Lecture Notes: Average Atomic Mass and Avogadro's Number

Introduction to Average Atomic Mass

• Average atomic mass is useful for understanding mass at the atomic/molecular level.
• In practical applications, we deal with masses in grams rather than individual atoms.

Connecting Atomic Mass to Lab Scale

• Chemistry uses average atomic mass to scale from atomic to practical lab masses.
• For example, lithium's average atomic mass is 6.94 unified atomic mass units per atom.

Avogadro's Number

• Definition: The number of atoms in a sample that yields a mass equal to its atomic mass in grams.
• Value: 6.02214076 x 10^23 atoms (commonly approximated as 6.022 x 10^23).
• Named after Amadeo Avogadro, a 19th-century Italian chemist.

Concept of the Mole

• A mole represents 6.022 x 10^23 units of a substance.
• Similar to a dozen (e.g., a dozen eggs = 12 eggs; a mole of atoms = 6.022 x 10^23 atoms).
• Originated from Wilhelm Ostwald in the late 19th century.

Application of Avogadro's Number and Moles

• Example Problem: Determine the number of atoms in 15.4 mg of germanium.
  • Step 1: Convert milligrams to grams:
    • 15.4 mg of germanium = 0.0154 grams (since 1 g = 1000 mg).
  • Step 2: Convert grams to moles of germanium:
    • Molar mass of germanium = 72.63 g/mol.
    • Moles = Grams / Molar Mass = 0.0154 g / 72.63 g/mol.
  • Step 3: Convert moles to atoms:
    • Multiply moles by Avogadro's number (6.022 x 10^23 atoms/mol).

Calculation Summary

• Grams of Germanium: 0.0154 g.
• Moles of Germanium: 0.0154 / 72.63 = small fraction of a mole.
• Atoms of Germanium: Moles x Avogadro's number = 1.28 x 10^20 atoms.
• Significant Figures: The final result is rounded to three significant figures based on the least precise measurement.

Conclusion

• Understanding moles and Avogadro's number aids in connecting atomic scale measures to practical lab-scale quantities.
• Calculations involve converting between units and using Avogadro's number for precise atomic measurements.