8-hydroxyquinoline (Q) can complex and precipitate many metals. A mixture of the aluminium and magnesium complexes weighed 1.0843g (MW=459.43; 312.61). When ignited, the mixture decomposed and left a residue of Al203 (MW 101.96) and MgO (MW 40.304) weighing 0.1344g. What is the weight percentage of the Al-complex in the mixture?

Question

8-hydroxyquinoline (Q) can complex and precipitate many metals. A mixture of the aluminium and magnesium complexes weighed 1.0843g (MW=459.43; 312.61).  When ignited, the mixture decomposed and left a residue of Al203 (MW 101.96) and MgO (MW 40.304) weighing 0.1344g. What is the weight percentage of the Al-complex in the mixture?

To be really honest, i don't have a single idea of how to start.
Should i calculate the no. of moles of Al2O3 and MgO and then assume it to be that of the Al complex and then find the mass of Al complex.

And finally use wt%=(mass Al complex)/(1.0843) * 100

Please guide me, im totally confused

DrBob222 · Accepted Answer

This is a two equation and solve simultaneously. Al reacts with 8-hydroxyqinoline (Q) to form Alq3. Mg reacts similarly to form Mgq2.
Let X = weight Alq3
and Y = weight Mgq2, then
--------------------
equation 1 is X + Y = 1.0843

The second equation is obtained from the Al2O3 residue.[MM stands for molar mass.]
Convert grams X to grams Al2O3 which will be
X*(MM Al2O3/2*MM Alq3)
Convert grams Y to grams MgO. That will be
Y*(MM MgO/MM Mgq2). When you add the Al2O3 and MgO you end up with 0.1344 g. so equation 2 is
X*(MM Al2O3/2*MM Alq3) + Y*(MM MgO/MM Mgq2) = 0.1344.

Solve for X and Y simultaneously, then
%Alq3 = (mass Alq3/1.0843)*100 = ?
I have for formula of Alq3 = Al(C9H6NO)3 and Mgq2 is Mg(C9H6NO)2