How Much CO2 is Produced from Brewing?

Off-week bonus post!  Ever wonder how much CO₂ is actually produced when you ferment your favorite bubbly beverage?  I’ve often found myself staring at my air-lock or blow-off bucket thinking it must be a lot.  I finally decided that I should just go ahead and calculate it; as turns out, I was right.

The results of my calculations are presented here, in both English and metric units.  Below I give a bit more detail on the actual calculations.  If you look at the plots, you’ll notice that not only is the volume of gas large (left axis), the weight of the gas (right axis) is also quite appreciable.  Yes, your fermentor looses weight over the course of the brew!  In fact, some people have proposed monitoring the attenuation of a batch of beer in real time by monitoring the weight of the fermentor.  I always assumed you would need an extremely precise scale, but it turns out the difference is so large, you actually wouldn’t.

CO₂ production for various specific gravities and apparent attenuations.

CO₂ production for various specific gravities and apparent attenuations.

For anyone interested in the math behind the plots, read on.

The first thing I did was to convert specific gravity (SG in 1.xxx format) to degrees Plato (°P) with the equation .  This is a third order polynomial fit derived from the American Society of Brewing Chemists gravity to Plato tables, given on en.wikipedia.org/wiki/brix.  This is a convenient unit to work in because every 1°P is 1% by weight sugar, so a 1.060 SG wort would be 14.741°P, or 14.741% sugar by weight.

To find the mass of sugar in each wort, I then found the weight of a 5 gal/ 18.9 L batch by multiplying this volume by its density (if working in metric, this amounts to simply multiplying by the specific gravity since water is 1 kg/L).  Then, its a simple matter of multiplying by the sugar weight percent.  The example 1.060 wort contains (44.23 lb wort)*(0.14741)= 6.52 lb or 2.96 kg sugar.  Note that I actually did all of my calculations in metric, then converted to English because the units work out so nicely.

Next, we must find how much sugar is actually consumed.  For this, remember that the “attenuation” used most often is “apparent attenuation” as measured by a hydrometer.  Hydrometers actually measure the density of the solution, so when alcohol is created (its density is lower than water), the hydrometer is tricked into thinking more of the sugar has been consumed than really was.  Actual attenuation can by found by multiplying the apparent attenuation by 0.814, as given in Greg Noonan’s New Brewing Lager Beer.  So the actual amount of sugar consumed in a 1.060 75% apparent attenuation beer is (6.52 lb)(0.75*0.814)= 3.98 lb or 1.81 kg.

Now we must bring in some very basic chemistry.  Lets assume this sugar is all (or could become) glucose.  A single mole (in units of mol) of glucose weighs 180.156 g or  6.35 oz.  A mole is simply a way to keep track of how many molecules of something there is so you can predict chemical reactions.  Basically, there are a different number of molecules of acid in vinegar in a gram than the number of baking soda molecules in a gram, but reactions happen to molecules.  So to have a complete reaction, you need to add the same number of molecules of acid in vinegar and baking soda, not the same number of grams of each.  The upshot is we have (1.81 kg)/(0.180156 kg/mol)=10.02 mol glucose consumed in this example wort.

For every 1 mol glucose consumed, 2 mol of ethanol and 2 mol of CO₂ are produced.  Thus we have 10.02*2=20.04 mol of total CO₂ produced from fermenting this 1.060 wort to 75% apparent attenuation.

Now we can put this in terms of volume or mass of gas.  For any “ideal” gas, its volume can be predicted by the ideal gas law at standard temperature and pressure, , where P is pressure (1 atm or atmospheres), V is volume (liters or L), n is the number of moles of gas (1 mol), R is the ideal gas constant (0.0821 (atm*L)/(mol*K)), and T is the temperature (273 Kelvin or K).  Solving for volume and plugging in the numbers, we get .  Thus every mol of gas occupies 22.41 L or 0.791 cubic feet.  This would be (20.04 mol)*(22.41 L/mol)=449.1 L or 15.86 ft³ of CO₂ produced for our example beer.  Put another way, from 5 gallons of 1.060 SG wort, you could fill the glass carboy its in almost 24 times with the CO₂!  In the most extreme case shown on the graph (1.110 wort, 85% attenuation), this more than doubles.

To find the mass of the gas, we simply use the mass of a mole of CO₂, 0.04401 kg/mol, to convert back.  We have (20.04 mol)*(0.04401 kg/mol)=0.88 kg or 1.94 lb of CO₂ produced.

– Dennis,
Life, Fermented

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