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GEOS 3010 Name: _
Homework #5 Due Date: _2/14/2023_________
Silicate Minerals and Phase Diagrams
Objective
As we have learned, silicate minerals are the most important mineral class in terms of their abundance and occurrence on the earth. They make up the vast majority of rocks exposed at the earth’s surface as well as in the crust and mantle. In this activity, you will practice normalization of a mineral formula for a common group of silicate minerals. You will also use a phase diagram to interpret temperature conditions of crystallization and melting for minerals of a given composition. Finally, you will plot silicate mineral compositions on a ternary diagram and interpret relationships among minerals in solid solution.
Introduction
The following is a chemical analysis of some silicate minerals.
What if we want to determine the formula for each of these mineral specimens? (Hint: They are all in the same mineral group!) These data illustrate a significant limitation with electron-beam analyses of oxygen-containing compounds compared to the analyses we worked with for non-silicate minerals. There is no convenient method of analyzing for oxygen (O), so that the analyses are expressed as weight percent (wt%) oxides of each element, instead of the weight percent of the elements themselves. In other words, the machine cannot measure oxygen, so we simply pretend that there are oxides rather than pure elements in order to do the calculations.
Instead of converting the analyses to atomic proportions, we instead convert to oxide proportions (OP). This is convenient because the various silicate mineral subclasses (isolated, paired, ring, single chain, double chain, sheet, and framework) have specific ratios of metal oxides to silica. The conversion to oxide proportions is accomplished by dividing the weight percent oxide by the atomic weight of each oxide. Here’s an example for Sample 1:
OP SiO2 = Sample 1 wt% SiO2 / atomic weight SiO2
OP SiO2 = (40.99) / (28.09 + (2*16.00)) = 0.682
After determining the proportions for each oxide, we can determine the ratio of oxide metals to silica for this mineral group. This will enable us to determine the group and write both a general formula, and formulas specific to each analysis.
Silicate Mineral Normalization Problem
1. Determine the proportions for each oxide (OP) for each sample in the table below. In the first column, calculate and enter the atomic weight for each oxide. Next, calculate the proportions for each oxide. Finally, add the metal oxides together (all except SiO2) and enter this information into the table in the row labeled “TOM.” Show your calculations in the table or on a separate sheet.
What is the ratio of total metal oxides to SiO2 for each sample? [HINT: the easiest way to do this is to normalize the smaller number, either TOM or SiO2, to 1 by dividing it by itself. Then divide the larger number by the smaller number. Round to the nearest whole number.] Show your work, below.
The ratio of total metal oxides to SiO2 for each sample is:
Sample 1: 1.358/0.682=1.9912 ~2 So: Metal oxides to SiO2 basically 2:1
Sample 2: 1.197/0.635 = 1.8850~2 So: Metal oxides to SiO2 basically 2:1
Sample 3: 1.119/0.561=1.9946~2 So: Metal oxides to SiO2 basically 2:1
Sample 4: 1.039/0.529=1.9640~2 So: Metal oxides to SiO2 basically 2:1
b. If these minerals are all in the same mineral group, then they should have the same ratio of total metals to silica. Do they? What accounts for the minor variations in the ratios?
These minerals do not have the same ratio of total metals to silica. The minor variations in the ratios may be due to different amounts of impurities, variations in the crystal structures, or differences in the methods used for sample preparation and analysis.
Using the ratio of metal oxides to SiO2, write a general formula for this mineral group. [HINT: group the solid solution elements together in parentheses. Then use the ratio to determine how many metal oxides should be in the formula. Next, add the SiO2. Last, move all of the oxygen to the end of the formula.]
The general formula for this mineral group can be written as:
(Fe,Mg,Mn)x(Si,Al)yOz
d. Name the mineral group and silicate subclass that these samples belong to. How did you determine this?
This mineral contains only Mn, Mg, Fe, Si, and O. So, it cannot be a member of any of the
following common silicate groups: quartz, feldspar, mica, or amphibole. That basically leaves
olivine, garnet, and pyroxene. Olivine and garnet have a 1:4 ratio of Si:O, so they are out.
Pyroxenes have a 1:3 ratio of Si:O and a 1:1 ratio of metals to Si, plus commonly contain Mg, Fe. The best match for our mystery mineral is pyroxene, a single chain silicate.
