1.

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

Answer»

`CHClO`
`CH_(3)ClO`
`C_(2)H_(2)Ocl`
`ClCH_(2)O`

Solution :`{:("ELEMENT"," NO. of moles"," Simple RATIO"),(C 18.5%,18.5//12=1.54,1),(H1.55%,1.55//1=1.55,1),(Cl55.04%,55.04//35.5=1.55,1),(O24.81%,24.81//16=1.55,1):}`
Hence,FORMULA `= CHClO`.


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