Question

The emperical formula of a compound is $C_2{H}_{5}O$ and its vapor density is $45$ . What is the molecular formula of the compound?

• A. $C_2{H}_{5}O$
• B. $C_6{H}_{13}{O}_3$
• C. $C_4{H}_{10}{O}_2$
• D. $C_6{H}_{15}{O}_3$

Answer 1

Answer: (C)

Given, empirical formula of compound $=C_2{H}_{5}O$
Vapour density $=45$
Molecular formula $=?$
Empirical formulas show the simplest whole number ratio of atoms in a compound.
Molecular formula show the number of each type of atom in a molecule.
Relation between empirical formula and molecular formula.
Molecular formula $=n$ (empirical formula)
$n=\frac{MM\text{ (molar mass) }}{EFM\text{ (empirical formula molar mass) }}$
Molecular weight $=2\times$ vapour density $=2\times 45=90$
Let, the molecular formula be $C_{2x}{H}_{5x}{O}_x$.
Therefore,
$n=\frac{90(\text{ molecular weight })}{47(EFM)}=1.47\approx 2$
Value of $x=2$
Molecular formula
$={\left(C_2{H}_{5}O\right)}_2=C_4{H}_{10}{O}_2$