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Son Mesaj Yazan *JujU*

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Nov 6, 2013
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A norm is essentially a set of idealized synthetic minerals that are calculated from a bulk chemical analysis of a rock for comparison purposes. Norms are usually calculated for volcanic rocks which have glass and/or exceedingly small crystals that make it difficult to determine a mode, and for metamorphosed igneous rocks that no longer have the original igneous mineralogy. The normative minerals can be thought of as representing the minerals that could potentially crystallize if the rock were cooled under perfect equilibrium dry conditions. The assemblages of normative minerals are based largely on reality. In real rocks the mineral pairs:
quartz and Mg-rich olivine
quartz and nepheline
Al-silicates and augite
orthopyroxene and nepheline,
are usually mutually exclusive. The mutually exclusive normative mineral pairs are similar:
quartz and olivine
quartz and nepheline
corundum and diopside
orthopyroxene and nepheline.

These mutually exclusive mineral pairs result in the four common normative assemblages:

[h=4]Norm assemblages[/h][h=4]Meaning[/h]
Quartz, corundum, OPX:
Silica-saturated, peraluminous, subalkaline
Quartz, diopside, OPX:
Silica-saturated, subaluminous, subalkaline
Olivine, diopside, OPX:
Silica-undersaturated, subaluminous, subalkaline
Olivine, diopside, nepheline:
Silica-undersaturated, subaluminous, peralkaline

NORM CALCULATION PROCEDURE
The following procedure has been abridged for the most common rocks, omitting many calculations that must be done if the rock is unusual. For example, the procedures to calculate normative leucite in strongly silica-undersaturated rocks, aegirine in alumina-undersaturated alkalic rocks, or hematite in oxidized rocks have been omitted. The procedure below also departs from the original CIPW norm in that the pyroxenes, plagioclase, and olivine are calculated as their solid solutions rather than as end member components. I have included four worked examples so you can see how calculations are done and how to keep track of the numbers. You must do all calculations to at least five decimal places or severe rounding errors will result.
1) Use only the eleven major element oxides, for which the gram formula weights are:
SiO[SUB]2[/SUB]​
60.0843​
TiO[SUB]2[/SUB]
79.8988​
Al[SUB]2[/SUB]O[SUB]3[/SUB]
101.9613​
Fe[SUB]2[/SUB]O[SUB]3[/SUB]
159.6922​
FeO
71.8464​
MnO
70.9374​
MgO
40.3044​
CaO
56.0794​
Na[SUB]2[/SUB]O
61.9789​
K[SUB]2[/SUB]O
94.1960​
P[SUB]2[/SUB]O[SUB]5[/SUB]
141.9445​

