Question

An unknown compound on elemental analysis, was found to contain 49.02% C, 2.74% H, and 48.24%CL by mass. its molecular mass was determined as 147.0 calculate the molecular formulat pf the compound.

Answer 1

C_xH_yCl_z;

x:y:z=(0.4902*147/12) : (0.0274*147/1) : (0.4824*147/35.5) = 6:4:2

Simplest formula - C₆H₄Cl₂

Mr(C6H4Cl2)= 147

Answer:  molecular formula of the compound is C6H4Cl2