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

consider the given equation

AlQ3 + MgQ2 = Al203 + MgO

let

weight of AlQ3 be z

now

moles = mass / molar mass

so

moles of AlQ3 = z / 459.441

now

weight of MgQ2 = 1.0843 - z

so

moles of MgQ2 = (1.0843 - z ) / 312.611

now

AlQ3 + MgQ2 = Al203 + MgO

we can see that

moles of Al203 = 0.5 x moles of AlQ3

so

moles of All203 = 0.5 x z / 459.441

so

mass of Al203 = 0.5 x z x 101.961 / 459.441 = 0.110962z

now

moles of MgO = moles of MgQ2 = (1.0843 - z ) / 312.611

so

mass of MgO = 40.304 x (1.0843 - z ) / 312.611 = 0.1397955 - 0.128927z

now

mass of Al203 + mass of MgO = 0.1344

so

0.110962z + 0.1397955 - 0.128927z = 0.1344

z = 0.3

now

we know that

% weight = mass of AlQ3 x 100 / total mass

so

% weight = 0.3 x 100 / 1.0843

% weight = 27.7

so

the weight percent of AlQ3 is 27.7 %