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Since the formation of the Earth as a planetary body, the lithophile part of the Earth, which we call the mantle, has continuously undergone modification by partial melting and the consequent haemorrhaging of basaltic melts to the surface under the influence of gravitational forces. Integrated over time this process has caused the mantle to become depleted in its low temperature melting fraction commensurate with the growth of an outer rim of basaltic material. The chemical effect of the melting can be represented in terms of the following mineralogical reactions (where the suffix(M) means 'phase entering the melt', and (R) means 'phase left as a residue'):

**
SPINEL**

**
Hercynite-Chromite-Magnetite**
**
Hercynite
Opx
Cpx
Anorthite
Olivine**
**1) (Fe,Mg)Al2O4
+ (Fe,Mg)2Si2O6 + Ca(Fe,Mg)Si2O6 -> CaAl2Si2O8(M)
+ 2(Fe,Mg)2SiO4(R)**

**
Chromite-Magnetite
Magnetite
Chromite**
**2) (Fe,Mg)(Fe,Cr)2O4
-> FeFe2O4(M)
+ (Fe,Mg)(Fe,Cr)2O4(R)**

**
where (Fe,Mg)Al2O4 and (Fe,Mg)(Fe,Cr)2O4 are solid
solution components of spinel. After melting the spinel is depleted
in Al and ferric iron and enriched in Cr. The chromite is refractory.**

** PYROXENES**

**
Clinopyroxene (Diopside-Hedenbergite, Jadeite, Ca-Tschermaks molecule)**

**3) Ca(Fe,Mg)Si2O6
-> Ca(Fe,Mg)Si2O6(M)**
**4) CaAl2SiO6
+ (Fe,Mg)2Si2O6 -> CaAl2Si2O8(M)
+ (Fe,Mg)2SiO4(R)**
**5) NaAlSi2O6
+ (Fe,Mg)2Si2O6 -> NaAlSi3O8(M)
+ (Fe,Mg)2SiO4(R)**
**6) 2CaTiAl2O6
+Fe2Si2O6 + 2(Fe,Mg)2Si2O6 ->
2FeTiO3(M)
+ 2CaAl2Si3O8(M)
+ 2 (Fe,Mg)2SiO4(R)**

**
Orthopyroxene (Enstatite - Ferrosilite)**

**7) (Fe,Mg)2Si2O6
-> (Fe,Mg)2Si2O6 (M)**
**8) (Fe,Mg)Al2SiO6
+ (CaMg)2Si2O6 -> CaAl2Si2O8(M)
+ (Fe,Mg)2SiO4(R)**

**At**
**high pressures garnet will appear as a component of the lherzolite assemblage,
replacing Ca-Al clinopyroxene and orthopyroxene, where,**

**
Ca-Al cpx Opx
Garnet**
**9)
CaAl2SiO6 + (Mg,Fe)2Si2O6 -> Ca(Mg,Fe)2Al2Si3O12**

** and where the normative basaltic
equivalents of the garnet would be anorthite and olivine,**

**
Garnet
Anorthite
Olivine**
**10)
Ca(Mg,Fe)2Al2Si3O12 = CaAl2Si2O8 + (Mg,Fe)2SiO4**

** AMPHIBOLE
AND PHOLOGOPITE**

**
Potassium feldspar is likely derived by the melting of small amounts of
either phlogopite (mica; KAlMg _{3}Si_{3}O_{10}(OH)_{2}
or amphibole (K_{2},Ca)_{2}Mg_{4}Al_{2}Si_{7}O_{22}(OH)_{2}.**

**
You will note that melting of the mantle causes the mantle to become enriched
in olivine, and to be depleted in potential
plagioclase,
pyroxene, magnetite, and ilmenite.
The melt is clearly enriched
in Ca, Na, Al, Ti, and Si, whereas the mantle residue is depleted
in these elements.
This is the first stage in crustal
fractionation.**

** How
much mantle is required to melt to produce granite?**

**
The second stage of crustal fractionation involves the formation of Si,
Na and K enriched continental crust. This may take place by two processes:**

1) crystal fractionation involving the continuous gravitational removal of the mafic minerals olivine and clinopyroxene, and the Ca-rich plagioclase anorthite as the basaltic melt cools, leaving a liquid enriched in Na, K, and Si components;2) alternatively, by reheating the solidified basaltic crust slighlty above its temperature of solidification and extracting the relatively small amounts of Na, K, and Si enriched melt that forms.

Let us look athe arithmetic of this proposition.Imagine that the initial liquid contains 0.05% K2O, the least refractory major element. If we remove 50% of olivine and clinopyroxene from the melt the concentration will increase to 0.1%. If we remove 50 % of what remains, the concentration will increase to 0.2% and the remaining liquid will be 25% of the original melt. This relationship is hyperbolic and is described by the simple equation:

C = C_{0}/x

where C is the present concentration of an element in the residual melt fraction x, and C_{0}is the original concentration of the element in the melt or source rock. (see Kd)

The graph of this simple equation represents a hyperbolic curve as represented in the following figure showing the behaviour of TiO2 as an incompatible element.

It also follows that if the mantle contains 0.04 wt% K20, and granite contains 3 wt% K2O, the granite would represent only (x = C_{0}/C; x = 0.04/3*100) 1.3 % of the mantle.

