1.

0.66g of an organic compound containing C, H and O gave on combustion 0.968g of CO_(2) and 0.792 g of H_(2)O. Calculate the percentage of O in the compound. (C= 12, H= 1, O= 16)

Answer»

Solution :Moles of C in `CO_(2)=1 XX` moles of `CO_(2)`
`=1 xx (0.968)/(44)= 0.022`
Wt. of `C= 0.022 xx 12= 0.264g`
Moles of H in `H_(2)O=2 xx` moles of `H_(2)O`
`=2 xx (0.792)/(18)= 0.088`
Weight of H= `0.088 xx1= 0.088g`
Total weight of C and H in the COMPOUND `=(0.264 + 0.088)G = 0.352g`
`therefore` weight of O in the compound `= (0.66- 0.352)g= 0.308g`
`therefore` % of O in the compound `= (0.308)/(0.66) xx 100= 46.67%`


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