48. Saccharomyces cerevisiae produces ethanol by fermentation.
The theoretical yield of ethanol from 2.5 g of glucose is __________ g.
Theoretical Yield of Ethanol from 2.5 g Glucose in Saccharomyces cerevisiae Fermentation
Saccharomyces cerevisiae fermentation converts glucose to ethanol anaerobically, yielding a precise theoretical amount from 2.5 g glucose. The answer is 1.22 g, based on the balanced equation and molar ratios. This calculation applies standard biochemical engineering principles for maximum ethanol output.
Fermentation Reaction
Glucose undergoes glycolysis to pyruvate, then alcoholic fermentation produces ethanol and CO2. The balanced equation is:
C6H12O6 → 2 CH3CH2OH + 2 CO2
One mole of glucose (180 g/mol) yields two moles of ethanol (46 g/mol each), giving a theoretical mass yield of (2 × 46)/180 = 0.511 g ethanol per g glucose.
Step-by-Step Calculation
Start with 2.5 g glucose.
- Moles of glucose = 2.5 / 180 = 0.01389 mol
- Ethanol moles produced = 2 × 0.01389 = 0.02778 mol
- Ethanol mass = 0.02778 × 46 = 1.22 g (rounded to three significant figures)
Alternatively, direct yield: 2.5 g × 0.511 = 1.22 g, confirming the result.
Option Analysis
The query resembles multiple-choice formats with options like 120, 12.0, 1.20, 0.120 g.
- 120 g ignores stoichiometry entirely.
- 12.0 g overestimates by a factor of 10, possibly from molar mass confusion.
- 1.20 g approximates closely but rounds prematurely (exact is 1.217 g).
- 0.120 g underestimates by confusing yield ratio or units.
Biotech Applications
In microbial biotechnology, S. cerevisiae achieves 90-95% of theoretical yield industrially due to minor biomass diversion. For bioreactor optimization, this calculation guides scaling from lab (2.5 g glucose) to production, aligning with your interests in fermentation kinetics and biochemical engineering. Real yields factor maintenance coefficients but theoretical values set benchmarks.
