Fosfatul de magneziu: Mg3(PO4)2
Masa moleculara:
[tex]M_{Mg3(PO4)2}=3*24+2*31+8*16=262[/tex]
Aplicam regula de 3 simpla pentru a afla compozitia procentuala:
72 g Mg..................62 g P................128 g O............262 g Mg3(PO4)2
x.............................y.........................z....................100 g Mg3(PO4)2
[tex]x= \dfrac{72*100}{262}=27,48 \ \% \ Mg[/tex]
[tex]y= \dfrac{62*100}{262}=23,67 \ \% \ P[/tex]
[tex]z= \dfrac{128*100}{262}=48,85 \ \% \ O[/tex]
