Chem 16 Supplementary Lecture Notes

Supplementary Lectures: Moles Concept, Empirical and Numerical Formula, Chemical Reactions and Stoichiometry

## University of the Philippines Diliman - Institute of Chemistry AY 2025-2026

Introduction
  • Subject: Chemistry 16 Supplementary Lectures.
  • Purpose: Understanding moles, empirical and molecular formulas, and stoichiometry in chemical reactions.
Copyright Notice
  • This material is copyrighted under the Republic Act 8293 (Intellectual Property Code of the Philippines). Reproduction without permission is prohibited.
Primary Objectives for This Chapter
  • Conversions: Between number of molecules, ions, formula units, atoms, moles, and mass.
  • Percent Composition: By mass of an element in a compound.
  • Empirical Formula: Determining the empirical formula of a compound.
  • Molecular Formula: Determining the molecular formula of a compound.
The Concept of the Mole and the Avogadro Constant
  • Definition of the Mole: A mole is an amount of substance that contains as many elementary entities as there are atoms in 12 grams of the Carbon-12 isotope.
  • Avogadro's Number ($N_A$): $6.02214179 imes 10^{23}$ entities/mol.
Atomic Mass
  • Atomic Mass: Dependent on the number of electrons, protons, and neutrons. The mass of electrons is negligible.
  • Actual Mass of Particles:
      - Proton: $1.67262192 imes 10^{-27}$ kg
      - Neutron: $1.675 imes 10^{-27}$ kg
      - Electron: $9.1093837 imes 10^{-31}$ kg (generally ignored in calculations).
  • Reference for Atomic Mass Unit (amu): 1 amu is defined as $ rac{1}{N_A} = 1.660538921 imes 10^{-27}$ kg.
Molar Mass and Conversions
  • Molar Mass (M): The mass of one mole of a substance, expressed in grams/mol. For example:
      - Molar Mass of Carbon (C): $12.011$ g/mol corresponding to $6.022 imes 10^{23}$ C atoms.
      - Molar Mass of Oxygen ($O$): $15.999$ g/mol.
  • Conversions to Moles: Mass to moles can be calculated using:
    extMoles=extMass(g)extMolarMass(g/mol)ext{Moles} = \frac{ ext{Mass (g)}}{ ext{Molar Mass (g/mol)}}
Empirical and Molecular Formulas
  • Empirical Formula: Represents the simplest whole number ratio of elements in a compound.
  • Molecular Formula: Represents the actual number of atoms of each element in a molecule.
  • Example Compound: Water ($H_2O$), having empirical and molecular formulas as the same.
  • Finding Empirical Formula:
      1. Convert mass percent to grams (assuming 100 g total for simplicity).
      2. Convert grams to moles using atomic mass.
      3. Divide by the smallest number of moles.
      4. Obtain whole number ratios if necessary.
  • Example Calculation: For a compound with percent composition of C, H, and O:
    extC:extmass/extmolarmass(C)ext{C: } ext{mass}/ ext{molar mass (C)}
Percent Composition Calculations
  • To find the mass percent composition for a compound such as $C_2H_6O$:
      1. Calculate molar mass of the compound.
      2. Divide individual element mass by compound mass and multiply by 100 to find percentage.
    extPercentComposition=extmassofelementextmolarmassofcompoundimes100ext{Percent Composition} = \frac{ ext{mass of element}}{ ext{molar mass of compound}} imes 100
Practice Exercises

  1. Given Molecular Formula: Acetylsalicylic acid (C9H8O4).
       - (a) Calculate its molar mass.
       - (b) If a typical tablet contains 500 mg of active ingredient, calculate the number of moles and molecules.
       - (c) Calculate the mass percentages of C, H, and O.

  2. Example for Combustion Analysis: For an organic compound producing CO2 and H2O, calculate:
       - Mass percent of each constituent.
       - Empirical formula.
       - Molecular formula based on molar mass information.

Conclusion
  • The understanding of moles, empirical and molecular formulas, and stoichiometry is essential to grasp chemical reactions quantitatively.
  • Use proper dimensional analysis and conversion methods to solve chemical problems effectively.