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

Calculation of percentage of carbon and hydrogen :
Percentage of carbon `=(12)/(44 ) xx ("Mass of "CO_2 )/( "Mass of compound") xx 100`
` = (12)/(44 ) xx ( 0.792 )/( 0.45 ) xx 100 =48`
Percentage of hydrogen `= (2)/(18) xx ("Mass of "H_2 O ) /( "Mass of compound ")xx 100`
`= (2)/(18) xx ( 0.324 )/( 0.45 ) xx 100 =8`
Calculation of percentage of nitrogen :
Mass of the compound = 0.24 g
Volume of `H_2 SO_4` taken = 50.0 `cm^3`
Normality of `H_2 SO_4 = 1/ 4 N`
Volume of NaOH required = 77.0 `cm^3`
Normality of `NaOH =1/(10) N`
Suppose V `cm^3` of `H_2 SO_4` react with NaOH.
` therefore 1/4 xx V = (1)/(10) xx 77 " " (N_1 V_1 = N_2 V_2)`
or` V=1/10 xx (77 xx 4 )/(1) = 30.8 cm^3`
` therefore `The volume of `H_2 SO_4` neutralised by `NH_3 `= 50.0 - 30.8 = 19.2 `cm^3`
Number of gram equivalents of `NH_3` evolved
Number of gram equivalents of `H_2 SO_4` neutralised by `NH_3`
` = (1/4 xx 19.2 )/( 1000 )`
One gram equivalent of `NH_3` contains 14 g of nitrogen.
` therefore ` Mass of nitrogen present in `NH_3` evolved
` = ( 1/4 xx 19.2 )/( 1000 ) xx 14 = 0.0672 g `
This mass of nitrogen is present in 0.24 g of compound.
` therefore ` Percentage of nitrogen `= (0.0672 )/( 0.24 ) xx 100 =28`
Calculation of percentage of oxygen :
Percentage of oxygen `= 100-(% `of `C+% of H+ % of N)`
`= 100 - (48 + 8 +28) = 16 `
Calculation of empirical formula :

` therefore ` Empirical formula `= C_4 H_8 N_2 O .`