When 3.761g of a hydrate of strontium chloride is heated to 60 degree C it forms strontium chloride dihydrate of mass 2.746g What is the formula of the original sample?

Answer 1

SrCl2.xH2O ==> SrCl2.2H2O + xH2O
3.761g.........2.746g.......1.015g
(3.761-2.746 = 1.015g).
I would then convert SrCl2.2H2O to SrCl2 which can be done as follows:
2.746g x (molar mass SrCl2/molar mass SrCl2.2H2O) = 2.746 x (158.53/194.56) = 2.237 g SrCl2.

Now convert 2.237 g SrCl2 to moles and convert 1.015 g H2O to moles.

moles SrCl2 = grams/molar mass = about 0.014 but you need to do it more accurately.
moles H2O = 1.015/about 18 = about 0.056

Now find the ratio of SrCl2 to H2O this way. Divide the smaller number by itself which assures that number of being 1.000.
0.014/0.014 = 1.000
0.056/0.014 = 4.00
When you do this more accurately you may not end up with such even numbers; if so, round to whole numbers. You would write the formula as
SrCl2.4H2O