Question

A substance was known by its mode of synthesis to contain 10 atoms of carbon per molecule along with unknown number of atoms of chlorine hydrogen and oxygen. Analysis showed 60.5% carbon, 5.55% hydrogen, 16.10%j oxygen and 17.9% chlorine. The Empirical formula of the compound is ______.

Options

• \[\ce{C10H8OCl2}\]

• \[\ce{C10H11O2Cl}\]

• \[\ce{C10H10OCl}\]

• \[\ce{C10H12O2Cl}\]

Answer 1

A substance was known by its mode of synthesis to contain 10 atoms of carbon per molecule along with unknown number of atoms of chlorine hydrogen and oxygen. Analysis showed 60.5% carbon, 5.55% hydrogen, 16.10%j oxygen and 17.9% chlorine. The Empirical formula of the compound is \[\underline{\ce{C10H11O2Cl}}\].

Explanation:

Element	Relative number of atom	Simplest ratio
C = 60.5%	`60.5/12 = 5`	`5/0.5 = 5`
H = 5.55%	`5.55/1 = 5.55`	`5.55/0.5 = 11`
O = 16.1%	`16.10/16 = 1.00`	`1.00/0.5 = 2`
Cl = 17.9%	`17.9/35.5 = 0.5`	`0.5/0.5 = 1`