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ENVS212 Practical 6: Triangular compatibility diagrams for pelites

In this practical we study the AKF and the Thompson AFM diagrams for pelites.

First, the necessary information for plotting on the diagrams is as follows.

AKF: plotting

If [Al2O3] means 'number of moles of Al2O3' in either a rock or mineral, and so on, then we

have:

A = [Al2O3] - [K2O] - [CaO] - [Na2O]

K = [K2O]

F = [FeO] + [MgO]

Normalise A + K + F to 100% and plot in the normal way. Conventionally A is at the top, K at

the lower left, and F at the lower right of the triangle. Note that the numeric value of A is

affected by the amount of K in the mineral or rock being plotted. This is because, out of the total

Al in the system, some is allocated to the K vertex. The numeric value of A is affected by Ca

and Na in a rather similar way - these are assumed to be in plagioclase, which also contains Al.

The Al bound in plagioclase is then not available to contribute to other phases in the AKF

triangle.

AKF: other information

In the system KFMASH, projection is from quartz and water onto this plane. If in addition Na

and Ca are present then they are assumed to be combined in plagioclase, and we project from

plagioclase. This changes the way the aluminium content of a rock is plotted, because some of

the Al is contained in plagioclase. If yet other chemical components are present then they are

ignored: actually this means that:

(a) they are assumed to be present as pure oxide phases. For example TiO2 might be present as

rutile; and/or

(b) they are present in such small amounts that they are unlikely to affect phase equilibria.

Note that the K vertex is not simply K2O. This is because no rock would plot close to this

corner, and thus space would be wasted in the diagram. Instead the K vertex is relocated to have

the composition of microcline. Note also that merging FeO and MgO obscures some of the

effects related to solid solution (the Thompson AFM plot solves this problem). It is useful to

remember that microcline plots at the K corner; sillimanite, kyanite and andalusite all plot at the

A corner; and, for instance, hypersthene at the F corner.

Thompson AFM: plotting

A = [Al2O3] - 3[K2O] - [CaO] - [Na2O]

F = [FeO]

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M = [MgO]

Note that the actual value of A is different to that in AKF. This is because of the extra projection

step involved. Put A at the top of your triangle and F and M half-way up the sides of the triangle

(F on the left). This is so we can plot negative A values which will be points below the line

joining F to M. Negative A values are plotted exactly as described in the previous practical,

except a negative value is plotted by selecting the FM line and counting away from A by the

required value.

Thompson AFM: other information

In the system KFMASH, projection is from quartz, water, and muscovite onto this plane. With

additional Na and Ca, we project from plagioclase (as for AKF). The diagram is more rigorous

than AKF as a depiction of actual phase relations, but can only be used in rocks with muscovite.

You should do parts (1) to (7) in order. Then, attempt any or all of (8) to (10) as time permits.

(1) Plot the following minerals on an AKF diagram.

mu muscovite KAl3Si3O10(OH)2

kf K-feldspar KAlSi3O8

alm almandine Fe3Al2Si3O12

cd cordierite Mg2Al4Si5O18

ky kyanite Al2SiO5

st staurolite Fe2Al9Si4O23(OH)

ch clinochlore Mg9Al6Si5O20(OH)16 (a chlorite)

Use Fig. 1. Each of these is a point on the AKF diagram. Now plot

phlogopite KMg3AlSi3O10(OH)2

eastonite K2Mg5Al4Si5O20(OH)4

Both of these are biotites (bt). The biotites show not only substitution of Fe for Mg (phlogopite

through to annite) but also substitution of Al for Mg and Si (giving rise to the eastonite -

siderophyllite series). There is complete solid solution here, so draw a thick line connecting the

phlogopite and eastonite points. This is not a tie-line but instead a line representing the range of

possible biotite compostions.

