Question

A 2.159 g sample of a leaves enjoyed by a humble gnu was collected. The sample was digested in sulfuric acid and the ammonia produced was passed through a solution of 25.00 mL of 0.1472 M HCl. What is the protein content of this sample of grass if 14.1 mL of 0.1112 M NaOH were required to titrate the leftover HCl? (use an N-to-protein conversion factor of 1.31 for Celtis integrifolia leaf protein)

Answer 1

Total moles of HCl added = Molarity*Volume = 0.1472*0.025 = 0.00368

Moles of NaOH required for titrating leftover HCl = Molarity*Volume = 0.1112*0.0141 = 0.00157

So, moles of HCl leftover = 0.00157

So,

Moles of HCl that reacted with ammonia = 0.00368-0.00157 = 0.00211

So,

Moles of ammonia produced = 0.00211

So, moles of nitrogen released by the digestion = 0.00211

Mass of nitrogen = Moles*MW = 0.00211*14 = 0.02954 g

Using the conversion factor:

Protein content = Nitrogen content*1.31 = 0.02954*1.31 = 0.0387 g

So,

Protein content = (0.0387/2.159)*100 = 1.792%

Hope this helps !