Question

What is the molecular formula of a compound with the empirical formula CaCl2 and with a molecular molar mass of 330g?

Answer 1

We need to find the Molecular formula of a Compound , which has an Empirical Formula of CaCl₂ . According to Question , the molar mass of the Compound is 330g.

• Firstly we need to find the Empirical Formula Mass of the Compound , Which will be 111g . As the individual masses of Ca is 40g/mol and Cl is 35.5 g/mol .
• The next step is to divide , the molecular mass by Empirical Formula mass , that will be ,

$\sf\dashrightarrow n =\dfrac{Molecular\ Mass}{Empirical\ Mass}= \dfrac{330g}{111g}= \red{3}$

Now , let us say if we have the empirical formula of Compound as , $\sf A_xB_y$ , and we got the value of n as t . Then the Compound will be ,

$\sf\dashrightarrow Compound =\pink{ (A_xB_y)_t }$

Similarly here , the Original formula of the Compound will be ,

$\sf\dashrightarrow Compound = (CaCl_2)_3 \\\\\sf\dashrightarrow Compound = (Ca_3Cl_{(2\times3)}) \\\\\sf\dashrightarrow \boxed{\pink{\sf Compound = Ca_3Cl_6 }}$