1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hydrated copper (II) sulfate mass calculation

  1. Apr 24, 2006 #1

    Aya

    User Avatar

    calculate the precent by mass of water in a sample of hydrated copper (II) sulfate.

    mass of beaker-165.3g
    mass of beaker with bluestone-173.9g
    mass of dehydrated bluestone-170.7g

    Would this be a ballanced equation for this reaction?
    CuSO4•5H2O+ heat= CuSO4+5H2O


    so would you do this

    %by mass=mass of part/mass of whole
    =90.1g/249.7g/molx100
    = 36%

    is this correct?
     
  2. jcsd
  3. Apr 24, 2006 #2

    Borek

    User Avatar

    Staff: Mentor

    No. But it is so wrong I have no idea how to help you :(

    Start calculating mass of the copper sulfate before roasting and after roasting.
     
  4. Apr 24, 2006 #3

    Aya

    User Avatar

    lol, its ok i figured that part out

    i have different questions now

    Suppose you heated a sample of hydrated ionic compound in a test tube. What might you expect to see inside the test tube, near the mount of the test tube? Explain

    You would see water droplets and steam. This is because a hydrate indicates the presents of water, when you heat a hydrated ionic compound the water will separate from the compound, and you will see some of the water collecting near the mount of the test tube. The steam could be a result of the water, or the compound evaporating

    would that be right?

    ****
    Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the molecular formula you determined?

    you would get more molecules of water???
     
  5. Apr 24, 2006 #4
    Ionic compounds generally are nonvolatile solids and have high boiling points. Stick to the water answer.

    No, think about what the question is saying here. Some water molecules are leaving the compound, just not all of them. When all the water has gone from the compound, then you will have the anhydrous product.
     
  6. Apr 24, 2006 #5

    Aya

    User Avatar

    ^ok thanks
     
  7. Apr 24, 2006 #6

    Aya

    User Avatar

    7. Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the calculated percent by mass of water in the compound?

    If the hydrate was not completely converted to the anhydrous compound, the mass of anhydrous Cu(II)SO4 would increase because it would not be pure anhydrous Cu(II)SO4, instead it would be CuSO4•5H2O, in other words you would be weighing the compound plus water. In turn, this would change the mass of H2O, making to higher (for example instead of 3.2g it might be 4.2g.) the results of the percent composition calculation would also change to a higher number, for example %= (4.2/8.6)100 would equal 48.8% of water, compared to 37.2% of water, which is what we got by completely convert the hydrate to the anhydrous compound.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?