Chemistry Reference
In-Depth Information
Cl
according to the theoretical structural formula. The quantities of OH
in the structural formula OH and H
2
O wt% in the analysis can then be
recalculated.
Example 3
: A biotite
Potassium in biotites (and more generally in micas) is often poorly meas-
ured, for several reasons:
III - Calculation the percentage of H
2
O
is by fixing the sum OH
+
F
+
K and Na are volatile components that evaporate under the beam of the
microprobe; it is preferable to make the determination of such elements
early in the cycle of dosage and possibly move the sample a little after
having measured them;
Table 3.
Example of a mineral with water a biotite X
2
Y
4-6
Z
8
O
20
(OH)
4
.
Wt %
Results of the
microprobe
Recalculated
analysis
Structural formula
SiO
2
39,01
39,01
Si
5,76
site Z (8)
TiO
2
1,20
1,20
AI IV
2,24
AI
2
O
3
15,78
15,78
AI total
2,75
FeO
14,58
14,58
AI VI
0,51
MnO
0,04
0,04
Ti
0,13
MgO
14,96
14,96
Cr
0,00
CaO
0,00
0,00
R
3
+
0,64
site Y (5,74)
Na
2
O
0,13
0,13
Fe
1,80
K
2
O
9,41
9,41
Mn
0,01
F
1,32
1,32
Mg
3,29
CI
0,13
0,13
R
2
+
5,10
H
2
O
3,40
Ca
0,00
96,56
99,96
Na
0,04
K
1,77
Ca
+
Na
+
K
1,81
site X (1,81)
OH
3,35
OH
+
F
+
CI
=
4
F
0,62
CI
0,03
Parameters of the calcuation.
It is assumed that all the iron is at ferrous state.
K, Na, Ca not taken into account.
Number of oxygens corresponding to the other cations
=
21.
OH
+
F
+
CI
=
4
This biotite is not strictly octahedral.
K, Na, Ca are probably under-estimated or this biotite is slightly chloritized.
Columns of intermediate calculations are omitted.
