Question

In a gravimetric estimation of Al, a sample of 0.1000 g AlCl₃ is precipitated with 8-hydroxyquinoline. The weight of the precipitate is _____ g.
[Given: atomic weight of Al is 26.98; molecular weight of AlCl₃ is 133.34; and molecular weight of 8-hydroxyquinoline is 145.16] 
(round off to 4 decimal places)

Answer 1

Correct Answer: 0.344 - 0.3452

To determine the weight of the precipitate formed from a 0.1000 g sample of AlCl₃ when precipitated with 8-hydroxyquinoline, we need to follow these steps:

Step 1: Determine moles of AlCl₃.

The molecular weight of AlCl₃ is given as 133.34 g/mol. The moles of AlCl₃ can be calculated as follows:

moles of AlCl₃=mass of AlCl₃molecular weight of AlCl₃=0.1000 g133.34 g/mol=0.000750 mol

Step 2: Relate Al in AlCl₃ to AlC₉H₆NO (precipitate).

The balanced equation for the formation of the aluminum-8-hydroxyquinoline precipitate (Al(C₉H₆ON)₃) is:

AlCl₃ + 3C₉H₆NO → Al(C₉H₆ON)₃ + 3HCl

One mole of AlCl₃ yields one mole of Al(C₉H₆ON)₃ as the precipitate. Hence, moles of AlCl₃ = moles of precipitate formed.

Step 3: Calculate weight of the precipitate.

The molecular weight of Al(C₉H₆ON)₃ is calculated as 26.98 + (3×145.16) = 462.46 g/mol.

The weight of the precipitate is calculated as follows:

Weight of precipitate=moles of precipitate×molecular weight of precipitate=0.000750 mol×462.46 g/mol=0.3468 g

Conclusion: The weight of the precipitate is 0.3468 g.