A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M `H_(2)SO_(4)`. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. What would be the percentage composition of nitrogen in the compound?

Explanation

In the present case,
Mass of organic compound (w) = 0.50 g
Volume of `H_2 SO_4 `taken `(V_1)` = 50 mL
Normality of `H_2 SO_4 (N_1)` = 1 N
`(0.5 M H_2 SO_4 = 1 N H_2 SO_4`,
Mol. mass of `H_2 SO_4 = 2 xx `Eq. mass)
Volume of NaOH required `(V_2) = 60 mL `
Normality of NaOH `(N_2) = 0.5 N`
`( therefore 0.5 M NaOH = 0.5 N NaOH`
Molar mass of NaOH = Eq. mass)
Suppose v mL of `H_2 SO_4 `react with alkali. Therefore,
`N_1 V_1= N_2 V_2`
` 1 xx V = 0.5 xx 60`
or ` v= (0.5 xx 60)/(1 ) = 30 mL`
` therefore ` The volume of `H_2 SO_4` that reacted with
`NH_3 = 50-30 = 20 mL `
Gram equivalents of `NH_3` evolved
= Gram equivalent of `H_2 SO_4` reacted with `NH_3 `
` =(N xx V)/(1000) xx (1 xx 20)/(1000 )`
` therefore `Mass of nitrogen present in NH3 evolved
`=(1 xx 20 )/( 1000) xx 14 = 0.28 g`
` therefore ` Percentage of N in the given compound
` = ( 0.28 )/(0.50 ) xx 100 =56.0`