Chemistry Reference
In-Depth Information
Table 1.
Detailed example of calculation of the structural formula of a mineral without water nor without ferric iron a plagioclase (Na,K)
(1
−
x)
Ca
x
AI
(1
+
x)
Si
(3
−
x)
O
8
.
In 99,2 g (total weight) of the
mineral
Calculated structural foumula
Molecular
weight
Number of
cations
Number of
oxygens
On a base of
5 cations
On a base of
8 oxygénes
Weight %
SiO
2
59,43
60,09
60,09
0,98902
1,97803
Si
2,68
2,67
TiO
2
0,03
79,90
79,90
0,00038
0,00075
Ti
AI
2
O
3
25,17
101,94
50,97
0,49382
0,74073
AI
1,34
1,33
Cr
2
O
3
0
152,02
76,01
0
0
Cr
Fe
2
O
3
0
159,70
79,85
0
0
Fe
3
+
FeO
0,02
71,85
71,85
0,00028
0,00028
Fe
2
+
MnO
0,03
70,94
70,94
0,00042
0,00042
Mn
MgO
0
40,32
40,32
0
Mg
CaO
7,47
56,08
56,08
0,13320
0,13320
Ca
0,36
0,36
% An
36,7
Na
2
O
6,93
61,09
30,55
0,22688
0,11344
Na
0,61
0,61
% Ab
62,6
K
2
O
0,12
94,20
47,10
0,00255
0,00127
K
0,01
0,01
% FK
0,7
BaO
0
153,36
153,36
0
0
Ba
F
0
19,00
19,00
0
0
F
CI
0
35,46
35,46
0
0
CI
H
2
O
0
18,02
9,01
0
0
OH
total
99,20
1,84654
2,96813
O
8,04
8
number of
cations
5,00
4,97
factor
2,70776
2,69530
Remarks
1-there are two atoms of AI in 101,94 g of AI
2
0
3
; to obtain the number of atoms you have thus to divide by 101,94/2
=
50,97.
2-it is enough for the structural formula to round the results to two decimal digits.
3-the quality of the analysis is estimated by concordance with the theoretical structural formula, the value of the parameter x must be identical for each cation and lesser
important, the total sum of weight % that should be 100%.
The main cations found in the compositions of common minerals are here indicated, even if they are not involved in this example, so that to give a table of their molecular weights.
