- #1
Mitchtwitchita
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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?
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?