00:03
So continuing on with the work surrounding the lattice structures.
00:08
So we have two tetrahedral holes for each anion in either our hexagonal closed -packed or cubic closed -packed structure for our anions.
00:17
That is the ratio of the octahedral holes to anions in either of these structures that is a one -to -one.
00:23
So for example, our oxide can form a close -pack array, whilst titanium will occupy one half of the holes where these holes are octahedral.
00:30
So for example, we let n use the number of oxide ions, where 2 n tetrahedral holes, n octahedral holes.
00:49
So one half of the octahedral holes are occupied by titanium.
00:55
So we take one half multiplied by n, we get n over 2 for our titanium.
01:00
And so the ratio for our titanium to oxide ions, we have n over 2 for titanium to n for oxygen.
01:13
That gives us a 1 to 2 ratios, therefore we have tio2.
01:25
So next, we can calculate the percent by mass of titanium...