Nitrification — Cell Mass Produced in a Septic Tank: 04-BS-13, December 2015, Q4

Question 4: Nitrification — Cell Mass Produced in a Septic Tank (20 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.

Quantity	Value
Wastewater processed	20 000 kg
NH₄⁺ content	5% by weight
Fraction of NH₄⁺ consumed	95%
MW NH₄⁺	18 g/mol
Biomass formula (no ash)	C₅H₇NO₂, MW = 113 g/mol
Stoichiometric ratio	55 mol NH₄⁺ → 1 mol biomass

Find. The mass of cells (biomass) produced.

Approach. Verify the given stoichiometry balances (C, H, N, O and charge), then use it as a direct mole-ratio bridge: mass of wastewater → mass of NH₄⁺ present → moles reacted (95% conversion) → moles of biomass (55:1 ratio) → mass of biomass via its molecular weight.

Confirm the equation balances (elemental + charge audit). C: $5=5$✓. H: $55(4)=220$ vs. $7+52(2)+109=220$✓. N: $55=1+54$✓. O: $5(2)+76(2)=162$ vs. $2+52(1)+54(2)=162$✓. Charge: $+55$ vs. $54(-1)+109(+1)=+55$✓. The equation is internally consistent as written. (The printed paper shows the last term as “log H+”; that is a typesetting slip for 109 H⁺, the only coefficient that closes both the H and charge balances above.)
Mass of NH₄⁺ available and reacted. $$m_{\text{NH}_4^+}=20\,000(0.05)=1000\ \text{kg}, \qquad m_{\text{reacted}}=1000(0.95)=\boxed{950\ \text{kg}}.$$
Moles of NH₄⁺ reacted. $$n_{\text{NH}_4^+}=\frac{950\,000\ \text{g}}{18\ \text{g/mol}}=52\,778\ \text{mol}.$$
Moles of biomass from the 55:1 stoichiometric ratio. $$n_{\text{biomass}}=\frac{52\,778}{55}=\boxed{959.6\ \text{mol}}.$$
Mass of biomass produced. MW of C₅H₇NO₂ (no ash) $=5(12)+7(1)+14+2(16)=113$ g/mol: $$m_{\text{biomass}}=959.6(113)=108\,434\ \text{g}=\boxed{108.4\ \text{kg cells}}.$$

Quantity	Result
NH₄⁺ reacted	950 kg = 52 778 mol
Biomass formed	959.6 mol
Mass of cells produced	108.4 kg

Topic: Using a fully-specified biological stoichiometric equation as a direct mass/mole conversion bridge.

Key relations: mole-ratio bridging via balanced stoichiometric coefficients; $n=m/\text{MW}$.

Why this works: once every species and coefficient in a reaction is known (unlike Q1–Q3, where an unknown coefficient had to be solved from elemental balances), the problem reduces to ordinary stoichiometry — convert the limiting/given reactant to moles, apply the fixed mole ratio, and convert the product back to mass.

Common pitfall: forgetting the 95%-conversion factor (using the full 1000 kg of NH₄⁺ instead of the 950 kg actually consumed), which would overstate the biomass yield by about 5%.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (nitrification stoichiometry, autotrophic biomass yield).

Question 4 of 10: Nitrification — Cell Mass Produced in a Septic Tank

Question 4: Nitrification — Cell Mass Produced in a Septic Tank (20 marks)