Record of Observation

 

Volume of base required for titration (8 points): 5.5 mL

 

 

 

Questions & Answers

 

 

1.  Calculate % nitrogen in the sample using the follow equation. (4 points)

 

         (mL acid x 0.1 N x 0.014 g nitrogen/milliequivalent) x 100 =

                           sample weight, g

 

         5.5 x 0.1 x 0.014 x 100 / 5.12 = 0.150% nitrogen

    

 

2.  The % nitrogen for this sample can be converted to % protein by using a factor of 6.25. That is 6.25 x % nitrogen = % protein. What is the % protein? (4 points)

 

    6.25 x 0.15% nitrogen = 0.94% protein

 

3.  A second sample weighed 6.3 grams and required 17.22 mL of acid for titration. What is the % nitrogen in this sample? (4 points)

 

    17.22 x 0.1 x 0.014 x 100 / 6.3 = 0.383% nitrogen

 

4.  For this sample, the conversion factor is 6.38. How much protein did it contain? (4 points)

 

    6.38 x 0.383% nitrogen = 2.44% protein

 

5.  A third sample weighed 7.32g and required 13.42 mL of acid for titration. What is the % nitrogen in this sample? (4 points)

 

    13.42 x 0.1 x 0.014 x 100 / 7.32 = 0.257% nitrogen

 

6.  For this sample, the conversion factor is 5.65. How much protein did it contain? (4 points)

 

    5.65 x 0.257% nitrogen = 1.45% protein

 

7.  Using the molecular weights (MWs) provided, derive the factor of approximately 6.25 used to convert the % nitrogen to % protein. (8 points)

        

         Example: GLY = [2C + 1N + 3H] / N = (28 + 14 + 32 + 3) / 14 = 5.5

 

Note: Value derived will vary depending on the amino acid used for the calculation so a range of values can be considered correct what is important is the logic.