Over the weekend I was working on my latest batch of typha(cattail) ethanol and I realized that I had forgotten my alcohol refractometer at school. I wanted to know the amount of ethanol I was producing. How could I figure this out? Let’s start with the chemical equation for fermentation:

**C**_{6}H_{12}O_{6} -> 2(C_{2}H_{5}OH) + 2(CO_{2})

If I know the atomic weights for each of the atoms in the various molecules I can calculate a ratio of masses(or weights if everything is at 1g sitting on a lab bench). Using closely rounded numbers:

C = Carbon = 12 grams/mol

H = Hydrogen = 1 gram/mol

O = Oxygen = 16 grams/mol

Therefore fermentation expressed as weights becomes:

(6*12+1*12+16*6) -> 2*(2*12+5*1+1*16+1*1) + 2*(1*12+2*16)

or:

(180g) -> (92g) + (88g)

So if 88 grams of CO2 is given off, 92 grams of ethanol should have been formed. (For the mathematically astute, I multiplied both sides of the equation by one mole.) Using a pure ethanol density of .789 grams/milliliter, (92grams/(.789 grams/ml)) = 116.6ml of pure ethanol is created for every 88grams of** **CO_{2 }given off. Lastly, we can use all this to create the ratio:

Milliliters of alcohol produced = (116.6ml of ethanol)/(88g of CO_{2}) * weight of CO_{2 }given off by the fermenting mash

Milliliters of alcohol produced = 1.325 * weight of CO_{2 }given off by the fermenting mash

My fermenting beaker off gassed 5g of weight therefore I made a vast 6.63ml of ethanol. That should run my tractor for about twenty feet!

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