Elemental & Electron Balance for Clostridium acetobutylicum Fermentation: 04-BS-13, May 2013, Q1

Question 1: Elemental & Electron Balance for Clostridium acetobutylicum Fermentation (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. The fully-specified reaction above: 100 mol glucose + 11.2 mol NH₃ forming 13 mol biomass (unknown formula) + 22 mol acetone + 0.4 mol butyrate (C₄H₈O₂) + 14 mol acetate (C₂H₄O₂) + 221 mol CO₂ + 135 mol H₂ + 0.7 mol ethanol (C₂H₆O).

Find. (a) α, β, γ, δ in the biomass formula C_αH_βN_γO_δ; (b) the total electrons transferred (available-electron balance) and where they end up.

Approach. Four unknowns (α,β,γ,δ) are pinned down exactly by four independent atom balances (C, H, N, O), since every other species in the equation is fully specified. The degree-of-reduction (available-electron) balance is then a linear combination of those same atom balances relative to the CO₂/H₂O/NH₃ reference state — it is not new information, but it is the standard cross-check (and the tool part (b) asks for) that confirms the elemental balance in part (a) is self-consistent.

Carbon balance. Left side: $100\times6=600$. Right side (known species): $22(3)+0.4(4)+14(2)+221(1)+135(0)+0.7(2)=66+1.6+28+221+1.4=318$. So $$13\alpha = 600-318=282\ \Rightarrow\ \alpha=\frac{282}{13}=21.69.$$
Hydrogen balance. Left side: $100(12)+11.2(3)=1200+33.6=1233.6$. Right side (known): $22(6)+0.4(8)+14(4)+221(0)+135(2)+0.7(6)=132+3.2+56+0+270+4.2=465.4$. So $$13\beta=1233.6-465.4=768.2\ \Rightarrow\ \beta=\frac{768.2}{13}=59.09.$$
Nitrogen balance. Only NH₃ and the biomass carry nitrogen: $13\gamma = 11.2\ \Rightarrow\ \gamma=\dfrac{11.2}{13}=0.862.$
Oxygen balance. Left side: $100(6)=600$. Right side (known): $22(1)+0.4(2)+14(2)+221(2)+135(0)+0.7(1)=22+0.8+28+442+0.7=493.5$. So $$13\delta=600-493.5=106.5\ \Rightarrow\ \delta=\frac{106.5}{13}=8.19.$$
Part (b): degree-of-reduction (available electrons) of each species. Using the standard convention relative to CO₂, H₂O and NH₃ (each defined as zero available electrons), a compound C_cH_hO_oN_n carries $\gamma=4c+h-2o-3n$ available electrons per mole (C contributes 4, H contributes 1, O withdraws 2, N withdraws 3). Glucose: $\gamma_{\text{glu}}=4(6)+12-2(6)=24$ per mole, so the 100 mol charged carry $$\text{AE}_{\text{glucose}} = 100(24) = \boxed{2400\ \text{mol e}^-\text{-equivalents}}.$$
Distribute the electrons among the products and verify closure. Substituting $\alpha,\beta,\gamma,\delta$ into $\gamma_{\text{bio}}=4\alpha+\beta-2\delta-3\gamma=126.9$ per mole gives $13(126.9)=1649.6$ routed to biomass. The remaining products carry: acetone $22(4\cdot3+6-2)=352$; butyrate $0.4(4\cdot4+8-4)=8$; acetate $14(4\cdot2+4-4)=112$; CO₂ $221(0)=0$ (fully oxidized, by definition of the reference state); H₂ $135(2)=270$; ethanol $0.7(4\cdot2+6-2)=8.4$. Summing: $$1649.6+352+8+112+0+270+8.4 = 2400\ \text{mol e}^-\text{-equivalents},$$ exactly matching the glucose electron content in Step 5 — the balance closes with zero discrepancy, confirming the part (a) elemental balance is internally consistent.

Quantity	Result
α (C)	21.69
β (H)	59.09
γ (N)	0.862
δ (O)	8.19
Total available electrons transferred from glucose	2400 mol e⁻-equivalents
Electrons routed to biomass	1649.6 (68.7%)
Electrons routed to H₂ (excess-reductant sink)	270 (11.3%)
Electron balance closure	exact (0 discrepancy)

Concept & Theory

Topic. Elemental mass balances and the degree-of-reduction (available-electron) method for fermentation stoichiometry.

Key relations. Atom balance: moles of each element in $=$ moles out. Degree of reduction relative to CO₂/H₂O/NH₃: $\gamma=4c+h-2o-3n$ per mole of C_cH_hO_oN_n; the overall electron balance $\sum(\text{moles}_i\cdot\gamma_i)_{\text{in}}=\sum(\text{moles}_i\cdot\gamma_i)_{\text{out}}$.

Why this works. With every species except the biomass formula fully specified, C/H/N/O balances give four independent linear equations in the four unknowns α,β,γ,δ — an exactly-determined system. The degree-of-reduction balance is a fixed linear combination of the C, H and O balances (it carries no N term when NH₃ is the reference), so it cannot fix a fifth unknown; its value lies in auditing the arithmetic of parts (a): if the elemental balances contain an error, the electron balance will not close.

Common pitfall. Treating the electron balance as an independent equation that could re-solve α–δ a second, inconsistent way — it cannot, because it is linearly dependent on the atom balances already used. A second pitfall is forgetting that CO₂, H₂O and NH₃ are defined to carry zero available electrons in this convention, so none of the 221 mol CO₂ contributes to the electron tally even though it is the single largest product by mole count.

Source. Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (Stoichiometry of Microbial Growth and Product Formation — elemental and available-electron balances).

Question 1 of 10: Elemental & Electron Balance for Clostridium acetobutylicum Fermentation

Question 1: Elemental & Electron Balance for Clostridium acetobutylicum Fermentation (20 marks)