Lecture Notes on Composition Stoichiometry

Introduction to Stoichiometry

• Definition: Stoichiometry is derived from Greek words:
  • Stoichion: meaning elements
  • Metron: meaning to measure
• Stoichiometry defines mass-to-mole relationships in chemical reactions, enabling quantitative analysis.
• Key concepts to be covered in this lecture:
  • Atomic mass
  • Mole
  • Avogadro's number
  • Chemical formula
  • Percent compositions
  • Chemical equations

Atomic Mass

• Represents the average mass of all naturally occurring isotopes of an element.
• Example: Hydrogen (H) has three isotopes:
  • H-1 (proton)
  • H-2 (deuterium)
  • H-3 (tritium)
• Average atomic mass for an element is based on the relative abundance of its isotopes.
• Units: Unified atomic mass (u) or grams per mole (g/mol).

Calculation Example: Chlorine Average Atomic Mass

• Isotopes:
  • Cl-35: 34.9689 u (75.76% abundance)
  • Cl-37: 36.9659 u (24.25% abundance)
• Calculation:
  • Average atomic mass = (0.7576 * 34.9689) + (0.2424 * 36.9659)
  • Result: 35.45 u

Practice Problem

• Calculate the average atomic mass of silicon using its isotopes.

Mole

• Mole defined as:
  • Amount of substance with the same number of particles as 0.012 kg of Carbon-12.
• Conversion between mass and moles is crucial in chemistry.

Avogadro's Number

• Value: 6.022 x 10^23
• Significance: One mole of any substance contains Avogadro's number of particles (atoms, molecules, etc.).
  • Example: 1 mole of sodium = 6.022 x 10^23 atoms.

Molecular Weight

• Defined as: mass of a substance divided by the number of moles.
• Units: grams per mole (g/mol).
• Conversion factor between mass and moles.

Mass, Mole, and Atom Relationships

• Chart Summary:
  • Mass to Mole: Use molecular weight.
  • Mole to Number of Atoms: Use Avogadro's number.
  • Mass to Number of Atoms: Two-step process (mass to mole, then mole to atoms).

Example Problems

1. 3.5 moles of Aluminum

   • Molecular weight of Al = 26.982 g/mol
   • Mass = 3.5 moles x 26.982 g/mol = 94.4 g

2. 200 grams of Osmium

   • Molecular weight of Os = 190.23 g/mol
   • Moles = 200g / 190.23 g/mol = 1.05 moles

3. 1 gram of Carbon

   • Molecular weight of C = 12.011 g/mol
   • Steps: Convert to moles, then to atoms using Avogadro’s number.

4. Single Oxygen Atom Weight

   • Exercise left for practice.

Chemical Formulas

• Chemical formulas represent elements and compounds.
• Governed by the law of definite proportions: A fixed ratio exists between constituent atoms of a compound.

Example: Molecular Weight of Water (H2O)

• H = 1.008 g/mol, O = 15.999 g/mol
• Calculation:
  • Molecular weight of water = (2 * 1.008) + (1 * 15.999) = 18.015 g/mol

Example: Molecular Weight of Sucrose (C12H22O11)

• Atomic masses: C = 12.011 g/mol; H = 1.008 g/mol; O = 15.999 g/mol
• Calculation yields: 342.297 g/mol

Percent Composition

• Determines the elemental composition of a compound by weight.
• Requires molecular weight calculation.

Example: Percent Composition of Aluminum Sulfate

• Chemical formula: Al2(SO4)3
• Percent composition calculations for Aluminum, Sulfur, and Oxygen.

Empirical and Molecular Formulas

• Empirical Formula: Simplest positive integer ratio of atoms.
• Molecular Formula: Complete atomic composition of a compound.
• Example: NO2 (empirical) vs N2O4 (molecular).

Problems on Empirical and Molecular Formulas

1. Given Percent Composition: 21.6% Na, 33.3% Cl, 45.1% O

   • Assume 100g of compound for simplicity.
   • Calculate empirical formula.

2. Vitamin C Calculation: 40.91% C, 4.59% H, 54.50% O

   • Determine empirical formula and convert to molecular formula using given molecular weight.

Conclusion

• Next lesson: Reaction stoichiometry and stoichiometric calculations for chemical reactions.

• Feel free to ask questions via course messages on Blackboard.