1.

60g ofcompound on analysis gave C= 24g, H= 4 g and O = 32g. If Empirical formula is

Answer»

`C_(2)H_(4)O_(2)`
`C_(2)H_(2)O`
`CH_(2)O_(2)`
`CH_(2)O`

Solution :`{:("Element"," NO. of moles"," Simple RATIO"),(C =24,24//12=2,1),(H=4,4//1=4,2),(O=32,32//16,1):}`
Therefore `CH_(2)O`.


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