Duma & Kjeldahl’s Method : Nitrogen

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Duma & Kjeldahl’s Method : Nitrogen

Estimation of Nitrogen- It is determined by two methods

1. 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.

V1=Vol. of N2                                               P2= 760 mm or 76cm

P1=Pressure of N2(mm)                            V2= ? Vol. of N2 at NTP

T1=Temp. of N2                                            T2= 273K

Numericals

1. An organic Compound was analysed by Duma’s Method. 0.45 gm. of compd on combustion give 48.6 ml N2 at 270C & 756mm pressure. calculate  % of  N2.

Ans.                                                             At N.T.P.

P1=756 mm                                                    P2= 760 mm

V1=48.6 ml                                                    V2=? ml

T1=27+273= 300K                                      T2= 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 N2 = =[28 x 43.99x 100] / [22400 x 0.45]

% of N2 = 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 150C=12.75mm)

P1 = 732.7-12.75 = 719.95 mm                   P2 = 760 mm

V1= 30.4 ml                                                    V2= ?ml

T1 = 15+273=288k                                        T2=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-

1. i)   0. 25 gm of organic compd on complete combustion gave 0.37 gm. of CO2 & 0.2 gm of H2O
2. ii) 0. 25 gm. of  organic compd on analysis by Duma’s Method gave 32 ml of N2 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. H2SO4 in presence of CuSO4 & K2SO4. The Nitrogen is converted into (NH4)2SO4. The (NH4)2SO4 obtained is decomposed with caustic soda or caustic potash then NH3 is evolved which is passed into H2SO4 of known quantity & strength. The remaining unreacted H2SO4 is titrated against NaOH Solution of known strength. Hence % of N2 is then calculated

% of N2=[ 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

1. 0.2 gm of an org. compd was analysed by Kjeldahl’s method .  NH3 evolved was absorbed in 60ml N/5 H2SO4. unused acid required 40ml N/10 NaOH for complete neutralisation find the % of N2

60 ml N/5 H2SO4 = 12 ml N H2SO4

40 ml N/10 NaOH=4ml N NaOH

Acid used with NH3= (12-4)

= 8 ml N H2SO4

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

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

1. 0.788 gm. of an org. compd was analysed by Kjeldahl’s method. NH3 liberated was absorbed in 100 ml N H2SO4. Excess of H2SO4 is neutralised by 73.7 ml of N NaOH .Calculate % of N2= ?

Ans.  Acid used with NH3 = (100-73.7) ml N H2SO4

= 26.3 ml N H2SO4

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

% of N2= 46.72 %

1. 1.029 gm of an organic compd on boiling with NaOH liberates some NH3, the complete neutralisation of which  requires 14 ml of N/2 H2SO4 . Calculate % of N2=?

Ans.  w=1.029 , V =14 ml

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

% of N2 =(1.4  x  0.5 x 14) /1.029

% of N2 =9.5 %