Lesson 3 - Gas Laws SAQ

**Determine the molecular mass of beer if the composition is ethanol 4% (w/w), CO₂ at 5 g/kg and the remainder is water.

Response:

Previously, we considered the molecular mass of beer to be the equivalent of pure water (H₂O) which would be 2 x 1 + 16 = 18 kg/kmol.

We now have three components (ethanol, CO_2, and water) to consider. The molecular mass of the mixture would be given by summing the molecular masses of the three components in a known mass of liquid.

Let us first work out the masses of the three components. To do this, we will consider 100 kg of the mixture.

• As ethanol is given at 4% (w/w), we know that there are 4/100 x 100 = 4 kg of ethanol.

• The molecular formula for ethanol is C₂H₅OH so the molecular mass is 2 x 12 + 5 x 1 + 16 + 1 = 46 kg/kmol.

• Hence, we have 4/46 = 0.087 kmol of ethanol.

Moving on to CO₂:

• CO₂ concentration is given as 5 g/kg. Consequently, there are 5 x 100 = 500 g CO₂.

• The molecular formula for CO₂ so the molecular mass is 12 + 2 x 16 = 44 kg/kmol.

• Therefore, we have 0.5/44 = 0.011 kmol of CO₂.

Finally, considering water:

• The remainder of the mixture is water with a mass of 100 – 4 – 0.5 = 95.5 kg.

• The molecular formula for water is H₂O, so the molecular mass is 2 x 1 + 16 = 18 kg/kmol.

• We have 95.5/18 = 5.3 kmol of water.

In 100 kg of model beer, the total molar quantity of ethanol, CO₂ and water is 0.087 + 0.011 + 5.3 = 5.4 kmol. The molar mass of the model beer is 100/5.4 = 18.52 kg/kmol.

You can appreciate that our previous assumption of beer’s molecular mass being approximated by water is quite accurate, i.e. out by 100 x 0.52/18 = 2.9%.