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I really need help with how to solve the following question:

A hydrate of zinc nitrate has the formula Zn(NO3)2 * xH2O. If the mass of 1 mol of anydrous zinc nitrate is 63.67% of the mass of 1 mol of the hydrate, what is the value of x?

Note: 3 and 2 in the formula are subscripts.

In the book the answer to the question is x= 6, but I can't seem to get this answer. I need to know how to do it.

2006-10-29 13:01:09 · 3 answers · asked by angelslove 1 in Science & Mathematics Chemistry

3 answers

Step 1: Treat the 63.67% as grams of zinc nitrate per 100 grams. If you do this, the other 36.33% is H2O and therefore you can treat that as grams per 100. At this point we have 63.67g Of zinc nitrate and 36.33g of H2O.

Step 2: Convert the mass in grams of each compound by using their respective molar masses. You should get .336mol of zinc nitrate and 2.02mol of water.

Step 3: All you have to do now is divide the 2.02 by .336 (just dividing the larger by the smaller) and you will get the value 6 (approximately 6, but as long as it is close, due to mathematically accepted error, it is considered the integer 6) which is the value you are looking for for H2O.

Note: Essentially what you are doing is looking at this as for every .336 moles of Zn(NO3)2 you have 2.02 moles of H2O. We just divide the smaller into the larger as a mathematical technique to get the integer of both terms. We are actually dividing .336/.336 = 1 for Zn(NO3)2 and 2.02/.336 = 6 for H2O. There may be other ways of finding these values, but I think this is most practical, and it is the approach I use with people I am tutoring here at the university I attend and it seems to help them on their tests. Good luck.

2006-10-29 13:17:36 · answer #1 · answered by Anonymous · 0 0

the mass of the hydrate is the sum of the mass of Zn(NO3)2 and the total mass of x moleculars of H2O

the anydrous zinc nitrate is Zn(NO3)2

=> mZn(NO3)2/ mZn(NO3)2 * xH2O = 189 / (189 + x*18) = 63.67/100

sovle this equation we will have x is nearly equal to 6 ( 5.99445 )

2006-10-29 21:29:08 · answer #2 · answered by James Chan 4 · 0 0

Mr of zinc nitrate=189.41
Mr of water of cystallisation=18x (or y, a dummy variable i will use now)

189.41/63.67=18x /36.33
18x=108.1
x=108.1/18
=6.004

2006-10-29 21:18:18 · answer #3 · answered by Anonymous · 0 0

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