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Identity of the metal in a particular sulfate salt

  1. Mar 6, 2005 #1
    Three Students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472-g sample of the salt in water adn treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After teh precipitate had been filtered and dried, it weighed 0.2327 g.
    Each student analyzed the data independently and came to different conclusions.
    The conclusions were that the metal is sodium, titanium or gallium. What further tests you suggest to determine which student is most likely correct.

    This is how I tried to tackle the problem. Since the precipitate is BaSO4 and weighs .2327 g, calculating the molar mass of BaSO4, the no. of moles of precipitate I get is 9.97129E-4 mol. Thus the sulfate ions are 9.97129E-4 mol. Since no. of moles of sulfate cannot change during reaction, when the sulphur is combined with the metal, then too it is in the quantity 9.97129E-4 mol which means it has mass 0.09579 g which means that mass of metal is 0.051405 g

    If the metal has valency 1, then no. of moles of metal is 2*9.97E-4 mol
    If the metal has valency 2, then no. of moles of metal is 9.97E-4 mol
    If the metal has valency 3, then no. of moles of metal is 9.97E-4*2/3 mol

    Taking each possibility into account, I get as molar mass
    for metal valency 1, 25.77 g/mol
    for metal valency 2, 51.55 g/mol
    for metal valency 3, 77.329 g/mol

    None of these molar masses correspond to any metal. Where am I going wrong?
  2. jcsd
  3. Mar 6, 2005 #2


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    I hope you realize that your calculation, while a good thing to do, is irrelevent to the question.

    valency=1 :
    [tex]M_2SO_4 + BaCl_2 \longrightarrow 2MCl + BaSO_4 [/tex]

    Mol. Wt. of BaSO4 is about 233g/mol, so there's about 0.001 moles of BaSO4 formed. Here, 1 mole of M2SO4 gives 1 mole of BaSO4, so there must have been about 0.001 moles of M2SO4. Hence the Mol. Wt. of M2SO4 should be about 147 g/mol. This sets the atomic wt. of M at about 25. The nearest monovalent metal is Na at 23.

    valency = 2 :

    [tex]MSO_4 + BaCl_2 \longrightarrow MCl_2 + BaSO_4 [/tex]
    1 mole of MSO4 gives 1 mole of BaSO4. So, there must be 0.001 moles of MSO4 again. This sets the atomic weight of M at about 50 g/mol. Ti is often bivalent and has atomic mass 49. V is rarely bivalent (usually trivalent or pentavalent).

    valency = 3 :
    [tex]M_2(SO_4)_3 + 3BaCl_2 \longrightarrow 2MCl_3 + 3BaSO_4 [/tex]
    1 mole of M2(SO4)3 gives 3 moles of BaSO4, so, there must have been about 0.00033 moles of M2(SO4)3. So, the mol. wt. of M2(SO4)3 would be about 426 g/mol. This requires the atomic weight of M to be about 69, and Ga, is the closest trivalent element at about 70.

    Now, as for the real question...it wants you to perform chemical tests to determine which of these possibilities is the more likely answer.
  4. Mar 6, 2005 #3
    I know it does not hold much relevance. I was just trying to analyze the data on my own and was surprised by the large discrepancy between my results and the book's. Like my valency 1 metal had molar mass more than 25 g/mol.
  5. Mar 7, 2005 #4


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    In the real world that will be perfectly normal. Nobody told that the sample was reagent grade so whole analysis is somewhat approximate. 1g out of 25g is 4% error - which is rather good considering the situation.

    Now, most textbooks authors 'forget' about the real world and numbers they give are absurdally accurate. In this particular case if the data was too accurate whole question will make no sense as the answer will be obvious :smile:
  6. Mar 7, 2005 #5


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    In "the real world" your answer would most likely be sodium...least expensive.
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