Calculation of alcohol content If the fermentation reaction is rewritten taking water into account, one has: x sugar + (1-x) water → 2xy alcohol + 2xy CO2 + x(1-y) sugar + (1-x) water where y represents the proportion of sugar turned into alcohol, called attenuation.

Sugar Water Alcohol Honey CO2 (gas)

Molar mass M (g/mol)

Density d

180 18 46

1,55 1 0,79 1,41 0,0018

44

(d-1)/d

0,355 0 -0,2595 0,2908 -554,6

Molar volume (L/mol) (= M/d) 0,116 0,018 0,058 24

Table 8: some properties of sugar, alcohol, etc. One can notice that

Ms d ≈ 1,96 and s ≈ 1,95 . 2M a da

d s 2M a ≈1, that is da Ms 2xyVa + x(1 - y)Vs + (1 − x)Vw =

So

xVs + (1 − x)Vw + x

Ms ds

the

 d s 2M a  − 1 y   da Ms 

Final density is by definition: V 2xyM a + x(1 - y)M s + (1 − x)M w df = w M w 2xyVa + x(1 - y)Vs + (1 − x)Vw This can be simplified using that the volume is constant:

 d -d  d f ≈ 1 + 1 − y s a  (di − 1) ds - 1   This expression can be modified to get the attenuation as a function of di and df: d − df d − 1 di − d f ≈ 0,73 i y≈ s di − 1 d s − d a di − 1 The alcohol by volume is equal to the volume of alcohol (in ml) per 100 ml mead: 2xyVa ABV = 2xyVa + x(1 - y)Vs + (1 − x)Vw

volume

is

constant. By definition the initial density (specific gravity) is the initial mass of the must (water +sugar) divided by the initial volume: V xM s + (1 − x)M w di = w . Which can be M w xVs + (1 − x)Vw rewritten: Ms ds −1 x Mw ds di = 1 + M 1  1 +  s - 1 x  M w ds  Mead made complicated: alcohol content

It is also possible to get x as a function of the M 1 ds − di 1 initial density: - 1 = s x M w ds di −1

≈

di − df ≈ 1,32 (d i − d f ) ds − da

So the attenuation and the alcohol by volume can be calculated, knowning initial and final densities,. The sugar by volume is: x(1 - y)Vs SBV = x(1 - y)Vs + 2xyVa + (1 − x)Vw ≈ (1 − y)

•

di −1 di −1 di − df − ≈ ds −1 ds −1 ds − da

If all the sugar has been consumed (y = 1) :

df = 1−

1− da (d i − 1) ≈ 1 − 0,37 (d i − 1) ds −1

TVa ≈

1

di −1 y ≈ 1,8 (d i − 1)y ds −1

Mathieu Bouville 2001-2002

•

•

ds − 1 ≈ 73 % . So when the ds − da density is 1, three quarters or so of the sugar have been turned into alcohol (whatever the initial density is). White labs publishes attenuations for their yeasts of 70-80 %.

di 1,052 1,092 1,132

If d = 1, y ≈

Examples (final density df is equal to 0.995 in all examples).

Mead made complicated: alcohol content

y 80,0 % 77,0 % 75,8 %

TVa 7,6 % 12,9 % 18,3 %

TVs 1,9 % 3,9 % 5,8 %

TMs 30 g/L 60 g/L 90 g/L

Table 9: some properties of mead as a function of original gravity (final gravity is .995). One can notice that there is a lot of sugar remaining but the sugar content does not depend on the final density only. So the final density should not be used, as some people do, as a criterion to determine whether the mead is dry or sweet. Is there really so much sugar at the end? If so, why do bacteria not attack all meads, not only the sweeter ones?

2

Mathieu Bouville 2001-2002