Problems & Solutions
Practice problems with solutions.
Chapter 1: Matter, Energy, and the Origins of the Universe
Problem 1: Classify the following as pure substances or mixtures: (a) O₂ (b) Fe (c) H₂O (d) Saltwater (e) Chocolate chip cookie
Answer:
- O₂ → Element → Pure substance
- Fe → Element → Pure substance
- H₂O → Compound → Pure substance
- Saltwater → Homogeneous mixture
- Chocolate chip cookie → Heterogeneous mixture
Explanation: Pure substances have fixed composition; mixtures can be separated physically.
Problem 2: Describe the states of matter in terms of particle arrangement.
Answer:
- Solid → tightly packed, vibrate in place
- Liquid → close together, flow past each other
- Gas → far apart, move rapidly
Explanation: Molecular motion increases from solid → liquid → gas.
Chapter 2: Atoms, Ions, and Compounds
Problem 1: Write isotopic notation for an atom with 26 protons and 30 neutrons.
Answer:
^56_26 Fe
Explanation: Mass number = protons + neutrons; atomic number = protons.
Problem 2: Calculate average atomic mass for Neon isotopes: 19.9924 (90.48%), 20.9940 (0.27%), 21.9914 (9.25%).
Answer:
Avg Mass = 0.904819.9924 + 0.002720.9940 + 0.0925*21.9914 Avg Mass ≈ 20.1797 amu
Explanation: Weighted average based on natural abundance.
Problem 3: Name the following compounds: (a) NaCl (b) MgO (c) FeCl₂ (d) N₂O₅
Answer:
- NaCl → Sodium chloride
- MgO → Magnesium oxide
- FeCl₂ → Iron(II) chloride
- N₂O₅ → Dinitrogen pentoxide
Explanation: Ionic → cation first, anion with -ide; Covalent → use prefixes.
Problem 4: Convert empirical formula CH₂O to molecular formula if molar mass = 180 g/mol.
Answer:
- Empirical mass = 30 g/mol
- n = 180 / 30 = 6
- Molecular formula = C₆H₁₂O₆
Explanation: Molecular formula = empirical × n.
Chapter 3: Chemical Reactions and Earth's Composition
Problem 1: How many molecules are in 5.32 moles of CaCO₃?
Answer:
5.32 mol × 6.022×10^23 = 3.20×10^24 molecules
Explanation: Use Avogadro’s number.
Problem 2: Percent Fe in Fe₂O₃
Answer:
- Molar mass Fe₂O₃ = 159.70 g/mol
- %Fe = (2×55.85 / 159.70) × 100 ≈ 69.94%
Problem 3: Limiting reactant: CH₄ + 2O₂ → CO₂ + 2H₂O Given 10.0 g CH₄ and 20.0 g O₂
Answer:
- Moles CH₄ = 10.0 / 16.04 ≈ 0.623 mol
- Moles O₂ = 20.0 / 32.00 = 0.625 mol
- H₂O from CH₄ = 1.246 mol, from O₂ = 0.625 mol → O₂ limiting
- Mass H₂O = 0.625 × 18.02 ≈ 11.3 g
Explanation: Limiting reactant produces least product.
Chapter 4: Solution Chemistry and the Hydrosphere
Problem 1: Molarity of 36.5 g BaCl₂ in 0.750 L solution
Answer:
- Moles BaCl₂ = 36.5 / 208.23 ≈ 0.175 mol
- Molarity = 0.175 / 0.750 ≈ 0.233 M
Problem 2: Titration: 25.0 mL HCl neutralized by 30.0 mL NaOH (1:1) Find [HCl] if [NaOH] = 0.100 M
Answer:
MaVa = MbVb Ma × 25.0 mL = 0.100 × 30.0 Ma = 0.120 M
Explanation: Use stoichiometry for neutralization.
Chapter 5: Thermochemistry
Problem 1: q for heating 50 g water from 25°C → 75°C, c = 4.18 J/g°C
Answer:
q = m c ΔT = 50 × 4.18 × (75-25) = 10450 J
Problem 2: For expansion of gas, ΔE = q + w, w = -PΔV. If q = +500 J, w = -150 J → ΔE?
Answer:
ΔE = 500 - 150 = 350 J
Chapter 6: Properties of Gases
Problem 1: Find molar mass of 0.495 g gas in 127 mL at 98°C, 754 torr
Answer:
- V = 0.127 L, T = 371 K, P = 0.992 atm
- n = PV / RT = 0.00413 mol
- M = 0.495 / 0.00413 ≈ 120 g/mol
Problem 2: Ideal Gas Law: 2.0 mol gas at 1.5 atm, 300 K → V?
Answer:
V = nRT / P = (2×0.08206×300)/1.5 ≈ 32.8 L
Chapter 7: Chemical Bonding
Problem 1: Draw Lewis structure for CO₂.
Answer:
- O=C=O with double bonds, formal charges = 0
Problem 2: Determine bond type for H-Cl (ΔEN = 0.9)
Answer: Polar covalent (0 < ΔEN < 2)
Problem 3: Formal charge of N in NH₄⁺
Answer:
FC = valence - (lone + ½ bonding) = 5 - (0 + ½*8) = 1
Extra Topics
Problem 1: pH of [H⁺] = 2×10⁻⁴ M
Answer:
pH = -log[H⁺] ≈ 3.70
Problem 2: Kc for aA + bB ⇌ cC + dD
Answer:
Kc = [C]^c [D]^d / [A]^a [B]^b
Problem 3: Kp = Kc(RT)^Δn, Δn = moles products - moles reactants (gases only)