Lepidolite

Chemistry

K2O7.300
Li2O3.000
Al2O326.900
SiO247.700

Volatiles

H2O4.700
LOI10.000

Links to Other Materials

Miscellaneous

Notes

Although this material is sometimes referred to as a lithium feldspar, it is not a feldspar, but rather a lithium mica. A natural source of Li2O, usually about 5%. It is a raw material for the production of lithium carbonate.

Like other feldspars, it softens over a range (1150-1300C). It has a lower fusion point than most high temperature feldspars.

The fluorine content can cause problems with glaze bubbles and surface pitting.

Its ideal formula is LiF.KF.Al2O3.3SiO2.
However it is tricky to establish a formula weight and analysis for this material. Here are some of the factors:
-LOI varies, but we have assumed 5% water in this example.
-Lepidolite does not actually contain Li2O, it has to absorb oxygen from the atmosphere or other materials during firing since the Li is bonded to F.
-Fluorine is bonded to potassium so oxygen has to be absorbed to create K2O.
-For both Li and K only 1/2 of a molecule per input molecule of lepidolite will be yeilded.
-In a complete oxidation firing all of the fluorine will volatilize and none will be in the final glass. However in real world conditions none of these is likely to happen completely, and in a reduction firing none may happen at all. In this case we are assuming it is all LOI.

Given the above assumptions, an input of LiF.KF.Al2O3.3SiO2 would yield:
1/2 Li2O
1/2 K2O
2 F
1 Al2O3
3 SiO2
5 LOI
Considering the F as LOI we would have:
3% Li2O
7.3% K2O
26.9% Al2O3
47.7% SiO2
15% LOI

To properly handle a material like this in ceramic calculations fluorine conversion factors have to be employed.

(Richard Willis)

Mineral, of the ideal form K(Li, Al)3[(OH, F)2(AlSi4O10)] , of a complex and variable empirical composition.
Hardness: 2.5-3 Density: 2.8-2.9 Soluble in acids
Important lithium ore. see spodumene

Data

URLs

Suppliers

Authors

Pictures

XML

<?xml version="1.0" encoding="UTF-8"?>
<material name="Lepidolite" descrip="" generic="1" rawmineral="1" searchkey="Lithium Feldspar" loi="0.00">
<families>
<family name="Feldspar"/>
</families>
<regions>
<region name="North America"/>
</regions>
<oxides>
<oxide symbol="Li2O" name="Lithium Oxide, Lithia" status="U" percent="3.000" tolerance=""/>
<oxide symbol="K2O" name="Potassium Oxide" status="U" percent="7.300" tolerance=""/>
<oxide symbol="Al2O3" name="Aluminum Oxide, Alumina" status="" percent="26.900" tolerance=""/>
<oxide symbol="SiO2" name="Silicon Dioxide, Silica" status="" percent="47.700" tolerance=""/>
</oxides>
<volatiles>
<volatile symbol="" name="" percent="4.700" tolerance=""/>
<volatile symbol="" name="" percent="10.000" tolerance=""/>
</volatiles>
<references>
<reference name="seealso" reason=""/>
<reference name="seealso" reason=""/>
<reference name="seealso" reason=""/>
<reference name="seealso" reason=""/>
</references>
<suppliers>
<supplier name="Generic" country="" url="" label=""/>
</suppliers>
<notes>
<note>Although this material is sometimes referred to as a lithium feldspar, it is not a feldspar, but rather a lithium mica. A natural source of Li2O, usually about 5%. It is a raw material for the production of lithium carbonate.

Like other feldspars, it softens over a range (1150-1300C). It has a lower fusion point than most high temperature feldspars.

The fluorine content can cause problems with glaze bubbles and surface pitting.

Its ideal formula is LiF.KF.Al2O3.3SiO2.
However it is tricky to establish a formula weight and analysis for this material. Here are some of the factors:
-LOI varies, but we have assumed 5% water in this example.
-Lepidolite does not actually contain Li2O, it has to absorb oxygen from the atmosphere or other materials during firing since the Li is bonded to F.
-Fluorine is bonded to potassium so oxygen has to be absorbed to create K2O.
-For both Li and K only 1/2 of a molecule per input molecule of lepidolite will be yeilded.
-In a complete oxidation firing all of the fluorine will volatilize and none will be in the final glass. However in real world conditions none of these is likely to happen completely, and in a reduction firing none may happen at all. In this case we are assuming it is all LOI.

Given the above assumptions, an input of LiF.KF.Al2O3.3SiO2 would yield:
1/2 Li2O
1/2 K2O
2 F
1 Al2O3
3 SiO2
5 LOI
Considering the F as LOI we would have:
3% Li2O
7.3% K2O
26.9% Al2O3
47.7% SiO2
15% LOI

To properly handle a material like this in ceramic calculations fluorine conversion factors have to be employed.</note>
<note>Mineral, of the ideal form K(Li, Al)&lt;FONT SIZE=2&gt;&lt;SUB&gt;3&lt;/SUB&gt;&lt;/FONT&gt;[(OH, F)&lt;FONT SIZE=2&gt;&lt;SUB&gt;2&lt;/SUB&gt;&lt;/FONT&gt;(AlSi&lt;FONT SIZE=2&gt;&lt;SUB&gt;4&lt;/SUB&gt;&lt;/FONT&gt;O&lt;FONT SIZE=2&gt;&lt;SUB&gt;10&lt;/SUB&gt;&lt;/FONT&gt;)] , of a complex and variable empirical composition.
Hardness: 2.5-3 Density: 2.8-2.9 Soluble in acids
Important lithium ore. &lt;FONT SIZE=2&gt;see &lt;B&gt;&lt;I&gt;spodumene&lt;/I&gt;&lt;/B&gt;&lt;/FONT&gt;

</note>
</notes>
<testdata>
<testitem testname="4" value="1200C"/>
<testitem testname="4" value="3.00"/>
<testitem testname="4" value="3.00"/>
<testitem testname="4" value="1200C"/>
</testdata>
<pictures>
<picture description="Lepidolite" filename="lepidolite.jpg"/>
<picture description="Lepidolite Stone" filename="lepidolite_stone.jpg"/>
</pictures>
</material>



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