Chemistry Lecture Notes: Moles and Stoichiometry

Introduction to Moles

• Moles: A way to count particles, similar to how a dozen equals 12.
• Avogadro's Number: 6.02 x 10^23 particles per mole.
• Moles serve as a bridge between the atomic scale (particles) and the macroscopic scale (mass/volume).

Stoichiometry in Reactions

• Balancing Equations: Coefficients in chemical equations indicate the ratio of moles needed for reactants and products.
• Example: Mg(OH)₂ + 2HCl → MgCl₂ + H₂O
  • 1 mole Mg(OH)₂ reacts with 2 moles HCl.
• Stoichiometry allows the prediction of reactant/product quantities.

Calculating Moles

• Formula: [ \text{moles} = \frac{\text{mass (g)}}{\text{RAM or RFM}} ]
  • RAM: Relative Atomic Mass (for elements).
  • RFM: Relative Formula Mass (for compounds).
• Example Calculation:
  • Carbon: 12 g / 12 (RAM of Carbon) = 1 mole
  • Water (H₂O): RAM of H is 1, O is 16; RFM = 18
    • 54 g H₂O / 18 = 3 moles

Example Problem: Reaction Mass Calculation

• Calculate mass of HCl needed to react with 2.9 g Mg(OH)₂:
  1. Find RFM: Mg(OH)₂ RFM = 24 (Mg) + 34 (OH) = 58
  2. Calculate moles of Mg(OH)₂: 2.9 g / 58 = 0.05 moles
  3. Stoichiometry: 1:2 ratio means 0.1 moles of HCl needed
  4. Convert moles to mass: 0.1 moles x 36.5 (HCl RFM) = 3.65 g

Concentration Calculations

• Concentration Formula: [ \text{Concentration} = \frac{\text{mass (g) or moles}}{\text{volume (dm}^3\text{)}} ]
• Units Conversion:
  • 1000 cm³ = 1 dm³
  • Example: 200 cm³ = 0.2 dm³
• Example Calculation:
  • 3.65 g HCl in 0.2 dm³: 18.25 g/dm³
  • 0.1 moles HCl in 0.2 dm³: 0.5 moles/dm³ (0.5 Molar)

Titration and Neutralization

• Titration: Determining unknown concentrations using neutralization.
• Example Problem:
  • 50 cm³ of 0.2 M Mg(OH)₂ neutralizes 25 cm³ HCl.
  1. Convert volume: 50 cm³ = 0.05 dm³
  2. Calculate moles: 0.2 M x 0.05 dm³ = 0.01 moles Mg(OH)₂
  3. Stoichiometry: 1:2 ratio means 0.02 moles HCl needed
  4. Concentration of HCl: 0.02 moles / 0.025 dm³ = 0.8 M

Conclusion

• Moles, stoichiometry, and concentration are key to understanding chemical reactions.
• Practice calculations and units conversion to master these concepts.