COD and TOC of an Ethylene Glycol Wastewater: 18-Env-B5 Industrial & Hazardous Waste Management, May 2014, Q7

Question 7: COD and TOC of an Ethylene Glycol Wastewater (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An ethylene glycol concentration of 150 mg/L, with atomic weights C = 12, H = 1, O = 16 supplied for computing the molecular weight and the stoichiometric oxygen demand.

Given data
Quantity	Symbol	Value
Ethylene glycol concentration	C_substrate	150 mg/L
Formula	—	C₂H₆O₂
Atomic weights	C, H, O	12, 1, 16

Find. The chemical oxygen demand (COD) and total organic carbon (TOC) of the 150 mg/L ethylene glycol solution.

Approach. Compute the molecular weight of C₂H₆O₂, balance its complete combustion to CO₂ and H₂O to get the stoichiometric O₂ demand per mole (which, for a fully-oxidizable organic with no COD-resistant fraction, equals its COD), then scale both the oxygen demand and the carbon content to the given 150 mg/L.

Molecular weight of ethylene glycol. $$MW = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\ \text{g/mol}$$
Balance the complete oxidation reaction. Complete combustion converts the carbon to CO₂ and the hydrogen to H₂O: $$C_2H_6O_2 + 2.5\,O_2 \rightarrow 2\,CO_2 + 3\,H_2O$$ Checking atoms: C: 2=2; H: 6=6; O (LHS): $2 + 2(2.5) = 7$, O (RHS): $2(2)+3(1)=7$ — balanced with 2.5 mol O₂ consumed per mol of substrate.
COD (theoretical oxygen demand). The mass of O₂ consumed per gram of ethylene glycol is $2.5 \times 32 / 62$, applied directly to the 150 mg/L concentration since ethylene glycol is fully biodegradable/oxidizable with no COD-resistant residue: $$COD = 150\ \text{mg/L} \times \dfrac{2.5 \times 32}{62} = 150 \times 1.290$$ $$\boxed{COD \approx 193.5\ \text{mg/L}}$$
TOC (total organic carbon). The mass fraction of carbon in the molecule is $2(12)/62$, applied to the same 150 mg/L: $$TOC = 150\ \text{mg/L} \times \dfrac{2(12)}{62} = 150 \times 0.3871$$ $$\boxed{TOC \approx 58.1\ \text{mg/L}}$$

Final results
Quantity	Value
Molecular weight of C₂H₆O₂	62 g/mol
COD	≈ 193.5 mg/L
TOC	≈ 58.1 mg/L

The COD/TOC ratio here ($193.5/58.1 \approx 3.33$ mg O₂ per mg C) is a useful cross-check: it is close to the ratio expected for a fully-oxidized, moderately-oxygenated small organic molecule, and is typical of why ethylene glycol — a common de-icing-fluid and antifreeze constituent — is treated as essentially 100% biodegradable in an activated-sludge system, rather than carrying any refractory (non-biodegradable) COD fraction.

Topic: Stoichiometric COD and TOC of a pure organic compound from its molecular formula

Key relations: balanced combustion $C_2H_6O_2 + 2.5\,O_2 \rightarrow 2\,CO_2 + 3\,H_2O$; $COD = C_{substrate}\times(n_{O_2}\times32)/MW$; $TOC = C_{substrate}\times(n_C\times12)/MW$

Why this works: COD is, by definition, the mass of oxygen required to fully oxidize a compound, so balancing the combustion equation and scaling by the molecular-weight ratio gives the theoretical (and, for a readily-oxidizable compound like ethylene glycol, the actual) COD directly; TOC is simply the compound's own carbon mass fraction applied to its concentration.

Common pitfall: forgetting to balance the oxygen atoms on both sides (a common slip is guessing "2" or "3" mol O₂ instead of the correct 2.5), or computing TOC from the whole molecular weight instead of only the carbon-atom mass fraction.

Source: Metcalf & Eddy, Wastewater Engineering, 5th ed. (COD/BOD/TOC relationships); Davis & Cornwell, Introduction to Environmental Engineering, 6th ed.

Question 7 of 18: COD and TOC of an Ethylene Glycol Wastewater

Question 7: COD and TOC of an Ethylene Glycol Wastewater (5 marks)