2) Take the oxide weight percents in the chemical analysis and divide them by their respective formula weights to give molar oxide proportions. Use these molar oxide proportions in all subsequent calculations.
3) Add MnO to FeO. MnO is now zero. The combined FeO and MnO will now be called FeO.
4) Apatite (Ap): Multiply P[SUB]2[/SUB]O[SUB]5[/SUB] times 3.33 and subtract this number from CaO. P[SUB]2[/SUB]O[SUB]5[/SUB] represents 2/3 of an apatite molecule, so multiply P[SUB]2[/SUB]O[SUB]5[/SUB] times 2/3, and put this number in Apatite. P[SUB]2[/SUB]O[SUB]5[/SUB] is now zero.
5) Ilmenite (Ilm): Subtract TiO[SUB]2[/SUB] from FeO. Put the TiO[SUB]2[/SUB] value in ilmenite. TiO[SUB]2[/SUB] is now zero.
6) Magnetite (Mt): Subtract Fe[SUB]2[/SUB]O[SUB]3[/SUB] from FeO. Put the Fe[SUB]2[/SUB]O[SUB]3[/SUB] value in magnetite. Fe[SUB]2[/SUB]O[SUB]3[/SUB] is now zero.
7) Orthoclase (Or): Subtract K[SUB]2[/SUB]O from Al[SUB]2[/SUB]O[SUB]3[/SUB]. Put the K[SUB]2[/SUB]O value in orthoclase. K[SUB]2[/SUB]O is now zero.
8) Albite (Ab) [Provisional]: Subtract Na[SUB]2[/SUB]O from Al[SUB]2[/SUB]O[SUB]3[/SUB]. Put the Na[SUB]2[/SUB]O value in albite. Retain the Na[SUB]2[/SUB]O value for possible normative nepheline.
9) Anorthite (An):
a) If CaO is more than Al[SUB]2[/SUB]O[SUB]3[/SUB], then subtract Al[SUB]2[/SUB]O[SUB]3[/SUB] from CaO. Put all Al[SUB]2[/SUB]O[SUB]3[/SUB] into anorthite. Al[SUB]2[/SUB]O[SUB]3[/SUB] is now zero.
b) If Al[SUB]2[/SUB]O[SUB]3[/SUB] is more than CaO, then subtract CaO from Al[SUB]2[/SUB]O[SUB]3[/SUB]. Put all CaO into anorthite. CaO is now zero.
10) Corundum (Cor): If Al[SUB]2[/SUB]O[SUB]3[/SUB] is not zero, put the remaining Al[SUB]2[/SUB]O[SUB]3[/SUB] into Corundum. Diopside and Al[SUB]2[/SUB]O[SUB]3[/SUB] are now zero.
11) Calculate the current ratio of Mg/(Mg+Fe[SUP]+2[/SUP]). This ratio is called Mg' and will be the Mg/(Mg+Fe[SUP]+2[/SUP]) ratio for all normative silicates.
Mg' = MgO/(MgO+FeO)
12) Calculate the mean formula weight of the remaining FeO and MgO. This combined Fe-Mg oxide called FMO will be used in subsequent calculations.
Formula weight of FMO = (Mg'*40.3044)+((1-Mg')*71.8464)
13) Add FeO and MgO to get FMO.
14) Diopside (Di): If CaO is not zero, subtract CaO from FMO. Put all CaO into diopside. CaO is now zero.
15) Hypersthene (Hy) [Provisional]: Put all remaining FMO into hypersthene. Retain the FMO value for the possible calculation of normative olivine.
16) Calculate the amount of SiO[SUB]2[/SUB] needed for all of the normative silicate minerals listed above, allotting SiO[SUB]2[/SUB] as follows:
Orthoclase * 6 = needed SiO[SUB]2[/SUB] for each Orthoclase
Albite * 6 = needed SiO[SUB]2[/SUB] for each Albite
Anorthite * 2 = needed SiO[SUB]2[/SUB] for each Anorthite
Diopside * 2 = needed SiO[SUB]2[/SUB] for each Diopside
Hypersthene * 1 = needed SiO[SUB]2[/SUB] for each Hypersthene

Sum the five SiO2 values just calculated, and call this number pSi for provisional SiO[SUB]2[/SUB].
17) Quartz (Qz): If there is enough silica to make all five minerals then the rock is quartz-normative. Otherwise there is no quartz in the norm and silica to make the rest of the silicates must come from other sources.
a) If pSi calculated in step 16 is less than SiO[SUB]2[/SUB], then there is excess silica. Subtract pSi from SiO[SUB]2[/SUB], and put excess SiO[SUB]2[/SUB] in quartz. SiO[SUB]2[/SUB], Nepheline, and olivine are now zero. Skip to step 20.
b) If pSi calculated in step 16 is more than SiO[SUB]2[/SUB], then the rock is silica deficient. Proceed to step 18.
18) Olivine (Ol), Hypersthene (Hy): If pSi calculated in step 16 is more than SiO[SUB]2[/SUB], then there is an SiO[SUB]2[/SUB] deficit. Quartz is now zero. Calculate a new pSi as in step 16, omitting hypersthene.
Orthoclase * 6 = needed SiO[SUB]2[/SUB] for each Orthoclase
Albite * 6 = needed SiO[SUB]2[/SUB] for each Albite
Anorthite * 2 = needed SiO[SUB]2[/SUB] for each Anorthite
Diopside * 2 = needed SiO[SUB]2[/SUB] for each Diopside

Sum the four SiO[SUB]2[/SUB] values just calculated to get a new value of pSi. Subtract the new pSi from SiO[SUB]2[/SUB] to get the amount of SiO[SUB]2[/SUB] available for olivine and hypersthene, called aSi.
a) If FMO is greater than or equal to 2 times aSi, then put all FMO in olivine. FMO and hypersthene are now zero. Proceed to step 19.
b) If FMO is less than 2 times aSi, then nepheline is zero. Calculate the amount of hypersthene and olivine as follows:
Hypersthene = ((2 * aSi) - FMO)
Olivine = (FMO - Hypersthene)
Skip to step 20.

