FeSO4, miu= 56+32+4×16= 154 g/mol
Ca(OH)2, miu = 40+2(1+16)= 74 g/mol
Raport masic
Fe:S:O = 56:32:64 impartite la cel mai mic multiplu comun
= 7:4:8
Ca:O:H = 40:32:2, impartim la 2
= 20:16:1
Raport atomic
Fe:S:O = 1:1:4
Ca:O:H =1:2:2
Compozitie procentuala
La 154g FeSO4 avem..56gFe...32gS...64gO
La 100% ........................a%Fe..........b%S......c%O
36,36% 20,78% 41,56%
La 74 g Ca(OH)2............ 40gCa......32gO........2gH
La 100%.............................a%................b%.............c%
54,05% 43,24% 2,7%
