Nitrogen can be estimated either by Dumas' method or Kjeldahl's method.

`text(Duma's Method)` : Known mass of an organic compound is heated with cupric oxide in an atmosphere of carbon dioxide. The carbon and the hydrogen get oxidised to carbon dioxide and water while the nitrogen is set free. Traces of oxides of nitrogen, which may be formed in some cases are reduced to elemental nitrogen by passing over heated copper spiral. The gaseous mixture is collected over an aqueous solution of potassium hydroxide. `CO_2` is absorbed by the caustic potash solution while `H_2O` condenses and nitrogen is collected by downward displacement of `KOH` solution. The volume of nitrogen is measured and this is equivalent to the nitrogen present in the compound. The volume of nitrogen is reduced to `STP`, its weight calculated and from this the percentage of nitrogen present in the organic compound is estimated.

The following reactions take place during Duma's process :

`C + 2CuO -> CO_2 + 2Cu`; `2H + CuO -> H_2O + Cu`

`2N + CuO -> N_2 (+)` `text(Oxides of nitrogen)`

`text(Oxides of nitrogen)` `+ Cu -> CuO + N_2`

Let the mass of organic compound `= w` `g`

Volume of nitrogen collected `= V` `cm^3`,

Pressure of moist `N_2 = P` mm of `Hg`,

Room ternperature `= t^oC`,

Aqueous tension at `t^oC = a` mm of `Hg`,

Pressure of dry nitrogen `= (P - a)` mm of `Hg`,

Let us first convert the volume of nitrogen at given temperature & pressure to the volume of `N_2` at `STP`.

`P_(N_2) xx V_(N_2) = n_(N_2) xx RT`

`n_(N_2) = ((P-a) xx V xx 10^(-3))/(760 xx 0.821 xx (273 + t))`

Volume of `N_2` at STP `= ((P-a) xx V xx 10^(-3) xx 22.4)/(760 xx 0.821 xx (273 + t)) = x L` (say)

Now, `22.4` `L` of `N_2` at `STP` weigh = `28` `g`

`x` `L` of `N_2` at `STP` will weigh `= (28/22.4 xx x)` `g`

Percentage Of nitrogen `= (text(Mass of nitrogen))/(text(Mass of organic Compound)) xx 100`

` = (28 x xx 100)/(22.4 xx W)`

`text(Kjeldahl's Method)`

This method is simpler and more convenient than Duma's method. This method is largely used for the estimation of nitrogen in food stuffs, drugs, fertilizers and many other organic compounds. However, this method cannot be used for

(i) organic compounds containing nitrogen in the ring such as pyridine, quinoline etc.

(ii) organic compounds containing nitro `(- NO_2)` and diazo `(- N=N-)` groups.

`text(Principle)` : A known weight of the organic compound is heated with concentrated `H_2SO_4` so that nitrogen is quantitatively converted into ammonium sulphate. The solution is then heated with excess of sodium hydroxide. The ammonia gas evolved is passed into a known but excess volume of standard acid (`HCl` or `H_2SO_4`). The acid left unused is estimated by titrating the solution with standard alkali. From the amount of acid left unused, the amount of acid used for neutralization of ammonia can be calculated. From this, percentage of nitrogen can be calculated. The chemical reactions involved are

`undersettext(From organic Compound)(C,H,S) overset[text(conc.)H_2SO_4]-> CO_2 + H_2O + SO_2`

`undersettext(From organic Compound)(N) overset[text(conc.)H_2SO_4]-> undersettext[(Ammonium sulphate)]((NH_4)_2SO_4)`

`(NH_4)_2SO_4 + 2NaOH -> Na_2SO_4 + 2NH_3 + 2H_2O`

`2NH_3 + H_2SO_4 -> (NH_4)_2SO_4` (`n = 1`)

`H_2SO_4 + 2NaOH -> Na_2SO_4 + 2H_2O`

Let the mass of organic compound = `w` `g`,

Total volume of standard acid taken `= V_1` `cm^3`,

Normality of acid `= N_1`

Now, the excess acid after dilution is titrated against standard alkali.

Volume of standard alkali used for neutralization of unused acid `= V_2` `cm^3`,

Normality of standard alkali `= N_2`

Equivalents of unused acid = Equivalents of alkali = `N_2V_2 xx10^(-3)`

Initial equivalents of acid `= N_1 V_1 xx 10^(-3)`

Equivalents of acid consumed by `NH_3 = (N_1V_1 - N_2V_2) xx10^(-3) =` Equivalents of `NH_3` reacted.

Moles of `NH_3` reacted `= (N_1 V_1 - N_2V_2) xx 10^(-3)`

Moles of `NH_3` liberated `= (N_1 V_1 - N_2V_2) xx 10^(-3) =` Moles of nitrogen in `NH_3` = Moles of nitrogen in organic compound.

Mass of nitrogen in the organic compound `= (N_1 V_1 - N_2V_2)xx10^(-3) xx14`

Percentage of nitrogen `= (text(Mass of nitrogen))/(text(Mass of organic Compound)) xx 100`

`= ((N_1V_1 - N_2V_2) xx 10^(-3) xx 14)/w xx 100 = 1.4 (N_1V_1 - N_2V_2)/w`

where `V_1` and `V_2` are the volume of standard acid and alkali respectively, in `cm^3`.