Stoichiometry and hydration of Mg nitrate

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?

chemisttree
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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?

Very close. Here is where you got off track...
0.009961 mol H2O x 1 mol Mg(NO3)2 / 2 mol H20 = 0.004981 mol Mg(NO3)2
This is actually the number of moles of the dihydrate of magnesium nitrate. Continue from here and you will get the correct answer.