An organic compound consists of C, H and N. 0.5g of the compound, on combustion, forms 1.44g of `CO_(2)` and 0.3785g of water, find the molecular fomrula of the compound if its vapour density is 53.35.

Explanation

Molecular mass of the compound=2`xx`vapour density off the compound=`2xx53.5=107`
As given in the question, .5g of the compound, on combustion, forms 1.44g of `CO_(2)` and 0.3785g of water.
`thereforeCO_(2)` produced on combustion`=(1.44)/(0.5)xx107=308.16g`
`thereforeH_(2)O` produced on combustion`=(0.3785)/(0.5)xx107=81g`
No. of moles of C-atoms in 308.16g of
`CO_(2)=(1xx308.16)/(44)=7`
No. of moles of H-atoms present in 81g of
`H_(2)O=(2xx81)/(18)=9`
`therefore` Number of C and H-atoms present in each molecule of the compound are 7 and 9 respectively.
Let, the number of N-atoms in each molecule of the compound be x, then molecular mass
`=7xx12+9xx1+x xx14=93+14x`
`therefore 93+14x=107` or, x=1
`therefore` Molecular formula fo the compound: `C_(7)H_(9)N`