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Mass of a hydrate before and after the hydrate was evaporated away

  1. Oct 28, 2004 #1
    If I have the mass of a hydrate before and after the hydrate was evaporated away, how can I find the ratio of molecules?

    For example: CoCl2 * xH2O (The * is meant to be a dot). I need to find "x".
     
  2. jcsd
  3. Oct 28, 2004 #2

    chem_tr

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    Yes, the hydrate formulae are found in this way, .by successive evaporations and subsequent weight measurements until a stable reading is achieved.

    Suppose that you started with [itex]a[/itex] grams of [itex]CoCl_2\cdot xH_2O[/itex] and after several steps, you get [itex]b[/itex] grams of [itex]CoCl_2[/itex]. Then it means that [itex]b-a[/itex] grams of water was evaporated in the process, just divide it by 18 (molar mass of water) to find the overall x. Note that you may not be very precise in this calculation, so if you find [itex]x=2,5[/itex], you may conclude that it can be regarded as [itex]x=3[/itex], etc.
     
    Last edited: Oct 28, 2004
  4. Oct 28, 2004 #3
    Are you sure it's not "a - b"? If I try it as you said then I get -0.041 (a = 1.62, b = 0.88). This doesn't seem correct. Am I doing something wrong?
     
  5. Oct 28, 2004 #4
    Yes, what chem_tr means that if a is total weight before evaporation, then b is to be the weight after evaporation... This difference gives the amount of water. (Definitely positive)

    Of course, we suppose the hydrate compound is pure...
     
  6. Oct 28, 2004 #5
    [tex]CoCl_2 \cdot xH_2O[/tex]

    1.62 is the mass before evaportation
    0.88 is the mass after evaportation

    [tex]1.62 - 0.88 = 0.74[/tex]
    [tex]0.74 / 18.02 = 0.041[/tex]

    How can this be right?
     
    Last edited by a moderator: Oct 28, 2004
  7. Oct 28, 2004 #6

    chem_tr

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    Oops, I should have written [itex]a-b[/itex] of course. The positive difference between these two measurements gives the amount of water evaporated.

    About the difference, [itex]0.041[/itex] moles of water is present in this compound, namely [itex]CoCl_2 \cdot xH_2O[/itex]. The molar amount of the initial compound is not known, but we may consider that [itex]0,88[/itex] grams of [itex]CoCl_2[/itex] is present, you can find the molar mass from [itex]Co:58.93[/itex] and [itex]Cl:35.45[/itex] grams/mol. You then set up a proportion equation to find how many moles of water are present in one mole of [itex]CoCl_2[/itex]. This will give [itex]\displaystyle x[/itex] you're looking for.

    I recommend that you use greater amounts of salt and multiple determinations to minimize errors. For example, do the analysis triplicate at one time and use at least 5 or 10 grams of sample, then average the findings you obtained. This will be better.
     
    Last edited: Oct 28, 2004
  8. Oct 31, 2004 #7
    Could you tell me a bit more about this proportion equation?
     
  9. Oct 31, 2004 #8

    chem_tr

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    Okay, first find how many moles are there in 0.88 grams of [itex]\displaystyle CoCl_2[/itex]. Then calculate this: "If there are n moles in 0.88 grams, 1 mole would be X". Then set up a second calculation, "if 0.041 moles of water is present in n moles of compound, how many moles of water are present in 1 mole of compound?"

    I think you'll be able to do these.
     
  10. Oct 31, 2004 #9
    First calculation:

    [tex]
    CoCl_2 = 129.83 g/mol
    [/tex]

    [tex]
    \frac{0.88}{129.83} = 0.0068 mol
    [/tex]

    Second Calculation:

    [tex]
    \frac{0.041}{0.0068} = 6.0
    [/tex]

    I'm not sure what you wanted for `X', and I don't think I did the second calculation right.
     
  11. Oct 31, 2004 #10

    chem_tr

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    Congrats, and you're right about X, we don't need to use this, as we know that [itex]\displaystyle CoCl_2[/itex] is 129.83 g/mol. I did the same calculation and found 6.0, and it is very characteristic for cobalt to coordinate six water molecules in the form [itex]\displaystyle [Co(H_2O)_6]Cl_2[/itex].
     
  12. Oct 31, 2004 #11
    So from all this we get that the formula must be [itex]CoCl_2 \cdot 6H_2O[/itex]? Thanks for all your help!
     
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