A sample of 0.50 gm of an organic compound was treated according to Kjeldahl's method the smmonia evolved was absorbed in 50 ml of `0.5MH_2SO_4`. The residual acid required 60 cm of 0.5M solution of `NaOH` for neutralisation. Find the percentage composition of nitrogen in the compound.

Explanation

Step 1. To determine the volume of `H_(2)SO_(4)` used.
Volume of acid taken = 50 mL of 0.5 M `H_(2)SO_(4) = 25 mL` of 1 M `H_(2)SO_(4)`
Volume of alkali used for neutralization of excess acid = 60 mL of 0.5 M NaOH = 30 mL of 1 M NaOH.
Now 1 mole of `H_(2)SO_(4)` neutralizes 2 moles of NaOH (i.e. `H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O`)
`:.` 30 mL of 1 M `NaOH -= 15 mL` of 1 M `H_(2)SO_(4)`
`:.` Volume of acid used by ammonia = 25 - 15 = 10 mL
Step 2. To determine percentage of nitrogen.
Again 1 mole of `H_(2)SO_(4)` neutralizes 2 moles of `NH_(3) :. 10 mL` of 1 M `H_(2)SO_(4) -= 20 mL` of 1 M `NH_(3)`
But 1000 mL of 1 M `NH_(3)` contain nitrogen = 14g
`:.` 20 mL of 1 M `NH_(3)` will contain nitrogen `= (14)/(1000) xx 20 g`
But this much amount of nitrogen is present in 0.5 g of the organic compound.
`:.` Percentage of nitrogen `= (14)/(1000) xx (20)/(0.5) xx 100 = 56.0`.
Alternatively, % of N can be determined by applying the following equation,
`% N = (1.4 xx "Molarity of the acid" xx "Basicity of the acid" xx "Vol. of the acid used")/("Mass of substance taken")`
Substituting the values of all the items in the above equation, we have, `% N = (1.4 xx 1 xx 2 xx 10)/(0.5) = 56.0`