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Spring 2025 (1st Sem)FinalChemistry

Suggestions

Study suggestions and review guidance.

1) Isotopes & Atomic Mass Calculation

Concept:

Atomic Mass = Sum of (isotope mass × % abundance) ÷ 100

Example

Chlorine has two isotopes: Cl-35 (75%) and Cl-37 (25%).

Step-by-step:

  1. Multiply each isotope by its abundance:
    • 35 × 75 = 2625
    • 37 × 25 = 925
  2. Sum = 2625 + 925 = 3550
  3. Divide by 100 → 3550 ÷ 100 = 35.5 u

Answer: 35.5 u

Practice Problems with Solutions

  1. Boron: B-10 (20%), B-11 (80%)

    • Mass = (10×20 + 11×80)/100 = (200+880)/100 = 10.8 u

  2. Magnesium: Mg-24 (78.6%), Mg-25 (10.1%), Mg-26 (11.3%)

    • Mass = 24×0.786 + 25×0.101 + 26×0.113 = 24.33 u

  3. Hydrogen: H-1 (99.985%), H-2 (0.015%)

    • Mass = 1×0.99985 + 2×0.00015 = 1.0002 u

  4. Carbon: C-12 (98.9%), C-13 (1.1%)

    • Mass = 12×0.989 + 13×0.011 = 12.011 u

  5. Neon: Ne-20 (90.5%), Ne-21 (0.3%), Ne-22 (9.2%)

    • Mass = 20×0.905 + 21×0.003 + 22×0.092 = 20.19 u

  6. Lithium: Li-6 (7.5%), Li-7 (92.5%)

    • Mass = 6×0.075 + 7×0.925 = 6.93 u

  7. Chlorine: Cl-35 (76%), Cl-37 (24%)

    • Mass = 35×0.76 + 37×0.24 = 35.48 u

  8. Silver: Ag-107 (51.8%), Ag-109 (48.2%)

    • Mass = 107×0.518 + 109×0.482 = 107.96 u

  9. Gallium: Ga-69 (60.1%), Ga-71 (39.9%)

    • Mass = 69×0.601 + 71×0.399 = 69.80 u

  10. Oxygen: O-16 (99.8%), O-18 (0.2%)

    • Mass = 16×0.998 + 18×0.002 = 16.00 u

2) % Composition & Mass of Elements in Compounds

Concept:

% element = (mass of element in formula / molar mass of compound) × 100

Example

In 18 g H2O, how much H is present?

  1. Molar mass: M(H2O) = 2×1.008 + 16 = 18.016 g/mol
  2. Mass of H = 2.016 g
  3. %H = 2.016 / 18.016 × 100 ≈ 11.19%
  4. Mass of H in 18 g water = 18 × 2.016 / 18.016 ≈ 2.0 g

Answer: 2.0 g H

Practice Problems

  1. %H and %O in H2O

    • M = 18.016, mH = 2.016, mO = 16
    • %H = 11.19%, %O = 88.81%

  2. %C in CO2

    • M = 12.01 + 32 = 44.01
    • %C = 12.01/44.01 ×100 = 27.29%

  3. %Na and %Cl in NaCl

    • M = 22.99 + 35.45 = 58.44
    • %Na = 39.34%, %Cl = 60.66%

  4. Mass of O in 90 g H2O

    • Mass = 90 × 0.8881 = 79.93 g

  5. Mass of Ca in 100 g CaCO3

    • M = 40.08 + 12.01 + 48 = 100.09
    • %Ca = 40.08%, Mass = 40.04 g

3) Empirical & Molecular Formula

Steps:

  1. Assume 100 g sample → convert % → grams
  2. Convert grams → moles: n = mass / atomic mass
  3. Divide each mole value by smallest → simplest ratio (empirical)
  4. Multiply by integer if needed → molecular formula

Practice Problems

  1. 52.14%C, 13.13%H, 34.73%O, M=46

    • Moles: C=4.341, H=13.027, O=2.171
    • Divide by smallest (2.171): C=2, H=6, O=1 → Empirical: C2H6O
    • Multiplier = 46/46 ≈1 → Molecular: C2H6O

  2. 85.7%C, 14.3%H, M=28

    • Moles: C=7.137, H=14.185 → divide by 7.137 → C=1, H≈2
    • Empirical: CH2, Multiplier = 28/14 ≈2 → Molecular: C2H4

  3. 27.3%C, 72.7%O → CO2

  4. 40%C, 6.7%H, 53.3%O, M=180

    • Moles: C=3.331, H=6.646, O=3.331 → divide by 3.331 → C=1, H≈2, O=1
    • Empirical: CH2O, Multiplier=180/30≈6 → Molecular: C6H12O6

  5. 75%C, 25%H, M=30 → Empirical: CH4 (Molecular inconsistent, note typo)

  6. 29.1%Na, 40.5%S, 30.4%O → Empirical: Na2S2O3

  7. 69.6%C, 30.4%O → Empirical: C3O

  8. 20%H, 80%O → Empirical: H4O

  9. 92.3%C, 7.7%H, M=26 → Empirical: CH, Molecular: C2H2

  10. 10.5%C, 27.9%O, 61.6%Cl → Empirical: CCl2O2

4) Kp and Kc Math

Formula: Kp = Kc × (RT)^Δn Δn = moles of gas products − moles of gas reactants R = 0.0821 L·atm/(mol·K)