AKF Thompson AFM

Mineral A K F A F M

muscovite 1 0.5 0 n/a

K-feldspar 0 0.5 0 n/a

almandine 1 0 3 1 3 0

cordierite 2 0 2 2 0 2

kyanite 1 0 0 1 0 0

staurolite 4.5 0 2 4.5 2 0

clinochlore 3 0 9 3 0 9

phlogopite 0 0.5 3 -1 0 3

eastonite 1 1 5 -1 0 5

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(2) Plot the following 3 rocks on your diagram.

Molecular Weight % oxide in rock ..

weight 1 2 3

SiO2 - 59.6 64.1 77.0

TiO2 - 0.93 0.88 0.61

Al2O3 102.0 22.6 18.1 10.98

Fe2O3 - 0.97 1.61 0.82

FeO 71.9 4.40 3.59 3.02

MgO 40.3 1.82 3.12 0.82

CaO 56.1 0.81 0.77 0.64

Na2O 62.0 1.12 1.76 2.64

K2O 94.2 3.90 3.70 2.0

H2O - 3.63 2.31 1.24

Do this by filling in the following tables, one for each rock. First calculate the relative molar

amounts by

[Al2O3] etc. = (Weight % in rock)/(Molecular weight)

Use 3 decimal places.Then calculate A, F and M. Use 3 decimal places.

ROCK 1

[Al2O3] 0.222

[FeO] 0.061

[MgO] 0.045

[CaO] 0.014

[Na2O] 0.018

[K2O] 0.041

AKF Thompson AFM

Value Percentage Value Percentage

A 0.149 50 A 0.067 39

K 0.041 14 F 0.061 35

F 0.106 36 M 0.045 26

Total 0.296 Total 0.173

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ROCK 2

[Al2O3] 0.177

[FeO] 0.050

[MgO] 0.077

[CaO] 0.014

[Na2O] 0.028

[K2O] 0.039

AKF Thompson AFM

Value Percentage Value Percentage

A 0.096 37 A 0.018 12

K 0.039 15 F 0.050 35

F 0.127 48 M 0.077 53

Total 0.262 Total 0.145

ROCK 3

[Al2O3] 0.108

[FeO] 0.042

[MgO] 0.020

[CaO] 0.011

[Na2O] 0.043

[K2O] 0.021

AKF Thompson AFM

Value Percentage Value Percentage

A 0.033 28.5 A -0.009 -17

K 0.021 18 F 0.042 80

F 0.062 53.5 M 0.020 38

Total 0.116 Total 0.053

(3) Mark the same rocks and minerals on a Thompson AFM diagram. Leave out K-

feldspar. Use a separate sheet of triangular graph paper (Fig. 2). Note that the mineral formulae

above are given as either Fe or Mg end-members. Mark them as such (so some are on the left,

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some are on the right) but bear in mind that they are all in fact solid solutions. Note also that,

since muscovite is being projected from, it cannot appear on the plot.

(4) Annotate the AFM diagram you are provided with. The diagram (Fig. 3) is representative

of the cordierite zone of the Buchan style of metamorphism (low-P, moderate T). Label all the

minerals, using the answer to part (3), assuming the stable Al2SiO5 phase is andalusite and that

chlorite is stable at this grade for appropriate rock compositions. Also label the fields of possible

assemblages. Some fields have 3 AFM phases (this means 6 phases altogether including

projecting phases). Label these as A+B+C where these are the three minerals. Other fields have

only 2 AFM phases: label these as A+B. You may find it helpful to colour in the two- and three-

phase fields differently. What AFM solid-solution minerals are completely unstable under these

conditions?

(5) Predict the assemblages in the three rocks using the diagram annotated in (4). Mark the

rocks on the diagram and read off the assemblages. Each assemblage has a mineral or minerals

which are a solid solution. These are often characterised by their F/M ratio which is the molar

ratio of Fe/Mg. Thus a mineral with no Fe, plotting at the right of the diagram, has F/M = 0. A

mineral with 50% Fe and 50% Mg has F/M = 1 and plots down the middle of the AFM diagram.

For each rock give the F/M ratio for the ferromagnesian phases.

