So we’ve gone through what mass balancing is and the mass balancing equations. Here is one of the harder parts: calculating the mass balance when there are reactions occurring. This is called stoichiometric mass balancing. This will be basic stoichiometric mass balancing (aka what I learnt in year 1). There is also elemental mass balancing, which is more cell-based and a bit more complex so I’ll go through that in another post.
Stoichiometry (Textbook Definition): The chemical and biochemical reactions involved in the rearrangement of atoms and molecules to form new groups.
Stoichiometry is basically calculating the amount of a reactant or product based on the related chemical equation.
To do this you must know the mass of at least one of the molecules, the moles of each molecule and the molecular weight of each molecule.
The best way to teach this is by example, so here is a chemical equation:
This is the chemical equation for alcoholic fermentation. In this equation, 1 mole of glucose (left) is broken down into 2 moles of ethanol (right) and 2 moles of carbon dioxide (right). Now, imagine we have 300 grams of glucose. How much ethanol would we produce?
Here are the 3 main rules:
- Matter cannot be created or destroyed. Therefore, the total mass on the left of the equation must be the same as the total mass on the right.
- Elements cannot be created or destroyed in a chemical equation. Therefore, both sides of the equation will have the same amount of each element.
- Stoichiometric balances work with moles, not mass – you must convert mass into moles when doing stoichiometric balances.
Firstly, since stoichiometry is done in moles, let’s convert our 300g of glucose into moles. From the basic calculations and equations post, we know that moles = mass / molecular weight. To find the molecular weight of glucose, we must add the molecular weight of all the elements in glucose together. If you google the periodic table (or just google the molecular weight of each element) you will find that:
- Carbon (C): Molecular weight of 12
- Hydrogen (H): Molecular weight of 1
- Oxygen (O): Molecular weight of 16
However, since there is more than 1 of each element (as indicated by the subscripts in the equation) we must multiply the molecular weight of each element by how much there is in glucose.
(C x 6) + (H x 12) + (O x 6) = (12 x 6) + (1 x 12) + (16 x 6) = 180
Hence the total molecular weight of glucose is 180. To find the molar mass:
Moles = Mass / MW = 300/180 = 1.67 gmol
Now that we know the molar mass of glucose, we can determine the molar mass of ethanol. If you look at the equation, you can see that there is 2 moles of ethanol produced for every 1 mole of glucose (if there is no number in front of a molecule it means there is only 1 mole). Therefore, the molar mass of ethanol would be 2 times the molar mass of glucose. So:
Molar mass of ethanol = molar mass glucose x 2
= 1.67 x 2 = 3.33 gmol ethanol
From this, we can find the mass of ethanol using the MW of ethanol and the molar mass calculated above:
Mass = MW x moles
= 46 x 3.33 = 153g
Where 46 is the molecular weight of ethanol. And voila! There we have the amount of ethanol produced with 300 grams of glucose. If you were to find the mass of carbon dioxide produced (which is roughly 147, you can calculate it yourself if you want practice). you’ll find that:
300 grams glucose = 153 grams ethanol + 147 grams carbon dioxide
Both sides of the equation equal 300 grams! This is a good way of checking if you’ve done the mass balance right, since the mass must be the same on both sides of the equation.
Now, lets try a harder one…
This LOOKS complicated, but the same method applies. How would we work out how much oxygen is needed to produce 40g of glutamic acid (C5H9NO4)?
Firstly, lets find the molecular weight:
(C x 5) + (H x 9) + (N x 1) + (O x 4)
= (12 x 5) + (1 x 9) + (14 x 1) + (16 x 4) = 147
Therefore the molar mass of L-glutamic acid is:
Molar mass = Mass/ MW
= 40/147 = 0.272 gmol
From the chemical equation, we know that 1.5 moles of oxygen is needed to make 1 mole of L-glutamic acid. Therefore we need 1.5 x more of oxygen:
0.272 x 1.5 = 0.408 gmol
To find the mass of oxygen needed:
Mass = MW x moles
= 0.408 x (16 x 2) = 0.408 x 32 = 13.06 grams
Therefore to make 40g of L-glutamic acid, we need 13.06 grams of oxygen.
And there you have it! It sounds complicated, but once you understand it its actually pretty simple! However I don’t know if its just simple to me, so if you have any questions about anything please leave a comment below and I will get back to you!
Bioprocessing Explained 