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General Chemistry: Mole, Stoichiometry and Chemical Reactions

Overview of Chemistry Concepts

  • General Chemistry (Course Code: 202-SN1-RE)

  • Chapter 3: Focus on Mole, Stoichiometry, and Chemical Reactions

  • Lectured by: Sylvain Vigier, Champlain College - Lennoxville

1. Measuring Atomic Mass

  • Challenges:

    • Atoms are extremely small (approximately $10^{-10}$ m) and very light, making it difficult to ascertain their mass directly.

    • Historical figures: Lavoisier, Dalton, Avogadro, Gay-Lussac contributed to understanding atomic masses by analyzing proportions in which elements combine to form compounds.

  • Dalton's Law of Multiple Proportions:

    • Examines ratios of masses of oxygen per gram of carbon in different compounds:

    • For CO: 1.33 ext{ g O} / 1 ext{ g C}

    • For CO$_2$: 2.66 ext{ g O} / 1 ext{ g C}

2. Unified Atomic Mass Unit

  • Definition: 1 atom of 12C is assigned exactly 12 unified atomic mass units (u), also known as daltons (Da).

  • Importance: All atomic masses are given relative to this standard.

  • Example Calculation: Mass of carbon-13 isotope:

    • ext{Ratio of masses} = 13C/12C = 1.0836129

    • m(13C) = 1.0836129 imes 12 ext{ Da} = 13.003355 ext{ Da}

3. Average Atomic Mass

  • Isotopes: Each element has isotopes with different masses and percentages:

    • Hydrogen: 1H (99.99%), 2H (0.01%)

    • Carbon: 12C (99%), 13C (1%)

  • Average Atomic Mass Formula:

    • ext{average atomic mass} = ext{sum of fractional abundance} imes ext{isotopic mass}

    • Example for Carbon:

    • ext{Average mass} = 0.9889 imes 12 ext{ Da} + 0.0111 imes 13.0034 ext{ Da} = 12.01 ext{ Da}

4. The Mole

  • Definition: The mole (mol) is defined as the number of carbon atoms in 12 grams of pure carbon-12:

    • N_A = 6.022 imes 10^{23} ext{ mol}^{-1}

  • Comparative Examples:

    • 1 mole of eggs = 6.022 x 10$^{23}$ eggs

    • 1 mole of playing cards = 6.022 x 10$^{23}$ cards

5. Molar Mass

  • Definition: Molar mass (M or MM) is the mass of one mole of a substance, measured in g/mol.

  • Relation to Atomic Mass: Molar mass of any substance is numerically equivalent to its atomic mass in Da.

    • 1 ext{ mol } 12C = 12 ext{ g}
      eq 12 ext{ Da}

  • Calculation Example:

    • Molar mass of H$_2$O:

    • = (2 imes 1.00794 ext{ g/mol}) + (16.00 ext{ g/mol}) = 18.01528 ext{ g/mol}

6. Chemical Reactions

  • Principles:

    • Atoms reorganize during a chemical reaction with no creation or destruction (Law of Conservation of Mass).

  • Chemical Equation Structure:

    • Reactants (left side) transform into products (right side).

    • Arrow indicates direction and reversibility of the reaction.

Balancing Chemical Equations

  • Goal: Make sure the same number of each element appears on both sides of the equation.

  • Example - Combustion of Ethanol:

    • C2H5OH(l) + O2(g) ightarrow 2 CO2(g) + 3 H_2O(g)

7. Stoichiometry

  • Definition: The quantitative relationship in a chemical reaction allowing calculations based on balances of reactants and products.

  • Methods:

    1. Write and balance the reaction equation.

    2. Convert masses to moles using the molar mass.

    3. Utilize the mole ratios from the balanced equation to find the number of moles needed or produced.

    4. Convert back to mass if required.

8. Limiting Reagents

  • Definition: The reactant that is completely consumed first in a reaction, thus limiting the amount of product formed.

  • Example - Combustion of Propane:

    • 2 C3H8(g) + 5 O2(g) ightarrow 3 CO2(g) + 4 H_2O(g)

9. Theoretical and Actual Yield

  • Theoretical Yield: The maximum amount of product obtainable; actual yield is usually less due to various reaction efficiencies.

  • Percent Yield Formula:

    • ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100

10. Exercises

  1. Calculate the number of moles in a given mass of a reactant.

  2. Identify limiting reagent in a given reaction.

  3. Calculate theoretical yield and actual yield based on given quantities.