Answer to Question #169385 in General Chemistry for thisrani

Question

0.7535 g sample of wheat flour was analysed by the Kjeldahl method. The ammonia (NH3) formed by addition of concentrated base after digestion with H2SO4 was distilled into 25.00mL of 0.06211 M HCl. The excess HCl was then back-titrated with 9.95 mL of 0.03512 M NaOH. Calculate the mass of nitrogen in the sample in grams. Given that proteins are 5.70 % nitrogen by mass, calculate the percentage protein in the flour.

Answer 1

1 Quantity of excess HCl (mol):

HCl+NaOH=NaCl+H2O

$$n(HCl)=n(NaOH)=Cm*V=0.03512*9.95/1000=0.00035 mol$$

2 Quantity of 25 ml 0.06211M HCl (mol):

$$n(HCl)=0.06211*25/1000=0.00155mol$$

3 Quantity of HCl that reacted with NH3:

$$n(HCl)=0.00155-0.00035=0.0012mol$$

$$HCl+NH3=NH4Cl$$

4 Quantity of NH3:

$$n(NH3)=n(HCl)=0.0012 mol$$

5 Mass of NH3:

$$m(NH3)=nM=0.0012*17=0.0204g$$

6 Massof N:

0.0204g - 17 g/mol

m(N) - 14 g/mol

$$m(N)=0.0204*14/17=0.0168g$$

7 Mass of protein:

$$m(protein)=0.0168/0.057=0.295g$$

8 The percentage protein in the flour:

% (protein) = 0.295/0.7535=0.391*100%=39.1%