Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate: H2PtCl6 + 2NH4+ = (NH4)2PtCl6 + 2H+ The percipitate decomposes upon ignition, yielding metallic platinum and gaseous products: (NH4)2PtCl6 = Pt(s) + 2Cl2(g) + 2NH3(g) + 2HCl(g) What is the percentage of ammonium in a sample of 0.2213g of sample gave rise to 0.5881g of platinum?

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