Chem 1045: Unit 5 Moles

Chem 1045: Unit 5 Moles

Formula Mass and the Mole Concept

  • Formula Mass

    • Definition: The formula mass of a substance is the sum of the average atomic masses of all the atoms in the substance’s formula.

    • Covalent substances: Exist as discrete molecules.

    • Molecular mass: For covalent substances, the formula mass may be correctly referred to as molecular mass.

Molecular Mass Examples

  • Chloroform (CHCl₃)

    • Average mass: 119.37 amu

    • Calculation: Sum of average atomic masses of constituent atoms.

  • Aspirin (C₉H₈O₄)

    • Average mass: 180.15 amu

    • Molecular structure depicted in accompanying model.

Example 3.1: Aspirin Molecular Mass Calculation

  • Components:

    • C: 13 atoms, average atomic mass = 12.01 amu

    • Subtotal = 13 × 12.01 = 156.13 amu

    • H: 18 atoms, average atomic mass = 1.008 amu

    • Subtotal = 18 × 1.008 = 18.144 amu

    • O: 2 atoms, average atomic mass = 16.00 amu

    • Subtotal = 2 × 16.00 = 32.00 amu

  • Total Molecular Mass: 206.27 amu

Formula Mass for Ionic Compounds

  • Ionic Substances:

    • Composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter.

    • Ionic compounds do not exist as molecules.

    • Formula mass for ionic compounds cannot be referred to as molecular mass.

    • Average atomic masses of the ions approximate the average atomic masses of the neutral atoms.

Example 3.2: Sodium Chloride (NaCl) Formula Mass Calculation

  • Formula Mass Computation:

    • Components:

    • Al: 2 atoms, average atomic mass = 26.98 amu

      • Subtotal = 2 × 26.98 = 53.96 amu

    • S: 3 atoms, average atomic mass = 32.06 amu

      • Subtotal = 3 × 32.06 = 96.18 amu

    • O: 12 atoms, average atomic mass = 16.00 amu

      • Subtotal = 12 × 16.00 = 192.00 amu

  • Total Formula Mass: 342.14 amu

The Mole

  • Definition: The mole is a unit of amount similar to a pair, dozen, gross, etc.

    • Defined as the amount of substance containing the same number of discrete entities as the number of atoms in a sample of pure carbon-12 weighing 12 g.

  • Importance: The mole provides a connection between the mass of a sample and the number of atoms, molecules, or ions present.

Avogadro’s Number

  • Constant: The number of entities in a mole is approximately 6.02214179 × 10²³ (Avogadro's Number, denoted as $N_A$).

  • Diverse Masses: Different elements have different molar masses despite each mole containing the same number of entities.

  • Example: 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, 63.5 g copper, and their corresponding quantities of other elements, all containing 1.00 mol of atoms.

Molar Mass

  • Definition: The molar mass of a substance is the mass in grams of 1 mole of that substance (units: g/mol).

    • Equivalence: Molar mass numerical values are equivalent to atomic or formula mass in amu.

  • Example:

    • A single carbon-12 atom has a mass of 12 amu, hence a mole of carbon-12 atoms weighs 12 g.

Example 3.7: Vitamin C Mass Calculation

  • Molecular Formula: C₆H₈O₆

  • Recommended Daily Allowance: 1.42 × 10⁻⁴ mol

  • Calculated Molar Mass: 176.124 g/mol.

    • Desired mass in grams: Formula via multiple.

Determining Empirical and Molecular Formulas

  • Percent Composition: By mass percentage of each element in a compound.

  • Example Calculation: A compound weighing 10.0 g containing 2.5 g hydrogen and 7.5 g carbon.

Determination of Empirical Formulas

  • Process:

    1. Convert masses to moles using molar masses.

    2. Divide each number of moles by the smallest number of moles.

    3. If needed, multiply by an integer for whole-number ratios.

  • Example: For a compound containing 1.71 g C and 0.287 g H.

  • Summary: Mass ratios can inform about empirical formula formulation.

Derivation of Molecular Formulas

  • Method: Determined by the empirical formula and its molar mass.

    • Formula: Molecular formula = empirical formula × n, where 'n' is a whole number that scales it to molecular mass.

  • Example: From an empirical formula CH₂O with a mass of 30 amu to a molecular mass of 180 amu giving C₆H₁₂O₆.

Miscellaneous Examples

  • Example 3.8: Saccharin quantity analysis from a 40.0 mg sample.

    • Apply quantities to figure molecular count.

  • Example 3.12: Determining the empirical formula from a gas composed of 27.29% C and 72.71% O.

    • Method involves assumption and calculation of moles from percent composition.

Conclusion

  • The study of moles, molar mass, and the relationship to empirical and molecular formulas is crucial in understanding chemical composition and stoichiometry in chemistry.