Chemistry Reference
In-Depth Information
II - Calcul of the distribution of iron between ferrous and ferric iron
1
The structural formula is calculated on a number of cations in all, or
part, of the unit cell;
2
The amount of ferrous iron and ferric iron is equal to the total iron:
Fe*
=
Fe
2+
+
Fe
3+
3
The number of oxygens corresponding to the number of cations is fixed:
=
2 (Si
+
Ti)
+
1,5 (Al,
+
Fe
3+
+
Cr)
+
(Fe
2+
+
Mn
+
Mg
+
Ca)
+
0,5
(Na
K).
So we have two equations to calculate Fe
2+
and Fe
3+
+
4
The contents of FeO and Fe
2
O
3
can then be recalculated in weight of
oxides in the analysis.
Example 2
: An aluminous diopside Ca (Mg, Fe
2+
)
1−x
(Al
VI
, Fe
3+
)
x
Si
2−x
,
Al
IV
x
O
6
number of cations
=
4
=
6 number of oxygens.
Table 2.
Example of a mineral with ferrous and ferric iron without water aluminous
diopside Ca (Mg, Fe
2
+
)
1
−
x
(AI
VI
, Fe
3
+
)
x
Si
2
−
x
AI
IV
x
O
6
.
Weight %
Results of the
microprobe
Recalculated
analysis
Structural formula
SiO
2
45,76
45,76
Si
1,72
Z site (2)
TiO
2
0,96
0,96
AIiv
0,28
AI
2
O
3
8,45
8,45
total AI
0,38
FeO*
9,77
AIvi
0,10
Fe
2
O
3
5,11
Ti
0,03
FeO
5,18
0,14
Fe
3
+
MnO
0,09
0,09
0,27
Y site (1)
R
3
+
MgO
10,15
10,15
0,16
Fe
2
+
CaO
24,01
24,01
Mn
0,00
Na
2
O
0,31
0,31
Mg
0,57
K
2
O
0,00
0,00
R
2
+
0,73
BaO
0,14
0,14
Ca
0,97
X site (1)
total
99,64
100,16
Na
0,02
K
0,00
Ba
0,00
Ca
+
Na
+
K
+
Ba
0,99
Parameters of the calculation.
Number of cations
=
4.
Number of oxygens
=
6.
Columns of intermediate calculations are omitted.
Total iron
=
Fe
2
+
+
Fe
3
+
=
Fe
=
0,31.
Number of oxygens calculated on the previous number of cations.
Regardless the distribution of ferrous iron
=
5.93.
