Lecture Notes: Finding Empirical and Molecular Formulas

Overview

• Objective: Learn how to determine the empirical and molecular formulas of a compound given either percent composition or grams of each element.
• Applications: Useful in chemistry for determining the composition of molecular structures.

Part A: Finding the Empirical Formula from Percent Composition

Problem Setup

• Given percentages:
  • Carbon: 52.14%
  • Hydrogen: 13.13%
  • Oxygen: 34.73%
• Assume 100 grams of compound:
  • Convert percentages to grams:
    • Carbon: 52.14g
    • Hydrogen: 13.13g
    • Oxygen: 34.73g

Conversion to Moles

• Carbon:
  • Molar Mass: 12.01 g/mol
  • Moles: ( \frac{52.14}{12.01} = 4.34 ) moles
• Hydrogen:
  • Molar Mass: 1.008 g/mol
  • Moles: ( \frac{13.13}{1.008} = 13.026 ) moles
• Oxygen:
  • Molar Mass: 16.00 g/mol
  • Moles: ( \frac{34.73}{16} = 2.171 ) moles

Determine Empirical Formula

• Identify smallest mole value: Oxygen 2.171 moles
• Divide each by smallest:
  • Carbon: ( \frac{4.34}{2.171} \approx 2 )
  • Hydrogen: ( \frac{13.026}{2.171} \approx 6 )
  • Oxygen: ( \frac{2.171}{2.171} = 1 )
• Empirical Formula: ( C_2H_6O )

Part B: Finding the Molecular Formula

Given

• Molar Mass of compound: 138.204 g/mol
• Empirical formula mass ( C_2H_6O ):
  • Calculation: ( 2(12.01) + 6(1.008) + 16 = 46.068 ) g/mol

Calculate Molecular Formula

• Divide molecular mass by empirical formula mass:
  • ( \frac{138.204}{46.068} = 3 )
• Multiply subscripts by 3:
  • Molecular Formula: ( C_6H_{18}O_3 )_

Part C: Finding Empirical Formula from Grams

Problem Setup

• Given grams:
  • Carbon: 20.32g
  • Hydrogen: 5.12g
  • Nitrogen: 7.9g

Conversion to Moles

• Carbon:
  • Molar Mass: 12.01 g/mol
  • Moles: ( \frac{20.32}{12.01} = 1.692 ) moles
• Hydrogen:
  • Molar Mass: 1.008 g/mol
  • Moles: ( \frac{5.12}{1.008} = 5.079 ) moles
• Nitrogen:
  • Molar Mass: 14.01 g/mol
  • Moles: ( \frac{7.9}{14.01} = 0.5639 ) moles

Determine Empirical Formula

• Smallest mole value: Nitrogen 0.5639 moles
• Divide each by smallest:
  • Carbon: ( \frac{1.692}{0.5639} \approx 3 )
  • Hydrogen: ( \frac{5.079}{0.5639} \approx 9 )
  • Nitrogen: ( \frac{0.5639}{0.5639} = 1 )
• Empirical Formula: ( C_3H_9N )

Part D: Finding the Molecular Formula from Empirical Formula

Given

• Molar Mass: 236.448 g/mol
• Empirical formula mass ( C_3H_9N ):
  • Calculation: ( 3(12.01) + 9(1.008) + 14.01 = 59.112 ) g/mol

Calculate Molecular Formula

• Divide molecular mass by empirical formula mass:
  • ( \frac{236.448}{59.112} = 4 )
• Multiply subscripts by 4:
  • Molecular Formula: ( C_{12}H_{36}N_4 )

Conclusion

• Empirical Formula: Reflects the simplest ratio of elements in a compound.
• Molecular Formula: Reflects the actual number of atoms of each element in a compound.
• Process involves converting mass/percent to moles, standardizing to smallest mole value, and using molar mass to find the molecular formula.