e. If we ignore the minor amounts of Mn in these samples, we can normalize the proportions of Mg and Fe to write a specific formula for each analysis. In other words, we want to write a formula that expresses the exact amount of Mg and Fe in each sample of this mineral. To do this, we calculate the relative proportions of Mg and Fe in each sample:
Proportion of MgO = OP MgO / (OP MgO + OP FeO)
Proportion of FeO = 1 – proportion of MgO
Write the formula here for :
Sample 1: (Fe0.1594Mg0.1459) xSi0.682Oy
Sample 2: (Fe0.0867Mg0.1079) xSi0.377Oy
Sample 3: (Fe0.1052Mg0.1135) xSi0.317Oy
Sample 4: (Fe0.0759Mg0.0481) xSi0.293Oy
The calculations for the relative proportions of MgO and FeO are as follows:
Sample 1:
Proportion of MgO = 0.1263 / (0.1263 + 0.0304) = 0.807
Proportion of FeO = 1 – 0.807 = 0.193
Sample 2:
Proportion of MgO = 0.0756 / (0.0756 + 0.1058) = 0.417
Proportion of FeO = 1 – 0.417 = 0.583
Sample 3:
Proportion of MgO = 0.0458 / (0.0458 + 0.1571) = 0.225
Proportion of FeO = 1 – 0.225 = 0.775
Sample 4:
Proportion of MgO = 0.0209 / (0.0209 + 0.2044) = 0.093
Proportion of FeO = 1 – 0.093 = 0.907
Phase Diagram Problems
2. Use the plagioclase feldspar T-X phase diagram below to answer the following questions.
a. What is the coexisting solid mineral and liquid compositions at 1200°C?
from reading right from 1200 C on the temperature axis to find where 1200
intersects the solidus (for the mineral composition) and liquidus (for the liquid composition).
The mineral has a composition of ~68% NaAlSi3O8 (Ab) [or ~33% CaAl2Si2O8 (An]. the coexisting liquid has a composition of ~94% NaAlSi3O8 (Ab) [or~ 6% CaAl2Si2O8 (An)].
b. Starting with a solid mineral of composition 80% An (80% CaAl2Si2O8), at what temperature will this mineral melt? What is the composition of the liquid that coexists with this mineral?
Finding 80% An (or 20% Ab) on the compositional axis. Read up to the solidus and then left over to the temperature axis to find the melting temperature. Find where this temperature intersects the liquidus and read down to the compositional axis to find the liquid
temperature.
Answers: (1) The mineral melts at 1400°C. (2) The co-existing liquid has a composition of about
62% NaAlSi3O8 (Ab) and 38% CaAl2Si2O8 (An).
c. Starting with a liquid of composition 40% Ab (40% NaAlSi3O8), at what temperature will this melt start to crystallize? What is the composition of the mineral that coexists with this liquid?
Reading down from the top compositional axis starting at 40% Ab (or 60% An) to the liquidus. Reading left over to the temperature axis to find the mineralization temperature. Reading right to the solids to find the coexisting minerals, then read down to find their composition.
Answers (1): the mineral first form approximately at 1500 C, (2) the coexisting mineral has a composition of 94~95% An and ~5% Ab
3. Use the alkali feldspar T-X phase diagram below to answer the following questions.
What are the coexisting mineral and liquid compositions at 1100°C? [HINT: there are TWO answers to this question.]
At 1100°C, the coexisting mineral and liquid compositions are 88% albite (NaAlSi3O8) and 12% orthoclase (KAlSi3O8) for one set of coexisting phases, and the reverse composition of 30% albite and 70% orthoclase for the other set.
Starting with a solid mineral of composition 80% Or (80% KAlSi3O8), at what temperature will this mineral melt? What is the composition of the liquid that coexists with this mineral?
At the temperature of melting, the solid mineral of 80% orthoclase will become a homogeneous liquid of the same composition. Based on the T-X phase diagram, the solid mineral of 80% orthoclase (80% KAlSi3O8) will melt at 1150 degrees Celsius. The composition of the liquid that coexists with this mineral at this temperature will also be 80% orthoclase.
c. If a mineral with composition 70% Ab (70% NaAlSi3O8) cools to 650°C and enters the miscibility gap, what will happen to it? What are the [HINT: two] compositions of intergrown minerals that will form? What name do we give this mineral?
When the mineral with composition 70% albite cools to 650°C and enters the miscibility gap, it will undergo a solid-state phase separation. Two intergrown minerals will form, one rich in albite and the other rich in orthoclase. The mineral resulting from this process is known as perthite.
4. Plot the following mineral formulas on the ternary diagram and label each point with the appropriate letter:
a. (Ca0.5Mg0.5)2Si2O6
b. (Ca0.5Fe0.5)2Si2O6
c. (Ca0.2Fe0.4Mg0.4)2Si2O6
d. (Ca0.02Fe0.50Mg0.48)2Si2O6
e. What mineral group(s) is/are represented on this phase diagram?
The mineral group represented on this phase diagram is silicate minerals.
f. Which mineral (a, b, c, or d) is most likely to be part of a solid solution? Why?
Mineral (c) is most likely to be part of a solid solution because it has intermediate composition between enstatite and ferrosilite, and solid solutions commonly form between two minerals with intermediate compositions.
g. Which mineral (a, b, c, or d) is most likely not found in naturally occurring rocks? Why?
Mineral (d) is most likely not found in naturally occurring rocks because it has an unusual composition with a very low concentration of calcium and a high concentration of iron and magnesium. This mineral may form because of some artificial or high-pressure, high-temperature conditions, but is unlikely to occur naturally in the Earth’s crust.

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