Lecture: Average Atomic Mass and Moles

Introduction to Average Atomic Mass

• Idea: Useful concept for understanding mass at atomic/molecular level.
• Objective: Connect atomic-scale masses to sample masses in chemistry labs.

From Atomic Mass to Lab Scale

• Context: Chemists deal with grams of substances, not individual atoms.
• Average Atomic Mass Example:
  • Lithium's average atomic mass: 6.94 unified atomic mass units per atom.

Avogadro's Number

• Definition: The number of atoms that equate to a substance's average atomic mass in grams.
• Value: 6.02214076 x 10^23 (approximated as 6.022 x 10^23).
• Named After: Amadeo Avogadro, 19th-century Italian chemist.

Concept of a Mole

• Definition: A unit representing 6.022 x 10^23 entities (atoms, molecules).
• Origin: Term coined by German chemist Wilhelm Ostwald, based on 'molecule'.
• Analogy: A dozen represents 12 items; a mole represents Avogadro's number of items.

Practical Example: Calculating Atoms in a Sample

• Problem: Determine the number of germanium atoms in a 15.4 mg sample.

Steps to Solve

1. Convert Milligrams to Grams:
   • 15.4 mg = 0.0154 g (divide by 1000).
2. Calculate Moles of Germanium:
   • Use Germanium's molar mass: 72.63 g/mol.
   • Formula: ( \text{Moles} = \frac{\text{grams}}{\text{grams per mole}} ).
3. Determine Number of Atoms:
   • Multiply moles by Avogadro's number.
   • Calculation: 15.4 mg → grams → moles → atoms.

Detailed Calculation

• Convert to Grams:
  • 15.4 mg to grams: 0.0154 g.
• Determine Moles:
  • ( 0.0154 \text{ g} \div 72.63 \text{ g/mol} ).
• Calculate Atoms:
  • Moles x Avogadro's number = atoms.
  • Final result: Approximately 1.28 x 10^20 atoms.

Conclusion

• Outcome: Understanding moles and Avogadro's number helps in converting atomic-scale measures to practical lab-scale measures.
• Significance: Provides a basis for calculating the number of atoms in a given mass of a substance, improving practical lab calculations.