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posted by  alana on 10/31/2007 1:25:36 PM  |  status: Live  

Percent water

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Question Details:
The following data were collected from a gravimeter analysis of a hydrated salt;
Mass of crucible and lid (g)              18.733
Mass of crucible, lid and hydrated salt (g)      21.171
Mass of crucible, lid and anhydrous salt (g)    20.122
 
Determine the percent water in the hydrated salt. What is the percent by mass of water in copper(II)sulfate pentahydrate, CuSO4 . 5H2O ? What mass due to waters of crystalization is present in a 50.0 g sample of CuSO4 . H2O ?
Please show all calculations. Thank you!!!!!!!!!!!!!!

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posted by io2chip on 10/31/2007 5:59:16 PM  |  status: Live
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alana's comment:
"Great job!!!!!!!!!!!!! Thanks!"
Response Details:
1)      
         Mass of crucible, lid and hydrated salt                             21.171(g)
      -
         Mass of crucible and lid                                                  18.733(g)        
---------------------------------------------------------------------------
--->   Mass of hydrated salt : (21.171 - 18.733)                  =    2.438 (g)
          
2)
          Mass of crucible, lid and hydrated salt (salt + water):      21.171(g)
      -       
          Mass of crucible, lid and anhydrous salt (salt only) :        20.122(g)
         --------------------------------------------------------------------
--->   Mass of water :   (21.171 - 20.122)                   =            1.049(g)
         
         The percent water in the hydrated salt:  (mH2O/mhydrate salt) * 100 = (1.049/2.438)*100 = 43.03o/o
         The percent by mass of water in copper(II)sulfate pentahydrate, CuSO4 . 5H2O:
              ( Molar mass of 5H2O/Molar mass of CuSO4 . 5H2O)*100 = [(5*18)/(5*18 + 63.55 + 32.07 + 16*4)]*100 = (90/249.62)*100 = 36.05o/o
         Mass due to waters of crystalization is present in a 50.0 g sample of CuSO4 . H2O
     Molar mass of CuSO4 . H2O : 63.55 + 32.07 +   (16*4) + (1.008*2)+16.00 = 177.64g/mol
     Molar mass of H2O: (1.008*2) + 16.00 = 18.02g/mol
 In 177.64g CuSO4 . H2O, there is 18.02g H2O
---->in a 50.0 g sample of CuSO4 . H2O, there is:
             50.0g CuSO4 . H2O * (18.02g H2O /  177.64g CuSO4 . H2O) = 5.07g H2O
 
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