Question

A student weighs out 3.23 g of Fe(NO3)3 • ?H2O and heats it. The remaining anhydrous salt has a mass of 1.93 g. Calculate the empirical formula for the hydrate.

Answer 1

The empirical formula is :
`Fe(NO_3)_3.9H_2O`

Step-by-step explanation:

Let us assume the moles of hydration in ferric nitrate is '
`x`'

Thus the formula of the compound becomes :

`Fe(NO_3)_3.xH_2O`

Now when its heated , the reaction proceeds like :

`Fe(NO_3)_3.xH_2O`⇒
`Fe(NO_3)_3+xH_2O`

Given ,

• The mass of anhydrated
  `Fe(NO_3)_3=1.93g`

Molar mass of
`Fe,N,O=56,14,16` respectively.

The mass of 1 mol of
`Fe(NO_3)_3 = 56+(14+16*3)*3=56+62*3=242g`

The number of moles =
`(1.93)/(242) \\=0.008mol`

Therefore the number of hydrated moles must also be 0.008

Given ,

• The mass of hydrated
  `Fe(NO_3)_3=3.23g`

`0.008=(3.23)/(242+18x) \\242+18x=(3.23)/(0.008)\\242+18x=403.75\\18x=161.75\\x=9`