Question

The empirical formula of a compound is $A{B_2}$. If its empirical formula weight is $\dfrac{2}{3}$ times of its vapor density, calculate the molecular formula of the compound.
A. ${A_3}{B_6}$
B. ${A_2}{B_4}$
C. ${A_6}{B_{12}}$
D. ${A_4}{B_8}$

Answer 1

Hint: The molecular formula of the compound is found by the empirical formula weight. It depends on the vapor density. Actual molecular formula is the whole number of ratios between the elements. It comes from the molecular mass and empirical formula. It is shown by the simplest ratio of elements of the compound.

Complete step by step solution:
Molecular formula describes the type and number of atoms in a single molecule of compound. It represents the simplest whole integer ratio of atoms. It indicates the ratio of all the atoms in a molecule.
Empirical formula is the simplest whole number ratio of elements. It is represented by the number of grams of each element, given the problem. Molecular formula calculates by the molecular weight and empirical weight.
The given values are as below:
Empirical weight $ = \dfrac{2}{3}$ (V.D)
Molecular weight= 2 (V.D)
Now we put the above values in the formula:
Therefore, $n = \dfrac{{2 \times V.D}}{{\dfrac{2}{3} \times V.D}} = 3$
Put the value in the empirical formula in the compound of $A{B_2}$,
Molecular formula $ = {(A{B_2})_3} = {A_3}{B_6}$

Thus, option (A) is the correct answer.

Note: The compound has the vapor density equal to its formula weight and then finds its molecular formula. A compound with formula of AB which has vapor density 3 times its empirical formula weight. We remember that the empirical formula is equal to vapor density. Vapor density is half of the molecular mass.