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Determing The Chemical Formula of a Hydrate

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    When copper (II) sulfate hydrate, a blue crystalline solid containing embedded water molecules (called a hydrate), is heated in air, it loses the water molecules and the blue solid is transformed to a white anyhydrous (no water) crystal known as copper (II) sulfate.


    2. Relevant equations
    CuSO4 .x H2O +heat --> CuSO4 (s) + H2O (g)


    Observations: Mass of copper (II) sulfate hydrate = 0.1534g
    Mass of anyhydrous copper (II) sulfate = 0.0980g

    1a) Calculate mass of H2O that was driven off.
    b) Calculate the number of moles of water present.
    c) Calculate the number of moles of CuSO4 present.

    My attempt to answer these questions:
    1a) Since anhydrous copper sulfate has no water the mass of H2O driven off is 0.1534g - 0.0980g =0.0554g
    b) n = m/M n= 0.0554/18.02 therefore, n= 3.07 x10 to the power of -3.
    c) Mcuso4 = 159.62g/mol so, n=m/M n= 0.0980g/159.62g/mol which = 6.13 x10 to the power of -4
     
    Last edited: Apr 29, 2012
  2. jcsd
  3. Apr 29, 2012 #2

    Borek

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    Staff: Mentor

    So far so good (that is, I have not checked the numbers, just the general idea).

    What is the ratio?
     
  4. Apr 29, 2012 #3
    The numbers just dont look right for some reason...
    No ratio given, thats a whole new question "Determine the ratio of moles of water to moles of CuSO4" that i have yet to attempt. If my answers are right then i can move on.
     
  5. Apr 30, 2012 #4

    Borek

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    Staff: Mentor

    Come on, "whole new question" means just divide two numbers you got so far.
     
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