Question

A carbon compound on analysis gives the following percentage composition. Carbon 14.5 %, hydrogen 1.8 % chlorine 64.6 %, oxygen 19.24 %. The empirical formula of the compound is:

• A. $C_2{H}_{3}C{l}_3{O}_2$
• B. $C_{2}HC{l}_3{O}_4$
• C. $C_2{H}_{2}C{l}_3{O}_3$
• D. $C_2{H}_{4}C{l}_3{O}_2$

Answer 1

The correct option is A $C_2{H}_{3}C{l}_3{O}_2$

Percentage compostion of the elements present in the compound,
C : H : Cl : O = 14.5 : 1.8 : 64.46 : 19.24

Dividing by molar mass we get the mole ratio as,
C : H : Cl : O = 1.21 : 1.8 : 1.81 : 1.2

Dividing by the smallest number and converting the same to whole numbers, we get the simplest whole number ratio.
C : H : Cl : O = 2 : 3 : 3 : 2