Question

(i) A compound containing carbon, hydrogen and chlorine contains 14.4% of carbon, 1.2% of
hydrogen and 84.4% of chlorine by mass. Determine its empirical formula. (C = 12, H = 1
and Cl = 35.5).​

Answer 1

Answer:

empirical formula is CHCl2. = 2. Therefore, molecular formula =(CHCl2)×2=C2H2Cl4

Explanation:

Answer 2

Answer:

Empirical formula :- CHCl2

Explanation:

carbon (c) =

%weight =14.4 ; atomic weight = 12

%weight /atomic weight = 14.4 /12 = 1.2

hydrogen (H) =

%weight = 1.2 ; atomic weight = 1

%weight / atomic weight = 1.2 /1 = 1.2

chlorine (cl) =

%weight = 84.4 ; atomic weight = 35.5

%weight / atomic weight = 84.4 /35.5 = 2.4 (approximately)

simple ratio :-

carbon = 1.2 / 1.2 = 1

hydrogen = 1.2 / 1.2 = 1

chlorine = 2.4 / 1.2 = 2

Empirical formula :- CHCl2