Aerobic Growth Stoichiometry of S. cerevisiae on Ethanol: 04-BS-13, May 2013, Q4

Question 4: Aerobic Growth Stoichiometry of S. cerevisiae on Ethanol (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. Basis: 1 mol ethanol reacted. Biomass empirical formula $\text{CH}_{1.704}\text{N}_{0.149}\text{O}_{0.408}$ (C-mole basis, MW $=22.32$ g/C-mol); $\text{RQ}=d/a=0.66$; MW ethanol $=46$; MW O₂ $=32$.

Find. (a) $a,b,c,d$ (and, needed to close the system, $e$); (b) $Y_{X/S}$ (kg biomass/kg ethanol) and $Y_{X/\text{O}_2}$ (kg biomass/kg O₂).

Approach. Write C, H, O and N atom balances on the reaction (basis 1 mol ethanol) together with the RQ constraint $d=0.66a$ — five equations in the five unknowns $a,b,c,d,e$, solved simultaneously. The two yield coefficients then follow from $c$ and $a$ converted to a mass basis via the molar masses.

Atom balances (basis 1 mol ethanol, C₂H₆O). $$\text{C: } 2=c+d,\qquad \text{H: } 6+3b=1.704c+2e,\qquad \text{O: } 1+2a=0.408c+2d+e,\qquad \text{N: } b=0.149c.$$
RQ constraint. $d=0.66a$.
Solve the 5×5 linear system. Substituting $d=0.66a$ and $c=2-d=2-0.66a$ into the N, H and O balances and eliminating $b$ and $e$ reduces the system to one equation in $a$, which resolves to $$a=2.917,\quad b=0.01115,\quad c=0.07484,\quad d=1.925,\quad e=2.953.$$
Cross-check via the electron balance. Ethanol's degree of reduction (relative to CO₂/H₂O/NH₃) is $\gamma_{\text{eth}}=4(2)+6-2(1)=12$ per mole; the biomass formula gives $\gamma_X=4(1)+1.704-2(0.408)-3(0.149)=4.44$ per C-mole. O₂ accepts 4 electrons per mole as the terminal acceptor, so the balance $\gamma_{\text{eth}}(1)=\gamma_X\,c+4a$ gives $4.44(0.07484)+4(2.917)=0.332+11.668=12.0$ — matches exactly, confirming the solution.
Part (b): mass-basis yield coefficients. Biomass molar mass $M_X=12+1.704(1)+0.149(14)+0.408(16)=22.32$ g/C-mol. $$Y_{X/S}=\frac{c\,M_X}{(1)(46)}=\frac{0.07484(22.32)}{46}=\boxed{0.0363\ \text{kg biomass/kg ethanol}},$$ $$Y_{X/\text{O}_2}=\frac{c\,M_X}{a(32)}=\frac{0.07484(22.32)}{2.917(32)}=\boxed{0.0179\ \text{kg biomass/kg O}_2}.$$

Quantity	Result
$a$ (O₂)	2.917
$b$ (NH₃)	0.01115
$c$ (biomass)	0.07484
$d$ (CO₂)	1.925
$e$ (H₂O)	2.953
$Y_{X/S}$	0.0363 kg X/kg ethanol
$Y_{X/\text{O}_2}$	0.0179 kg X/kg O₂

Concept & Theory

Topic. Aerobic growth stoichiometry via combined atom and respiratory-quotient balances.

Key relations. C/H/O/N atom balances on the growth reaction; $\text{RQ}=\text{mol CO}_2/\text{mol O}_2$; degree-of-reduction balance $\gamma_S n_S=\gamma_X c+4a$ as an independent check on the atom-balance solution; mass-basis yield $Y_{X/S}=cM_X/(n_S M_S)$.

Why this works. Complete combustion of ethanol alone ($\text{C}_2\text{H}_5\text{OH}+3\text{O}_2\rightarrow2\text{CO}_2+3\text{H}_2\text{O}$) has $\text{RQ}=2/3=0.667$. The given $\text{RQ}=0.66$ sits just below this ceiling, which is the signature of a fermentation where only a small slice of the substrate carbon is diverted into biomass and the overwhelming majority is respired — hence the small but strictly positive value of $c$; an RQ any higher than $2/3$ is not physically achievable by this reaction and would return a negative (unphysical) $c$.

Common pitfall. Reading a small $c$ as an arithmetic error rather than the correct consequence of an RQ close to the full-combustion value — always sanity-check a computed $c$ against the RQ-vs-$2/3$ gap before assuming a mistake. A second pitfall: using the biomass's *molecular* formula weight instead of its *C-mole* formula weight (22.32 g/C-mol here) when computing $Y_{X/S}$, since the formula is deliberately written on a per-carbon-atom basis.

Source. Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (Yield coefficients, respiratory quotient, elemental balances for growth on ethanol).

Question 4 of 10: Aerobic Growth Stoichiometry of S. cerevisiae on Ethanol

Question 4: Aerobic Growth Stoichiometry of S. cerevisiae on Ethanol (20 marks)