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General ChemistryHard

Determining molecular formula

A compound contains: C 54.52%, H 9.15%, O 36.33%. Molar mass is 132.16 g/mol. Determine empirical and molecular formulas.
Atomic masses: C=12.01, H=1.008, O=16.00.
Given data
C: 54.52%H: 9.15%O: 36.33%M = 132.16 g/mol
Review the theory: Chimica Generale
Steps
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  1. Calculate moles of C in 100 g: n(C) = 54.52/12.01.
  2. Calculate moles of H in 100 g: n(H) = 9.15/1.008.
  3. Calculate moles of O in 100 g: n(O) = 36.33/16.00.
  4. Divide by smallest (2.271). C ratio: 4.539/2.271.
  5. H ratio: 9.077/2.271.
  6. O ratio: 2.271/2.271.
  7. Empirical formula mass C₂H₄O = 2×12.01 + 4×1.008 + 16.00.
  8. n = Molecular / Empirical mass = 132.16/44.05.
Full worked solution
  1. Calculate moles of C in 100 g: n(C) = 54.52/12.01.
    54.52/12.0154.52/12.01
    n(C) = 4.539 mol.
  2. Calculate moles of H in 100 g: n(H) = 9.15/1.008.
    9.15/1.0089.15/1.008
    n(H) = 9.077 mol.
  3. Calculate moles of O in 100 g: n(O) = 36.33/16.00.
    36.33/16.0036.33/16.00
    n(O) = 2.271 mol.
  4. Divide by smallest (2.271). C ratio: 4.539/2.271.
    4.539/2.2714.539/2.271
    C ratio = 2.00.
  5. H ratio: 9.077/2.271.
    9.077/2.2719.077/2.271
    H ratio = 4.00.
  6. O ratio: 2.271/2.271.
    Empirical formula: C₂H₄O.
  7. Empirical formula mass C₂H₄O = 2×12.01 + 4×1.008 + 16.00.
    2⋅12.01+4⋅1.008+16.002\cdot12.01 + 4\cdot1.008 + 16.00
    M(C₂H₄O) = 44.05 g/mol.
  8. n = Molecular / Empirical mass = 132.16/44.05.
    132.16/44.05132.16/44.05
    n = 3.00 → molecular: C₆H₁₂O₃.
Result:Empirical: C₂H₄O — Molecular: C₆H₁₂O₃.
TheoryChimica GeneraleFormula sheetChemistry

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