<p>I'd like to add to @SabreTooth correct answer and add the following:</p>
<ol>
<li><p>Plagioclase in nature commonly has a K-feldspar component: <span class="math-container">$\ce{KAlSi3O8}$</span>. So a plagioclase crystal may be 48% anorthite, 48% albite and 4% K-feldspar. This component is usually written as the <span class="math-container">$\ce{Or}$</span> component (after the mineral orthoclase). This introduces a new problem. Let's say you have a plagioclase with the Composition <span class="math-container">$\ce{An49Ab46Or5}$</span>. Normally you would think that the mineral is an andesine because it has 49% <span class="math-container">$\ce{An}$</span>. However, if you want to project the mineral to the plagioclase side of the ternary feldspar diagram, you need to normalise it to <span class="math-container">$\ce{An52Ab48}$</span>. Now it is a labradorite! So which one is it?</p>
<p>Well, there are no defined rules for determining it. However, it doesn't really matter. Remember that these names are artificially divided by us humans. Nature is not so nicely divided but rather quite continuous. Another reason is doesn't really matter is...</p>
</li>
<li><p>Plagioclase in igneous rocks (which is where you are most likely to find it) is rarely homogenous. It is commonly <strong>zoned</strong>, meaning that its composition varies across the crystal. Here are maps of <span class="math-container">$\ce{Ca^2+}$</span> and <span class="math-container">$\ce{Na+}$</span> distributions inside plagioclase crystals:</p>
</li>
</ol>
<p><img src="https://i.sstatic.net/A7C03.jpg" alt="Plagioclase X-ray maps" />
<sup>source: <a href="http://serc.carleton.edu/details/images/8598.html" rel="noreferrer">http://serc.carleton.edu/details/images/8598.html</a></sup></p>
<p>Brigher colours correspond to higher calcium contents. Each crystal has different composition in different zones! It may be labradorite in the cores whereas the rims would be andesine or oligoclase. Sometimes you may encounter a single crystal exhibiting the entire range from bytownite to oligoclase, which are the compositions that currently occur in igneous rocks. So the exact composition of the plagioclase is meaningless.</p>
<p>What about albite and anorthite? These two extreme cases deserve a special treatment.</p>
<ol>
<li>Albite is technically not even a plagioclase, but rather an alkali feldspar. It occurs in settings which are not your usual igneous rock: low grade metamorphic rocks, metasomatic rocks, pegmatites, etc. In this case the identity of the crystal is indeed interesting, and it is rarely called &quot;plagioclase&quot;, but rather &quot;albite&quot;.</li>
<li>Anorthite is extremely rare on Earth. Even the so called &quot;anorthosites&quot; (rocks composed almost entirely of plagioclase) contain labradorite and not true anorthite. Conversely, anorthite (as <span class="math-container">$\ce{An_{95-98}}$</span>) is the main constituent of Lunar anorthosites. These unique rocks and minerals truly deserve to be called anorthites.</li>
</ol>

CC BY-SA 3.0
2015-01-23T16:38:29.750Z
Thank you so much for the very, very interesting details!
Self-teaching worker
CC BY-SA 3.0
2015-01-23T15:31:30.473Z
2015-01-23T15:31:30.473Z
2020-06-18T08:25:19.090Z
Gimelist
<p>Actually, according to IMA classification rules, only albite and anorthite are valid placioglace group mineral names. The content can be easily constructed from Ca/Na ratio. There are more ways to do that but the most simple is the use of <em>apfu (atoms per formula unit)</em>.</p>

<p>Let say we have theoretical feldspar with composition K0.1Na0.3Ca0.6Al1.2Si3O8 - so you have 0.1 K, 0.3 Na and 0.6 Ca in one feldspar molecule. Together its 1 so we can directly recalculate that into %. So it has (0.1 x 100 =) 10% K-feldspar, 30% albite and 60% anorthite. By modern classification this is anorthite. By the old classification (still 
 used in petrology) its labradorite. The 50 % rule applies so basically what is over 50% ir right. Over 0.5 Na is albite, over 0.5 is anorthite. If there is also some K present, rarely both Na and Ca might be less then 0.5 - then the bigger number wins. </p>

<p>If you have real data, you rarely have such ideal composition. But you can get the % of each component from equation (apfu of alkali metal/sum of all alkali metals apfu) x 100. So e.g. [0.256 Na/(0.256 Na + 0.035 K + 0. 723 Ca)] x 100 gives you the % of albite component. </p>

<p>The miscibility of Ca-plagioclase with K-feldspar is quite poor while albite (Na-plagiocalse) can contain some more K-feldspar component. More about miscibility and origin of feldspars <a href="http://www.mineralexpert.org/overview-plagioclases-feldspars.php" rel="nofollow">here</a>. The miscibility strongly depends also on temperature. Also rare Ba-feldspars exist, more in the reference.   </p>

CC BY-SA 3.0
2016-03-16T20:13:13.797Z
2016-03-16T20:13:13.797Z
Zbynek Burival
<p>The fractions of albite $\ce{NaAlSi_3O_8}$ and anorthite $\ce{CaAl_2Si_2O_8}$, which I call $\ce{Ab}$ and $\ce{An}$ in the following list, present in a plagioclase are often used to define it in the following way:</p>

<ul>
<li>Anorthite: $\ce{An}=90\%-100\%$, $\ce{Ab}=0\%-10\%$;</li>
<li>Bytownite: $\ce{An}=70\%-90\%$, $\ce{Ab}=10\%-30\%$;</li>
<li>Labradorite: $\ce{An}=50\%-70\%$, $\ce{Ab}=30\%-50\%$;</li>
<li>Andesine: $\ce{An}=30\%-50\%$, $\ce{Ab}=50\%-30\%$;</li>
<li>Oligoclase: $\ce{An}=10\%-30\%$, $\ce{Ab}=70\%-90\%$;</li>
<li>Albite: $\ce{An}=0\%-10\%$, $\ce{Ab}=90\%-100\%$.</li>
</ul>

