HOMEWORK
IÕd like you to try to work through the process of ÒnormalizingÓ a mineral analysis (that is, converting a chemical analysis to a mineral formula) between now and ThursdayÕs class. I will not require you to hand your work in Ð instead, bring it with you and I will go over it at the start of class.
Analytical instruments can be used to obtain chemical analyses. The resulting data are generally reported in weight percent of the major oxides in the mineral. Use the following chemical analysis and the instructions below to determine the specific mineral formula and the identity of the unknown mineral. Be sure to show all your work! Using a spreadsheet program like Excel makes these calculations easier, although this particular analysis can be done pretty quickly by hand as well. A similar set of instructions appears in Box 1.5 (p. 22) of your text.
SiO2 38.05%
MgO 41.57%
(Assume the known total number of oxygen atoms per formula unit is 4)
Mineral formula_______________________________________
What mineral is this?________________________________
Steps for determining a specific mineral formula:
1) Convert oxide wt.% into molecular proportion of each oxide. This is done by dividing the wt. % of each oxide by the molecular weight of the oxide. This gives the molecular proportion of each oxide. (The molecular weight for each is calculated from their atomic weights.)
2) Multiply the molecular proportion for each oxide by the # of oxygen atoms present in each oxide. This gives the O atomic proportion.
3) Sum the O atomic proportion column.
4) Divide the known total # of Oxygen atoms per unit cell in the mineral by the sum of the O atomic proportion. (e.g. Olivine is know to have 4 oxygen atoms in its mineral formula, Feldspars have 8). This operation gives you a normalization factor.
5) Next, normalize the O atomic proportions from each oxide by multiplying each entry by this normalization factor. This gives the number of anions based on the known number in the mineral formula.
6) Determine the number of cations associated with the oxygens by dividing the number of anions determined in step 5 by the number of oxygens in the reported oxide. (e.g. SiO2 has 2 O per 1 Si, Al2O3 has 1.5 O per 1 Al).
7) The number you obtain after doing step six is the number of cations that are in the final mineral formula.
EXAMPLE:
Olivine:
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step 1 |
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step 2 |
step 4 |
step 5 |
step 6 |
step 7 |
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Element |
Oxide wt. % |
Molec. Wt. |
Molec. Prop. |
# Oxygen |
O atomic |
Normaliza. |
# of Anions |
Oxygen |
Cations |
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of oxides |
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proportion |
factor |
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per cation |
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SiO2 |
31.85 |
60.074 |
0.530179 |
2 |
1.060368 |
1.90543 |
2.02043 |
2 |
1.0102 |
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FeO |
58.64 |
71.841 |
0.816247 |
1 |
0.816247 |
1.90543 |
1.5553 |
1 |
1.5553 |
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MnO |
0.85 |
70.937 |
0.011982 |
1 |
0.011982 |
1.90543 |
0.02283 |
1 |
0.02283 |
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MgO |
8.49 |
40.299 |
0.210675 |
1 |
0.210675 |
1.09543 |
0.04142 |
1 |
0.40142 |
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(step 3) |
Total |
2.09926 |
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Normalization factor= |
4Oxygens/2.09926 |
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Formula: (Fe 1.555 Mg 0.401 Mn 0.023) Si 1.01 O 4