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Question

A gaseous hydrocarbon contains 82.76% of Carbon. Given that Its vapor density Is 29. Find its molecular formula. [C = 12, H =1].


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Solution

Determination of the molecular formula from vapour density

  • Step 1: Determination of the percentage of each element in hydrocarbon
    Percentage Cabon present in gaseous hydrocarbon = 82.76%
    Percentage of Hydrogen present in the gaseous hydrocarbon = 100 - 82.76%
    Percentage of Hydrogen present in the gaseous hydrocarbon = 17.24%
  • Step 2: Determination of Empirical formula
S.No. ElementMolecular massPercentage compositionNumber of atomsSimplest ratioNearest whole number rounding off
1C (Carbon)1282.76%82.7612=6.896.896.89=11×2=2
2H (Hydrogen)117.24%17.241=17.2417.246.89=2.52.5×2=5

The empirical formula of gaseous hydrocarbon is = C2H5

  • Step 3: Determination of Empirical formula mass
    The empirical formula mass of C2H5 = 2×12+5×1
    The empirical formula mass of C2H5 = 29
  • Step 4: Determination of molecular mass
    The vapour density = 29
    Molecular mass = 2×Vapourdensity
    Molecular mass = 2×29
    Molecular mass = 58
  • Step 5 Relation between empirical mass and molecular mass
    Molecular mass = n×Empiricalformulamass
    n = MolecularmassEmpiricalformulamass
    n = 5829
    n = 2
  • Step 6: Determination of Molecular formula
    Molecular formula = n×Empiricalformula
    Molecular formula = C4H10

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