Growth Stoichiometry and Yield Coefficients on Glucose: 04-BS-13, May 2016, Q4

Question 3: Growth Stoichiometry and Yield Coefficients on Glucose (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
Biomass formula (whole molecule, ash neglected)	C_4.4H_7.3N_0.86O_1.2, MW = 91.34 g/mol
Glucose	C₆H₁₂O₆, MW = 180 g/mol (6 C atoms)
Carbon converted to biomass	2/3 (w/w of substrate carbon)

Find. (a) coefficients $a,b,c,d,e$; (b) $Y_X$ (g biomass/g glucose) and $Y_{X/O_2}$ (g biomass/g O₂).

Approach. The "2/3 of substrate carbon to biomass" statement is the fifth piece of information needed alongside the four C/H/N/O atom balances (5 unknowns, only 4 balances). Fix $c$ from the carbon-conversion fraction first, then solve the remaining four balances in sequence (N, C, H, O).

Biomass coefficient $c$ from the carbon-conversion fraction. Per mole glucose, carbon supplied $=6(12)=72$ g. Carbon routed to biomass $=\tfrac23(72)=48$ g $=4$ mol C atoms. Each biomass molecule carries 4.4 C atoms, so $$c=\frac{4}{4.4}=\boxed{0.909\ \text{mol biomass/mol glucose}}.$$
Nitrogen balance → $b$. NH₃ is the only N source: $b=0.86c=0.86(0.909)=\boxed{0.782\ \text{mol NH}_3\text{/mol glucose}}.$
Carbon balance → $e$. $6=4.4c+e\Rightarrow e=6-4.4(0.909)=\boxed{2.000\ \text{mol CO}_2\text{/mol glucose}}$ (the carbon not fixed in biomass is fully respired).
Hydrogen balance → $d$. $12+3b=7.3c+2d\Rightarrow d=\dfrac{12+3(0.782)-7.3(0.909)}{2}=\boxed{3.855\ \text{mol H}_2\text{O/mol glucose}}.$
Oxygen balance → $a$. $6+2a=1.2c+d+2e\Rightarrow a=\dfrac{1.2(0.909)+3.855+2(2.000)-6}{2}=\boxed{1.473\ \text{mol O}_2\text{/mol glucose}}.$
Part (b): yield coefficients. $$Y_X=\frac{c\,(\text{MW}_{\text{biomass}})}{1\,\text{mol}\times\text{MW}_{\text{glucose}}}=\frac{0.909(91.34)}{180}=\boxed{0.461\ \text{g biomass/g glucose}},$$ $$Y_{X/O_2}=\frac{c\,(\text{MW}_{\text{biomass}})}{a\,(\text{MW}_{O_2})}=\frac{0.909(91.34)}{1.473(32)}=\boxed{1.762\ \text{g biomass/g O}_2}.$$

Quantity	Result
$c$ (biomass)	0.909 mol/mol glucose
$b$ (NH₃)	0.782 mol/mol glucose
$e$ (CO₂)	2.000 mol/mol glucose
$d$ (H₂O)	3.855 mol/mol glucose
$a$ (O₂)	1.473 mol/mol glucose
$Y_X$	0.461 g biomass/g glucose
$Y_{X/O_2}$	1.762 g biomass/g O₂

Topic: Elemental balances for aerobic growth stoichiometry when a mass-fraction (not molar) split is given.

Key relations: C/H/N/O atom balances (4 equations); an experimentally measured carbon-partition fraction supplies the 5th equation needed to fix all 5 unknown coefficients.

Why this works: a written stoichiometric equation with an unspecified empirical biomass formula always has one more unknown than the atom balances can resolve; any independent measured quantity — a yield, a partition fraction, or a degree-of-reduction ratio — supplies exactly the missing equation, and it does not matter which one is given as long as it is expressed on a consistent (molar or mass) basis.

Common pitfall: applying the 2/3 split to moles of glucose instead of moles (or mass) of carbon — the statement is explicitly "2/3 (w/w) of the substrate carbon", i.e. a carbon mass balance, not a molecule count, and misreading it as "2/3 of the glucose molecules become biomass" gives a badly wrong $c$.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 5 (elemental balances, yield coefficients).

Question 4 of 9: Growth Stoichiometry and Yield Coefficients on Glucose

Question 3: Growth Stoichiometry and Yield Coefficients on Glucose (20 marks)