Question

A mixture of 8-hydroxyquinoline complexes of aluminum and magnesium weigh 1.0843 g. When ignited in a furnace open to the air, the mixutre decomposed, leaving a residue of Al₂O₃ and Mg weighing 0.1344. What is the weight percent of Al(C₉H₆NO₃) in the original mixture?

AlQ₃ + MgQ₂ → Al₂O₃ + MgO

Where:

Q represents	8-hydroxyquinoline
AlQ3	F.W. = 459.441 g
MgQ2	F.W. = 312.611 g
Al2O3	F.W. = 101.961 g
MgO	F.W. = 40.304 g

Answer 1

You have a balanced reaction:

2 AlQ3 + MgQ2 = Al2O3 + MgO

You have the expressions:

i) g AlQ3 + g MgQ2 = 1.0843

ii) g Al2O3 + g MgO = 0.1344

The expression ii becomes the elements of the expression i:

g Al2O3 = g AlQ3 * (1 mol AlQ3 / 459.441 g) * (1 mol Al2O3 / 2 mol AlQ3) * (101.961 g Al2O3 / 1 mol) = 0.111 g AlQ3

g MgO = g MgQ2 * (1 mol MgQ2 / 312.611 g) * (1 mol MgO / 1 mol MgQ2) * (40.304 g MgO / 1 mol) = 0.129 g MgQ2

They are replaced in expression ii:

ii) 0.111 g AlQ3 + 0.129 g MgQ2 = 0.1344

System of equations between i and ii is applied and you have:

g AlQ3 = 0.3042 g

g MgQ2 = 0.7802 g

AlQ3 percentage:

% AlQ3 = g AlQ3 * 100 / g sample = 0.3042 * 100 / 1.0843 = 28.05%

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