1.

0.16g of an organic compound, on complete combustion, produced 0.44g of CO_(2) and 0.18g of H_(2)O. Calcuate the percentage of carbon and hydrogen in the organic compounds

Answer»

Solution :Moles of C in `CO_(2)=1 xx` moles of `CO_(2)`
`=1 xx (0.44)/(44)= 0.01`
Weight of C= moles of C `xx` at. Wt . of C
`=0.01 xx 12 = 0.12g`
Moles of H in `H_(2)O=2 xx` moles of `H_(2)O`
`=2 xx (0.18)/(18)= 0.02`
weight of H= moles of `H xx` at. wt. of H
`= 0.02 xx 1 = 0.02g`
`THEREFORE` % of `C= (0.12)/(0.16)xx 100= 75%`
and % of `H= (0.02)/(0.16) xx 100= 12.5%`


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