Question

The molecular formula of a compound gives the actual number of atoms of each element present in a molecule of the compound. It is either the same as the empirical formula or a simple multiple of it.
Molecular formula $=(\text{ Empirical formula }{)}_n$
Where, $n$ is a whole number and its value is obtained by dividing the molecular mass of the compound by its empirical formula mass.
Structural formula is that formula which shows in what way different kinds of atoms in a molecule are united or linked together
$\underset{\text{ethyl alcohol}}{H-\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}-\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}-O-H}\underset{\text{dimethyl ether}}{H-\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}-O-\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}-H}$
What is the molecular formula of a compound, its empirical formula is $C{H}_{2}O$ and its molecular weight is $90^{\circ }$ ?

• A. $C_3{H}_6{O}_3$
• B. $C_6{H}_6{O}_3$
• C. $C_4{H}_8{O}_4$
• D. $C_2{H}_6{O}_2$

Answer 1

Answer: (A)

Empirical formula $=C{H}_{2}O$
Molecular weight $=90$
Empirical formula weight of $C{H}_{2}O$
$=12+(1\times 2)+16=30$
$n=\frac{\text{ molecular formula }}{\text{ empirical formula weight }}$
$=\frac{90}{30}=3$
Molecular formula $={\left(C{H}_{2}O\right)}_3=C_3{H}_6{O}_3$