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Chemistry

A compound with empirical formula AB has vapour density three times its empirical formula weight. Find the molecular formula.

Mole Concept

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Answer

Given,

Empirical formula = AB

V.D. = 3 x Empirical formula weight

Hence, Empirical formula weight = V.D.3\dfrac{\text{V.D.}}{3}

and we know, Molecular weight = 2 x V.D.

Substituting in the formula for n we get,

n=Molecular weightEmpirical formula weightn= 2 x V.D.V.D.3n= 3 x 2 x V.D.V.D.n=6\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \dfrac{\text{ 2 x V.D.}}{\dfrac{\text{V.D.}}{3}} \\[0.5em] \text{n} = \dfrac{\text{ 3 x 2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 6

∴ Molecular formula = n[E.F.] = 6[AB] = A6B6

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