19) Nepheline (Ne), Albite (Ab): If you reached this step, then turning hypersthene into olivine in step 18a did not yield enough silica to make Or, Ab, An, Di, and Ol. Calculate a new pSi value as in step 16, omitting hypersthene and albite.
Orthoclase * 6 = needed SiO[SUB]2[/SUB] for each Orthoclase
Anorthite * 2 = needed SiO[SUB]2[/SUB] for each Anorthite
Diopside * 2 = needed SiO[SUB]2[/SUB] for each Diopside
Olivine * 0.5 = needed SiO[SUB]2[/SUB] for each Olivine

Sum the three SiO[SUB]2[/SUB] values just calculated to get a new value of pSi. Subtract this pSi from SiO[SUB]2[/SUB] to get a new value of aSi, which is the amount of SiO[SUB]2[/SUB] available for albite and nepheline.
Albite = (aSi-(2*Na[SUB]2[/SUB]O))/4
Nepheline = Na[SUB]2[/SUB]O-Albite

20) Multiply orthoclase, albite, and nepheline by two. Divide olivine by two.
21) Calculate An', which is the Ca/(Ca+Na) ratio in normative plagioclase:
An' = Anorthite/(Anorthite+Albite)
22) Plagioclase (Plag): Add albite to anorthite to make plagioclase. Retain the albite value.
23) Calculate the formula weight of plagioclase, using the An' value from step 21.
Formula weight Plag = (An'*278.2093)+((1-An')*262.2230)
24) Multiply all of the normative mineral values by their respective formula weights. The formula weight of FMO is from step 12.
Qz
60.0843
SiO[SUB]2[/SUB]
Or
278.3315
KAlSi[SUB]3[/SUB]O[SUB]8[/SUB]
Plag
Calculated in step 23
(Ca,Na)(Al,Si)[SUB]4[/SUB]O[SUB]8[/SUB]
Cor
101.9613
Al[SUB]2[/SUB]O[SUB]3[/SUB]
Ne
142.0544
NaAlSiO[SUB]4[/SUB]
Di
176.2480 + formula weight of FMO
Ca(Fe,Mg)Si[SUB]2[/SUB]O[SUB]6[/SUB]
Hy
60.0843 + formula weight of FMO
(Fe,Mg)SiO[SUB]3[/SUB]
Ol
60.0843 + (2 * formula weight of FMO)
(Fe,Mg)[SUB]2[/SUB]SiO[SUB]4[/SUB]
Mt
231.5386
Fe[SUB]3[/SUB]O[SUB]4[/SUB]
Ilm
151.7452
FeTiO[SUB]3[/SUB]
Ap
504.3125
Ca[SUB]5[/SUB](PO[SUB]4[/SUB])[SUB]3[/SUB]F

25) This is your weight norm. This would usually be published along with four other useful parameters:
Total:
The sum of all mineral weights in the norm, calculated in step 24. Within rounding error the Total should be the same as the original total of the weight % of the oxides from the chemical analysis.
Mg'%:
100*Mg' From step 11.
An'%:
100*An' from step 21.
DI:
The Differentiation Index of Thornton and Tuttle (1960). This is the weight ratio: (Qz+Or+Ab+Ne)/Total. Ab is albite from step 20 times 262.2230.

26) The weight norm is the way norms are usually expressed. If you want to compare norms to modes (which are volume fractions) you should divide each normative mineral by its density and recalculate the Total back to 100%. Mineral densities can be gotten from mineralogy texts, the CRC Handbook of Chemistry and Physics, or elsewhere. The density of plagioclase and the Fe-Mg silicates with intermediate solid solutions can be found in tables or figures in mineralogy texts, or equations derived from them


منبع:
egeology.blogfa.com
 
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