Question

10. What is the empirical formula of a compound containing 31.0% Ti and 69.0% CI?

Answer 1

Solution :

Let us consider that we have taken 100 g of the compound.

According to question,

Amount of Ti in the compound = 31 g

Amount of Cl in the compound = 69 g

$Number\ of\ moles\ of\ Ti = \frac{Amount\ of\ Ti\ in\ the\ compound }{Molar\ mass\ of\ Ti}$

$or, Number\ of\ moles\ of\ Ti = \frac{31}{47.87} = 0.6476\ mol$

$Number\ of\ moles\ of\ Cl = \frac{Amount\ of\ Cl\ in\ the\ compound }{Molar\ mass\ of\ Cl}$

$or, Number\ of\ moles\ of\ Cl = \frac{69}{35.45} = 1.9464\ mol$

Here we can observe that Ti has less number of moles. So, dividing by least molar quantity for each case.

The Emperical formula we get is : $Ti_{\frac{0.6476}{0.6476}}Cl_{\frac{1.9464}{0.6476}} = TiCl_{3}$

Therefore, the Emperical formula of the compound is TiCl₃