Question

Calculate the ultimate biochemical oxygen demand (BOD) and the total organic carbon (TOC) concentration of an industrial wastewater containing 434 mg/L of organic matter that has been analyzed and found to have the empirical formula C9H15O5N and to be 100% biodegradable. Assume that no other organic, biodegradable or oxygen-demanding materials are present in the wastewater; and also assume that the nitrogen is converted to ammonia, NH3. Hints: a) although the molecular weight of the organic matter is not known, and may vary from one molecule to another, the formula weight can be used in similar fashion to convert mg/L to M; b) for completely biodegradable organic matter, the ultimate BOD is equal to the theoretical oxygen demand; and c) it is best to begin by writing the appropriate reaction and balancing it.

P.S. I know this is supposed to go in the Chemical Engineering subject but I can't find it. I have tried Civil Engineering but I have received no replies. Please help as much as you can.

Answer 1

Calculation of BOD:

The balanced equation will be...

C₉H₁₅O₅N + 19/2 O₂ = 9CO₂ + NH₃ + 6H₂O

According to the equation, 19/2 mol of oxygen is required to completely dissociate 1 mol of the organic compound.

Molar mass of the compound = [9*12 +15*1 + 5*16 + 14] g/mol = 217g/mol

The amount of the organic compound in water

= 0.434 g/L

= 0.434/217 M

=0.002 M

Oxygen required to completely dissociate 0.002 mol of the compound

= [0.002 * (19/2)] g/L

= 0.608 g/L

= 608 mg/L

So, ultimate BOD = 608 mg/L

Calculation of TOC:

Total Organic Carbon (TOC) is an indirect measure of organic molecules present in water and measured as carbon. To measure TOC the organic molecules in the sample water is completely oxidized to carbon dioxide (CO₂). The resultant CO₂ concentration is as carbon concentration.

C₉H₁₅O₅N + 19/2 O₂ = 9CO₂ + NH₃ + 6H₂O

So we see that, 9 mol of CO₂ is produced by complete oxidization of the compound. So, 0.002 mol of the compound will produce

= (9 * 0.002) mol of CO₂

=0.018 mol of CO₂

= 0.792 g/L of CO₂

= 792 mg/L of CO₂

So, TOC = 792 mg/L