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. The miscibility strongly depends also on temperature. Also rare Ba-feldspars exist, more in the reference.