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:
C6H12O6 -> 2(C2H5OH) + 2(CO2)
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 CO2 given off. Lastly, we can use all this to create the ratio:
Milliliters of alcohol produced = (116.6ml of ethanol)/(88g of CO2) * weight of CO2 given off by the fermenting mash
Milliliters of alcohol produced = 1.325 * weight of CO2 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!