Question

A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a mass of 23.875 g for the crucible, lid, and sample. The mass of the empty crucible and lid was found earlier to be 22.652 g. He then heats the crucible to expel the water of hydration, keeping the crucible at red heat for 10 minutes with the lid slightly ajar. On colling, he finds the mass of crucible, lid, and contents to be 23.403 g. The sample was changed in the process to very light clue anhydrous CuSO4. How many moles of water are present per mole of CuSO4?

Answer 1

There are present 5,5668 moles of water per mole of CuSO₄.

Step-by-step explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g ×
`(1mol)/(159,609g)` = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g ×
`(1mol)/(18,02g)` = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have 5,5668 moles of water per mole of CuSO₄.

I hope it helps!