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

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1 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 1 of 10 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]

2 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 2 of 10 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 n/a K-feldspar n/a almandine cordierite kyanite staurolite clinochlore phlogopite eastonite

3 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 3 of 10 (2) Plot the following 3 rocks on your diagram. Molecular Weight % oxide in rock.. weight SiO TiO Al2O Fe2O FeO MgO CaO Na2O K2O H2O 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] [FeO] [MgO] [CaO] [Na2O] [K2O] AKF Thompson AFM Value Percentage Value Percentage A A K F F M Total Total 0.173

4 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 4 of 10 ROCK 2 [Al2O3] [FeO] [MgO] [CaO] [Na2O] [K2O] AKF Thompson AFM Value Percentage Value Percentage A A K F F M Total Total ROCK 3 [Al2O3] [FeO] [MgO] [CaO] [Na2O] [K2O] AKF Thompson AFM Value Percentage Value Percentage A A K F F M Total Total (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,

5 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 5 of 10 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 threephase 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. As well as musc and qz we have: Rock1 and, chl, bi; Rock 2 cord, chl, bi; Rock 3 and, bi. For a bulk rock composition or a single mineral composition you can calculate F/M Either by F/M directly where numbers are known (e.g. for rocks in tables above) Or by F/M ratio = (horizontal distance to right hand side of AFM) / (horizontal distance to left hand side of AFM) (e.g. for mineral positions on solid solution lines in this exercise) Rock 1 F/M = / = 1.36 Rock / = 0.65 Rock / = 2.1 (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

6 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 6 of 10 K chl-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: K staurolite-biotite = 6.5 K staurolite-chlorite = 10.0 K garnet-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). (F/M in st) = = 8.45 This means the st is rich in Fe and should plot near the Fe end of the st solid solution line. The F/M ratio is used to calculate the actual F percentage F% = (F/M)/(F/M + 1) 100% = 89% M% = 1/(F/M + 1) 100% = 11% The M% tells you how far (starting from the left) along the st solid solution line to plot this particular st composition For garnet (F/M in ga) = = 5 M% = 17% for ga in this assemblage You are told (F/M in bi) = 1.3 M% = 43% for bi in this assemblage 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 K biotite-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). K biotite-chlorite = K staurolite-chlorite / K staurolite-biotite = 10.0/6.5 = 1.54 so (F/M in st) = = 6.5 M% = 13% for st in this assemblage (F/M in chl) = 1.0 / 1.54 = 0.65 M% = 61% for chl in this assemblage You are told (F/M in bi) = 1.0 M% = 50% for bi in this assemblage

7 LIVERPOOL UNIVERSITY EARTH SCIENCE ENVS212 page 7 of 10 (7) Deduce the assemblages for rocks 1, 2, 3 in the staurolite zone by marking them on the answer to (6). As well as musc and qz we have: Rock1 st, chl, bi; Rock 2 chl, bi; Rock 3 ga, bi. (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 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. The AFM diagram with cordierite fits with field labelled L-m because it has to be to the right of St + Chl = Bt + As but to the left of Chl = Crd + Bt + As The AFM diagram with staurolite fits with field labelled M-k because it has to be to the right of Grt + Chl = St + Bt but to the left of St + Chl = Bt + As 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