CH 4: Stoichiometry: Quantitative Information about Chemical Reactions

Overview of Stoichiometry

• Definition: Stoichiometry is the study of amount relationships in chemical reactions, providing quantitative information about chemical reactions.

• Basis: It is founded on the law of conservation of matter, stating that mass is conserved in a chemical reaction.

Law of Conservation of Matter

• Mass Conservation Equation:

  • Total mass of reactants = Total mass of products

• Balanced Chemical Equations: Must balance for mass:

  • Numbers of atoms on the reactant side = Numbers of atoms on the product side

Stoichiometric Coefficients

• Definition: Stoichiometric coefficients are the numbers in front of chemical formulas or symbols in a chemical equation.

• Function: They provide information about the numbers of atoms, molecules, and ions involved in the reaction, as well as the number of moles of pure substances (or species) involved.

• Example Equation:

  • CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

  • Indicates that 1 molecule (or mole) of methane reacts with 2 molecules (or moles) of oxygen, yielding 1 molecule (or mole) of carbon dioxide and 2 molecules (or moles) of water.

Moles and Stoichiometric Ratios

• To analyze amounts:

  • Molecular Interpretation:

    • One molecule of methane reacts with two molecules of oxygen yielding products as indicated.

  • Mole Interpretation:

  • One mole of methane reacts with two moles of oxygen yielding one mole of carbon dioxide and two moles of water.

• Relation to Avogadro’s Number: Assumes each stoichiometric coefficient multiplied by Avogadro’s number gives the corresponding number of entities involved.

Conversion Factors in Stoichiometry

• Utilization of Stoichiometric Coefficients:

  • They help establish conversion factors between reactants and products, allowing for calculations to convert between amounts (number of moles or atoms) of reactants and products.

• Example Calculation: Given the complete combustion of ethane, we can calculate how many moles of water are produced from a specific amount of oxygen.

Example Problem: Ethane Combustion

1. Given Reaction: C₂H₆ + 7 O₂ → 6 CO₂ + 6 H₂O

2. Determine Moles of Water Produced from Moles of Oxygen: 3.0 moles O₂

   • From the reaction, 7 moles of O₂ produce 6 moles of H₂O.

   • 
     $ext{Moles of H}<em>2 ext{O} = (3.0 ext{ moles O}</em>2) imes rac{6 ext{ moles H}<em>2 ext{O}}{7 ext{ moles O}</em>2} = 2.57 ext{ moles H}_2 ext{O}$

Stoichiometry Problems

• General Approach: Mass (reactant) → # of moles (reactant) → # of moles (product) → mass (product)

• Steps to solve:

  1. Convert mass of the reactants to moles using molar mass.

  2. Use molar ratios from the balanced equation to get moles of product.

  3. Convert moles of product back to mass using its molar mass.

Reaction Limiting Reactant Concept

• Definition: The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Identification: Often necessary in reactions involving two or more reactants to determine the maximum yield of products.

• Example:

  1. If 50.0 g of Al reacts with 50.0 g of O₂ to form Al₂O₃, determine limiting reactant and theoretical yield by molar calculations.

  2. Balanced Equation: 4 Al + 3 O₂ → 2 Al₂O₃

Stoichiometry Calculations with Limiting Reactants

• An example involves a unit operation:

  • Determining Product Yield from Reactant Mass:

  • Molar mass of reactants must be calculated.

  • Stoichiometric ratios apply to convert initial reactant moles to products.

Theoretical Yield and Percent Yield

• Theoretical Yield: Maximum amount of product expected from stoichiometric calculations.

• Actual Yield: Amount obtained experimentally after the reaction.

• Percent Yield Calculation:
  ext{Percent Yield} ( ext{in C}) = rac{ ext{Actual Yield (g or mol)}}{ ext{Theoretical Yield (g or mol)}} imes 100

Example: Aluminum and Oxygen Reaction

1. Calculating Theoretical Yield: If 50.0 g of Al provides a theoretical yield of 94.3 g of Al₂O₃ (given, with 48.5 g collected), what is the % yield?

   • ext{Percent Yield} = rac{48.5 ext{ g}}{94.3 ext{ g}} imes 100 = 51.4 ext{%}

Troubleshooting Stoichiometry Problems

• Apply the limiting reactant theory by:

  • Calculating potential product yield from both reactants.

  • Selecting the limiting reactant based on which produces less product.

Chemical Analysis and Stoichiometry

• Chemical Analysis: Methods to quantitatively measure the composition of solutions or compounds.

• Common Approaches: Reaction of substances with known amounts, as in titrations.

• Example scenario: Measuring Na₂SO₄ in samples.

Combustion Analysis of Hydrocarbons

• Procedure: Analyze products of combustion to determine empirical formulas.

  • Use CO₂ and H₂O recorded from combustion to find mass composition of original compounds.

• Empirical and Molecular Formulas: Determine molecular formula using empirical formula relationship and molecular weight.

Measuring Concentrations of Compound in Solutions

• Concentration Definitions: Molarity (concentration in mol/L) and others like mass percent.

• Concentration influences stoichiometric calculations, often requiring transformations for solution preparations.

Conducting Dilutions and Molarity Calculations

• Use formula: $M<em>1V</em>1 = M<em>2V</em>2$ for dilutions.

• Convert concentrations and volumes to find new concentrations post-dilutions.

Example Problem: Diluting Solutions

1. Given 50.0 mL of 0.515 M HCl, find new molarity after dilution to 500.0 mL:

   • $V<em>1 = 50 ext{ mL}, M</em>1 = 0.515 ext{ M}$

   • Using $M<em>1V</em>1 = M<em>2V</em>2$ calculate $M_2$ and verify.