(6) Draw an AFM diagram for conditions of the Barrovian staurolite zone. Modify Fig. 2.

for this purpose. The following 3-phase assemblages which we are concerned with are stable at

this grade:

garnet-staurolite-biotite

chlorite-staurolite-biotite

This is the basic information we require, but we also need the compositions of these phases

when in equilibrium: specifically the F/M ratios. From the last question we see that minerals in

equilibrium need not have the same F/M. Some prefer more Fe than others. However the

partitioning of Fe and Mg is controlled by a constant distribution coefficient K for any pair of

minerals. For instance, we write

Kchl-crd

= (F/M in chlorite) / (F/M in cordierite)

etc. Although F/M for the minerals is a function of the overall rock F/M, the distribution

coefficient is not: it thus allows us to calculate F/M in sets of minerals given the F/M of one. Use

the following information:

Kstaurolite-biotite

= 6.5

Kstaurolite-chlorite

= 10.0

Kgarnet-biotite

= 3.85

For the ga - st - bi field, (F/M in bi) = 1.3. Calculate F/M for st and ga coexisting here, and mark

this triangle on your answer to (3). (Try to work out for yourself how to turn an F/M ratio into a

position on a solid-solution line).

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For the ch - st - bi field, (F/M in bi) = 1.0. Calculate F/M for each of st and ch coexisting here.

Note that you have not been given the value of

Kbiotite-chlorite

Use the definitions of the K values to see how you can deduce this and thus solve the problem.

Mark this other triangle on and draw in the solid-solution limits for garnet, chlorite and biotite

(the latter is stable for all F/M ratios).

(7) Deduce the assemblages for rocks 1, 2, 3 in the staurolite zone by marking them on the

answer to (6).

(8) Compare you answers so far with the corresponding equilibria displayed on the AKF

diagram.

(9) Study the attached P,T grid. (from Spear, F. S. 1995. Metamorphic Phase Equilibria and

Pressure-Temperature-Time Paths. Mineralogical Society of America, Washington). Despite

its apparent complexity, we can relate our two AFM diagrams to this grid (Fig. 4). Specifically

we can note the reaction lines in P,T space which must limit the conditions under which our

AFM diagrams will apply. There are enough of these to allow us to determine and colour in the

P,T conditions for the Buchan cordierite zone and for the Barrovian staurolite zone. For the

latter, assume that kyanite would be stable in rocks of appropriate composition.

Key to PT grid (Mg or Fe before name indicates Mg or Fe end member)

And Andalusite

As Unspecified aluminium silicate

Bt Biotite Phl Phlogopite = MgBt

Chl Chlorite

Cld Chloritoid

Crd Cordierite

Grt Garnet Prp Pyrope = MgGrt Alm Almandine = FeGrt

Kfs K-feldspar

Ky Kyanite

Ms Muscovite

Prl Pyrophyllite

Qtz Quartz often involved, but not explicit!

Sil Sillimanite

St Staurolite

Tlc Talc

Solid black lines – reactions in KFMASH

Solid grey lines – reactions in KFASH (pure Fe system)

Dashed grey lines – reactions in KMASH (pure Mg system)

When you have finished this practical correctly you should be able to:

Plot minerals and bulk rock compositions on triangular compatibility diagrams

Identify possible reactions using the marked positions of minerals

Given a compatibility diagram for a specific PT, decide on the equilibrium assemblages

Relate the compatibility diagrams to PT grids

Comment on strengths, weaknesses and assumptions behind different types of compatibility diagram

0 40 80 120 160 200 Scale marked

for M

A

F M

Fig. 2. AFM

+ quartz

+ muscovite

+ water

Biotite

Kyanite etc.

Clinochlore

Mg-Cordierite

2

1

3

Almandine

Phlogopite

Eastonite

Fe-Staurolite

2

1

3

A

F M

Fig. 2. AFM

+ quartz

+ muscovite

+ water

ChloriteGarnet

Staurolite

2

1

3