Chemistry Labs

Problem 1

A solution was prepared from 0.5284 g of an alloy containing aluminium. The aluminium was precipitated as aluminium 8-hydroxyquinolate, Al(CX9HX6NO)X3\ce{Al(C9H6NO)3}. The precipitate was separated, dissolved in hydrochloric acid, and the liberated 8-hydroxyquinoline was titrated with a standard potassium bromate solution containing potassium bromide; 17.40 cm3^3 of the standard solution were required, and the resultant product is a dibromo derivative of 8-hydroxyquinoline. The relative atomic mass of aluminium is 26.98. 1.1 Write the equation for the reaction of AlX3+\ce{Al^{3+}} with 8-hydroxyquinoline. 1.2 Give the name of the type of compound formed in the precipitation. 1.3 Write the equation in which bromine is produced from bromate and bromide. 1.4 Write the equation for the reaction of bromine with 8-hydroxyquinoline. 1.5 Calculate the molar ratio of aluminium ions to bromate ions. 1.6 Calculate the percentage by weight of aluminium in the alloy.
Step 3 of 4: Al : bromate equivalence chain
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Intuition

Count electrons, not molecules: oxidation of Br− to Br2 costs 5 e− per BrO3−, while each Al 'holds' 6 Br2 = 12 e−.

Al≐Al(CX9HX6NO)X3≐3 CX9HX7NO≐6 BrX2≐12 e−⇒n(AlX3+)n(BrOX3X−)=512\ce{Al} \doteq \ce{Al(C9H6NO)3} \doteq 3\,\ce{C9H7NO} \doteq 6\,\ce{Br2} \doteq 12\,e^- \quad\Rightarrow\quad \frac{n(\ce{Al^{3+}})}{n(\ce{BrO3^-})} = \frac{5}{12}
Analysis

The chain Al≐3 oxine≐6 BrX2\ce{Al} \doteq 3\,\text{oxine} \doteq 6\,\ce{Br2} corresponds to 12 electron equivalents per Al; the oxidant BrOX3X−\ce{BrO3^-} is a 5-electron reagent, so the chemical equivalent of Al is 26.98/12=2.24826.98/12 = 2.248 g eq−1^{-1} and the molar ratio n(Al):n(BrOX3X−)=5:12n(\ce{Al}):n(\ce{BrO3^-}) = 5:12.

Common pitfall. A frequent error is to count BrOX3X−\ce{BrO3^-} as equivalent to one BrX2\ce{Br2}: each bromate generates THREE BrX2\ce{Br2} molecules (5 Br− + 1 BrO3− → 3 Br2), so the factor linking titre to Al is 12 equivalents per Al, not 6.