<p>What I am not sure of is whether these fractions are intended as applying to mass, molarity, volume.</p>

<geology><geochemistry><mineralogy>
The different types of plagioclase
Self-teaching worker
# The different types of plagioclase

(Asked by: username_0 on 2015-01-23)

The fractions of albite $\ce{NaAlSi\_3O\_8}$ and anorthite $\ce{CaAl\_2Si\_2O\_8}$, which I call $\ce{Ab}$ and $\ce{An}$ in the following list, present in a plagioclase are often used to define it in the following way:

* Anorthite: $\ce{An}=90\%-100\%$, $\ce{Ab}=0\%-10\%$;
* Bytownite: $\ce{An}=70\%-90\%$, $\ce{Ab}=10\%-30\%$;
* Labradorite: $\ce{An}=50\%-70\%$, $\ce{Ab}=30\%-50\%$;
* Andesine: $\ce{An}=30\%-50\%$, $\ce{Ab}=50\%-30\%$;
* Oligoclase: $\ce{An}=10\%-30\%$, $\ce{Ab}=70\%-90\%$;
* Albite: $\ce{An}=0\%-10\%$, $\ce{Ab}=90\%-100\%$.

What I am not sure of is whether these fractions are intended as applying to mass, molarity, volume.

**Tags:** geology, geochemistry, mineralogy


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## 2 answers

### Answer by username_1 on 2015-01-23. Score: 6

I'd like to add to @SabreTooth correct answer and add the following:

1. Plagioclase in nature commonly has a K-feldspar component: $\ce{KAlSi3O8}$. So a plagioclase crystal may be 48% anorthite, 48% albite and 4% K-feldspar. This component is usually written as the $\ce{Or}$ component (after the mineral orthoclase). This introduces a new problem. Let's say you have a plagioclase with the Composition $\ce{An49Ab46Or5}$. Normally you would think that the mineral is an andesine because it has 49% $\ce{An}$. However, if you want to project the mineral to the plagioclase side of the ternary feldspar diagram, you need to normalise it to $\ce{An52Ab48}$. Now it is a labradorite! So which one is it?

   Well, there are no defined rules for determining it. However, it doesn't really matter. Remember that these names are artificially divided by us humans. Nature is not so nicely divided but rather quite continuous. Another reason is doesn't really matter is...
2. Plagioclase in igneous rocks (which is where you are most likely to find it) is rarely homogenous. It is commonly **zoned**, meaning that its composition varies across the crystal. Here are maps of $\ce{Ca^2+}$ and $\ce{Na+}$ distributions inside plagioclase crystals:

![Plagioclase X-ray maps](https://i.sstatic.net/A7C03.jpg)
source: <http://serc.carleton.edu/details/images/8598.html>

Brigher colours correspond to higher calcium contents. Each crystal has different composition in different zones! It may be labradorite in the cores whereas the rims would be andesine or oligoclase. Sometimes you may encounter a single crystal exhibiting the entire range from bytownite to oligoclase, which are the compositions that currently occur in igneous rocks. So the exact composition of the plagioclase is meaningless.

What about albite and anorthite? These two extreme cases deserve a special treatment.

1. Albite is technically not even a plagioclase, but rather an alkali feldspar. It occurs in settings which are not your usual igneous rock: low grade metamorphic rocks, metasomatic rocks, pegmatites, etc. In this case the identity of the crystal is indeed interesting, and it is rarely called "plagioclase", but rather "albite".
2. Anorthite is extremely rare on Earth. Even the so called "anorthosites" (rocks composed almost entirely of plagioclase) contain labradorite and not true anorthite. Conversely, anorthite (as $\ce{An\_{95-98}}$) is the main constituent of Lunar anorthosites. These unique rocks and minerals truly deserve to be called anorthites.

> **Comments:**
> **username_0**: Thank you so much for the very, very interesting details!  

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### Answer by username_2 on 2016-03-16. Score: 2

Actually, according to IMA classification rules, only albite and anorthite are valid placioglace group mineral names. The content can be easily constructed from Ca/Na ratio. There are more ways to do that but the most simple is the use of *apfu (atoms per formula unit)*.

Let say we have theoretical feldspar with composition K0.1Na0.3Ca0.6Al1.2Si3O8 - so you have 0.1 K, 0.3 Na and 0.6 Ca in one feldspar molecule. Together its 1 so we can directly recalculate that into %. So it has (0.1 x 100 =) 10% K-feldspar, 30% albite and 60% anorthite. By modern classification this is anorthite. By the old classification (still
used in petrology) its labradorite. The 50 % rule applies so basically what is over 50% ir right. Over 0.5 Na is albite, over 0.5 is anorthite. If there is also some K present, rarely both Na and Ca might be less then 0.5 - then the bigger number wins.

If you have real data, you rarely have such ideal composition. But you can get the % of each component from equation (apfu of alkali metal/sum of all alkali metals apfu) x 100. So e.g. [0.256 Na/(0.256 Na + 0.035 K + 0. 723 Ca)] x 100 gives you the % of albite component.

The miscibility of Ca-plagioclase with K-feldspar is quite poor while albite (Na-plagiocalse) can contain some more K-feldspar component. More about miscibility and origin of feldspars [here](http://www.mineralexpert.org/overview-plagioclases-feldspars.php). The miscibility strongly depends also on temperature. Also rare Ba-feldspars exist, more in the reference.


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