1.

0.465g of an organic compound upon combustion produced 1.32g of CO_(2) and 0.315g of H_(2)O. 0.2325g of the same compound gave 27.8 mL of dry N_(2) at NTP. If the compound contained C, H and N find its formula

Answer»

Solution :Moles of C in `CO_(2)=1 xx` moles of `CO_(2) = (1.32)/(44)= 0.03`
moles of H in `H_(2)O=2xx` moles of `H_(2)O=2 xx (0.315)/(18)= 0.035`
Moles of N in `N_(2)=2xx` moles of `N_(2)=2 xx (27.8)/(22400)= 0.0025`
Now, 0.0025 mole of N is contained in 0.2325g of the compound.
`therefore` moles of N in 0.465g of the compound `=0.0025 xx (0.465)/(0.2325)=0.005`
`therefore` moles of `C: H: N= 0.030: 0.035 : 0.005`
`=30: 35: 5`
`=6: 7: 1`
`therefore` empirical FORMULA is `C_(6)H_(7)N`


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