# What Carbohydrates remain in brewed beer?

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#### aamcle

##### Well-Known Member

At the moment I'm on a Keto diet (don't ask ) and will be for the next couple of months then I'm going to trade up to a Low Carb diet and maybe have a beer or two.

Now I know most beers run something like 10 to 20g of carbs per pint but I don't know if they are resistant carbs (that is fibre) or the more usual easily digested carbs.

If you know the details or can post a reference I'd appreciate it.

Note. I am NOT asking How Many Calories a beer contains.

Many thanks. Aamcle

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#### aamcle

##### Well-Known Member
I've just found Bamforths paper relating to this.

Nearly all (2/3) the carbs are soluble fibre and can therefore be deducted from the total carbs to give a " net carb" value. Dropping a pint 6-7g of carbs max!

I'll be brewing a lot lager and pale ale.

#### ajdelange

##### Supporting Member
HBT Supporter
I can't help you with the fraction of the residual carbs which are 'good' and which aren't but I guess you have a number from Charley for that. You can estimate the amount of carbs in a beer from the real degree of fermentation (RDF) which the yeast supplier should be able to give you. Or you can WAG it from the apparent degree of attenuation (ADF) which the yeast suppliers aways give. RDF seems to be 10 - 15% less than ADF. Thus if you see ADF of 75% the RDF may be about 60%. Of course you can determine RDF yourself. Measure out 100 mL of beer as accurately as you can, transfer it quantititatively to a beaker and then evaporate it down to 1/3 its original volume. You can do this gently (water bath) or by boiling. Now make back up to the original volume 100 mL with DI water and measure the SG. Convert the SG to Plato. That's the number of grams of extract per 100 grams of the reconstituted beer. Multiply the volume of reconstituted beer (100) mL by 0.998203*SG which is the density of the reconstituted sample to get the mass of the reconstituted beer, 100*0.998203*SG. As the Plato value corresponding to SG is the grams of extract in 100 grams of reconstituted beer you have 100*Plato*0.998203*SG/100 grams of extract in the beaker before you. That came from 100*0.9982038*SG_u (in which SG_u is the specific gravity of the untreated beer sample) grams of beer so the grams of extract per gram of beer is
(100*Plato*0.998203*SG/100)/100*0.9982038*SG_u = Plato*(SG/SG_u)/100 . The grams of extract in the untreated beer per 100 grams of unfinished beer is thus Plato*(SG/SG_u). This is the true extract (TE) of the beer and the RDF is then (OE - TE)/OE.

The TE is the grams of extract in 100 grams of the beer. 100 grams of the beer is 100/(SG_u*.998302) ml of beer so Plato*(SG/SG_u)/100/(SG_u*.998302)= Plato*(SG/SG_u)*(SG_u*.998302)/100 = Plato*SG*.998203 is the number of grams of extract in 100 mL of beer. You can apply Charly's fraction in order to determine how much of this is of which type. You should knock of a wee bit (fraction of a percent) for ash (minerals).

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Thanks

Aamcle

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