Mass fractions of iron and sulfur nbsp;of compounds are, respectively, 46.67%

given:
(Fе) = 46,67 nbsp;%
(S) = 53,33 nbsp;%
find :
Fех Sу nbsp;-?
Solution:
m(Fе)= (m(Fе_х S_у )(Fе))/(100 %)= (100 г46,67 %)/(100 %)=46,67 g;
m(S)= (m(Fе_х S_у )(S))/(100 %)= (100 г53,33 %)/(100 %)=53,33 g;
М(Fе) = 56 g/mol,
(Fе)= (46,67g)/(56 g/mol) nbsp;nbsp;=0,83 mol ;
М(S) = 32 g/mol,
(S)= (53,33 g )/(32 g/mol) nbsp;nbsp;=1,66 mol ;
(Fе) nbsp;: nbsp;(S) nbsp;= nbsp;х nbsp;: nbsp;у = nbsp;0,83 : 1,66
ху= (0,83)/(0,83) (1,66)/(0,83)=12 .
formula iron compound FеS2.
Answer: FеS2.