Practice Problems

  1. N2 + O2 ⇌ 2NO → Δn=0 → Kp = Kc

  2. Kc=0.1 at 400 K, Δn=2

    • RT = 0.0821 × 400 = 32.84
    • Kp = 0.1 × (32.84)^2 ≈ 107.8

  3. Kc=2.5, 298K, Δn=-1 → Kp = 2.5 / 24.45 ≈ 0.102

  4. 2SO2 + O2 ⇌ 2SO3, Kc=3, 600K → Δn=-1

    • RT=0.0821×600=49.26 → Kp = 3/49.26 ≈0.061

  5. A + B ⇌ C → Δn=0 → Kp = Kc

  6. Kc=4, 273K, Δn=2 → RT=22.4 → Kp=4×(22.4)^2≈2008

  7. 2H2 + O2 ⇌ 2H2O → Δn=-1 → Kp = Kc / RT

  8. Kc=1.2×10^-3, 700K, Δn=-1 → Kp ≈ 2.09×10^-5

  9. Kc=10, 500K, Δn=1 → Kp ≈ 410

  10. Kc=5, 1000K, Δn=-3 → Kp = 5 / (82.057^3) ≈ 9.0×10^-7

5) pH Math

Concepts:

  • pH = -log[H+]
  • pOH = -log[OH-]
  • pH + pOH = 14 (at 25°C)
  • Strong acids/bases: [H+] or [OH-] = concentration
  • Weak acids/bases: use Ka or Kb

Practice Problems

  1. 0.01 M HCl → Strong acid

    • [H+] = 0.01
    • pH = -log(0.01) = 2

  2. 0.001 M NaOH → Strong base

    • [OH-] = 0.001
    • pOH = -log(0.001) = 3 → pH = 14 - 3 = 11

  3. 0.1 M H2SO4 → Strong diprotic acid (first proton fully dissociates)

    • [H+] ≈ 0.2 M
    • pH = -log(0.2) ≈ 0.7

  4. Weak acid: HA, 0.1 M, Ka = 1×10^-5

    • [H+] = √(Ka × C) = √(1×10^-5 × 0.1) = √1×10^-6 = 0.001
    • pH = -log(0.001) = 3

  5. Weak base: NH3, 0.1 M, Kb = 1.8×10^-5

    • [OH-] = √(Kb × C) = √(1.8×10^-5 × 0.1) ≈ 0.00134
    • pOH = -log(0.00134) ≈ 2.87 → pH ≈ 11.13

  6. Neutral solution: [H+] = 1×10^-7 → pH = 7

  7. pH = 4 → [H+] = 10^-4 = 0.0001 M

  8. pOH = 5 → [OH-] = 10^-5 = 0.00001 M → pH = 14 - 5 = 9

  9. 0.05 M NaOH → pOH = -log(0.05) ≈ 1.3 → pH ≈ 12.7

  10. 0.02 M HCl → pH = -log(0.02) ≈ 1.7

6) Bond Order

Concept: Bond order = (Number of bonding electrons − Number of antibonding electrons)/2

Practice Problems

  1. H2 → (2-0)/2 = 1
  2. He2 → (2-2)/2 = 0 (unstable)
  3. O2 → (10-6)/2 = 2
  4. N2 → (10-4)/2 = 3
  5. F2 → (8-4)/2 = 2
  6. C2 → (8-4)/2 = 2
  7. B2 → (6-2)/2 = 2
  8. Li2 → (2-0)/2 = 1
  9. O2^- → (11-6)/2 = 2.5
  10. NO → (7-3)/2 = 2

7) Effective Atomic Number (EAN)

Concept: EAN = Number of valence e⁻ in metal + Number of e⁻ donated by ligands

Practice Problems

  1. [Fe(CO)5] → Fe = 8 valence e⁻ (Fe0), CO donates 2 e⁻ each ×5 =10 → EAN = 18
  2. [Ni(CO)4] → Ni = 10 valence e⁻, 4×2=8 → EAN=18
  3. [Cr(CO)6] → Cr=6, 6×2=12 → EAN=18
  4. [Co(NH3)6]^3+ → Co3+ = 6 e⁻, NH3 donates 2×6=12 → EAN=18
  5. [PtCl2(CO)2] → Pt=10, Cl2+CO2 = 8 e⁻ → EAN=18
  6. [Mn(CO)5] → Mn=7, 5×2=10 → EAN=17
  7. [Fe(CN)6]^4- → Fe2+ = 6, CN- = 12 → EAN=18
  8. [Ru(NH3)5Cl]^2+ → Ru2+ = 6, NH3×5=10, Cl=2 → EAN=18
  9. [Mo(CO)6] → Mo=6, CO×6=12 → EAN=18
  10. [V(CO)6] → V=5, CO×6=12 → EAN=17

8) Notes & Cheat Sheet

  • Δn = moles gas products − moles gas reactants
  • Kp = Kc × (RT)^Δn
  • pH + pOH = 14
  • Bond order = (bonding − antibonding)/2
  • EAN = metal valence e⁻ + ligand e⁻

Quick Revision

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On this page

1) Isotopes & Atomic Mass CalculationExamplePractice Problems with Solutions2) % Composition & Mass of Elements in CompoundsExamplePractice Problems3) Empirical & Molecular FormulaPractice Problems4) Kp and Kc MathPractice Problems5) pH MathPractice Problems6) Bond OrderPractice Problems7) Effective Atomic Number (EAN)Practice Problems8) Notes & Cheat Sheet