A granulated sample of aircraft alloy(Al,Mg,Cu) weighing 8.72 g was first treated with alkali to dissolved Al, then with very dilute HCl to dissolve Mg,leaving a residue of Cu. The residue after alkali boiling weighed 2.1 g and the acid insoluble residue from this weighed 0.69g. What is the composition of the alloy?

We know that the amount lost during alkali boiling is the total amount of aluminum: subtract 2.1g from 8.72g to get 6.62g Al. The residue after acid is the mass of the copper: .69g Cu. The mass of Mg is the amount that dissolved in the acid: 2.1g - .69g = 1.41g.

Divide each mass by the total amount of alloy used and multiply by 100% to get the percentage of each element in the alloy.

(6.62/8.72) x 100% = 75.9% Al
(1.14/8.72) x 100% = 13.1% Mg
(0.69/8.72) x 100% = 7.9% Cu