Stoichiometry and Mole Concept Lecture Notes

Introduction to Stoichiometry

• Stoichiometry involves writing chemical equations to determine how much volume, mass, or molecules are formed.
• Understanding the relationship between moles, mass, and volume.

Key Formulas for Moles

Formula 1: Mass to Moles

• Number of moles = Mass given / Molar mass
• Example: Find the number of moles in 32 grams of methane (CH₄)
  • Carbon atomic mass = 12; Hydrogen atomic mass = 1
  • Molar mass of CH₄ = 12 + (1 x 4) = 16
  • Moles of CH₄ = 32 / 16 = 2 moles

Formula 2: Volume to Moles

• Number of moles = Volume given / Molar volume
• Standard molar volume at STP (Standard Temperature and Pressure) = 22.4 L
• Example: Find the volume of 3 moles of carbon dioxide at STP
  • Volume = 3 x 22.4 L = 67.2 L

Problem Solving Examples

Example 1: Mass to Moles and Volume

• Problem: If 16g of methane is burned, find the mass of CO₂ produced
• Solution:
  • Moles of CH₄ = 16/16 = 1 mole
  • Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O
  • 1 mole of CH₄ gives 1 mole of CO₂
  • Mass of CO₂ = (Number of moles) x (Molar mass of CO₂)
  • Molar mass of CO₂ = 44
  • Mass of CO₂ = 1 x 44 = 44g

Example 2: Find Volume of CO₂ at STP

• Problem: Find the volume of CO₂ produced when 50g of CaCO₃ is heated
• Solution:
  • Number of moles of CaCO₃ = 50/100 = 0.5 moles
  • Balanced equation: CaCO₃ → CaO + CO₂
  • 1 mole of CaCO₃ gives 1 mole of CO₂
  • Volume of CO₂ = 0.5 x 22.4 = 11.2 L

Advanced Stoichiometry Problems

Example 3: Oxygen Requirement

• Problem: Find mass of O₂ required to produce 44.8 L of CO₂
• Solution:
  • 44.8 L of CO₂ = 2 moles
  • Equation: 2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
  • 4 moles of CO₂ requires 5 moles of O₂
  • 2 moles of CO₂ requires 2.5 moles of O₂
  • Mass of O₂ = 2.5 x 32 = 80g

Example 4: Mass of Acetylene Required

• Problem: Find mass of acetylene (C₂H₂) required to produce 44.8 L of CO₂
• Solution:
  • 44.8 L of CO₂ = 2 moles
  • 4 moles of CO₂ from 2 moles C₂H₂
  • 2 moles of CO₂ from 1 mole C₂H₂
  • Mass of C₂H₂ = 26g (molar mass)

Conclusion

• Stoichiometry is essential for understanding the quantitative aspects of chemical reactions.
• By converting between mass, moles, and volume, we solve diverse chemical problems efficiently.
• Always balance chemical equations to accurately calculate the reactants and products involved.

Tip: Break down complex stoichiometry problems into smaller steps using mole concepts for easier calculation.