03_Lecture_2021

Chapter 3: Chemical Reactions and Reaction Stoichiometry

Types of Chemical Reactions

• Combination or Synthesis Reaction

  • General form: A + B → AB

• Decomposition Reaction

  • General form: AB → A + B

  • Example: 2 H₂(g) + O₂(g) → 2 H₂O(l) (2 moles H₂ + 1 mole O₂ = 2 moles H₂O)

• Single-replacement Reaction

  • General form: A + BC → AC + B

• Double-replacement Reaction

  • General form: AB + CD → AD + CB

• Combustion Reaction

  • Involves oxygen and produces CO₂ and H₂O

Conservation of Mass

• The total mass of reactants equals the total mass of products in a chemical reaction.

  • Example: 4.0 g H₂ + 32.0 g O₂ = 36.0 g H₂O

Stoichiometry

• Definition: Study of the quantities of substances consumed and produced in chemical reactions.

• Based on the Law of Conservation of Mass:

  • "Nothing is created or destroyed; equal amounts are present before and after the reaction." — Antoine Lavoisier

Chemical Equations

• Representation of Chemical Reactions:

  • Reactants (left) → Products (right)

  • "+" is used for multiple reactants or products.

Balancing Equations

• Follow the Law of Conservation of Mass:

  • Balance one element at a time by changing coefficients (not subscripts).

Why Use Coefficients?

• Changing subscripts alters the identity of the substance.

• Example: H₂O vs H₂O₂ (used in hydrogen peroxide).

States of Matter in Chemical Equations

• Indicate states:

  • (g) = gas, (l) = liquid, (s) = solid, (aq) = aqueous (dissolved in water).

Reaction Conditions

• Symbols used to indicate conditions during reactions:

  • Example: Δ over the arrow indicates heat requirement.

Practice

• Balance the following chemical equations and predict products.

Simple Patterns of Chemical Reactivity

• Combination, Decomposition, and Combustion reactions can be predicted based on reactants.

Combination Reactions

• Definition: Two or more substances react to form a single product.

Predictions in Combination Reactions

• Ability to predict products when metals react with nonmetals based on common charges.

Decomposition Reactions

• Definition: One substance breaks down into two or more new substances.

• Example: Solid sodium azide decomposes to release nitrogen gas.

Heating Metal Carbonate

• Metal carbonates decompose upon heating to give carbon dioxide and metal oxide.

Combustion Reactions

• Definition: Rapid reactions that produce flames, usually involving oxygen.

Products of Combustion of Hydrocarbons

• Typically yield CO₂ and H₂O when burning compounds containing C and H.

Formula Weight (FW)

• Definition: Sum of atomic weights in a chemical formula. Significance in quantifying substance.

• Example: For sulfuric acid (H₂SO₄):FW = 2(1.0 amu) + 32.1 amu + 4(16.0 amu) = 98.1 amu

Molecular Weight (MW)

• Equivalent to formula weight when dealing with molecules.

• Example: For glucose (C₆H₁₂O₆): MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180.0 amu.

Percent Composition

• Formula: % Element = (Number of Atoms × Atomic Weight) / Formula Weight of Compound × 100.

• Example for glucose (C₆H₁₂O₆):%C = (6 × 12.0 amu) / 180.0 amu × 100 = 40.0%.

Avogadro’s Number

• One mole = 6.022 x 10²³ particles (atoms/molecules).

Molar Mass

• Definition: Mass of 1 mole of a substance (g/mol).

• Example of diatomic elements: If diatomic, molar mass = 2 × atomic weight.

Mole Relationships

• 1 mole of atoms or molecules = Avogadro's number of particles.

• Number of atoms is the subscript multiplicatively adjusted by Avogadro's number.

Converting Amounts

• Moles provide conversion between the molecular scale and real-world mass.

• Example calculation of atoms in 3 g of copper.

Determining Empirical Formulas

• Definition: Simplest whole-number ratio of atoms in a compound.

Example: Empirical Formula of PABA

• Steps include:

  • Convert percentages to moles,

  • Calculate mole ratio,

  • Obtain empirical formula (C₇H₇NO₂).

Determining a Molecular Formula

• The molecular formula is a multiple of the empirical formula based on molar mass.

Determining Percent Yield

• Percent yield = (actual yield / theoretical yield) × 100.