# chemistry

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?

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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.

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

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