Question 3 of 8: Single-Cell Protein from Hexadecane — Stoichiometric Coefficients
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved calculator allowed). Format: Part I offers 5 questions (any 3 constitute a complete answer, 20 marks each) and Part II offers 3 questions (any 2 constitute a complete answer, 20 marks each) — a full paper is 5 questions. All 8 numbered questions are solved below for completeness. Q2–Q5 and Q7 are calculation/derivation questions; Q1, Q6, and Q8 are essay questions.
Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, fermenter mass and energy balances, growth kinetics; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial/eukaryotic cell structure, fungi, protozoa/algae, Gram-stain cell envelope; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — plant tissue structure and cereal grain morphology.
Question 3: Single-Cell Protein from Hexadecane — Stoichiometric Coefficients (20 marks)
Find. The five stoichiometric coefficients $a$ (O2), $b$ (NH3), $c$ (biomass), $d$ (CO2), $e$ (H2O).
Approach. Balance carbon, hydrogen, nitrogen, and oxygen atoms across the reaction, then close the system with the given RQ, giving exactly five independent equations for the five unknowns.
Element
Balance equation
C
$16 = c + d$
H
$34 + 3b = 1.66c + 2e$
N
$b = 0.20c$
O
$2a = 0.27c + 2d + e$
RQ
$d/a = 0.43$
Reduce to two unknowns using the N and RQ equations. From the N balance, $b=0.20c$. From RQ, $d=0.43a$. Substituting these (and $d=16-c$ from the C balance) leaves the H and O balances as two linear equations purely in $a$ and $c$.
Eliminate $e$ between H and O. Solving the H balance for $e$: $e=\tfrac12(34+0.6c-1.66c)=17-0.53c$. Solving the O balance (with $d=16-c$) for $a$ in terms of $c$ and $e$, then substituting $e$, gives a single linear equation in $c$ alone once $d=0.43a$ is also imposed.
Solve the resulting linear system. Carrying the elimination through (four linear equations in $a,c$ after substituting $b,d$) gives:
$$\boxed{a=12.49,\quad b=2.13,\quad c=10.63,\quad d=5.37,\quad e=11.37}$$
(coefficients per mole of hexadecane, rounded to 2 decimal places).
Check all four elemental balances plus RQ. C: $c+d=10.63+5.37=16.00\checkmark$. N: $b=2.13=0.20(10.63)=2.13\checkmark$. H: $34+3(2.13)=40.38$ and $1.66(10.63)+2(11.37)=17.65+22.73=40.38\checkmark$. O: $2(12.49)=24.98$ and $0.27(10.63)+2(5.37)+11.37=2.87+10.74+11.37=24.98\checkmark$. RQ: $5.37/12.49=0.430\checkmark$.
