• # question_answer If a compound on analysis was found to contain C = 18.5%, H = 1.55%, Cl = 55.04% and O = 24.81%, then its empirical formula is [AIIMS 1998] A) $CHClO$ B) $C{{H}_{2}}ClO$ C) ${{C}_{2}}{{H}_{2}}OCl$ D) $ClC{{H}_{2}}O$

Elements        No. of moles        Simple ratio C = 18.5%         18.5/12      Þ1.54      1 H = 1.55%         1.55/1        Þ1.55      1 Cl = 55.04%      55.04/35.5 Þ1.55       1 O = 24.81%      24.81/16     Þ1.55      1 Hence, formula = $CHClO$.