In a sample of the non-stoichiometric crystal Fe0.96O,the % of cationic sites occupied by Fe3+ ions is?

ans.8

please tell how?

let x be wt of fe3+ and then 0.96-x will be fe2+ 

the molecule / crystal is neutral therefore net positive = net negative

3x+ 2(0.96-x) = 2

x = 0.08

or 8 percent

96 Fe atom are associated with 100 O atoms . out of 96 Fe atoms suppose Fe present as Fe+2 = x and Fe present as Fe+3 = 96 -x

total charge on x Fe+2 ions and (96-x) Fe+3 ions should be equal to charge on 100 O-2 ions...this is because overall molecule is neutral

so x X 2 + (96-x) X 3 = 100 X 2

2x + 3(96-x) = 200

2x + 288 - 3x = 200

-x = 200 - 288

x = 88

so out of 96 atoms 88 are Fe+ and 8 are Fe+3...

so fraction of Fe present as Fe+2 = 88/96 X 100 = 91.67 %

and fraction of Fe present as Fe+3 = 8/96 X 100 = 8.33 %

PLZ  APPROVE!

THNX!

:)

i can write it as Fe96O100
Let Fe3+ be x and Fe2+ be 96-x
by neutralising the charge
3x + 2(96-x) = 200
=> x= 8
and 96-x = 88
so percentage of Fe(3+) is 8.33 %

Please approve