Single-Cell Protein from Hexadecane — Stoichiometry and Yield Coefficients: 04-BS-13, December 2019, Q4

Question 4: Single-Cell Protein from Hexadecane — Stoichiometry and Yield Coefficients (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.

Quantity	Value
Substrate	hexadecane, $C_{16}H_{34}$, MW = 226 g/mol
Biomass formula (per mole)	$C_{4.4}H_{7.3}N_{0.86}O_{1.2}$, MW = 91.34 g/mol
Carbon conversion to biomass	2/3 (w/w, mass basis)

Find. (a) $a,b,c,d,e$; (b) $Y_{x/s}$, $Y_{x/O_2}$ (g/g).

Approach. Basis 1 mol hexadecane. The C, H, N, O atom balances give four of the five equations needed; the fifth comes from the stated mass-carbon-conversion fraction, which fixes $c$ directly (it constrains only the ratio of carbon atoms in biomass to carbon atoms in substrate, both multiplied by the same atomic mass of carbon, so the atomic-mass factor cancels).

(a) Fixing $c$ from the carbon-conversion fraction. Mass of carbon in 1 mol substrate $=16(12)=192$; mass of carbon converted to biomass $=c(4.4)(12)$. Setting the ratio to 2/3: $$\frac{c(4.4)(12)}{16(12)}=\frac{2}{3}\ \Longrightarrow\ \boxed{c=\dfrac{80}{33}=2.424\ \text{mol biomass/mol hexadecane}}.$$
Remaining atom balances. With $c$ fixed, the N balance gives $b$ directly, the C balance gives $e$, and the H and O balances solve simultaneously for $d$ and $a$: $$\text{N: } b=0.86c=\boxed{2.085},\qquad \text{C: } 16=4.4c+e\Rightarrow e=\boxed{5.333\ (=16/3)},$$ $$\text{H: } 34+3b=7.3c+2d\ \Rightarrow\ d=\boxed{11.279},\qquad \text{O: } 2a=1.2c+d+2e\ \Rightarrow\ a=\boxed{12.427}.$$ Check (all four elemental balances closed exactly by construction): $a=1367/110,\ b=344/165,\ c=80/33,\ d=1861/165,\ e=16/3$.
(b) Yield coefficients. Biomass MW $=4.4(12)+7.3(1)+0.86(14)+1.2(16)=91.34$ g/mol; hexadecane MW $=16(12)+34(1)=226$ g/mol. $$Y_{x/s}=\frac{c\,(91.34)}{1(226)}=\frac{2.424(91.34)}{226}=\boxed{0.980\ \text{g biomass/g substrate}},$$ $$Y_{x/O_2}=\frac{c\,(91.34)}{a\,(32)}=\frac{2.424(91.34)}{12.427(32)}=\boxed{0.557\ \text{g biomass/g O}_2}.$$ A yield above 0.9 g biomass per g substrate is plausible here specifically because hexadecane is a highly reduced hydrocarbon (no oxygen atoms, high available-electron content per gram), so a substantial fraction of its mass can end up as biomass even though nearly the same fraction of carbon (2/3) is diverted as for far less reduced substrates like glucose.

Coefficient	Value
$a$ (O₂)	12.427
$b$ (NH₃)	2.085
$c$ (biomass)	2.424
$d$ (H₂O)	11.279
$e$ (CO₂)	5.333
$Y_{x/s}$	0.980 g/g
$Y_{x/O_2}$	0.557 g/g

Topic: Elemental-balance fermentation stoichiometry with a mass-fraction carbon-conversion constraint supplying the fifth equation.

Key relations: C/H/N/O atom balances on the empirical biomass formula; a mass-basis carbon-conversion fraction $f_C=\dfrac{c\cdot n_{C,biomass}\cdot 12}{n_{C,substrate}\cdot12}$ fixes $c$ before the remaining balances are solved.

Why this works: five stoichiometric unknowns ($a$–$e$) need five independent equations; the four atom balances supply four, and the stated carbon-partitioning fraction (a piece of given experimental data, not derivable from the atom balances alone) supplies the fifth, letting $c$ be solved first and the rest to follow directly.

Common pitfall: forgetting that the 12 g/mol carbon atomic mass cancels out of the carbon-conversion-fraction equation (since it appears identically on both substrate and biomass sides) — the fraction is really just a ratio of carbon atom counts weighted by the biomass coefficient, not a full-molecule mass ratio.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (elemental balances and yield coefficients in single-cell-protein stoichiometry).

Question 4 of 9: Single-Cell Protein from Hexadecane — Stoichiometry and Yield Coefficients

Question 4: Single-Cell Protein from Hexadecane — Stoichiometry and Yield Coefficients (20 marks)