A Compound on analysis gave the following percentage composition C=54.55%, H=9.09%, O=36.36%. Determine the empirical formula of the compound.
suvamswagatpanpb7045:
C6H12o6
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HEYA!!!
HERE IS YOUR ANSWER,
=> We can deduce the empirical formula of the organic compound = C4H3O2.
=> The empirical Formula mass = 4 x 12 + 3 x 1 + 2 x 16 = 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.
HOPE IT HELPS YOU,
THANK YOU.☺️
HERE IS YOUR ANSWER,
=> We can deduce the empirical formula of the organic compound = C4H3O2.
=> The empirical Formula mass = 4 x 12 + 3 x 1 + 2 x 16 = 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.
HOPE IT HELPS YOU,
THANK YOU.☺️
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