Mole Concepts and Calculations

Overview of Mole Concept

• Mole (mol):
  • Unit of measurement for the amount of a substance.
  • Used to measure large amounts of tiny particles.
• Avogadro’s Number:
  • 1 mole contains 6.022 \times 10^{23} particles (atoms, molecules, etc.).
  • Example: 1 mole of gold = 6.022 \times 10^{23} atoms of gold.

Conversion Between Moles and Atoms

• Calculation Example:
  • 4.0 moles of He:
  • 4.0 \text{ moles He} \times 6.022 \times 10^{23} \text{ atoms/mole} = 2.4 \times 10^{24} \text{ atoms}.

Molar Mass

• Definition:
  • Mass of one mole of a pure substance, expressed in g/mol.
  • Same numerical value as atomic mass on the periodic table but different in meaning.
  • Atomic mass = mass of one atom (amu).
  • Molar mass = mass of one mole of a substance (g/mol).
• Conversion Factor:
  • Molar mass can serve as a conversion factor in calculations.
  • Example: The molar mass of aluminum (Al) is 26.98 g/mol, meaning:
  • 26.98 \text{ g of Al} \equiv 1 \text{ mole of Al}.

Molar Mass Calculations

• Example 1: 3.50 moles of gold.
  • Molar mass of gold (Au) = 196.97 g/mol:
  • 3.50 \text{ mol} \times 196.97 \text{ g/mol} = 689.40 \text{ g of Au}.
• Example 2: Given 12.6 g of Al, find moles.
  • Molar mass of Al = 26.98 g/mol:
  • \frac{12.6 \text{ g}}{26.98 \text{ g/mol}} = 0.467 \text{ mol of Al}.

Molar Mass of Compounds

• Formula mass: Sum of average atomic masses of elements in a compound, same numerically as molar mass.
• Example: Find molar mass of H2O:
  • \text{H molar mass} = 1.008 \text{ g/mol}
  • \text{O molar mass} = 16.00 \text{ g/mol}
  • \text{Molar mass of H2O} = 2(1.008) + 1(16.00) = 18.02 \text{ g/mol}.

Percent Composition

• Definition: Percentage of each element in a compound by mass.
• Formula:
  • \% \text{ of an element} = \left( \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \right) \times 100.
• Example: Percent composition of CO2:
  • Molar mass of CO2 = 44.01 g/mol:
  • \% \text{C} = \left( \frac{12.01 \text{ g}}{44.01 \text{ g}} \right) \times 100 = 27.29\%.
  • \% \text{O} = \left( \frac{32.00 \text{ g}}{44.01 \text{ g}} \right) \times 100 = 72.71\%.

Determining Chemical Formulas

• Steps:

  1. Calculate grams of each element based on % composition.
  2. Calculate moles of each element using their molar masses.
  3. Find the mole ratio by dividing by the smallest number of moles.

• Example:

  • Unknown compound: 110.0 g, 18.8% Na, 29.0% Cl, 52.2% O.
  • Find grams: Na = 20.7 g; Cl = 31.9 g; O = 57.4 g.
  • Moles: Na = 0.900 mol; Cl = 0.900 mol; O = 3.59 mol.
  • Determine ratio: 1:1:4 gives formula NaClO4.

• Empirical vs. Molecular Formulas:

  • Empirical formula: lowest whole-number ratio (e.g., BH3).
  • Molecular formula: actual amounts (e.g., B2H6).

• Finding Molecular Formula:

  1. Calculate empirical formula mass.
  2. Compare to actual molar mass to find the multiplication factor.
  • Example: For \text{CH}_2 with molar mass 42 g/mol:
  • Empirical mass = 14.026 g/mol; factor = 3 leads to \text{C}3\text{H}6.

• Another Example: \text{CH}_2 with molar mass 70 g/mol:

  • Factor = 5 leads to \text{C}5\text{H}{10}.