** Duma & Kjeldahl’s Method : Nitrogen**

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**Duma & Kjeldahl’s Method : Nitrogen**

**Estimation of Nitrogen-** It is determined by two methods

- Duma’s Methods ( ii) Kjeldahl method

**Duma’s Method**

% of Nitrogen =[28 x volume of nitrogen at N.T.P x 100] / [22400 x weight of organic compound

volume at N.T.P. can be calculated as-

P1V1/T1 =P2 V2/ T2

At N.T.P.

V_{1}=Vol. of N_{2} P_{2}= 760 mm or 76cm

P_{1}=Pressure of N_{2}(mm) V_{2}= ? Vol. of N_{2} at NTP

T_{1}=Temp. of N_{2} T_{2}= 273K

**Numericals**

- An organic Compound was analysed by Duma’s Method. 0.45 gm. of compd on combustion give 48.6 ml N
_{2}at 27^{0}C & 756mm pressure. calculate % of N_{2}.

**Ans.** At N.T.P.

P_{1}=756 mm P_{2}= 760 mm

V_{1}=48.6 ml V_{2}=? ml

T_{1}=27+273= 300K T_{2}= 273K

P1V1/T1 =P2 V2/ T2

( 756 X48.6)/300 =(760 x V2) /273

V2=43.99 ml

% of Nitrogen =[28 x volume of nitrogen at N.T.P x 100] / [22400 x weight of organic compound]

**% ** of N_{2} = =[28 x 43.99x 100] / [22400 x 0.45]

% of N_{2} = 12.22%

(2) 0.2046 gm of an org. compd give 30.4ml of N2 at 150C and 732.7mm pressure. calculate the % of N2 (aq. Tension at 15^{0}C=12.75mm)

P_{1} = 732.7-12.75 = 719.95 mm P_{2} = 760 mm

V_{1}= 30.4 ml V_{2}= ?ml

T_{1} = 15+273=288k T_{2}=273 k

P1V1/T1 =P2 V2/ T2

( 719.95 x 30.4)/288 = (760 x V2)/273

V2 =27.3 ml

% of Nitrogen =[28 x volume of nitrogen at N.T.P x 100] / [22400 x weight of organic compound]

=[28 x 27.3 x 100] / [22400 x 0.2046)

% of Nitrogen =16.68 %Ans.

Q. An org. Compd on analysis gave the following data-

- i) 0. 25 gm of organic compd on complete combustion gave 0.37 gm. of CO
_{2}& 0.2 gm of H_{2}O - ii) 0. 25 gm. of organic compd on analysis by Duma’s Method gave 32 ml of N
_{2}at N.T.P. Calculate % composition of org. Compd.

**Solve. **% of C=( 12 x 0.37 x 100) / (44 x 0.25) =40.36%

% of H=( 2 x 0.2 x 100) / (18 x 0,25 ) =8.89 %

% of Nitrogen =[28 x volume of nitrogen at N.T.P x 100] / [22400 x weight of organic compound]

% of Nitrogen =[28 x 32 x 100] / [22400 x 0.25] =16 %

% of O=100–(% of C+H+N)

% of O=100–(40.36+8.89+16) = 34.75% Ans.

**Kjeldahl’s Method-**

This method is based on the fact that many org. compd when heated with conc. H_{2}SO_{4} in presence of CuSO_{4} & K_{2}SO_{4}. The Nitrogen is converted into (NH_{4})_{2}SO_{4}. The (NH_{4})_{2}SO_{4} obtained is decomposed with caustic soda or caustic potash then NH_{3} is evolved which is passed into H_{2}SO_{4} of known quantity & strength. The remaining unreacted H_{2}SO_{4} is titrated against NaOH Solution of known strength. Hence % of N_{2} is then calculated

% of N_{2}=[ 1.4 x normality of acid x volume of acid used with NH3] /weight of organic compound

** **% of N2 =( 1.4 x N x V)/w

- 0.2 gm of an org. compd was analysed by Kjeldahl’s method . NH
_{3}evolved was absorbed in 60ml N/5 H_{2}SO_{4}. unused acid required 40ml N/10 NaOH for complete neutralisation find the % of N_{2}

60 ml N/5 H_{2}SO_{4} = 12 ml N H_{2}SO_{4}

40 ml N/10 NaOH=4ml N NaOH

Acid used with NH_{3}= (12-4)

= 8 ml N H_{2}SO_{4}

% of N2 =( 1.4 x N x V)/w

% of N2 =( 1.4 x 1 x 8)/0.2 =56 % Ans.

- 0.788 gm. of an org. compd was analysed by Kjeldahl’s method. NH
_{3}liberated was absorbed in 100 ml N H_{2}SO_{4}. Excess of H_{2}SO_{4}is neutralised by 73.7 ml of N NaOH .Calculate % of N_{2}= ?

**Ans.** Acid used with NH_{3} = (100-73.7) ml N H_{2}SO_{4}

= 26.3 ml N H_{2}SO_{4}

% of N2 =( 1.4 x 1 x 26.3)/0.788

% of N_{2}= 46.72 %

- 1.029 gm of an organic compd on boiling with NaOH liberates some NH
_{3}, the complete neutralisation of which requires 14 ml of N/2 H_{2}SO_{4 .}Calculate % of N_{2}=?

**Ans**. w=1.029 , V =14 ml

% of N2 =( 1.4 x N x V)/w

% of N_{2 =(1.4 x 0.5 x 14) /1.029 }

% of N2 =9.5 %