Question

What will be the mass of anhydrous Alum left over if one starts with 100.00g of Hydrated Alum?

Answer 1

Mass of hydrated alum = 100 g

The chemical formula of hydrated alum is KAl(SO₄)₂ 12H₂O.

The molar mass of hydrated alum = 474.39 g/mol

Determine the number of moles of alum using the mass of hydrated alum and molar mass of hydrated alum as follows:

= 100 g hydrated alum x ( 1 mol hydrated alum / 474.39 g hydrated alum)

= 0.2108 mol KAl(SO₄)₂ 12H₂O

Determine the moles of water in hydrated alum using the moles of KAl(SO₄)₂ 12H₂O as follows:

1 mol KAl(SO₄)₂ 12H₂O = 12 mol H₂O

Thus,

= 0.2108 mol KAl(SO₄)₂ 12H₂O x (12 mol H₂O/1 mol KAl(SO₄)₂ 12H₂O)

= 2.5296 mol H₂O

Convert moles of H₂O to grams as follows:

The molar mass of water = 18.02 g/mol

= 2.5296 mol H₂O x ( 18.02 g H₂O/1 mol H₂O)

= 45.583 g H₂O

Thus, 100 g hydrated alum contains 45.583 g water.

Determine the mass of anhydrous alum as follows:

Mass of anhydrous alum = Mass of hydrated alum - Mass of water

Substitute 100 g for the mass of hydrated alum and 45.583 g for the mass of water. Determine the mass of anhydrous alum as follows:

Mass of anhydrous alum = 100.00 g - 45.583 g

Mass of anhydrous alum = 54.417 g​​​​​​​