Question

An organic compound contains 4.7% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96% find its,
(a) Empirical formula (b) Molecular formula

Answer 1

Let total mass of the molecule = 100g

Mass of C in the molecule = 24.27g

Mass of H in the molecule = 4.07g

Mass of Cl in the molecule =71.65 g

Number of moles of C in molecule = $\frac{24.27}{molarmassofcarbon}=\frac{24.27}{12}=2.02$

Number of moles of H in molecule = $\frac{4.07}{molarmassofhydrogen}=\frac{4.07}{1}=4.07$

Number of moles of Cl in molecule = $\frac{71.65}{molarmassofnitrogen}=\frac{71.65}{35.45}=2.02$

Ratio of the number of moles of C, H & Cl

i.e.

C : H : Cl = 2.02 : 4.07 : 2.02 = 1 : 2 : 1

So the empirical formula will be: $C{H}_{2}Cl$

Empirical formula mass = 12.01+2+35.45 = 49.5 g

Now

$n=\frac{Molecularmass}{empiricalformulamass}=\frac{98.96}{49.5}=2$

$Molecularformula=n\times empiricalformula=2\left(C{H}_{2}Cl\right)=C_2{H}_{4}C{l}_2$