Question

Calculate the % Al in the sample.

Preparation of a stock solution of aluminum Accurately weigh out approximately 1 g of the aluminum foil. Tear it up into small pieces, place the pieces into a covered 250 mL beaker and add 60 mL of 6M HCl. This should be carried out in the fume hood because the reaction is exothermic and leads to violent frothing and emission of vapors. Cool, quantitatively transfer to a 250 mL flask, rinsing with deionized water, fill up to the mark and mix thoroughly. Label this stock solution accordingly.

Procedure for Method 1. Pipette 20 mL of the Al stock solution into a 400 mL beaker and dilute to about 150 mL with deionized water. Add 2 g of NH4Cl and a few drops of methyl red indicator. Heat the solution to about 90°C and add dilute ammonia solution slowly with constant stirring until the color changes from red to yellow. Keep the liquid at 90 °C for 5-10 minutes and filter hot through fluted coarse ashless filter paper (Whatman 541) using a glass funnel. Wash the precipitate with hot 0.5% NH4Cl solution made neutral with dilute ammonia solution against methyl red. Place the paper and precipitate in a weighed porcelain crucible, and heat gently with a Bunsen burner until dry, and then more strongly until all the paper has been burned off. Place the crucible and residue in a furnace at 1200 °C for 30 minutes, remove, cool and weigh.

Calculate the % Al in the sample.

Mass of the Al foil used for the stock solution is 1.0241 g

Mass of precipitate = 0.1502 g

The following equations may be helpful?

Al(OH)3 ⇌ Al(OH)4^-

NH4Cl → NH4 + + Cl−

NH4 + + H2O ⇌ NH4OH + H+

2 Al(OH)3 → Al2O3.xH2O → γ Al2O3 → αAl2O3

Answer 1

The related equations for the formation of Al2O3 from Al is given below.

Al (s) ----------> Al3+ (aq) + 3 e- ……..(1)

Al3+ (aq) + 3 OH- (aq) ----------> Al(OH)3 (s) ………(2)

2 Al(OH)3 (s) --------> Al2O3.3H2O (s) ……..(3)

Al2O3.3H2O (s) ---------> Al2O3 (s) + 3 H2O (g) …….(4)

As per the stoichiometric equations above,

1 mol Al = 1 mol Al(OH)3.

2 mols Al(OH)3 = 1 mol Al2O3.

The atomic masses are

Al: 26.981 g/mol.

O: 15.999 g/mol.

The gram molar mass of Al2O3 = (2*26.981 + 3*15.999) g/mol

= 101.959 g/mol.

Mass of the residue = 0.1502 g.

Mol(s) Al2O3 corresponding to 0.1502 g = (0.1502 g)/(101.959 g/mol)

= 0.001473 mol.

Mol(s) Al corresponding to 0.001473 mol Al2O3 =

(0.001473 mol Al2O3)*(2 mols Al)/(1 mol Al2O3)

= 0.002946 mol.

Mass of Al in the foil corresponding to 0.002946 mol = (0.002946 mol)*(26.981 g/mol)

= 0.07948 g

≈ 0.0795 g.

Mass percent Al in the foil = (0.0795 g)/(1.0241 g)*100

= 7.7629

≈ 7.763 (ans, correct to 4 sig. figs).