Question

How can I calculate Molar Ratio for struvite precipitation?

How can I calculate the molar ratio of Mg:P and N:P from the concentrations available in the attached table? Thank you

Mahmoud Elshahawy
Tanta University
To calculate the molar ratio of Mg:P and N:P from the given concentrations, we need to convert the concentrations from mg/L to mol/L.
The molar mass of Mg2+ is 24.31 g/mol, NH4+ is 18.04 g/mol and PO4 3- is 94.97 g/mol.
Let's take the first row of your table as an example:
So, the concentration of Mg2+ in mol/L is:
9363 mg/L * (1 g/1000 mg) * (1 mol/24.31 g) = 0.3852 mol/L
The concentration of NH4+ in mol/L is:
2138 mg/L * (1 g/1000 mg) * (1 mol/18.04 g) = 0.1185 mol/L
The concentration of PO4 3- in mol/L is:
18725 mg/L * (1 g/1000 mg) * (1 mol/94.97 g) = 0.1972 mol/L
The molar ratio of Mg:P is:
0.3852 mol/L / 0.1972 mol/L = 1.95
The molar ratio of N:P is:
0.1185 mol/L / 0.1972 mol/L = 0.60
I hope this helps!
1 Recommendation
University of Monastir
calculation example
- NH4+: 0.02 M
- Mg: 0.01 M
- N: 0.03 M
To calculate the Mg:P molar ratio, divide the Mg concentration by the P concentration (which comes from NH4+):
Mg:P Molar Ratio = Mg Concentration / P Concentration = 0.01 M / 0.02 M = 0.5
Similarly, for the N:P molar ratio, divide the N concentration by the P concentration:
N:P Molar Ratio = N Concentration / P Concentration = 0.03 M / 0.02 M = 1.5
So, in this example, the Mg:P molar ratio is 0.5 and the N:P molar ratio is 1.5.

Top contributors to discussions in this field

Got a technical question?