Library to solve stoichiometric balance problems with a concise and familiar syntax.
The API reference is available here.
Download /stoich to your project folder and
from element import Element
The best way to appreciate the syntax is to see it in action. The following is an example with the combustion of Octane.
Firstly, create a series of elements:
H = Element("H")
O = Element("O")
C = Element("C")
The necessary molecules can be created straightforwardly from the elements, as follows:
C8H18 = C*8-H*18
O2 = O*2
CO2 = C-O*2
H20 = H*2-O
The molecule creation syntax works as follows:
- Multiplication
- By Integer: single-species molecule (do not mistake for monoatomic)
- Subtraction
- By Element: diatomic molecule
- By Molecule: composed molecule
Once the molecules have been created, a mixture can be obtained from different molecules:
reactants = C8H18 + O2
products = CO2 + H20
Finally, the stoichiometric balance can be formulated and the coefficients solved for:
reactants >> products
The balanced stoichiometric equilibrium will be displayed in the terminal as follows:
1.0*C_8H_18 + 12.5*O_2 -> 8.0*H_2O + 9.0*H_2O
Note: as an avid chemist might have told, in this equilibrium there is one too many degrees of freedom (ie: there is one more molecule than there are atoms to conduct a balance with): the system is ill determined! A solution is found in this case however, as follows:
- If there is 1 molecule more than there are atoms, the stoichiometric coefficient of the first reactant (C8H18 in this case) is set to 1, and the system is solved.
- If there are more than 1 molecules more than there are atoms in the balance, the system cannot be solved and the program will stop.
Lastly, the balance can be output direcly in LaTeX code, by calling
balance = reactants >> products
print(balance.latex())
Which will output
1.0*C_{8}H_{18} + 12.5*O_{2} \longrightarrow 8.0*H_{2}O + 9.0*H_{2}O
Which is to say:
