COD of C 5 H 7 NO 2: 18-Env-B9 Environmental Chemistry and Microbiology, May 2016, Q3

Question 3: COD of C₅H₇NO₂ (3 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. Cell-mass empirical formula $\text{C}_5\text{H}_7\text{NO}_2$ (the standard biomass formula used throughout wastewater engineering); atomic weights as stated.

Find. Chemical oxygen demand (COD) of the compound, in g $\text{O}_2$ per g compound.

Approach. Write the balanced complete-oxidation reaction (carbon to $\text{CO}_2$, hydrogen to $\text{H}_2\text{O}$, nitrogen released as $\text{NH}_3$ — the standard biological-oxidation convention, since nitrogen in cell mass is not itself oxidized on complete combustion of the carbonaceous fraction), then take the mass ratio of $\text{O}_2$ consumed to compound oxidized.

Balance the oxidation reaction. $$\text{C}_5\text{H}_7\text{NO}_2+5\text{O}_2\longrightarrow 5\text{CO}_2+2\text{H}_2\text{O}+\text{NH}_3$$ Check: C 5=5; H $7=2(2)+3$; O $2+10=10+2$; N $1=1$ — balances with 5 mol $\text{O}_2$ per mol compound.
Molecular weight of C₅H₇NO₂. $$M=5(12)+7(1)+14+2(16)=60+7+14+32=113\ \text{g/mol}$$
COD as a mass ratio. Each mole of compound consumes 5 mol $\text{O}_2$ (32 g/mol): $$\text{COD}=\dfrac{5(32)}{113}=\dfrac{160}{113}=\boxed{1.42\ \text{g O}_2/\text{g}}$$

Quantity	Value
Molecular weight, C₅H₇NO₂	113 g/mol
O₂ demand per mole	5 mol (160 g)
COD	1.42 g O₂/g compound

Concept & Theory

Topic. Theoretical/chemical oxygen demand from an empirical formula.

Key relations. $\text{C}_c\text{H}_h\text{O}_o\text{N}_n+x\,\text{O}_2\to c\,\text{CO}_2+\tfrac{h-3n}{2}\text{H}_2\text{O}+n\,\text{NH}_3$, with $x=c+\tfrac{h-3n}{4}-\tfrac{o}{2}$.

Why this works. Complete oxidation to $\text{CO}_2$/$\text{H}_2\text{O}$ is a mass/electron balance independent of pathway, so COD can be computed directly from the stoichiometric formula without any experimental data — this is the same 1.42 g/g figure used industry-wide to convert MLVSS to its oxygen-equivalent COD, and it recurs in Question 17 below as the potential carbonaceous BOD of cell mass.

Common pitfall. Oxidizing the nitrogen on to $\text{NO}_3^-$ (that is the nitrogenous, not carbonaceous, demand) instead of stopping at $\text{NH}_3$.

Source. Metcalf & Eddy, Wastewater Engineering, ch. 8 (biological stoichiometry).

Question 3 of 20: COD of C 5 H 7 NO 2

Section 1: Chemistry (8 questions, 50 marks)

Question 3: COD of C₅H₇NO₂ (3 marks)

Question 3 of 20: COD of C 5 H 7 NO 2

Section 1: Chemistry (8 questions, 50 marks)

Question 3: COD of C5H7NO2 (3 marks)

Question 3: COD of C₅H₇NO₂ (3 marks)