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A tablet weighing 2.074 g and containing aluminium salts was dissolved and diluted to 100.0 mL...

A tablet weighing 2.074 g and containing aluminium salts was dissolved and diluted to 100.0 mL in a volumetric flask. A 25.00 mL aliquot was removed and treated with enough ammonia to precipitate the aluminium (Al) present as aluminium hydroxide (Al(OH)3). After ignition at 800 oC, this precipitate was converted to aluminium oxide (Al2O3) weighed 0.1967 g. Calculate the Al% in the tablet.

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Answer #1

In 25 ml aliquot,

mass of Al2O3 collected = 0.1967 g

moles of Al2O3 = 0.1967 g/101.96 g/mol = 1.93 x 10^-3 mol

moles Al = 2 x 1.93 x 10^-3 mol = 3.86 x 10^-3 mol

mass Al in 25 ml aliquot = 3.86 x 10^-3 mol x 27 g/mol = 0.1042 g

mass of Al in 100 ml solution = 0.1042 g x 100 ml/25 ml = 0.4168 g

mass% Al in tablet = 0.4168 g x 100/2.074 g = 20.10%

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