Elemental-Balance Stoichiometry of Aerobic Glucose-to-Biomass Growth: 20-Bio-A5 Systems Analysis & Control, December 2017, Q4

Question 4: Elemental-Balance Stoichiometry of Aerobic Glucose-to-Biomass Growth (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	Symbol	Value
Initial glucose supplied	$S_0$	15 mmol
Biomass (dry cell weight) formed	—	0.3 g
Biomass molecular weight	$MW_x$	25.2 g/mol
Oxygen consumed	—	15 mmol
Ammonia (nitrogen source)	—	supplied in excess

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

Approach. Balance the C, H, N, O elements of the stated growth reaction (per 1 mol glucose consumed) to express every coefficient in terms of the single biomass coefficient $c$; then use the measured ratio of O$_2$ consumed to biomass formed — both proportional to the same reaction extent — to pin down $c$, and hence every coefficient and the reaction extent itself.

Biomass formed, in moles. $$n_x=\frac{0.3\ \text{g}}{25.2\ \text{g/mol}}=0.011905\ \text{mol}=11.905\ \text{mmol}.$$
Elemental balances (per 1 mol glucose consumed). Nitrogen: $b=0.20c$. Carbon: $d=6-c$. Hydrogen ($12+3b=1.6c+2e$): $e=6-0.5c$. Oxygen ($6+2a=0.55c+2d+e$): $a=6-0.975c$.
Tie the coefficients to the measured data. O$_2$ consumed and biomass formed are both proportional to the reaction extent $\xi$ ($a\xi$ and $c\xi$ respectively), so their measured ratio fixes $a/c$ directly: $$\frac{a}{c}=\frac{15\ \text{mmol O}_2}{11.905\ \text{mmol biomass}}=1.260.$$
Solve for the coefficients. Combining $a=1.260c$ with $a=6-0.975c$: $$c=\frac{6}{1.260+0.975}=2.685,\quad a=3.383,\quad b=0.537,\quad d=3.315,\quad e=4.658.$$
(a) Yield coefficient. $$Y_{x/s}=c\times MW_x=2.685(25.2)=\boxed{67.7\ \text{g DCW/mol glucose}}.$$
Reaction extent and glucose remaining. Since the glucose coefficient is 1, the extent equals the moles of glucose consumed: $$\xi=\frac{\text{O}_2\text{ consumed}}{a}=\frac{15}{3.383}=4.435\ \text{mmol}, \qquad S_{\text{remaining}}=S_0-\xi=15-4.435=\boxed{10.57\ \text{mmol}}.$$
(b) CO$_2$ produced. $$n_{\text{CO}_2}=d\,\xi=3.315(4.435)=\boxed{14.7\ \text{mmol}}.$$

Result	Value
Stoichiometric coefficients $a,b,c,d,e$	3.38, 0.537, 2.68, 3.32, 4.66
(a) Yield $Y_{x/s}$	67.7 g DCW/mol glucose
(a) Glucose remaining	10.57 mmol
(b) CO$_2$ produced	14.7 mmol

Concept & Theory

Topic. Macroscopic elemental (C, H, N, O) balances for aerobic biomass growth.

Key relations. Elemental balance on a formal biomass formula $CH_pN_qO_r$; every stoichiometric coefficient scales with a single reaction extent $\xi$, so any one measured extensive quantity (here O$_2$ consumed) converts directly to all the others once the coefficients are known.

Why this works. Four elemental balances (C, H, N, O) leave one coefficient free after normalizing the glucose coefficient to 1; a single independent experimental measurement (the O$_2$/biomass ratio) supplies exactly the missing piece of information, making the system fully determined without needing a degree-of-reduction or respiratory-quotient assumption.

Common pitfall. Using the total moles of O$_2$ or biomass directly as if they were per-mole-of-glucose coefficients, instead of first forming their RATIO (which cancels the unknown reaction extent) — and forgetting that NH$_3$ supplies nitrogen only, so it never appears in the oxygen balance.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 4 (stoichiometry and elemental balances of microbial growth).

Question 4 of 6: Elemental-Balance Stoichiometry of Aerobic Glucose-to-Biomass Growth

Question 4: Elemental-Balance Stoichiometry of Aerobic Glucose-to-Biomass Growth (a. 15 marks; b. 5 marks)