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2.51g Mg

Explanation:

The empty crucible had a mass of 35.84. With the magnesium added, it had a mass of 38.45. This means that you should subtract the mass of the empty crucible from the mass of the crucible with Mg to find how much was added.

38.35g - 35.84g = 25.1g

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the answer is 0.8m.

by: nicholas