Molasses Fermentation: Yield, Losses and a Yeast Sales Claim
A small distillery ferments molasses. In a trial, 500 kg of sucrose (C₁₂H₂₂O₁₁) is hydrolysed with water and fermented according to the overall equations: C₁₂H₂₂O₁₁ + H₂O → 2 C₆H₁₂O₆ C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂ After fermentation and distillation, 200 kg of ethanol is collected. The distillery is considering a new yeast advertised to give '95% of theoretical yield'. The current yeast tolerates up to 12% ethanol by volume, and the fermentation broth reaches about 10% ethanol before distillation. The new yeast costs US$8,000 per year and the ethanol sells for US$1.20 per kg. (Aᵣ: C=12, H=1, O=16; density of ethanol = 0.789 g cm⁻³.)
1. Calculate the theoretical mass of ethanol obtainable from 500 kg of sucrose, and the percentage yield actually obtained.
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Molar mass of sucrose = (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g mol⁻¹.
n(sucrose) = 500 000 g / 342 g mol⁻¹ = 1.462 × 10³ mol = 1.462 kmol.
From the equations: 1 mol sucrose → 2 mol hexose → 4 mol ethanol. (Each hexose gives 2 ethanol.)
n(ethanol) = 4 × 1.462 = 5.848 kmol.
Molar mass of ethanol = 46.0 g mol⁻¹.
theoretical mass = 5.848 × 46.0 = 269.0 kg.
Percentage yield = (200 / 269.0) × 100 = 74.3%.
So the trial is about 74% of theoretical, not 95%.
2. Explain two reasons why the actual yield is below the theoretical yield.
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Fermentation is not complete. The yeast converts sugar to ethanol and CO₂, but some sugar remains, and some is used for yeast growth and by-products such as glycerol. The reaction is also inhibited by ethanol.
Ethanol is lost during processing. Some evaporates with the CO₂ produced, and some remains in the stillage after distillation. The 200 kg figure is measured after distillation, so it includes these losses.
Other valid reasons: the yeast is inhibited as ethanol approaches 10–12% by volume (the broth reaches about 10%), so the last part of the fermentation is slow; the molasses may contain non-fermentable solids; and some ethanol is lost in sampling and handling. The theoretical calculation assumes 100% conversion and perfect recovery, which never happens.
3. Evaluate the new yeast's claim that it gives '95% of theoretical yield'. Include a calculation and a commercial judgement.
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95% of the theoretical 269.0 kg is 0.95 × 269.0 = 255.6 kg ethanol per 500 kg sucrose.
The current trial gives 200 kg, so the claimed improvement is 55.6 kg per batch.
At US$1.20 per kg, extra revenue = 55.6 × 1.20 = US$66.7 per batch.
The yeast costs US$8,000 per year, so the break-even number of batches is 8000 / 66.7 ≈ 120 batches per year. That is about one batch every three days, before allowing for any extra fermentation time, larger vessels or distillation cost.
Chemically, 95% is very high for a fermentation that must stop at about 10–12% ethanol, where yeast inhibition and product loss are significant. The claim is not impossible, but it is an advertising claim and must be tested in a controlled trial using the distillery's own molasses, temperature and distillation. The decision should not be based on the advert alone: compare the yield before distillation and after distillation, check the yeast's ethanol tolerance, and calculate the payback period with the real batch frequency.