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Stoichiometry and hydration of Mg nitrate

  1. Oct 1, 2007 #1
    Hey Guys, I'm having a little trouble with this problem:

    A mixture of Mg(NO3)2 and its hydrate Mg(NO3)2*2H2O has a mass of 1.7242 g. After heating to drive off all the water, the mass is only 1.5447 g. What is the percentage of Mg(NO3)2*H2O in the original mixture? Report your answer to 4 significant figures.

    Molar Masses: Mg(NO3)2=148.31 g/mol; Mg(NO3)2*2H2O = 184.34 g/mol.

    So, 1.7242g - 1.5447g = 0.1795g

    0.1795 g H2O x 1mol H2O/18.02g H2O = 0.009961mol H2O

    0.009961 mol H2O x 1 mol Mg(NO3)2 / 2 mol H20 = 0.004981 mol Mg(NO3)2

    0.004981 mol Mg(NO3)2 x 148.31g Mg(NO3)2/1 mol Mg(NO3)2 = 0.07387 g Mg(NO3)2

    Therefore, 0.7387 g Mg(NO3)2 / 1.7242 g x 100% = 42.84%

    Thus, % mass of Mg(NO3)2*2H20 = 57.16%

    However, 0.004981 mol Mg(NO3)2*2H2O x 184.34g Mg(NO3)2*2H2O / 1 mol Mg(NO3)2*2H2O = 0.9181g Mg(NO3)2*2H2O

    0.9181g Mg(NO3)2*(2)H2O / 1.7242g x 100% = 53.25%

    Can anybody tell me where I'm going wrong?
  2. jcsd
  3. Oct 1, 2007 #2


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    Very close. Here is where you got off track...
    This is actually the number of moles of the dihydrate of magnesium nitrate. Continue from here and you will get the correct answer.
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