Chapter 6

Chapter 6: Quantities in Chemical Reactions

The Mole

Formula Weight

• Understanding the mass of a compound in reactions is critical.

• Formula weight is determined by summing the average atomic masses of its constituent atoms.

• Different terms:

  • Formula Weight: For ionic compounds.

  • Molecular Weight: For molecular (covalent) compounds.

• Example calculations:

  • Calculate formula or molecular weights for: H₂O, Fe(OH)₂, C₁₂H₂₂O₁₁.

The Mole Explained

• Individual atomic weights are too small for practical use.

• Mole (mol): Defined as the equivalent of the number of Carbon-12 atoms in 12 grams, which is approximately 6.022 x 10²³ atoms.

• Example question: What is the mass of 6.022 x 10²³ Magnesium atoms?

Avogadro’s Number

• Named after physicist Amadeo Avogadro.

• Avogadro’s Number: 6.022 x 10²³ items/mole, applicable regardless of the mass or nature of the item (atoms, molecules, etc).

• Example:

  • 6.022 x 10²³ hydrogen atoms in 1 mole of H₂.

  • 6.022 x 10²³ sucrose molecules in 1 mole of sucrose.

The Concept of the Mole

• Analogy:

  • 1 dozen = 12

  • 1 gross = 144

  • 1 mole = 6.022 x 10²³

Conversion Factors

• Usage of conversion factors helps in transitioning between moles and actual quantities.

  • Example using dimensional analysis:

    • 1.45 moles CH₄ → 8.71 x 10²³ CH₄ molecules.

    • Conversion: 1 mole of any substance contains 6.022 x 10²³ entities.

Practice Problems

1. Determine the number of oxygen atoms in 1.35 moles of oxygen.

2. How many moles of N₂H₄ are in 7.20 x 10²¹ molecules?

Molar Mass Concept

• Molar Mass: The mass in grams of one mole of a substance, numerically equivalent to its formula weight.

• Examples:

  • 1 mole of Oxygen = 15.9994 grams.

  • 1 mole of Lithium = 6.941 grams.

  • 1 mole of water = 18.0152 grams.

Molar Mass Calculations

• Example calculation:

  • Determine molecules in a sample: For 23.3 g of CH₄:

  • Calculate molar mass of CH₄: 12.011 g/mol + 4(1.0079 g/mol) = 16.043 g/mol.

  • Calculation steps for number of moles yielded by mass of compound.

Practice Problems

1. Determine the mass of 1.25 moles of silicon.

2. Moles of CF₄ in 52.99 g of CF₄.

3. Number of Argon atoms in 20.5 g of Argon.

Chemical Formulas and Moles

• Example: Composition of H₂O in terms of moles.

  • 1 molecule of H₂O: 2 H atoms and 1 O atom.

  • Scaling up to multiple molecules and applying mole relationships.

Relating Moles of Compounds to Elements

• Use of compound formulas to determine moles of individual elements:

  • Mg(OH)₂ yields moles of oxygen per mole of compound.

  • E.g., For every mole of Mg(OH)₂, there are two moles of oxygen.

Practice Problem Example

• Determine ethanol molecules in 21.6 g of ethanol (C₂H₆O).

• Calculate grams of oxygen in 27.39 g of C₆H₅NO₃.

Stoichiometry

Stoichiometry Fundamentals

• Balanced chemical equations indicate which reaction substances and their molar ratios.

• Stoichiometry studies mass relationships in chemical reactions.

Grams and Moles Conversion

• Use of balanced equations for molar conversion factors.

  • For example: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g).

  • 1 mole of C₃H₈ yields 3 moles of CO₂.

Grams Stoichiometry Example

• To find grams of N₂ needed for 7.50 g of NH₃:

  • Molar mass of NH₃ calculated.

  • Include steps from grams of A → moles A → moles B → grams B.

Practice Problems

1. Using Chromium metal with oxygen to determine mass yields from given chromium mass.

2. PbS with oxygen reaction to find required masses.

Limiting Reagents

Concept of Limiting Reagents

• Reactants can be mixed in proportions differing from a balanced equation.

• Example: 5 moles N₂ with 1 mole O₂. Oxygen is limiting, thus restricting product yield.

Determining Limiting Reagents

• Pancake analogy to understand limiting reactants.

• Stoichiometric calculations assess which reactants limit product creation.

Example Problem

• Calculation methods for determining limiting reagents based on given quantities.

Theoretical Yield and Percent Yield

Theoretical Yield

• Amount of product predicted from complete reaction of the limiting reactant.

  • Real-world issues (leftover material, side reactions) affect actual yield.

Percent Yield Calculation

• Percent Yield defined as:

  • (Actual yield / Theoretical yield) × 100.

Practice Problem

• Find percent yield from the production of calcium nitride given initial conditions.