54. During anaerobic growth, an organism coverts glucose (P) into biomass (Q), ethanol (R),
acetaldehyde (S), and glycerol (T). Every mole of carbon present in glucose gets distributed among the
products as follows:
1 (C–mole P) → 0.14 (C–mole Q) + 0.25 (C–mole R) + 0.3 (C–mole S) + 0.31 (C–mole T)
From 1800 grams of glucose fed to the organism, the ethanol produced (in grams) is ___________.
Given data: Atomic weights (Da) of C = 12, H = 1, O= 16, and N – 14
Glucose (C6H12O6, MW 180 g/mol) provides 6 C-moles per mole. From 1800 g glucose, 10 moles are fed, yielding 60 C-moles total.
Carbon Distribution
Every C-mole from glucose distributes as 0.14 C-mole biomass (Q), 0.25 C-mole ethanol (R, C2H5OH), 0.30 C-mole acetaldehyde (S, C2H4O), and 0.31 C-mole glycerol (T, C3H8O3). Verify: 0.14 + 0.25 + 0.30 + 0.31 = 1.00 C-mole, fully accounting for all carbon.
Ethanol Calculation
Ethanol receives 0.25 × 60 = 15 C-moles. Each C-mole ethanol (MW 46 g/mol, 2 C atoms) equals 46/2 = 23 g. Thus, ethanol mass = 15 × 23 = 345 g.
Detailed Yield Breakdown
| Product | C-mole fraction | Total C-moles | MW per C-mole (g) | Mass (g) |
|---|---|---|---|---|
| Q (biomass) | 0.14 | 8.4 | Unknown (not needed) | N/A |
| R (ethanol) | 0.25 | 15 | 23 | 345 |
| S (acetaldehyde) | 0.30 | 18 | 22 (44/2) | 396 |
| T (glycerol) | 0.31 | 18.6 | 23.33 (70/3) | 434 |
Atomic weights confirm MWs: ethanol C2=24, H6=6, O=16 → 46 g/mol.
In anaerobic growth, organisms like yeast convert glucose (P) to biomass (Q), ethanol (R), acetaldehyde (S), and glycerol (T) via metabolic pathways such as glycolysis and alcohol fermentation. This carbon mole distribution—1 C-mole P → 0.14 Q + 0.25 R + 0.3 S + 0.31 T—models real bioprocesses in biochemical engineering and microbial biotechnology, key for ethanol production optimization.
The primary anaerobic glucose ethanol yield hinges on C-mole fractions. Glucose (180 g/mol, 6 C-atoms) from 1800g equals 10 mol or 60 C-moles. Ethanol’s 0.25 fraction yields 15 C-moles. With ethanol MW 46 g/mol (2 C), mass per C-mole is 23g, so 15 × 23 = 345g ethanol—exact for GATE 2017 BT Q54.
Step-by-Step Anaerobic Fermentation Calculation
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Moles of glucose: 1800g / 180 g/mol = 10 mol.
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Total C-moles: 10 × 6 = 60 C-moles.
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Ethanol C-moles: 0.25 × 60 = 15.
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Ethanol MW breakdown: C2H6O = (24 + 6 + 16) = 46 g/mol; per C-mole = 23 g.
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Final yield: 15 × 23 = 345 g. No atomic weights for N needed, as biomass unspecified.
This mirrors Saccharomyces cerevisiae fermentation, balancing carbon without oxygen.
SEO-Optimized Insights for Biochemical Engineers
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Carbon balance verification: Fractions sum to 1, ensuring mass conservation.
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Practical applications: Scale-up for bioethanol from 1800g glucose feedstock.
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Common pitfalls: Forget C-moles vs. moles—always normalize to carbon atoms.
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Related queries: Glycerol yield (18.6 C-moles × 70/3 g/C-mole ≈ 434g); acetaldehyde (18 × 22g = 396g).
For students in biotechnology or GATE preparation, this anaerobic growth glucose ethanol yield problem tests stoichiometry fundamentals.
