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:

  • A
    $CHClO$
  • B
    $CH_{2}ClO$
  • C
    $C_{2}H_{2}OCl$
  • D
    $ClCH_{2}O$

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