Accuracy

[wedobetter]

chemlib==2.2.3

In [15]: r = Reaction.by_formula('KOH + H3PO4 --> K2HPO4 + H2O')

In [16]: r.balance()

In [17]: r.formula
Out[17]: '2K₁O₁H₁ + 1H₃P₁O₄ --> 1K₂H₁P₁O₄ + 2H₂O₁'

In [18]: dict(zip(r.constituents, [c["grams"] for c in r.get_amounts(3, grams=1)]))
Out[18]: {'K₁O₁H₁': 0.673, 'H₃P₁O₄': 0.588, 'K₂H₁P₁O₄': 1, 'H₂O₁': 0.216}

In [19]: dict(zip(r.constituents, [c["grams"] for c in r.get_amounts(3, grams=1000)]))
Out[19]: {'K₁O₁H₁': 644.255, 'H₃P₁O₄': 562.561, 'K₂H₁P₁O₄': 1000, 'H₂O₁': 206.906}

I'd expect KOH to be something like 673.xxx

[harirakul]

1 gram of the third compound in the formula (K₂HPO₄) is equivalent to 0.006 moles of K₂H₁P₁O₄, using its molar mass (174.17 g/mol). Using the 2:1 stoichiometric ratio of KOH : K₂HPO₄ in the balanced equation, this would be 0.012 mol KOH. This results in 0.673 grams of KOH, using its molar mass (56.1 g/mol).

1000 grams of K₂HPO₄ is equivalent to 5.741 moles of K₂HPO₄, using its molar mass (174.17 g/mol). In this case, that would give us 11.482 mol of KOH using the 2:1 stoichiometric ratio. This results in 644.25 grams of KOH, using its molar mass (56.1 g/mol).

Simply using 1000 times more mass of a product in a reaction won't mean that 1000 times more mass of a reactant will be needed. It depends also on the molar masses of the compounds in question, and stoichiometric ratios. I don't see any problems with the current calculation or its accuracy.

[wedobetter]

The problem is that all the round(value, 3) should be left to the users to do at the end of their calculations.
I'd love to see that 1 gram of K2HPO4 equivalent to 0.005741516908767 moles

By the way I forgot to mention that I really love this library and it's coming up really interesting, thank you!

[harirakul]

Oh ok, thanks for the clarification. Perhaps I will add an option for the user to override the default number of decimal values in the calculations.