If the mantle does not melt entirely but produces a partial melt of basalt representing 20% of the mantle, the K2O content of the basalt would be (C = C_{0}/x; C = 0.04/0.2) 0.2 wt%, and to derive a granite with 3 wt% K2O from a basalt with 0.2 wt% K2O the basaltic melt would have to be reduced to a residue of (x = C_{0}/C; x = 0.2/3 = 0.066) c. 7%. That is, 93% of the basalt would have to be returned to the mantle, and the granite would be still equivalent to only 1.3% of the mantle (i.e. 6.66% of 20%).(Note that if the mantle contains as little as 0.01 wt% K2O and the granite melt or residue contains as much as 4.3 wt% K2O, the granite would represent as little as (x = C_{0}/C; x = 0.01/4.3*100) 0.23% of the mantle.)

Crystal Fractionation of Basalt, x = C_{0}/C

The following is a sample calculation showing how one might use normative compositions to calculate the proportions of minerals that would crystallize in the during the conversion of basalt to granite by mineral fractionation. The compositions are represented as molar % proportions.

It is normal to present the chemical composition of rocks as 'oxide percent' in the order:

SiO2 TiO2 Al2O3 Fe2O3 FeO MnO MgO CaO K2O P2O5;

or as molar percents of the standard normative minerals:

Ilmenite (Il) Magnetite (Mt) Orthoclase (Or) Albite (Ab) Anorthite (An)Clinopyroxene (Cpx) Orthopyroxene (Opx) Olivine (Ol) Quartz (Qtz).

If the normative amount of orthoclase is 2.54 mol% in the primary basaltic liquid and 12.41 mol% in the residual granite, and if K has NOT been fractionated by any of the crystallizing mineral phases, that is, it is a perfectly incompatible element, then the granite fraction must represent (x = C_{0}/C; x = 2.54/12.41) 0.2047 of a unit amount of primary liquid (i.e. 20.47 mol%), and the crystallized phases must represent (100 - 20.47) 79.53 mol% of the primary liquid. In other words, the increase in concentration of orthoclase is purely the effect of the proportional removal of 79.53% of the liquid as non- K bearing solid phases.On the basis of this assumption we can therefore determine the proportions that all the other mineral phases in the residual 20.47% would have if they also had acted as incompatible elements. If the values calculated are the same as the concentration in the primary liquid, then the assumption of incompatibility must be true. If the calculated values are less than the primary proportion, then the difference must represent the proportion of the mineral crystallized from the primary melt.

Measured Measured Calculated CalculatedPrimary Residual Proportion of Proportion ofBasalt, % Granite, % Residual Granite Crystallized SolidIl 02.70 - ( 01.32 * 0.2047 = 00.27) = 02.43 IlMt 02.17 - (03.48 * 0.2047 = 00.71) = 01.46 MtOr 02.54 - (12.41 * 0.2047 = 02.54) = 00.00 OrAb 21.49 - (21.15 * 0.2047 = 04.33) = 17.16 AbAn 32.90 - (23.50 * 0.2047 = 04.81) = 28.09 AnCpx 22.89 - (08.87 * 0.2047 = 01.82) = 21.07 CpxOpx 09.86 - (14.82 * 0.2047 = 03.03) = 06.83 OpxOl 06.00 - (00.00 * 0.2047 = 00.00) = 06.00 OlQtz 00.00 - (13.76 * 0.2047 = 02.82) = -02.82 Qtz (negative)Total 100.00 - (100.00 * 0.2047 = 20.47) = 79.53 Total

We can account for the negative 2.82 mols of quartz by assuming that 2.82 mols of Opx have crystallized as 2.82 mols of quartz and an equivalent amount of olivine. (You will need to think about this!) That is, (6.00 + 2.82) 8.82 mols of Olivine and only (6.83-2.82) 4.01 mols of Opx would have crystallized. These calculations imply that 4 times as much material has been returned to the mantle as is now resident in granitic continental crust.

BIPOLAR GRAPHS

When attempting to examine the chemical variation in a suite of rock samples, it is sometimes more convenient to determine the fractionation vectors by using bipolar graphs.Theoretically, in a plot of one incompatible element against another, the value of the 'initial ratio' of the two elements will be reflected in the slope of the line, since in a plot of the concentration of element 'a', C_{a}, against the concentration of element 'b', C_{b}, and where C_{a}^{o}and C_{b}^{o}represent the initial concentrations of the two elements:

C_{a}=C_{b}.(C_{a}^{o}/C_{b}^{o})

The 'initial ratios' can therefore be used to distinguish between two chemically different basalt sequences.

If lon or log values are plotted the slope of the line will be 45 degrees and the initial ratios will be the intercept value of the line, since:

ln C_{a}= ln C_{b}+ ln (C_{a}^{o}/C_{b}^{o})

RATIO PLOTS

Ratio plots take away the effect of a simple proportional increase in concentration due to total mass fractionation. The fractionation effects of specific minerals can then be recognised by comparing the fractionation trend with the expected mineral vector variation.In this case:

C_{b}/C_{a }= C_{c}/C_{a }* (C_{b}^{o}/C_{c}^{o})*(f(K_{c},X)/f(K_{b},X))

where C_{a}represents the concentration of a known incompatible element 'a'; K_{c}, K_{b}are partition coefficents of the elements 'c' and 'b', respectively, and X is the melt proportion.If a, b, and c are all incompatible elements, then all data values will plot at a single point, and no trend will be observed on the ratio plot.

* Draw plots of Ca/Zr v Al/Zr plot when olivine, orthopyroxene, clinopyroxene and plagioclase crystallize successively?

Figure
1. Structural Provinces of North America.

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