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.

0.42g of an organic compound containing C, H, O and N gave on combusti

1.	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.
Answer» SOLUTION :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