Question

Will Fan and Medal!
A mixture of Mg(NO3)2 and its hydrate Mg(NO3)2x2H20 has a mass of 1.9151g. After heating to drive off all the water, the mass is only 1.7376g. What is the weight percentage of Mg(NO3)2xH2O in the original mixture. report your answer to 4 sig figs. Molar masses: Mg(NO3)2->148.31 g/mol; Mg(NO3)2xH2O 184.34 g/mol

Answer 1

Since heating gets rid of all the water, the mass difference before and after heating is the mass of the water:

1.9151 - 1.7376 = 0.1775 g

Converting this to moles will allow you to find how much Mg(NO3)2 it was associated with (note - molar mass of water is 18 g/mol):

0.1775 / 18 = 0.009861 mol

There are 2 water molecules for every 1 unit of Mg(NO3)2x2H2O. That means for however many moles of water you have, there are HALF as many moles of Mg(NO3)2x2H2O. This means you have 0.009861 / 2 = 0.004930 mol of Mg(NO3)2x2H2O. The mass of 0.004930 mol of Mg(NO3)2x2H2O is:

0.004930 x 184.34 = 0.9088 g

As a percentage of the mass of the original mixture, it is:

0.9088 / 1.9151 x 100 = 47.45%

Let me know if you have any questions!