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Gravimetric Analysis, Volatilization

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider an unknown solid containing BaCl2. 2H2O (MW = 244.26296). When the unknown is heated to 160 °C for 1 h, the water of crystallization is driven off:

    BaCl2.2H2O(s) ---> BaCl2(s)+ 2 H2O(g)

    A sample originally weighing 1.9226 g weighed 1.5647 g after heating. Calculate the weight percent of Ba (MW=137.327) in the original sample. BaCl2: MW=208.2324; H2O: MW=18.0153


    2. Relevant equations



    3. The attempt at a solution

    I think I'm missing something with the water, but I'm not sure how to work it in.

    I figured out mol of BaCl2*2H20
    1.9226 g / 244.26296 g/mol = 7.8710 mmol

    Calculated moles of BaCl2
    1.5647 g/208.2324 g/mol = 7.514 mmol

    Converted this to moles Ba

    7.514 mmol * 137.327 g/mol = 1.0319026 g

    Divided this by original weight to obtain w/w %

    1.0319 g/1.9226 g = 53.67 %

    This is not the right answer.
     
  2. jcsd
  3. Mar 21, 2009 #2

    symbolipoint

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    If you already know the compound's composition, and you want to find the percent Barium in the sample, was the sample also wet with more than just water of crystalization? If it only had the water of crystalization, then you need none of the experiemental weighings. Just use formula weights of the compound.

    You might have misworded your question, so I will take a thought path according to this idea. After heating, you only have weight of DRY, no water, BaCl2.
    Percent BaCl2 = (1.5647g)*100/(1.9226g).

    Now, can you simply find percent Ba in BaCl2 and finish your question?
     
  4. Mar 21, 2009 #3
    That question is exactly word for word from the book.
     
  5. Mar 22, 2009 #4

    symbolipoint

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    Based on post #3, I also suggest you calculate the theoretical percent water of crystalization to your experimentally determined loss of mass from your sample.
     
  6. Mar 22, 2009 #5

    Borek

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    Seems to me like there is something wrong with the question.

    Unknow solid seems to be suggesting it is a mixture containing hydrated barium chloride and something else.

    Heated sample lost 0.3579 g. Assuming it lost only water, that was 0.01987 moles of water. Assuming all water was from hydrated barium chloride, that means 9.933x10-3 moles of chloride. That in turn means 2.426 g of hydrated chloride, more than initial mass.

    I have no idea what other assumptions can be made. If sample lost water and something else, amount of barium chloride can be anything.

    I can be missing something, but I have no idea what else can be tried.
     
  7. Mar 24, 2009 #6
    One thing I was just thinking. Would amount of Ba always be the same in the total sample because of moles?

    Does it make sense to just take MW Ba/MW BaCl2*2H2O = 137.327/244.26296 = 56.22%
     
  8. Mar 24, 2009 #7

    Borek

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    In the pure dihydrate of barium chloride - yes. But you are not told it's a pure substance, quite the opposite - you are told it is "unknown soild containing". That suggests other compounds are present as well.
     
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