Formula chimica:
[tex]Na^{I}_{3}PO_{4}^{III}[/tex]
Raport atomic:
3:1:4
Raport de masa:
69:31:64
Masa molara:
[tex]M_{Na_{3}PO_{4}}=3M_{Na}+M_{P}+4M_{O}=3*23+31+4*16=69+31+64=164g/mol[/tex]
Compozitie procentuala:
69g Na......31g P......64g O........164g Na3PO4
x...................y.................z...................100g Na3PO4
x=69*100/164=42,07% Na
y=31*100/164=18,9% P
z = 100 - 42,07 - 18,9 = 39,03% O
