Question

compute the total organic carbon in mg/L for a wastewater containing.
a) 100mg/L C9N2H6O2
b) 100mg/L C6N2H4O
c) 100mg/L C12N4H6O2
d) 100mg/L C10N2H8O2

Answer 1

Total organic carbon in

a) 100mg/L C9N2H6O2 :

Molecular weight of C9N2H6O2 = (12*9 +14*2 +1*6 + 16*2) = 174mg

Thus , 174mg of C9N2H6O2 contains 108mg of Carbon.

100 mg will contain = (108/174)*100 = 62.06mg of Carbon.

Thus, 100mg/L C9N2H6O2 will contain 62.06mg/L of organic carbon.

b) 100mg/L C6N2H4O :

Molecular weight of C6N2H4O =(12*6 +14*2 + 1*4 +16) = 120mg

Thus , 120mg of C6N2H4O contains 72mg of Carbon.

100 mg will contain = (72/120)*100 = 60mg of carbon

Thus, 100mg/L C6N2H4O contains 60mg/L of organic carbon.

c) 100mg/L C12N4H6O2 :

Molecular weight of C12N4H6O2 = (12*12 + 14*4 + 1*6 + 16*2)= 238mg

Thus , 238mg of C12N4H6O2 contains 144mg of Carbon.

100 mg will contain = (144/238)*100 = 60.50 mg of carbon.

Thus, 100mg/L C12N4H6O2 contains 60.50mg/L of organic carbon.

d) 100mg/L C10N2H8O2:

Molecular weight of C10N2H8O2 = (12*10 + 14*2 + 1*8 + 16*2)= 188mg

Thus , 188mg of C10N2H8O2 contains 120mg of Carbon.

100 mg will contain = (120/188)*100 = 63.82 mg of carbon.

Thus, 100mg/L C10N2H8O2 contains 63.82mg/L of organic carbon.