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0.42g of an organic compound containing ...

0.42g of an organic compound containing C, H, O and N gave on combustion 0.924g of `CO_(2)` and 0.243g of water. 0.208 g of the substance when distilled with NaOH gave `NH_(3)`, which required 30 mL of (N/20) `H_(2)SO_(4)` solution for neutralization. Calculate the amount of each element in 0.42g of the compound.

Text Solution

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Carbon Moles of C in `CO_(2)=1 xx` moles of `CO_(2)`
`=1 xx (0.924)/(44)= 0.021`
Weight of `C= 0.021 xx 12= 0.252g`
Hydrogen Moles of H in `H_(2)O = 2 xx` moles of `H_(2)O`
`=2 xx (0.243)/(18)= 0.027`
Weight of `H= 0.027 xx 1= 0.027g`
Nitrogen m.e. of `H_(2)SO_(4)= (1)/(20) xx 30= 1.5`
`therefore` m.e. of `NH_(3)= 1.5`
Eq of `NH_(3)= (1.5)/(1000)= 0.0015`
moles of `NH_(3)= 0.0015`
Now moles of N in `NH_(3)=1 xx` moles of `NH_(3)= 0.0015`
Weight of N in 0.208g of the compound = `0.0015 xx 14`
=0.021g
`therefore` weight of N in 0.42g of the compound `=(0.021)/(0.208) xx 0.42` = 0.042g
Oxygen `therefore` total weight of C, H and `N= (0.252 + 0.027 + 0.042)g`
=0.321g
`therefore` weight of `O= 0.42- 0.321`
= 0.099g
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