Understanding Molecular and Empirical Formulas

In chemistry, an empirical formula represents the simplest whole-number ratio of elements in a compound, while a molecular formula shows the actual number of each atom in a molecule. Determining molecular formulas from empirical formulas is an essential skill in chemical analysis. Below are three diverse examples that demonstrate this concept in practical contexts.

Example 1: Common Table Sugar (Sucrose)

Understanding the composition of everyday substances can lead to insights in nutritional science. Sucrose, commonly known as table sugar, is a widely used carbohydrate in food. Its empirical formula is C(_{12})H(_{22})O(_{11}). To find its molecular formula:

• Step 1: Calculate the molar mass of the empirical formula:

  • C: 12.01 g/mol × 12 = 144.12 g/mol
  • H: 1.008 g/mol × 22 = 22.176 g/mol
  • O: 16.00 g/mol × 11 = 176.00 g/mol
  • Total = 144.12 + 22.176 + 176.00 = 342.296 g/mol

• Step 2: Determine the molar mass of the molecular formula:

  • The molecular formula for sucrose is C(_{12})H(_{22})O(_{11}) which has a molar mass of approximately 342.3 g/mol.

• Conclusion: Since the empirical formula represents the actual number of atoms, the molecular formula for sucrose is C(_{12})H(_{22})O(_{11}).

Notes:

• Sucrose is a disaccharide composed of glucose and fructose.
• It’s important for energy supply in the human body.

Example 2: Benzene

Benzene is a fundamental organic chemical compound used in various industries, including pharmaceuticals and plastics. Its empirical formula is C(_{6})H(_{6}). To derive its molecular formula:

• Step 1: Determine the empirical formula’s molar mass:

  • C: 12.01 g/mol × 6 = 72.06 g/mol
  • H: 1.008 g/mol × 6 = 6.048 g/mol
  • Total = 72.06 + 6.048 = 78.108 g/mol

• Step 2: Benzene’s known molar mass is approximately 78.11 g/mol.

• Conclusion: The empirical formula C(_{6})H(_{6}) is also the molecular formula for benzene.

Notes:

• Benzene is known for its aromatic properties and is a carcinogen.
• Its structure consists of a hexagonal ring with alternating double bonds.

Example 3: Ethylene Glycol

Ethylene glycol is a key ingredient in antifreeze and is used in the production of plastics. Its empirical formula is C(_{2})H(_{6})O. To find the molecular formula:

• Step 1: Calculate the empirical formula’s molar mass:

  • C: 12.01 g/mol × 2 = 24.02 g/mol
  • H: 1.008 g/mol × 6 = 6.048 g/mol
  • O: 16.00 g/mol × 1 = 16.00 g/mol
  • Total = 24.02 + 6.048 + 16.00 = 46.068 g/mol

• Step 2: The molecular formula of ethylene glycol is C(_{2})H(_{6})O, which has a molar mass of approximately 62.07 g/mol.

• Conclusion: To obtain the molecular formula:

  • Determine the ratio (62.07 g/mol / 46.068 g/mol) = 1.35, indicating the empirical formula needs to be multiplied by 2.
  • Therefore, the molecular formula is C(_{4})H(_{12})O(_{2}).

Notes:

• Ethylene glycol is toxic and should be handled with care.
• Its properties make it useful in various industrial applications.

By understanding these examples of determining molecular formulas from empirical formulas, you can better grasp the relationship between the two and apply this knowledge in various chemical contexts.