Lecture on Percent Composition, Empirical, and Molecular Formulas

Percent Composition

• Percent Composition refers to the percentage by mass of each element in a compound.

• Formula:

  $\text{Percent} = \left( \frac{\text{Part}}{\text{Total}} \right) \times 100%$

Example: Carbon Dioxide (CO2)

1. Molar Mass Calculation:
   • Carbon (C): 12.01 g/mol
   • Oxygen (O): 16 g/mol
   • Total Molar Mass of CO2 = 12.01 + 16 + 16 = 44.01 g/mol
2. Percent Calculation:
   • Carbon: $\left( \frac{12.01}{44.01} \right) \times 100 % = 27.29% $
   • Oxygen: $\left( \frac{32}{44.01} \right) \times 100% = 72.71% $

• Note: Percentages should sum to 100%.

Empirical and Molecular Formulas

Definitions

• Empirical Formula: Smallest whole number ratio of elements in a compound.
• Molecular Formula: Actual ratio of elements in a compound.

Practical Example of Each

• Benzene (C6H6): Empirical formula is CH
• Acetylene (C2H2): Empirical formula is CH
• Carbon Dioxide (CO2): Empirical formula remains CO2
• Empirical formulas are particularly useful for quick calculations without expensive equipment.

Relationships

1. Different compounds can have the same empirical formula (e.g., Benzene and Acetylene both have CH).
2. Molecular Formula = Whole Number Multiple of Empirical Formula (e.g., Benzene empirical formula CH, molecular formula $6 \times \text{CH}$).
3. An individual compound can have the same empirical and molecular formula (e.g., CO2).

Determining Empirical Formula from Percent Composition

1. Given Percent Composition: Assume 100g for easy calculation.
2. Convert Percentages to Grams:
   • Example: 40.92% C = 40.92g C
3. Convert Grams to Moles using molar masses.
4. Divide by the Smallest Mole Value to get smallest whole-number ratios.
5. If necessary, multiply by a common factor to get whole number subscripts.
   • Example: 40.92% C, 4.58% H, 54.50% O (C3H4O3)
   • Convert grams of each to moles, divide by smallest mole value, adjust to whole numbers.

Determining Molecular Formula

1. Known Empirical Formula and Molar Mass Range.
2. Calculate Empirical Formula Mass.
3. Determine the Molecular Formula:
   • Find the multiple by approximating molar mass.
   • Adjust empirical formula by multiple to get molecular formula.

Example: Compound with Empirical Formula C4H5ON2

• Mass: 97.11 g/mol, given molar mass between 180-200 g/mol.
• Double empirical mass to fit the molar mass range: $2 \times 97.11 = 194.22$ g/mol.
• Molecular Formula = C8H10O2N4