Chapter 3: Mass Relationships in Chemical Reactions

Chapter 3: Mass Relationships in Chemical Reactions

Stoichiometry
  • Definition: Area of study focusing on quantities of substances in chemical reactions.
  • Basis: Law of Conservation of Mass (Antoine Lavoisier, 1789)
  • No matter is created or destroyed in a chemical reaction.
  • Mass before equals mass after; foundational for chemical experiments.
Chemical Equations
  • Purpose: Represent chemical reactions on paper.
  • Components:
  • Reactants: Starting materials (left side of the equation).
  • Products: Ending materials (right side of the equation).
  • Symbols: "+" for multiple reactants/products, "→" for the direction of the reaction.
Writing and Balancing Chemical Equations
  1. Write correct formulas for reactants (left) and products (right).
  • Example: Ethane reacts with oxygen: C₂H₆ + O₂ → CO₂ + H₂O
  1. Adjust coefficients to balance atoms on each side.
  • Keep subscripts unchanged.
  • Example: C₂H₆ + O₂ → 2CO₂ + 3H₂O
  1. Steps to balance:
  • Start with elements found in only one reactant and one product (e.g., C or H).
  • Balance all elements gradually by adjusting coefficients.
Types of Reactions
  • Combination reactions: Two or more substances form one product.
  • Decomposition reactions: One substance breaks into two or more.
  • Combustion reactions: Rapid reactions producing flames, commonly involving C and H, yielding CO₂ and H₂O.
Mass Relationships and Molar Mass
  • Formula Weight: Sum of atomic masses in a formula unit.
  • Example: NaCl = 22.99 amu (Na) + 35.45 amu (Cl) = 58.44 amu.
  • Molecular Weight: Sum of atomic weights in a molecule.
  • Example: CO₂ = 12.01 amu (C) + 2 × 16.00 amu (O) = 44.01 amu.
Percent Composition
  • Calculation: Percentage of mass from each element in a compound.
  • Example: For glucose C₆H₁₂O₆, % Carbon = (6 × 12.0 amu / 180.0 amu) × 100 = 40.0%.
Avogadro’s Number (Nₐ)
  • Definition: Number of particles in one mole, approximately 6.022 × 10²³.
  • Example: 100 helium atoms weigh four times the weight of 100 hydrogen atoms.
  • Molar Mass: Mass of one mole of a substance in grams equals its atomic mass in amu.
The Mole Concept
  • 1 mole = Nₐ = 6.022 × 10²³.
  • Conversions: m (mass in g) = MM (molar mass in g/mol) × n (number of moles).
Empirical and Molecular Formulas
  • Empirical Formula: Lowest whole-number ratio of elements in a compound.
  • Procedure to calculate: Convert mass to moles, divide by the smallest number of moles.
  • Molecular Formula: Obtained by determining the molar mass and multiplying empirical formula subscripts by an integer.
Limiting Reactants
  • Definition: Reagent that is consumed first, limiting the amount of product formed.
  • Example scenario: In ammonia formation, identify limiting reactant based on reactant quantities.
Theoretical and Percent Yield
  • Theoretical Yield: Maximum product that can be formed based on stoichiometry.
  • Percent Yield: Comparison of actual yield to theoretical yield.
  • Formula: Percent Yield = (Actual Yield / Theoretical Yield) × 100.
Summary of Stoichiometry Problems
  1. Write a balanced equation.
  2. Identify given and sought quantities.
  3. Note molar masses needed.
  4. Determine limiting reactant if multiple reactants are involved.
  5. Execute calculations to find quantities of reactants/products.
Calculation Procedure
  • Convert grams to moles using molar masses.
  • Apply balanced equation coefficients for conversions between substances.
  • Convert moles back to grams as needed. Check reasonableness of results, ensure correct units and accuracy in calculations.