strontinum nitrous degree of hydration?

Question

A 4.372 g sample of hydrated strontium nitrate forms 3.26 g of anhydrous strontium nitrate Sr(NO3)2, when its water of hydration is removed. Find its degree of hydration.

Answer 1

MM Sr(NO3)2 = 211.62 g/mol

3.26 g / 211.62 g/mol= 0.0154 moles Sr(NO3)2

4.372-3.26= 1.11 g H2O

1.11g/18 g/mol=0.0618 moles of H2O

We divide for the smaller

0.0618/0.0154=4

The formula is Sr(NO3)2 4H2O

Answer 2

1. subtract 3.26 from 4.372 and divide the answer by 18.
2. divide 4.372 by the Mr of strontium nitrate.
3. Divide the answer to 1 by the answer to 2.