Question 4(ii)SolveA compound has the following percentage composition by mass, carbon 14.4%. hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. (H=1, C=12 and Cl=35.5)

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon14.41214.412\dfrac{14.4}{12}1214.4​ = 1.21.21.2\dfrac{1.2}{1.2}1.21.2​ = 1Hydrogen1.211.21\dfrac{1.2}{1}11.2​ = 1.21.21.2\dfrac{1.2}{1.2}1.21.2​ = 1chlorine84.535.584.535.5\dfrac{84.5}{35.5}35.584.5​ = 2.382.381.2\dfrac{2.38}{1.2}1.22.38​ = 1.98 = 2Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2Hence, empirical formula is CHCl2

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	14.4	12	$\dfrac{14.4}{12}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
Hydrogen	1.2	1	$\dfrac{1.2}{1}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
chlorine	84.5	35.5	$\dfrac{84.5}{35.5}$ = 2.38	$\dfrac{2.38}{1.2}$ = 1.98 = 2

Solved Sample Paper 1 — Question 6

Question 4(ii)