Question 3 of 10: Methylophilus methylotrophus — Maximum Yield and Oxygen Demand

Question 3: Methylophilus methylotrophus — Maximum Yield and Oxygen Demand (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.

Find. (a) $c_{\max}$, the maximum molar biomass yield (mol biomass/mol methanol); (b) $a$, the O₂ demand at the actual (42%-of-maximum) yield.

Approach. Methanol supplies only 1 carbon per mole, so $c\le1$ purely from the carbon balance — the true ceiling here is carbon-limited, not electron-limited (a naive $\gamma_S/\gamma_B$ ratio would over-predict $c_{\max}>1$, which is impossible with a single-carbon substrate). The thermodynamic maximum is therefore the case where all substrate carbon is fixed into biomass ($c=1$, no CO₂ by-product, $e=0$); the electron and atom balances then jointly fix how much O₂ is still needed to dispose of the substrate's surplus available electrons. Part (b) repeats the same balance machinery at $c=0.42(1)$.

1. Part (a): carbon-balance ceiling. With 1 C atom per mole methanol and the biomass formula normalized per C-atom, $1=c+e$. Maximum $c$ occurs at $e=0$: $$\boxed{c_{\max}=1.0\ \text{mol biomass/mol methanol}}.$$
2. Solve $b$, $d$, $a$ at $c=c_{\max}=1$, $e=0$. N: $b=0.22(1)=0.22$. H: $4+3b=1.68c+2d\Rightarrow d=\dfrac{4+3(0.22)-1.68(1)}{2}=1.49$. O: $1+2a=0.36c+d+2e\Rightarrow a=\dfrac{0.36(1)+1.49-1}{2}=\boxed{0.425\ \text{mol O}_2\text{/mol methanol}}$.
3. Cross-check via the electron balance. $\gamma_S(1)=\gamma_B c+4a\Rightarrow a=\dfrac{6(1)-4.3(1)}{4}=0.425$ — exact match, and the O-balance in step 2 closes exactly (1.850 both sides), confirming $c_{\max}=1$ with $a=0.425$, $b=0.22$, $d=1.49$ is a fully self-consistent stoichiometry.
4. Part (b): actual yield at 42% of maximum. $$c_{\text{actual}}=0.42\,c_{\max}=0.42(1.0)=\boxed{0.42\ \text{mol biomass/mol methanol}}.$$ Now $e\ne0$ (some carbon is respired to CO₂ instead of fixed): C: $e=1-c=0.58$. N: $b=0.22(0.42)=0.0924$. H: $d=\dfrac{4+3(0.0924)-1.68(0.42)}{2}=1.786$.
5. Oxygen demand at the actual yield. O: $1+2a=0.36c+d+2e=0.36(0.42)+1.786+2(0.58)=3.097\Rightarrow$ $$a=\boxed{1.049\ \text{mol O}_2\text{/mol methanol}}.$$ Electron-balance cross-check: $a=[6(1)-4.3(0.42)]/4=1.049$ — identical. Since MW methanol $=$ MW O₂ $=32$, the mass ratio equals the same numeric value: $1.049$ g O₂/g methanol.