Question 18 of 19: COD and TOC of an Ethylene Glycol Wastewater

Question 18: 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 industrial wastewater contains ethylene glycol, C₂H₆O₂, at a concentration of 200 mg/L. Atomic weights: C = 12, H = 1, O = 16.

Find. The theoretical chemical oxygen demand (COD) and total organic carbon (TOC) of the wastewater, in mg/L.

Approach. Write and balance the complete oxidation reaction of ethylene glycol to CO₂ and H₂O to find the stoichiometric oxygen demand per mole, then scale by the molar concentration for COD; find TOC directly from the carbon mass fraction of the molecule.

1. Molecular weight of ethylene glycol. $$MW = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\ \text{g/mol}$$
2. Balance the complete oxidation equation. Complete oxidation converts the compound entirely to carbon dioxide and water: $$\text{C}_2\text{H}_6\text{O}_2 + x\,\text{O}_2 \rightarrow 2\,\text{CO}_2 + 3\,\text{H}_2\text{O}$$ Carbon and hydrogen balance directly (2 C on each side; 6 H on each side, since 3 H₂O carries 6 H). Balancing oxygen: $$2 + 2x = 2(2) + 3(1) = 7 \quad\Rightarrow\quad x = 2.5$$ So one mole of ethylene glycol requires 2.5 mol O₂ for complete oxidation.
3. Convert the stoichiometric oxygen demand to a mass ratio. $$\text{COD ratio} = \frac{x \times 32\ \text{g O}_2/\text{mol}}{62\ \text{g glycol/mol}} = \frac{2.5 \times 32}{62} = \frac{80}{62} = 1.290\ \text{g O}_2/\text{g glycol}$$
4. Scale by the given concentration to find COD. $$\text{COD} = 200\ \text{mg/L} \times 1.290 = \boxed{258.1\ \text{mg/L}}$$
5. Find the carbon mass fraction of the molecule. Each mole of ethylene glycol contains 2 mol of carbon: $$\text{Carbon fraction} = \frac{2 \times 12}{62} = \frac{24}{62} = 0.3871\ \text{g C/g glycol}$$
6. Scale by the given concentration to find TOC. $$\text{TOC} = 200\ \text{mg/L} \times 0.3871 = \boxed{77.4\ \text{mg/L}}$$