An organic compound on analysis give the following percentage composition. C-57.8% H-3.6% o-? The vepure density of the compound was found to be 83.find the molecular formula of the compound
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So, we can deduce the empirical formula of the organic compound = C4H3O2.
The empirical Formula mass = 4x12 + 3x1 + 2x16 = 83 u.
The molecular mass of compound can be calculated as:
Molecular mass = 2 x Vapour Density = 2 x 83 = 166 u.
n = Molecular mass / Empirical Formula Mass = 166/83 = 2.
Molecular Formula of the compound = n(Empirical Formula) = 2(C4H3O2) = C8H6O4.
The empirical Formula mass = 4x12 + 3x1 + 2x16 = 83 u.
The molecular mass of compound can be calculated as:
Molecular mass = 2 x Vapour Density = 2 x 83 = 166 u.
n = Molecular mass / Empirical Formula Mass = 166/83 = 2.
Molecular Formula of the compound = n(Empirical Formula) = 2(C4H3O2) = C8H6O4.
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