Here's the problem from my homework. The amount of iron and manganese in an alloy can be determined by precipitating the metals with 8-hydroxyquinoline, C9H7NO (FW 145.16). After weighing the mixed precipitate, the precipitate is dissolved and the amount of 8-hydroxyquinoline is determined by another method. In a typical analysis, a 0.1273 g sample of an alloy containing iron, manganese, and other metals was dissolved in acid and treated with appropriate masking agents to prevent interference from other metals. The iron and manganese were precipitated and isolated as Fe(C9H6NO)3 (FW 488.29) and Mn(C9H6NO)2 (FW 343.24), yielding a total mass of 0.8678 g. The amount of 8-hydroxyquinoline in the mixed precipitate was determined to be 5.276 mmol. Calculate the %(m/m) Fe and %(m/m) Mn in the alloy.
I'm not sure where to start because I was out of town due to family things. Any help is appreciated!