Anonymous
Anonymous asked in Science & MathematicsChemistry · 4 weeks ago

chemistry help?

When 11.4 g of an organic compound known to be 57.15% C, 4.8% H, and 38.06% O by mass is dissolved in 859.2 g of water, the freezing point is −0.196 ∘C. The normal freezing point of water is 0 ∘C. What is the molecular formula for the organic compound? Assume that the organic compound is a molecular solid and does not ionize in water.

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  • 4 weeks ago

    In the compound, mole ratio C : H : O

    = (57.15/12) : (4.8/1) : (38.06/16)

    = 4.8 : 4.8 : 2.4

    = (4.8/2.4) : (4.8/2.4) : (2.4/2.4)

    = 2 : 2 : 1

    The empirical formula of the compound = C₂H₂O

    Consider the freezing point depression of the solution:

    Freezing point depression, ΔTf = [0 - (-0.196)] °C = 0.196 °C

    van't Hoff factor, i = 1

    Freezing point depression constant of water, Kf= 1.86 °C/m

    ΔTf = i Kf m

    0.196 = 1 × 1.86 × m

    Molality, m = 0.196/1.86 = 0.105 mol/kg

    Molar mass of the organic compound

    = 11.4 g / [(0.105 mol/kg) × (859.2/1000 kg)]

    = 126 g/mol

    Let the molecular formula of the compound = (C₂H₂O)ₙ

    Its molar mass in g/mol:

    (12×2 + 1×2 + 16)n = 126

    42n = 126

    n = 3

    The molecular formula of the compound = C₆H₆O₃

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