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Hydrated calcium sulfate - problem determining mass of H2O

  1. Oct 15, 2007 #1
    I just finished a lab and I was wondering if anyone could point out where I'm going wrong.

    Here are my findings for CaSO4:

    After First Heating After Second heating

    Wt. of crucible, lid, and hydrate 35.0537g 34.9779g
    Wt. of dry crucible and lid 34.0226g 34.0226g
    Wt. of hydrate 1.0331g 0.9553g

    Wt. of crucible, lid, and anhydrate 34.9779g 34.9781g
    Wt. of crucible and lid 34.0226g 34.0226g
    Wt. of anhydrate 0.9553g 0.9555g


    1. Calculate the % by weight of water in your unknown hydrate. Report the % to four digits.

    1.0331g CaSO4*H2O - 0.9555g CaSO4 = 0.0756 g H2O
    0.0756g/1.0311g x 100% = 7.332%

    2. Using the % water from Part 1 above, calculate the moles of water. (Hint: Assume a 100g sample of hydrate.) Report to four digits.

    7.332g H2O x 1mol. H20/18.02 g H2O = 0.6806 mol H2O

    3. Using the & anhydrate, calculate the moles of anhydrate. Report four digits.

    100% - 7.332% = 92.668%
    92.668g CaSO4 x 1 mol CaSO4/136.15g CaSO4 = 0.6806 mol CaSO4

    4. Using the anhydrate formula (CaSO4) given to you in lab, and the values from Parts 2 and 3 above, give the formula of your unknown hydrate. Show your calculation. Round the water to the nearest 1/2 or whole number of moles.

    H2O = 0.4069 mol/0.4609 mol =1
    CaSO4 = 0.6806 mol/0.4609 mol = 1.673
    x 3 = 5CaSO4*3H2O

    These calculations don't seem to be right, can anybody please help me understand?
  2. jcsd
  3. Oct 16, 2007 #2
    first, 1.0331 - 0.9555 = 0.0776g of water in 1.0331g of the hydrated salt. The % by weight of H2O is 7.511%

    i find this method weird. this is what i;d have done:

    weigh crucible + lid = 34.0226 g
    weigh crucible + lid + hydrated salt = 35.0537 g
    mass of hydrated salt = 1.0311 g

    heat first time:

    weigh crucible + lid + salt = 34.9779 g

    heat second time:

    weigh crucible + lid + salt = 34.9781 g

    therefore, mass of water of crystallisation evaporated = 35.0537 - 34.9781 = 0.0756 g
    and mass of anhydrous salt = 34.9781 - 34.0226 = 0.9555 g

    this is all i will be needing.

    i then calculate the number of moles of water of crystallisation = 0.0756/18 = 4.2*10^-3 mol

    this amount is present in 1.0311 g of the hydrated salt.

    i know the formula of the salt is CaSO4.xH2O

    1 mol CaSO4.xH2O contains 1 mol CaSO4

    i know the mass of CaSO4 which is 0.9555 g, and the molecular mass is 136
    the number of moles of CaSO4 thus present is 7.026*10^-3.

    this implies that 7.026*10^-3 mol CaSO4.xH20 is present.

    now coming back to the water,

    7.026*10^-3 mol CaSO4.xH20 contains 4.2*10^-3 mol water of crystallisation
    1 mol CaSO4.xH2O contains 0.59778 mol water

    you can approximate this value to 0.5 and you will get bassanite, CaSO4.(0.5)H2O
  4. Oct 16, 2007 #3
    i think i found the error, in part 2.

    (1) first the % water is wrong, it is 0.0776/1.0311 x 100% = 7.526 %

    (2) you are saying 100g of hydrate contains 7.526 g water, which is 0.418 mol
    (3) then 100g hydrate contains 92.474 g anhydrate, which is 0.680 mol

    (4) but then, CaSO4.xH2O contains 1 mol CaSO4.
    this implies that 100g of hydrate contains 0.680 mol of hydrate.

    you know that 0.680 mol hydrate contains 0.418 mol water
    therefore, 1 mol hydrate contains approximately 0.5 mol water (it is not the exact value obtained)

    same as above ;)
    Last edited: Oct 16, 2007
  5. Nov 21, 2008 #4
    Re: Hydrates

    I want to know about chemistry process to produce gypsum hemihydrate ( CaSO4. 1/2 H2O )
    and what should be controlled in the process to get stable CaSO4.1/2H2O:
    a. Temperature
    b. Moisture
    c. Pressure
    d. Air/powder ratio
    e. Any others parameter that needed
  6. Nov 21, 2008 #5


    User Avatar

    Staff: Mentor

    Re: Hydrates

    Please do not hijack the threads.

    Do you know how the hemihydrate is made?
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