

(a) Al³⁺ + NO₃⁻ → cross-multiply: Al(NO₃)₃ (brackets needed; subscript 3 outside scopes the NO₃ group).

(b) NH₄⁺ + SO₄²⁻ → cross-multiply: (NH₄)₂SO₄ (brackets around NH₄ because subscript 2 affects all 4 H and the N).

(c) Mg²⁺ + OH⁻ → cross-multiply: Mg(OH)₂.

✦ Answer: (a) Al(NO₃)₃, (b) (NH₄)₂SO₄, (c) Mg(OH)₂.

Rule: any time a polyatomic ion is taken more than once, bracket it.

Question 10

In an experiment, 1.375 g of cupric oxide on reduction yields 1.097 g of copper. In another, 1.179 g of copper, on dissolving in nitric acid and then heating, gave 1.476 g of cupric oxide. Show that the law of constant proportions is obeyed.

Accepted Answer

Step 1 — In experiment 1, find Cu : O ratio.
  Mass of O = 1.375 − 1.097 = 0.278 g.
  Ratio Cu : O = 1.097 : 0.278 = 1.097 / 0.278 ≈ 3.946 : 1.
  Simplify: ~ 4 : 1 (round).

Step 2 — In experiment 2, find Cu : O ratio.
  Mass of O in 1.476 g of CuO = 1.476 − 1.179 = 0.297 g.
  Ratio Cu : O = 1.179 : 0.297 ≈ 3.97 : 1.
  Simplify: ~ 4 : 1.

Step 3 — Compare.
  Both experiments give Cu : O ≈ 4 : 1 (matching the known atomic masses 63.5 : 16 ≈ 3.97 : 1).

✦ Answer: Both experiments give the same Cu : O mass ratio (~ 4 : 1) — verifying the law of constant proportions. CuO has a fixed composition regardless of how it's made.

Question 11

12 g of magnesium reacts with oxygen to form magnesium oxide. (a) How many moles of Mg is this? (b) How many moles of MgO are formed? (c) What is the mass of MgO produced? (M(Mg) = 24, M(O) = 16)

Accepted Answer

Step 1 — Write balanced equation.
  2 Mg + O₂ → 2 MgO.

Step 2 (a) — Moles of Mg.
  n = 12 / 24 = 0.5 mol.

Step 3 (b) — Moles of MgO.
  Stoichiometry: 2 mol Mg → 2 mol MgO (1 : 1 ratio).
  So 0.5 mol Mg → 0.5 mol MgO.

Step 4 (c) — Mass of MgO.
  M(MgO) = 24 + 16 = 40 g/mol.
  Mass = n × M = 0.5 × 40 = 20 g.

✦ Answer: (a) 0.5 mol Mg, (b) 0.5 mol MgO, (c) 20 g MgO.

Sanity check via conservation of mass: O used = 20 − 12 = 8 g. Moles of O = 8/16 = 0.5 mol O atoms = 0.25 mol O₂. From 2 Mg + O₂, ratio Mg:O₂ = 2:1 = 0.5:0.25 ✓.

Question 12

Find the number of (a) molecules and (b) atoms in 1.7 g of NH₃. (M(NH₃) = 17 g/mol, Nₐ = 6.022 × 10²³)

Accepted Answer

Step 1 — Moles of NH₃.
  n = m/M = 1.7 / 17 = 0.1 mol.

Step 2 (a) — Molecules.
  N = n × Nₐ = 0.1 × 6.022 × 10²³ = 6.022 × 10²².

Step 3 (b) — Total atoms.
  Each NH₃ has 4 atoms (1 N + 3 H).
  Atoms = 4 × 6.022 × 10²² = 2.4088 × 10²³.

✦ Answer: (a) 6.022 × 10²² molecules; (b) 2.41 × 10²³ atoms.

Lesson: 'atoms' is always 'molecules × atomicity'. Read the question carefully — 'how many atoms' vs 'how many molecules' is a common trap.

Question 13

An oxide of nitrogen contains 30.4 % nitrogen and 69.6 % oxygen by mass. Find its empirical formula. (N = 14, O = 16)

Accepted Answer

Step 1 — Take 100 g of the compound for convenience.
  Mass of N = 30.4 g. Mass of O = 69.6 g.

Step 2 — Convert to moles.
  n(N) = 30.4 / 14 = 2.17 mol.
  n(O) = 69.6 / 16 = 4.35 mol.

Step 3 — Find simplest ratio.
  Divide both by the smaller (2.17):
  N : O = 2.17/2.17 : 4.35/2.17 = 1 : 2.005 ≈ 1 : 2.

Step 4 — Write empirical formula.
  N₁O₂ = NO₂.

✦ Answer: Empirical formula = NO₂ (nitrogen dioxide).

Class 11 note: empirical formula = simplest whole-number ratio. The molecular formula could be a multiple (e.g., N₂O₄ has the same empirical formula NO₂).

Question 14

(a) Calculate the molar mass of ZnO. (b) A mineral sample of zinc oxide has mass 100 g. How many zinc atoms does it contain? (c) Comment on whether this is a small or large number. (Zn = 65, O = 16)

Accepted Answer

Step 1 (a) — Molar mass.
  M(ZnO) = 65 + 16 = 81 g/mol.

Step 2 (b) — Moles in 100 g.
  n = 100 / 81 ≈ 1.235 mol.

Step 3 — Each ZnO formula unit has 1 Zn atom.
  Moles of Zn atoms = 1.235.
  Number of Zn atoms = 1.235 × 6.022 × 10²³ ≈ 7.44 × 10²³.

Step 4 (c) — Comment.
  Comparison: world population ~ 8 × 10⁹. The number of Zn atoms in 100 g of ZnO is roughly 10¹⁴ times the world population — utterly enormous.

✦ Answer: (a) 81 g/mol, (b) 7.44 × 10²³ Zn atoms, (c) Enormous — it would take 100,000 years to count them at 1 per second × 100 billion humans counting in parallel.

Question 15

Convert 23 g of sodium into (a) moles of Na, (b) moles of Na⁺ ions, (c) number of Na⁺ ions, (d) total charge in coulombs (charge per ion = 1.6 × 10⁻¹⁹ C). (M(Na) = 23 g/mol, Nₐ = 6.022 × 10²³)

Accepted Answer

Step 1 (a) — Moles of Na atoms.
  n = 23 / 23 = 1 mol.

Step 2 (b) — Each Na atom becomes 1 Na⁺ ion on oxidation.
  So moles of Na⁺ ions = 1 mol.

Step 3 (c) — Number of Na⁺ ions.
  N = 1 × Nₐ = 6.022 × 10²³.

Step 4 (d) — Total charge.
  Q = N × charge per ion = 6.022 × 10²³ × 1.6 × 10⁻¹⁹ = 9.6352 × 10⁴ C ≈ 96,500 C.

✦ Answer: (a) 1 mol, (b) 1 mol, (c) 6.022 × 10²³ ions, (d) ≈ 96,500 C.

This '96,500 C' has a name: it's the Faraday constant (charge of 1 mole of electrons). You'll meet it again in Class 12 electrochemistry.

Practice — Atoms and Molecules

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