Question

An organic compound contains 4.075 % of H, 24.27% of C, 71.65% of Cl. calculate the empirical formula and molecular formula.the molecular mass of the compound is 98.96g​

Answer 1

Solution :-

Given :-

Percentage composition of carbon ( C ) = 24.27%

Percentage composition of hydrogen ( H ) = 4.075%

Percentage composition of chlorine ( Cl ) = 71.65%

We know :-

Atomic mass of carbon ( C ) = 12g

Atomic mass of hydrogen ( H ) = 1g

Atomic mass of chlorine ( Cl ) = 35.5g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{24.27}{12}= 2.022 \\\\ \bullet \ No \ of \ moles \ of \ hydrogen \ (H)= \dfrac{4.075}{1}= 4.075 \\\\\bullet \ No \ of \ moles \ of \ chlorine \ (Cl) = \dfrac{71.65}{35.5}= 2.018$

Simplest ratio :-

$\longrightarrow \sf \dfrac{2.022}{2.018} \ : \ \dfrac{4.075}{2.018} \ : \ \dfrac{2.018}{2.018} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1$

$\boxed{\bf Empirical \ formula = CH_2Cl}$

Empirical formula mass = 12 + 2 + 35.5

                                       = 49.5g

We know :-

$\sf Molecular \ formula = n \times Empirical \ formula$

Here,

$\longrightarrow \sf n = \dfrac{Molecular \ mass }{Empirical \ formula \ mass} \\\\\longrightarrow n = \dfrac{98.96}{49.5} \\\\\longrightarrow n = 1.99 \approx 2$

Molecular formula = $\sf 2 \times (CH_2Cl)$

                              = $\sf C_2H_4Cl_2$

$\boxed{\bf Molecular \ formula = C_2H_4Cl_2}$