Mole Concept and Stoichiometry — Question 20

Given,

Empirical formula = AB

V.D. = 3 x Empirical formula weight

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

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

Substituting in the formula for n we get,

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

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