Monday

November 30, 2015
Posted by **Gaby** on Wednesday, April 11, 2007 at 3:23pm.

whose general formula is

Mx(A)y * zH2O

Ignore the effect of interionic attractions in the solution.

a. A"-is a common oxyanion. When 30.0 mg of the anhydrous

sodium salt containing this oxyanion (Na,,A, where n = 1, 2,

or 3) is reduced, 15.26 mL of 0.02313 M reducing agent is

required to react completely with the Na,A present. Assume

a 1: 1 mole ratio in the reaction.

b. The cation is derived from a silvery white metal that is

relatively expensive. The metal itself crystallizes in a

- body-centered cubic'unit cell and has an atomic radius of

198.4 pm. The solid, pure metal has a density of 5.243 g/cm3.

The oxidation number of M in the strong electrolyte in question

is +3.

c. When 33.45 mg of the compound is present (dissolved) in

10.0 mL of aqueous solution at 25"C, the solution has an

osmotic pressure of 558 tor.

I already got a. A=NO3

I'm trying to do the rest, but I have no clue,

I used the density and the radius to get the volume of the cell and then the mass of the cell, but I don't know if that's what I'm supposed to do, or what to do. Please help! Thanks

I have worked this before.

From a), we can identify the anion from the titration.

0.01526 L *0.02313 M = 0.0003529 mols.

molar mass= g/mol = 0.030/0.0003529 = 84.99 for the sodium salt of the anion.

This is an oxyanion; therefore, we can do a little educated guessing here. There are a limited number of oxyanions; e.g.,

carbonate

nitrate

nitrite

phosphate

phosphite

chlorate

chlorite

hypochlorite

bromate

iodate

We don't know if the anion is -1, -2, or -3 so the sodium salt may be NaA, Na

So, let's assume the anion is trivalent. That would make the sodium salt Na

3*Na + 1*O = 68.97 + 16 = 84.97 which leaves only 94.99-84.97 = 0.02 available for the X part. I don't know of any element with an atomic mass of 0.02. So trivalent is out.

Try divalent. Na

2*Na + 1*O + X = 45.98 + 16 + X = 61.98 + X = 84.99 so that leaves 84.99-61.98 = 23.01 for X. Check out the periodic table and there aren't any non-metals in the list above with an atomic mass of 23.01.

Try y = 2.

2*Na + 2*O + X = 45.98 + 32 + X = 77.98 + X = 84.99 and that leaves 84.99-77.98 = 7.01 for the non-metal. Again, no takers.

Try y = 1.

2*Na + 3*O + X = isn't possible because 2*Na + 3*O addes to more than 84.99 leaving nothing available for X.

So Na

Now we try NaXO

First, try y = 1

Na + 1*O + X = 22.99 + 16 + X = 38.99 + X which leaves 84.99-38.99 = 46 for X with no takers in the periodic table.

Try y = 2.

Na + 2*O + X = 22.99 + 32.00 + X = 54.99 + X leaving 84.99 - 54.99 = 30 for X. P is 30.98 which is the closest but the titration uses 4 significant figures so "off" by almost 1.00 isn't good enough. Let's forge ahead.

Try y = 3

Na + 3*O + X = 22.99 + 48 + X = 70.99 + X which leaves 84.99-70.99 = 14.00 for X. That MUST be N so the salt is NaNO

All of this is using just the a) part of your question. But you now know the anion is NO

Finally, as one last hint, the osmostic pressure part of the problem will give you the molar mass of the COMPOUND and that will allow you to identify z, the number of mols of water in the crystal.

Post any work if you require further assistance. Also, please be detailed about any questions related to this problem.

In the original problem of

MxAy*zH2O. you now know x = 1 and y = 3. M is still unidentified and z is not known at this point. But 2 out of four isn't bad.

I understand the first part perfectly. Part b is what I'm having trouble with. I tried using the radius and density to get the mass of the atom, but it doesn't seem to work: I did the following:

because it's a body centered cubit unit cell:

r=0.43301a

a= 458.189pm

Volume=(458.189)^3 = 9.62e7pm^3=

9.62e-23cm^3

M=D*V

M= 5.243*9.62e-23 = 5.04e-22

and that's where I am at..

Go to www.webelements.com.

Scroll down the menu on the left side of the screen and click on density (I think its a sub topic--I think the main topic is bulk properties or something like that) and look for a density VERY CLOSE to 5.423 g/cc (I don't believe the site uses g/cc but you can convert. There is only one element with a density of that value. Check out if it has the correct crystal structure. You can check them out one at a time, which is tedious, or scan with graphs, bars, etc. Let me know how you do.

Notice I made a typo above. That density in the original post is 5.243 g/cc.

The website gives Europium it gives it in kg/m^3, but I converted it and it gives 5.244g/cc which is good enough and the crystal structure checks out, correct radius and everything.

ok. Eu is correct. You have the volume correct. But there are two particles per unit cell which makes the molar mass come out 151.9 or right at that. I can show you how to do that if you wish. And you omitted multiplying by 6.022E23 I believe but didn't look closely. Anyway, good work and let me know if there is anything else. You still need to determine z, the H2O of crystallization.

okay so it would be:

5.04e-22 * 6.022e23

= 303.75/2=151.9g/mol

Thank you.. I will try to work the rest from here..

Start from your 5.04E-22 g.

2 particles/unit cell.

[2*molar mass/6.022E23] = 5.04E-22

solve for molar mass = 151.8 amu.

So you were almost there and it would have taken less time to look up the molar mass and confirm with density and crystal structure rather than the other way around.

right. I got ahead of myself and showed you how to do it but you had already posted by the time I finished. Good work.

Don't forget to use the "i" in the osmotic pressure formula to account for the four particles that are formed in the solution.

okay.. I used "pi"= iMRT

M= "pi"/RTi

then I converted Molarity to moles by * the volume and I got 7.5e-5moles

then 0.03345g/7.5e-5moles

a molar mass of 446g/mol of COMPOUND

molar mass of Eu(NO3)3= 337.93g/mol

446-337.93= 108.07= molar mass of xH2O

(16)(2.016)x=108.07

x=3.35 .. round off to 3??

Eu(NO3)3*3H2O

Not quite.

You forgot the + sign.

[(16)+(2.016)]x = 108.07

18x=108

x = 6.00

You might have guessed something was not quite right since all the titrations are to four significant figures and the atomic mass of Eu was so close comparing calculated and published values--so rounding 3.35 to 3.00 makes one wonder about it. As it is,

108.07/18.016 = 5.999 which rounds to 6 much better.

By the way, this post is now on page 4 and most of us get lazy and don't look back this far to see about possible follow ups; therefore, if you have anything else to ask about this problem, please post a new question on page 1. Thanks, and I think you did very well on this problem, especially on the crystallography part. That was interesting how you progressed through that part. Good luck and thanks for using Jiskha.