Question 11 of 25: COD of the Standard Biomass Formula C₅H₄NO₂

Question 11: COD of the Standard Biomass Formula C₅H₄NO₂ (4 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. Empirical cell formula $\text{C}_5\text{H}_7\text{NO}_2$ (the standard biomass composition used throughout wastewater engineering); atomic weights C=12, H=1, N=14, O=16.

Find. The chemical oxygen demand (COD) of the compound, in g $\text{O}_2$ per g of $\text{C}_5\text{H}_7\text{NO}_2$.

Approach. Write the balanced complete-oxidation half-reaction of the compound to $\text{CO}_2$, $\text{H}_2\text{O}$, and $\text{NH}_3$ (the standard convention for biomass COD, since biological oxidation does not typically carry organic-N through to nitrate), find the stoichiometric moles of $\text{O}_2$ consumed per mole of compound, and convert to a mass ratio using each species' molecular weight.

1. Balanced oxidation reaction. $$\text{C}_5\text{H}_7\text{NO}_2 + 5\text{O}_2 \longrightarrow 5\text{CO}_2 + 2\text{H}_2\text{O} + \text{NH}_3$$ Checking atoms: C: 5=5; H: $7=2(2)+3$; N: $1=1$; O: $2+5(2)=5(2)+2(1)+0=12$. The equation balances with exactly 5 mol $\text{O}_2$ consumed per mole of biomass.
2. Molecular weight of the compound. $$MW = 5(12)+7(1)+14+2(16) = 60+7+14+32 = 113\ \text{g/mol}$$
3. COD per unit mass. $$\text{COD} = \frac{5\times MW_{\text{O}_2}}{MW_{\text{cells}}} = \frac{5\times32}{113} = \boxed{1.42\ \text{g O}_2\text{/g cells}}$$