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 neutralization. Find the percentage composition of nitrogen in the compound.
Strategy:
Step `1`. Convert molarity into normality using the relation
Normality `(N) = n xx Molarity (M)`
where `n` factor is either the acidity of base or basicity of acid.
Step `2`. Calculate the milliequivalents of `NaOH` which is equal to the milliequivalents of unreacted `H_(2)SO_(4)`.
Step `3`. Calculate the milliequivalents of total `H_(2)SO_(4)` and subtract the milliequivalents of unreacted `H_(2)SO_(4)` to get the milliequivalents of `NH_(3)` evolved.
Step `4`. Calculate the equivalents of `NH_(3)`, moles of `NH_(3)`, and moles of `N`.
Step `5`. Calculate the mass of `N` in the organic compound.
Step `6`. Finally, calculate `%` of `N` or directly apply Eq. `(13.8)` or `(13.9)` to get the `%` of `N` in the organic compound.

Explanation

Step 1. `N_(H_(2))SO_(4) = 2 xx M_(H_(2))SO_(4) = 2 xx 0.5 N = 1.0 N`
`N_(NaOH) = 1 xx M_(NaOH) = 1 xx 0.5 N = 0.5 N`
Step 2. Milliequivalents of `NaOH = 60 xx 0.5 = 30`
`:.` Milliequivalents of unreacted `H_(2)SO_(4) = 30`
Step 3. Milliequivalents of total `H_(2)SO_(4) = 50 xx 1 = 50`
`:.` Milliequivalents of `NH_(3) = 50 - 30 = 20`
Step 4. Equivalents of `NH_(3) = (20)/(1000) = 0.02`
Moles of `NH_(3) = 0.02`
Moles of `N = 0.02`
Step 5. Mass of `N` in organic compound
`= ("Moles of" N) xx ("Molar mass of" N)`
`= (0.02 mol)(14 g mol^(-1))`
`= 0.28 g`
Step 6. `%` of `N` in organic compound `= (0.28 g)/(0.50 g) xx 100%`
`= 56%`
Alternatively,
`%` of `N = ((N_(1)V_(1)) - (N_(2)V_(2)) xx 14)/(1000 xx m_(o . c.)) xx 100%`
`= ((1.0 xx 50 - 0.5 xx 60) xx14)/(1000 xx 0.50) xx 100%`
`= 56%` or
`%` of `N = (M_(ac id) xx "Basicity of acid xx Volume of acid")/(Mass_(o . c.)) xx 100%`
`= (14)/(1000) xx (0.5 xx 2 xx 20)/(0.50) xx 100%`
`= 56%`