Question

A compound contains 10.0 4% of carbon, 0.84 % of hydrogen and 89.1 2% chlorine. Determine its empirical formula .

Answer 1

Given :-

Percentage composition of carbon (C) = 10.04%

Percentage composition of hydrogen (H) = 0.84%

Percentage composition of chlorine (Cl) = 89.12%

To Find :-

Empirical formula

Solution :-

We know :-

Atomic mass pf carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of chlorine (Cl) = 35.5g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{10.04}{12}$

                                           $= \sf 0.84$

$\bullet \ \sf No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{0.84}{1}$

                                               $= \sf 0.84$

$\bullet \sf \ No \ of \ moles \ of \ chlorine \ (Cl) = \dfrac{89.12}{35.5}$

                                              $= \sf 2.51$

Simplest ratio :-

$\longrightarrow\sf \dfrac{0.84}{0.84} \ : \ \dfrac{0.84}{0.84} \ : \ \dfrac{2.51}{0.84} \\\\\longrightarrow 1 \ : \ 1 \ : 3$

$\underline{\boxed{\bf Empirical \ formula = CHCl_3}}$