Question

: A 2.800 g sample of meat is subjected to Kjeldahl analysis. The liberated NH3 (g) is absorbed by adding 50.00 mL of H2SO4 (aq) , which is more than enough. The excess acid requires 19.90 mL of 0.5510 M of NaOH for its complete neutralization. What is the percentage of protein in the meat?

Answer 1

According to the question ;

As we know that ;

19.90 mL of 0.5510 M NaOH

=19.90/1000*0.5510

=0.01096 mole

It will neutralize

=0.01096/2

= 0.00548 mole of sulphuric acid.

So,

50.00 mL of 0.2496 M sulphuric acid corresponds to =50.00/1000*0.2496

= 0.01248 mole

Thus,

The amount of sulphuric acid neutralized by ammonia is

=0.01248 mole -0.00548 mole

=0.007

0.007 mole of sulphuric acid will neutralize 0.014 mole of ammonia.

Now 1 mole of ammonia contains 14 g of nitrogen.

Therefore, 0.014 mole of ammonia will contain

=0.014*14

= 0.196g of nitrogen.

So,
3.100 g of organic compound contains 0.196 g of nitrogen.
Therefore , 100 g of compound will contain ;

=0.196*100/2.800

= 7% of nitrogen.