1.6 Empirical and Molecular Formula

Review of Empirical and Molecular Formulas

Percent Composition

• Percent composition is useful for determining the proportions of different elements in a compound.

• It is calculated using the formula:

  Percent by mass = (Mass of part / Total mass) × 100.

• Example Calculation:

  • For phosphoric acid (H₃PO₄):

    • Mass of hydrogen = 3 (since it's H₃)

    • Total mass = 3 (H) + 31 (P) + 4 (O) = 38

    • Percent by mass of H = (3 / 38) × 100 ≈ 3.02%

• Another Example: Calcium hydroxide (Ca(OH)₂)

  • Mass of calcium = 40 g

  • Total mass = 40 (Ca) + 16 (O) × 2 + 1 (H) × 2 = 74 g

  • Percent by mass of calcium = (40 / 74) × 100 ≈ 54%

• Mass calculation from percentage:

  • If a sample contains 74.5% chlorine, total mass = 190.2 g:

  • Mass of Cl = 0.745 × 190.2 g = 141.7 g.

Definition of Empirical Formulas

• Empirical formula is the simplest ratio of the elements in a compound, reduced to the lowest whole numbers.

• Examples of Reduction:

  • C₄H₈ can be reduced to C₂H₄ or CH₂

  • C₆H₁₂O₆ can be reduced to CH₂O

  • For sodium hydroxide (NaOH), dividing by two gives the same formula since it already is in simplest form.

Molecular Formulas

• Molecular formula provides the actual number of each atom in a molecule and can be a multiple of the empirical formula.

• Examples:

  • CH₄ and C₂H₈ are molecular formulas derived from the empirical formula CH₄.

• Example Calculation:

  • Given NH₃ with a mass of 34 g/mol:

    • Molar mass of NH₃ = (14 + (3 × 1)) = 17 g/mol.

    • 34 g / 17 g/mol = 2 moles → Molecular Formula = N₂H₆.

• Another example:

  • CH₃ has a molar mass = 15 g/mol, and if given mass = 45 g:

    • 45 / 15 = 3 → Molecular formula is C₃H₉.

Calculating the Empirical Formula from Percent Composition

• Steps:

  1. Assume a 100 g sample to convert percent to grams.

  2. Convert grams to moles by dividing by atomic masses.

  3. Find the simplest whole number ratio of moles.

• Example Calculation:

  • For a compound with 40.92% Carbon, 4.58% Hydrogen, and 54.5% Oxygen:

    • Assume 100 g sample: 40.92 g C, 4.58 g H, 54.5 g O.

    • Convert to moles:

      • C: 40.92 g / 12 g/mol = 3.41 moles

      • H: 4.58 g / 1 g/mol = 4.58 moles

      • O: 54.5 g / 16 g/mol = 3.41 moles.

  4. Divide each by smallest # of moles (3.41), to get a cleaner ratio:

  • C: 1, H: 1.5, O: 1.

  5. To eliminate the fraction, multiply through by 2: Empirical Formula is C₂H₃O₂.

Finding Molecular Formulas from Empirical Formulas

• To find the molecular formula, compare the empirical formula's mass with the molecular mass provided.

• Example:

  • If empirical formula is NO₂ and the calculated empirical mass = 46 g/mol, molar mass is 90 g/mol, so 90/46 = about 2 → Molecular formula is N₂O₄.

Practice Example

• Given 1.52 g of nitrogen and 3.47 g of oxygen:

  • Molar mass of N = 14 g/mol, O = 16 g/mol.

  • N: 1.52 g / 14 g/mol = 0.1086 moles; O: 3.47 g / 16 g/mol = 0.217 moles.

  • Ratio → NO₂ (Empirical).

Final Remarks

• Example question chosen in the video was to determine the empirical and molecular formulas based on given percentages and the molar mass:

  • 64.8% C, 13.62% H, 21.58% O, molar mass = 74.14 g/mol:

    1. Convert % to moles (refer to steps above).

    2. Determine ratios and find empirical formula through simplifications.

    3. Check if empirical and molecular formulas match given molar mass.