Question

Na2SO4.nH2O is Glauber’s salt, 16.1 g of Glauber’s salt is heated to remove water of crystals Atkin completely. Mass of remaining solid sample is 7.1g. What is value of n?

Answer 1

`0.5`

Explanation:

We have hydrated sodium sulfate or Glauber's salt with a mass of `16.1 \ "g"`.

When we heat it up, all the water is removed, and the remaining dehydrated sodium sulfate has a mass of `7.1 \ "g"`.

So, the mass of the water removed was `16.1 \ "g"-7.1 \ "g"=9 \ "g"`.

Water has a molar mass of `18.01528 \ "g/mol"`. So here, there will exist:

`(9color(red)cancelcolor(black)"g")/(18.01528color(red)cancelcolor(black)"g""/mol")`

`~~0.5 \ "mol"`

`:.n=0.5`

Therefore, Glauber's salt is a hemihydrate.

Answer 2

We address the stoichiometric equation....

`Na_2SO_4*nH_2O(s)stackrel(Delta)rarrNa_2SO_4(s)+nH_2O(g)uarr`

Explanation:

With all these types of dehydration reaction it is necessary to write the stoichiometric equation in order to inform our sense of the stoichiometric equivalence. Given the above, we can only assume that the MASS LOSS the hydrate undergoes is due to the loss of the so-called `"waters of crystallization"`, and that the final mass, `7.10*g` constitutes ONLY `"sodium sulfate"`

And so `n_"sodium sulfate"=(7.10*g)/(142.04*g*mol^-1)=0.0500*mol`

And `n_"water"=(16.10*g-7.10*g)/(18.01*g*mol^-1)=0.500*mol`

To get to the formula of the hydrate, we divide thru by the smallest molar quantity...i.e.

`(Na_2SO_4)_((0.050*mol)/(0.050*mol))*(OH_2)_((0.500*mol)/(0.050*mol))-=Na_2SO_4*10H_2O`

And so we gots `"sodium sulfate decahydrate.."` and clearly `"n=10"`...

Just on a cautionary note, I seem to remember that you can have several crystalline hydrates of sodium sulfate…`Na_2SO_4*2H_2O` is one such, but we really cannot address the degree of solvation, and the assumption that we make an anhydrous salt here is fully justified...