How can I get 1000ppm of cyanide in pottasium hexaferrocyanide with molar mass of 329.24g/mol
Bahati Lucas 1000ppm = 1g/l (the MW of the compound does not matter)
Bahati - Just one minor remark: The correct name of this complex is potassium hexacyanoferrate. Its'a a "ferrate" because we have here an anionic coordination compound. Potassium should be written with one "t". It should be "hexacyano" instead of "hexaferro" because we have 6 cyanide ligands and not six iron atoms. Furthermore, it must be specified it it's divalent or trivalent iron, which is not clear from your question. Please also note that in potassium hexacyanoferrate(II), K4[Fe(CN)6], the cyanide ligands are quite tightly bound to iron, so that in solution there are no free cyanide ions. If you require free cyanide ions in solution, you should perhaps use potassium cyanide right away.
Dear Bahati Lucas ,
Looking at the molecular weight you mean C6FeK3N6 (tripotassium;iron(3+);hexacyanide ) better known as potassium hexacyanoferrate (III) with formula K3[Fe(CN)6].
As indicated by Frank T. Edelmann dissolved in water you have [Fe(CN)6]3− so no cyanide-ions, only under for example (strong) acidic circumstances the ion ‘break’ and you might get HCN. Anyway, if I understand your question correctly you want to know how much gram of your salt you need to get a 1000 ppm equivalent of cyanide.
Strictly speaking ppm is a ratio and although frequently used it is better to use μL/L, μg/g, mg/kg. Since water has a density of 1 you can say 1 mg/L. So 1000 ppm CN(-) means 1000x1 mg = 1 g/L.
1 mol of K3[Fe(CN)6] gives 329.24 g salt of which (6x26,02) 156.12 g is cyanide. So, 1 gram of cyanide corresponds to 329.24/156.12 = 2.11 g salt. So, if you find 2.11 g salt in for example 1 L water or in 1 kg soil you may say you have found 1000 ppm cyanide in water or 1 kg of soil.
All this with the remark made by Frank T. Edelmann that it is questionable whether this is chemically relevant (perhaps in toxicological terms it is).
Best regards.
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