The value of $k *\delta_e$

In a previous post, I showed that
$$\begin{equation}ABV \approx ( \delta_i - \delta_f  ) ( k * \delta_e )^{-1} \end{equation}$$
Where $\delta_e$ is the density of ethanol
$$\begin{equation} \delta_e = 0.7892 \text{ g/cm}^3 \end{equation}$$
And $k$ is the proportionality constant between the weights of $CO_2$ and $\text{ethanol}$ that are produced during the fermentation process.
$$W_{CO_2} = k*W_e$$
We can derive the value of $k$ by having a look at the fermentation reaction:
$$C_6H_{12}O_{6} \rightarrow 2 C_2H_5OH + 2 CO_2$$

In other words, for each molecule of carbon dioxide released in fermentation, one ethanol molecule is produced. Since the molar mass of ethanol is 46.07 g/mol, compared to 44.01 g/mol for carbon dioxide; we have 

$$\begin{equation}k =  \frac{W_{CO_2}}{W_e}= \frac{44.01 \text{g/mol}}{ 46.07 \text{g/mol}} \approx  0.9553 \end{equation}$$

And now it is time to plug the numbers...

Using the values of (2) & (3) into (1), we finally get

$$ABV \approx ( \delta_i - \delta_f  ) ( 0.9553*0.7892\text{ g/cm}^3)^{-1}$$
$$\Rightarrow ABV \approx ( \delta_i - \delta_f  ) *1.326\text{ cm}^3/ \text{g}$$
In case the hydrometer readings for $\delta_i$ and $\delta_f$ are given in kg/m$^3$ instead of g/cm$^3$, we need to divide by 1000; and if we want the Alcohol by Volume to be expressed as a percentage, we have to multiply by 100. In such case,

$$\boxed{  \%ABV \approx ( \delta_i - \delta_f  ) *0.1326  \text{  [m}^3/  \text{kg]}   }$$

Alcohol content formula... Where does it come from?

One of the most used equations by homebrewers is the formula to calculate the alcohol content by volume (ABV).  Hereby I explain how to derive it.

During the fermentation process, glucose is converted into ethanol and carbon dioxide. Since carbon dioxide escapes from the fermentation bucket, the relation between the initial and final weights is given by:
$$W_i - W_f = W_{CO_2}$$
We know that ethanol and carbon dioxide are always produced in the same proportion. This means that there is a fixed ratio among their weights:  $W_{CO_2} = k*W_e$ , with $k$ a constant. Thus,
$$W_i - W_f = k*W_{e}$$
By expressing the weights in terms of volumes and densities $(W=V*\delta)$, we get:
$$V_i * \delta_i - V_f  * \delta_f = k*V_e * \delta_e$$

And here comes the trick... Let us assume that $V_i \approx V_f$. Thus,

$$V_f *( \delta_i - \delta_f ) \approx  k*V_e * \delta_e \Rightarrow \boxed{( \delta_i - \delta_f  ) ( k \delta_e )^{-1} \approx  \frac{V_e}{ V_f}  = ABV}$$

In the following posts, I will show that it is safe to assume that $V_i \approx V_f$, and determine the value of $k \delta_e$.

Calculating alcohol content in your homebrew beer

The method most employed by homebrewers involves the use of a hydrometer.

The process is very simple:

1. Using a hydrometer, measure the initial density of the must before starting fermentation: $\delta_i$
2. Then, after the fermentation has finished, take a second density reading: $\delta_f$
3. Now subtract both readings and multiply by a given factor:
   $$(\delta_i - \delta_f)* f= ABV$$   

The value obtained is the ABV (alcohol by volume) often expressed as a percentage.

This is simple to calculate, but not many homebrewers know where this formula comes from.

In a later post, I will explain how to derive this equation.