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Chemistry Unit 5 Worksheet 3: Empirical & Molecular Formulas 2025

Master empirical and molecular formulas with our Chemistry Unit 5 Worksheet 3. Includes practice problems and step-by-step solutions, updated for July 2025!

Chemistry Unit 5 Worksheet 3: Empirical & Molecular Formulas

Understanding empirical and molecular formulas is crucial for analyzing chemical compositions. Our Chemistry Unit 5 Worksheet 3, updated for July 2025, provides comprehensive practice to differentiate and calculate these formulas. Perfect for high school and college students, this resource includes detailed explanations and real-world applications.

Key Concepts

  • Empirical Formula: The simplest whole-number ratio of atoms in a compound (e.g., CH2OCH_2O for glucose).
  • Molecular Formula: The actual number of atoms in a molecule (e.g., C6H12O6C_6H_{12}O_6 for glucose).
  • Relationship: Molecular formula = n × empirical formula, where ( n ) is a whole number derived from molar mass ratios.

Steps to Calculate Empirical Formula

  1. Convert masses (or percentages) to moles.
  2. Divide by the smallest number of moles to get ratios.
  3. Adjust to whole numbers.
  4. If given molar mass, find ( n ) to determine the molecular formula.

Practice Problems

Problem 1: Empirical Formula

A compound contains 40% C, 6.7% H, 53.3% O by mass. Calculate its empirical formula.

  • Steps:
    • Assume 100 g: 40 g C, 6.7 g H, 53.3 g O.
    • Moles: C = ( \frac{40}{12} \approx 3.33 ), H = ( \frac{6.7}{1} \approx 6.7 ), O = ( \frac{53.3}{16} \approx 3.33 ).
    • Ratio: Divide by 3.33 → C: 1, H: 2, O: 1.
  • Answer: CH2OCH_2O.

Problem 2: Molecular Formula

The compound above has a molar mass of 180 g/mol. Find the molecular formula.

  • Steps:
    • Empirical formula mass (CH₂O): 12 + 2×1 + 16 = 30 g/mol.
    • ( n = \frac{180}{30} = 6 ).
    • Molecular formula: 6 × (CH₂O).
  • Answer: C6H12O6C_6H_{12}O_6.

Problem 3: Combustion Analysis

A 1.50 g sample of a hydrocarbon produces 4.40 g CO₂ and 2.70 g H₂O. Find the empirical formula.

  • Steps:
    • Moles C: ( \frac{4.40}{44} \times 1 = 0.1 ) mol C.
    • Moles H: ( \frac{2.70}{18} \times 2 = 0.3 ) mol H.
    • Ratio: C: 0.1 ÷ 0.1 = 1, H: 0.3 ÷ 0.1 = 3.
  • Answer: CH3CH_3.

Real-World Applications

  • Pharmaceuticals: Determining drug compositions (e.g., aspirin, C₉H₈O₄).
  • Forensics: Identifying substances in crime scene analysis via empirical formulas.
  • Materials Science: Designing polymers with specific molecular structures.

Tips for Success

  • Use a Calculator: Ensure precision in mole calculations.
  • Double-Check Ratios: Small errors in division can lead to incorrect formulas.
  • Practice Combustion Analysis: Common in lab settings and exams.

Download Now

Click here to download the worksheet (Note: Link to be added). Includes 15 problems, answer keys, and a guide to combustion analysis.

Interactive Element: Try It Yourself

Analyze a compound with 26.7% C, 2.2% H, 71.1% O, and molar mass 90 g/mol. Calculate its empirical and molecular formulas. Share your solution on X with #ChemistryUnit52025!


Keywords: chemistry unit 5 worksheet, empirical formulas, molecular formulas, chemistry practice, 2025 chemistry resources.

Last Updated: July 12, 2025, 03:14 PM +04

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