Question

Empirical formula of a compound $A_2{B}_4$. If its empirical formula weight is half of its vapour density. Determine the molecular formula of the compound.

• A. $A_4{B}_8$
• B. $A_8{B}_{16}$
• C. $A_2{B}_4$
• D. $A_3{B}_6$

Answer 1

The correct option is B $A_8{B}_{16}$

For the compound, $A_2{B}_4$,

Empirical weight $=\frac{1}{2}\times V.D$

where, $V.D=$ Vapour density

Also, molecular weight $=2\times V.D$

$\therefore$ Molecular weight $=2\times 2\times$ Empirical weight $=4\times$ Empirical weight

$\therefore$ $n=4$

Hence, the molecular formula is $(A_2{B}_4{)}_4=$$A_8{B}_{16}$.