Question

The molecular mass of a compound having empirical formula ${C_2}{H_5}O$ is 90. The molecular formula of the compound is:
A.${C_6}{H_{15}}{O_3}$
B.${C_4}{H_{10}}{O_2}$
C.${C_2}{H_5}O$
D.${C_3}{H_6}{O_3}$

Answer 1

Empirical formula is basically the simplest mathematical representation of the ratios of the constituent elements of any given compound.
Formula Used:
$ratio = \dfrac{{Mol.{\text{ }}mass{\text{ }}of{\text{ }}compound}}{{Mol.{\text{ }}mass{\text{ }}of{\text{ }}empirical{\text{ }}compound}}$

Complete step by step answer:
In order to find the molecular formula of the given compound with an empirical formula of ${C_2}{H_5}O$ and having a molecular mass of 90, we need to follow the following steps:
Calculate the total mass of the empirical formula of the given compound. The empirical formula is ${C_2}{H_5}O$, which means that there are 2 carbon atoms, 5 hydrogen atoms and 1 oxygen atom. The molecular mass of this empirical formula can be calculated as:
${m_e}$= 2(mol. mass of carbon) + 5(mol. mass of hydrogen) + 1(mol. mass of oxygen)
= 2 (12) + 5 (1) + 1(16)
= 45
Hence the molecular weight of the empirical formula is 45.
Calculating the ratio between molecular weight of the empirical compound and the actual compound can be found to be:
$ratio = \dfrac{{Mol.{\text{ }}mass{\text{ }}of{\text{ }}compound}}{{Mol.{\text{ }}mass{\text{ }}of{\text{ }}empirical{\text{ }}compound}}$= 90/45 = 2
In order to find the actual molecular formula, we must multiply the individual atomic significance of the atoms in the empirical formula with the ratio found above. Hence, the molecular formula of the actual compound is ${C_2}{H_5}O$x 2 = ${C_4}{H_{10}}{O_2}$

Hence, Option B is the correct option.

Note:
Sometimes, the value of the ratio found may not be a whole number. In such cases, we must approximate the value of the ratio to the nearest decimal.