<p>When it comes to igneous rocks classification, the bible is <a href="https://doi.org/10.1017/CBO9780511535581" rel="nofollow noreferrer">Igneous Rocks: A Classification and Glossary of Terms</a> (Le Maitre et al., 2002). It says (p.4):</p>
<blockquote>
<p>A = alkali feldspar, including orthoclase, microcline, perthite,
anorthoclase, sanidine, and albitic plagioclase (An<span class="math-container">$_0$</span> to An<span class="math-container">$_5$</span>)</p>
<p>P = plagioclase (An<span class="math-container">$_5$</span> to An<span class="math-container">$_{100}$</span>) and scapolite</p>
</blockquote>
<p>So I guess everyone is partially right: if your albite is An<span class="math-container">$_{0-5}$</span>, then you should consider it on the A side; if your albite is An<span class="math-container">$_{5-10}$</span>, then you should consider it on the P side.</p>

CC BY-SA 4.0
2022-08-19T17:35:04.093Z
Thanks. That definitely confirms some information and diagrams I got are wrong or too vague.
Hanatake Yurii
CC BY-SA 4.0
2022-08-20T13:49:33.047Z
Also, just to clarify something in your original post - albite is the sodium feldspar endmember, not the calcium endmember. Anorthite is the calcium endmember.
desander
CC BY-SA 4.0
2022-08-20T20:05:13.727Z
Yes my bad, you're right.
Hanatake Yurii
CC BY-SA 4.0
2022-08-18T19:41:18.547Z
2022-08-18T20:04:42.873Z
2022-08-18T20:04:42.873Z
Jean-Marie Prival
<p>QAPF diagrams are double ternary diagrams used for magmatic rock classification based on mineralogy.</p>
<p>Quartz and Feldspathoids are opposed on the pointy ends of the diamond, with Alkali feldspar and Plagioclase (feldspar) on the intermediary vertex. That direction indicates from more to less silicate (SiO2) content, leading to formation of one or the other.</p>
<p>The other direction differentiates feldspar types, from Alkali feldspar (Na-K) and Plagioclase (Na-Ca). My problem lies here: Albite (Sodium mineral) is the end of the ranges of BOTH of them, thus I don't know which side it should be considered.</p>
<p>I made some searches on the web but cannot find anything conclusive as some geologists claim one or the other, while most simply don't talk about it. Does anybody have any answer, if possible with reliable sources?</p>

<geology><mineralogy><rocks><minerals><petrology>
On QAPF (Steckeisen) diagrams, should Albite be considered on the Alkali feldspar or Plagioclase side? Or in-between?
Hanatake Yurii
# On QAPF (Steckeisen) diagrams, should Albite be considered on the Alkali feldspar or Plagioclase side? Or in-between?

(Asked by: username_0 on 2022-08-18)

QAPF diagrams are double ternary diagrams used for magmatic rock classification based on mineralogy.

Quartz and Feldspathoids are opposed on the pointy ends of the diamond, with Alkali feldspar and Plagioclase (feldspar) on the intermediary vertex. That direction indicates from more to less silicate (SiO2) content, leading to formation of one or the other.

The other direction differentiates feldspar types, from Alkali feldspar (Na-K) and Plagioclase (Na-Ca). My problem lies here: Albite (Sodium mineral) is the end of the ranges of BOTH of them, thus I don't know which side it should be considered.

I made some searches on the web but cannot find anything conclusive as some geologists claim one or the other, while most simply don't talk about it. Does anybody have any answer, if possible with reliable sources?

**Tags:** geology, mineralogy, rocks, minerals, petrology


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## 1 answers

### Answer by username_1 on 2022-08-18. Score: 4 | Accepted answer

When it comes to igneous rocks classification, the bible is [Igneous Rocks: A Classification and Glossary of Terms](https://doi.org/10.1017/CBO9780511535581) (Le Maitre et al., 2002). It says (p.4):

> A = alkali feldspar, including orthoclase, microcline, perthite,
> anorthoclase, sanidine, and albitic plagioclase (An$\_0$ to An$\_5$)
>
> P = plagioclase (An$\_5$ to An$\_{100}$) and scapolite

So I guess everyone is partially right: if your albite is An$\_{0-5}$, then you should consider it on the A side; if your albite is An$\_{5-10}$, then you should consider it on the P side.

> **Comments:**
> **username_0**: Thanks. That definitely confirms some information and diagrams I got are wrong or too vague.  
> **username_2**: Also, just to clarify something in your original post - albite is the sodium feldspar endmember, not the calcium endmember. Anorthite is the calcium endmember.  
> **username_0**: Yes my bad, you're right.  

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