tuna 0.917 g Kjeldahl NH3 0.1249 M HCl 20.59 mL

tuna 0.917 g Kjeldahl NH3 0.1249 M HCl 20.59 mL

 

 

A 0.917 g sample of canned tuna was analyzed by the Kjeldahl method.

A volume of 20.59 mL of 0.1249 M HCl was required

to titrate the liberated ammonia.

Calculate the percentage of nitrogen in the sample.

 

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(0.1249 mol/L) (20.59/1000 L) = 0.0025717 mol HCl

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

HCl + NH3 → NH4Cl

 

HCl : NH3 = 1 : 1 계수비(= 몰수비) 이므로,

NH3의 몰수 = 0.0025717 mol

 

 

 

NH3 : N = 1 : 1 계수비(= 몰수비) 이므로,

N의 몰수 = 0.0025717 mol

 

 

 

N의 몰질량 = 14.01 g/mol 이므로,

0.0025717 mol × (14.01 g/mol) = 0.03603 g N

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

(0.03603 g / 0.917 g) × 100 = 3.93%

 

 

 

답: 3.93%