Imidazolidinyl urea (Mw of 388.29 g/mol) can release the 3-fold amount of formaldehyde. A solution of 100 mg/ml Imidazolidinyl urea in DMSO contains:
100mg/ml divided by 388.29 g/mol = 258 µmol/ml Imidazolidinyl urea
So, if we take the concentration of the Imidazolidinyl urea in DMSO 20mg/ml then divided by 388.29 g/mol = 52 µmol/ml Imidazolidinyl urea
which is by a factor of 3 = 156 µmol/ml formaldehyde
Considering a spray coating solution containing 80 % (Vol.) Polymer in Acetonitrile and 20 % (Vol.) Imidazolidinyl urea in DMSO. So we have 20% of Imidazolidinyl urea solution in the total solution. There is a concentration of 0.2 time 156 µmol/ml formaldehyde that is roughly 31.2 µmol formaldehyde per ml of the spray coating solution mixture.
Considering a minimal formaldehyde (30,03 g/mol) concentration of 2.7 ppm (2.7 mg/L) then one liter will have:
2.7 mg divided by 30.03 g/mol =90 µmol.
which is 0.09 µmol/ml
When there is a sample volume of 10 ml, then we need 10 ml times 0.09 µmol/ml = 0.9 µmol
for this number of molecules, we have to spray coat at least a volume of:
0,9 µmol divided by 31.2 µmol/ml = 29 µl (spray coating solution)
Now, How do I calculate back to the amount of Imidazolidinyl urea (in ppm or any unit) for 2.7 ppm of Formaldehyde in that 20 % of the solution?
It's very complicated. Please make it simpler to make easy calculations....
Ok. My Bad.
Imidazolidinyl Urea is a compound that can release formaldehyde.
So, If we assume that Imidazolidinyl urea ( Mw 388.29 g/mol) can release 3 mol of Formaldehyde( Mw 30,03 g/mol) then to get 2.7 ppm of Formaldehyde how much Imidazolidiny Urea we need to get 2.7 ppm of Formaldehyde from that amount of Imadazolidinyl Urea?
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