Estimation of C&H

Estimation of Elements (Quantitative Analysis)

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Estimation of C&H

Carbon & Hydrogen are estimated simultaneously by Liebig’s combustion method

Principle

When a known weight of an organic compound is strongly heated with cupric oxide. C& H present are oxidised to CO₂ & H₂O respectively. Lime water or KOH is used to absorb obtained CO₂ while Anhy. CaCl₂ or Anhy. CuSO₄ is used to absorb H₂O. Increase in the wt. of lime water or KOH is the wt. of CO₂ while increase in the wt. of Anhy. CaCl₂ or Anhy. CuSO₄ is the weight of H₂O.

CxHy +O2 —–> xCO2 +y/2[H2O]

Calculation of  % of C

Let the mass of 0rganic compound is w gm.

Increase in the mass of KOH = Amount of CO₂=x gm

44 gms CO₂ contain C = 12 gm

x gm CO₂ contain C   =12x /44 gm

12x /44 gm of carbon is obtained from w gm of o.c.

w gm. 0rganic compound contain C = 12x /44 gm

100gm  o.c. contain C  =  ( 12x /44 w)  . 100

% of c=( 12 /44 )[( wt. of CO2) /( wt. of o.c.)100

Calculation of % of Hydrogen

Let the mass of o.c. is w gm.

Increase in the mass of Anhy. CaCl_{2 or Anhy . CuSO4 = wt of H2O= y gm.}

18 gm of water contain H= 2 gm

y gm  of water contain H = 2y/18 gm.                                     =

2y/18 gm of H are present in w gm of  o.c

w gm o.c contain H= 2y/18 gm

100 gm o.c contain H =  ( 2y /18w)  . 100                     =

% of H= ( 2 /18 )[( wt. of H2O) /( wt. of o.c.)]100

% of O =100 – (sum of % of all elements present in o.c)

Q 1)   0.2475 gm of an o.c on combustion gave 0.4950 gm of CO₂ & 0.2025 gm of H₂O. Determine the % composition of compound.

Ans .

wt.of.CO2=0.4950 gm.

wt.of H2O= 0.2025 gm.

wt.of o.c. =0.2475

% of C=( 12 /44 )[( wt. of CO2) /( wt. of o.c.)100

                     =( 12 /44 )[(.4950) /(.2475)]100  

                      =54.54%                      

% of H =( 2 /18 )[( wt. of H2O) /( wt. of o.c.)]100

percentage of H =( 2 /18 )[( .2025) /(.2475)]

    % of H  = 9.09%

% of O=100 – (54.54+9.09)=36.41%

Q 2) 0.39gm. hydrocarabon on combustion gave 1.32 gm of CO₂ & 0.27 gm of H₂O. Calculate percentage composition of hydrocarbon?

Solution:

weight of CO2 =1.32 gm.

wt. of H2O =0.27 gm.

wt. of o.c. =0.39 gm.

% of C=( 12 /44 )[( wt. of CO2) /( wt. of o.c.)100

                     =( 12 /44 )[(1.32) /(.39)]100  

                      =92.3%                      

% of H =( 2 /18 )[( wt. of H2O) /( wt. of o.c.)]100

=( 2 /18 )[( 0.27) /(.39)] 100

    % of H  = 7.7%

% of C =92.3, % of H=7.7%

Q 3) 0.46 gm. of an 0.c containing C, H & O was analysed by combustion method. The increase in mass of Anhy. CaCl₂ & KOH are 0.54gm & 0.88gm respectively. Determine % composition of the compound ?

Ans.

weight of CO2 =0.88 gm.

wt.of H2O=0.54 gm.

weight of o.c.=0.46

% of C=( 12 /44 )[( wt. of CO2) /( wt. of o.c.)]100

                     =( 12 /44 )[(0.88) /(0.46)]100  

                      =52.17%                      

% of H =( 2 /18 )[( wt. of H2O) /( wt. of o.c.)]100

percentage of H =( 2 /18 )[( 0.54) /(0.46)] 100

    % of H  = 13.04%

% of O=100 – (52.17+13.04)=34.79%

% of C=52.17, % of H= 13.04, % of O=34.79%

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