April 18, 2024, 06:38:06 AM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1]   Go Down

Topic: X ray diffraction and theoretical density  (Read 5266 times)

0 Members and 1 Guest are viewing this topic.

Offline elt2112

  • Regular Member
  • ***
  • Posts: 15
  • Mole Snacks: +0/-0
X ray diffraction and theoretical density
« on: May 17, 2015, 05:30:24 AM »
Struggling with a past paper question if anyone can help. So to find out the composition:
31.57 mg of V2O5 RMM = 181.88 
0.03157/181.88 = 1.7357 x 10-4
23.21 mg of V2Ox  RMM = (0.02324/0.03157) x 181.88 = 133.88
133.72 - (2 x 50.94) = 32.01
31.837 = 16 = 2
So the composition is V2O2

I've indexed the peaks and have a unit cell parameter of 4.06Å and determined that the unit cell is FCC which has 4 atoms in the unit cell so 1 formula unit.
The theoretical density is then (133.88/NA)/(4.06 x 10-8)3 and this is 3.322 g/cm3
But then it says that the actual density is 5.76 g/cm3, and so I must have gone wrong but can't figure out where because I'm assuming that my theoretical density should be higher than the actual density and it has defects lowering the density. Any help would be much appreciated!
Logged

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2053
  • Mole Snacks: +296/-12
Re: X ray diffraction and theoretical density
« Reply #1 on: May 17, 2015, 05:29:55 PM »
If the symmetry is FCC, what can you say about the ions at the vertices and the face centres of the cube? What does that imply about the stoichiometry of the unit cell?
Logged

Offline elt2112

  • Regular Member
  • ***
  • Posts: 15
  • Mole Snacks: +0/-0
Re: X ray diffraction and theoretical density
« Reply #2 on: May 20, 2015, 09:42:43 AM »
Well the ions at the corners contribute a quarter and the ones on the face contribute a half to the unit cell. So overall we have 5 ions?
Logged

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2053
  • Mole Snacks: +296/-12
Re: X ray diffraction and theoretical density
« Reply #3 on: May 20, 2015, 11:15:42 AM »
Quote
the ions at the corners contribute a quarter
Try again. You were right the first time. 8*1/8 + 6*1/2 = 4. But this can't be V2O2 because you have 1 of one type and 3 of the other - not 2 and 2. In fact the two ions must be the same (also for symmetry reasons - if they were different it wouldn't be FCC). So the unit cell must contain 4V and 4O (the other ions are on a displaced FCC lattice, at the centre and on the edges of the cube - like NaCl) and the theoretical density is twice what you calculated.
Logged

Offline elt2112

  • Regular Member
  • ***
  • Posts: 15
  • Mole Snacks: +0/-0
Re: X ray diffraction and theoretical density
« Reply #4 on: May 21, 2015, 02:40:14 AM »
Oh yes sorry, thanks for your *delete me*
Logged

Offline elt2112

  • Regular Member
  • ***
  • Posts: 15
  • Mole Snacks: +0/-0
Re: X ray diffraction and theoretical density
« Reply #5 on: May 21, 2015, 01:30:32 PM »
OK sorry just one more question, when it says discuss the defect structure and comment on its metallic character - would I say something like, where the oxygen defects occur vanadium is reduced to 4+ and is therefore metallic due to the partial population of the 3d orbitals in the reduced vanadium ions. These overlap with the O 2p orbital to give partially filled bands? Does defect structure mean what kind of defects?
Logged

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2053
  • Mole Snacks: +296/-12
Re: X ray diffraction and theoretical density
« Reply #6 on: May 22, 2015, 04:39:42 AM »
Quote
vanadium is reduced to 4+
What is the oxidation state of V in VO?
Logged
Pages: [1]   Go Up
« previous next »
Sponsored Links