Question

A compound of $X$ and $Y$ has the empirical formula $X{Y_2}$. Its vapor density is equal to its empirical formula weight. Determine its molecular formula.

Answer 1

Hint:Molecular weight is related to vapor density according to the relation :
$Molecular{\text{ Weight = 2}} \times {\text{ Vapour density}}$

Complete step by step answer:
The empirical formula is the simple whole – number ratio of the elements that are present in the compound. Empirical formula does not mention the arrangement or the number of atoms in the molecule. For example, the empirical formula of sulfur monoxide is ${\text{SO}}$and the empirical formula of disulfur dioxide will also be ${\text{SO}}$. Whereas, the molecular formulas of both the compounds are different.
The molecular formula represents the number of each type of atom present in the molecule. As you inferred above that the molecular formula of sulfur monoxide is ${\text{SO}}$and for disulfur dioxide is ${{\text{S}}_{\text{2}}}{{\text{O}}_{\text{2}}}$but the empirical formula for both of the compounds is the same. The molecular formula of the compound can be the empirical formula or it can be the multiple of the empirical formula. So, from the example of sulfur monoxide and sulfur dioxide, we can say that the empirical formula for different types of compounds can be the same as in this case.
Vapor density is defined as the ratio of weight of a certain volume of gas or vapor compound to the weight of an equal volume of hydrogen.
As we know,
$\operatorname{Molecular} {\text{ }}\operatorname{weight} {\text{ = 2}} \times {\text{ }}\operatorname{vapour} density{\text{ }}$
$ = 2 \times \operatorname{empirical} formula weight$
It is given that vapor density is equal to empirical formula weight.
That is, $\operatorname{vapour} density{\text{ = }}\operatorname{empirical} formula weight $ .
Now, as
$\operatorname{molecular} weight{\text{ = }} empirical{\text{ }} \times \operatorname{n} formula weight $
$2 \times \operatorname{empirical} formula weight{\text{ = }}\operatorname{empirical} formula weight{\text{ }} \times {\text{ n }}$
Comparing , we get $n = 2$
Hence, molecular formula
$ = \left( {\operatorname{empirical} formula} \right) \times n$
$ = \left( {X{Y_2}} \right) \times 2$
$ = {X_2}{Y_4}$
Therefore , the answer is ${X_2}{Y_4}$.

Note:
Empirical formulas are the simplest form of notation. They provide the lowest whole number ratio of the elements of a compound whereas molecular formulas do not provide information about the absolute number of atoms in a compound.