As an example of the recalculation procedure, consider the following microprobe data for a phlogopite (biotite) from the volcanic rocks of the Hickey Formation of central Arizona. Note that more digits than are significant are retained throughout the calculations and rounding only occurs at the end.
|
Oxide |
Oxide |
Single- |
Moles |
Oxygen |
Moles |
|
SiO2 |
37.7 |
60.0848 |
0.6275 |
2 |
1.2549 |
|
TiO2 |
2.26 |
79.8988 |
0.0283 |
2 |
0.0566 |
|
Al2O3 |
14.45 |
50.9806 |
0.2834 |
1.5 |
0.4252 |
|
Cr2O3 |
0.60 |
75.9951 |
0.0079 |
1.5 |
0.0118 |
|
FeO |
10.1 |
71.8464 |
0.1406 |
1 |
0.1406 |
|
MnO |
0.08 |
70.9374 |
0.0011 |
1 |
0.0011 |
|
MgO |
20.4 |
40.3044 |
0.5061 |
1 |
0.5061 |
|
CaO |
n.d. |
56.0794 |
0.0000 |
1 |
0.0000 |
|
BaO |
0.50 |
153.3394 |
0.0033 |
1 |
0.0033 |
|
Na2O |
0.71 |
30.9895 |
0.0229 |
0.5 |
0.0115 |
|
K2O |
9.16 |
47.0977 |
0.1945 |
0.5 |
0.0972 |
|
F |
3.02 |
|
|
|
|
|
Cl |
0.00 |
|
|
|
|
|
Total |
98.9 |
|
1.8155 |
|
2.5083 |
We want to recalculate atom concentrations on a 11-oxygen basis (10 oxygens and 2 OH); the factor needed to convert 2.5083 into 11 is C.F. = 11 / 2.5083 = 4.3854. Next we multiply all the cations by this conversion factor:
|
Si |
2.752 |
|
Ti |
0.124 |
|
Al |
1.243 |
|
Cr |
0.035 |
|
Fe |
0.617 |
|
Mn |
0.005 |
|
Mg |
2.220 |
|
Ba |
0.014 |
|
Na |
0.100 |
|
K |
0.853 |
First we fill the T-site with Si and enough Al to bring its sum to 4.00. If the amount of Si exceeds 4.00 or if Si and Al are insufficient to fill the T site (within a 2% error), the analysis is bad. We get 3.994.
(Si2.752 Al1.243)T
Next, we fill the M-site with the remaining Al, Ti and Fe and add Cr, Mg, and Mn. This should bring the total to 3.0000. The acceptable range is about 2.94 to 3.06 (±2% error); the data sum to 2.9650.
(Ti0.124 Fe0.617 Cr0.035 Mg2.220 Mn0.005)M
Finally, we fill the A-site with Na, K, Ca and Ba. The A-site should total to near 1.00; the data yield 0.9679.
(Na0.100 K0.853 Ba0.014)A
Next we divide F by its atomic weight (18.9984) and multiply by the conversion factor. Subtracting the result from 2.00 determines the OH occupancy in the OH-site:
F = (3.02 / 18.9984) x 4.3854 = 0.6971
OH = 2.0000 - 0.6971 = 1.3029
So the site occupancy is:
(F0.697 Cl0.000 OH1.303)OH
To back-calculate the H2O content, multiply the OH value by 9.0076 to get moles of H and divide by the conversion factor:
H2O = (1.3029 * 9.0076) / 4.3854 = 2.68%
Finally, add this to the total above and subtract the oxygen-equivalents for the F:
(OH = F) = 3.02 * 0.4211 = 1.27
Total = 98.9 + 2.68 - 1.27 = 100.3
The site occupancies should be reported to no more than 3 significant digits:
(Na0.10 K0.85 Ba0.01)A (Ti0.12 Fe0.62 Cr0.03 Mg2.22 Mn0.01)M (Si2.75 Al1.24)T (F0.70 OH1.30)OH
As another example, consider an alkali feldspar (#1 in Table 37, The Rock-Forming Minerals, 2nd ed., Deer, Howie & Zussman, 1992). Here is a summary table for the calculation:
| Wt % | Cation | Oxygen | Cations | |
| oxide | props | props | p.f.u. | |
| SiO2 | 65.76 | 1.0945 | 2.1889 | 2.929 |
| Al2O3 | 20.23 | 0.3968 | 0.5952 | 1.062 |
| FeO | 0.16 | 0.0023 | 0.0034 | 0.006 |
| CaO | 1.19 | 0.0212 | 0.0212 | 0.057 |
| BaO | 0.63 | 0.0041 | 0.0041 | 0.011 |
| Na2O | 8.44 | 0.2724 | 0.1362 | 0.729 |
| K2O | 3.29 | 0.0699 | 0.0349 | 0.187 |
| Totals | 99.82 | 2.9895 | 5.010 |
The formula is (Na0.73 K0.19 Ca0.06 Ba0.01)A (Si2.93 Al1.06 Fe0.01)T O8. There are several internal checks we can make:
The A site should total 1.00 ® it's 0.99.
The T site should total 4.00 ® it's 4.00.
Because of charge balance constraints, Al should equal (1.0 + Ba + Ca) = 1.07 ® it's 1.06.
Si should equal (2.0 + Na + K) = 2.92 for the same reasons ® it's 2.92.
We can conclude that this is a very good feldspar analysis.
Back: 6.2.3. Assigning Cations to Crystallographic Sites | Home: Course Overview
Copyright 1997-2003, James H. Wittke
Last update: 01/18/2006 01:47 PM.