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A carbon compound on analysis gave the f...

A carbon compound on analysis gave the following percentage composition, carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.

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`{:("Symbol of","% com-","Atomic","Atomic","Simple"),("the element","position","weight","ratio","ratio"),(C,14.5,12,(14.5)/(12) = 1.21,(1.21)/(1.2) = 1 xx 2 = 2),(H,1.8,1,(1.8)/(1) = 1.8,(1.8)/(1.2) = 1.5 xx 2 = 3),(Cl,64.46,35.5,(64.46)/(35.5) = 1.81,(1.81)/(1.2) = 1.5 xx 2 = 3),(O,19.24,16,(19.24)/(16) = 1.2,(1.2)/(1.2) = 1 xx 2 = 2):}`
`therefore` Empirical formula of the compound `= C_(2)H_(3) Cl_(3) O_(2)`
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