Question

An organic compound was analysed by dumas method. 0.45 gm of the compound on combustion gave 48.6 ml nitrogen at `27^@C` and 756 mm pressure. Calculate the percentage composition of the compound.

Answer 1

`{:(V_1=48.6ml,,V_2=?),(P_1=756mm,,P_2=760mm):}`
`T_1=27+273=300K,T_2=273K`
Applying the general gas equation `(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`
we get The volume of nitrogen at STP,
`V_2=(P_1V_1)/(T_1)(T_2)/(T_1)`
`=(756xx48.6)/(300)xx(273)/(760)`
`=43.99ml`
Mass of organic compound `=0.45gm`
Precentage of nitrogen in the compound
`=(28)/(22400)xx(V_2)/(W)xx100`
`=(28)/(22400)xx(43.99)/(0.45)xx100`
`=12.22`