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Question: A compound of \(X\) and \(Y\) has the empirical formula \(X{Y_2}\). Its vapor density is equal to it...

A compound of XX and YY has the empirical formula XY2X{Y_2}. Its vapor density is equal to its empirical formula weight. Determine its molecular formula.

Explanation

Solution

Molecular weight is related to vapor density according to the relation :
Molecular Weight = 2× Vapour densityMolecular{\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 SO{\text{SO}}and the empirical formula of disulfur dioxide will also be SO{\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 SO{\text{SO}}and for disulfur dioxide is S2O2{{\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,
Molecular⁡ weight⁡ = 2× vapour⁡density \operatorname{Molecular} {\text{ }}\operatorname{weight} {\text{ = 2}} \times {\text{ }}\operatorname{vapour} density{\text{ }}
=2×empirical⁡formulaweight= 2 \times \operatorname{empirical} formula weight
It is given that vapor density is equal to empirical formula weight.
That is, vapour⁡density = empirical⁡formulaweight\operatorname{vapour} density{\text{ = }}\operatorname{empirical} formula weight .
Now, as
molecular⁡weight = empirical ×n⁡formulaweight\operatorname{molecular} weight{\text{ = }} empirical{\text{ }} \times \operatorname{n} formula weight
2×empirical⁡formulaweight = empirical⁡formulaweight × n 2 \times \operatorname{empirical} formula weight{\text{ = }}\operatorname{empirical} formula weight{\text{ }} \times {\text{ n }}
Comparing , we get n=2n = 2
Hence, molecular formula
=(empirical⁡formula)×n= \left( {\operatorname{empirical} formula} \right) \times n
=(XY2)×2= \left( {X{Y_2}} \right) \times 2
=X2Y4= {X_2}{Y_4}
Therefore , the answer is X2Y4{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.