Aerobic Growth Stoichiometry — Yield and CO 2 Production: 20-Bio-A5 Systems Analysis & Control, Undated paper, Q4

Question 4: Aerobic Growth Stoichiometry — Yield and CO₂ Production (a. 15 marks; b. 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.

Quantity	Value
Glucose supplied initially	15 mmol
Dry cell weight formed	0.32 g
Biomass molecular weight	25.2 g/mol (CH$_{1.6}$N$_{0.20}$O$_{0.55}$)
O$_2$ consumed	16 mmol
NH$_3$, O$_2$ supplied	in excess

Find. (a) $Y_{X/S}$ (g DCW/mol glucose consumed) and the glucose remaining in the medium; (b) the CO$_2$ produced.

Approach. The reaction as written has four elemental balances (C, H, N, O) but five unknown coefficients ($a,b,c,d,e$); the measured $O_2$-consumed/biomass-formed ratio supplies the missing (fifth) equation, closing the stoichiometry so every coefficient — and hence glucose consumed — can be solved for explicitly.

Moles of biomass formed. $$n_X=\frac{0.32\ \text{g}}{25.2\ \text{g/mol}}=0.012698\ \text{mol}=12.698\ \text{mmol}.$$
Elemental balances (per mol glucose reacted, coefficients $a,b,c,d,e$). $$\text{C: }6=c+d,\quad \text{N: }b=0.2c,\quad \text{H: }12+3b=1.6c+2e,\quad \text{O: }6+2a=0.55c+2d+e.$$ Solving H and O in terms of $c$: $e=6-0.5c$ and $a=6-0.975c$.
Close the system with the measured ratio. Since $O_2$ consumed $=a\xi$ and biomass formed $=c\xi$ for extent of reaction $\xi$ (mmol glucose reacted), $$\frac{a}{c}=\frac{n_{O_2}}{n_X}=\frac{16}{12.698}=1.2600.$$ Combined with $a=6-0.975c$: $$1.2600c=6-0.975c\ \Rightarrow\ c=\frac{6}{1.2600+0.975}=2.6846.$$
Back-solve all coefficients. $$b=0.2(2.6846)=0.5369,\quad d=6-2.6846=3.3154,\quad e=6-0.5(2.6846)=4.6577,\quad a=1.2600(2.6846)=3.3826.$$ (All four elemental balances check to within rounding.)
(a) Glucose consumed and $Y_{X/S}$. Extent of reaction from the $O_2$ balance: $$\xi=\frac{n_{O_2}}{a}=\frac{16}{3.3826}=4.7302\ \text{mmol glucose consumed}.$$ $$Y_{X/S}=\frac{X_{\text{formed}}}{\xi}=\frac{0.32\ \text{g}}{4.7302\times10^{-3}\ \text{mol}}=\boxed{67.65\ \text{g DCW/mol glucose}}.$$ Glucose remaining: $$S_{\text{final}}=15-4.7302=\boxed{10.27\ \text{mmol}}.$$
(b) CO$_2$ produced. $$n_{CO_2}=d\,\xi=3.3154(4.7302\times10^{-3}\ \text{mol})=\boxed{15.68\ \text{mmol}}.$$

Result	Value
Stoichiometric coefficients	$a$=3.383, $b$=0.537, $c$=2.685, $d$=3.315, $e$=4.658
Glucose consumed	4.730 mmol
(a) $Y_{X/S}$	67.65 g DCW/mol glucose
(a) Glucose remaining	10.27 mmol
(b) CO$_2$ produced	15.68 mmol

Check: assumes ALL biomass formed and O$_2$ consumed at the sampling instant is attributable to the reaction above (negligible inoculum, as stated), and that the C/H/N/O elemental balances close exactly with the ideal biomass formula CH$_{1.6}$N$_{0.20}$O$_{0.55}$ (its molecular weight, $12+1.6+0.2(14)+0.55(16)=25.2$ g/mol, matches the given 25.2 exactly, confirming the formula is used consistently).

Concept & Theory

Topic. Elemental-balance stoichiometry for aerobic microbial growth.

Key relations. Four elemental balances (C, H, N, O) constrain five stoichiometric coefficients; a measured yield ratio (here O$_2$/biomass) supplies the missing degree of freedom, equivalent to fixing the respiratory quotient or a degree-of-reduction balance.

Why this works. The empirical biomass formula CH$_{1.6}$N$_{0.2}$O$_{0.55}$ converts the "black box" of cell composition into an ordinary chemical species so mass (elemental) balances can be applied exactly as for any reaction stoichiometry.

Common pitfall. Computing $Y_{X/S}$ against the initial glucose supplied (15 mmol) instead of the glucose actually consumed (4.73 mmol) at the sampling instant — the culture is still growing (excess O$_2$/NH$_3$, glucose not exhausted), so the yield must use the consumed amount from the stoichiometry, not the initial charge.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 6 (stoichiometry of microbial growth, elemental balances, yield coefficients).

Question 4 of 6: Aerobic Growth Stoichiometry — Yield and CO 2 Production

Question 4: Aerobic Growth Stoichiometry — Yield and CO2 Production (a. 15 marks; b. 5 marks)

Question 4: Aerobic Growth Stoichiometry — Yield and CO₂ Production (a. 15 marks; b. 5 marks)