Mass fractions of iron and sulfur of compounds are, respectively, 46.67% and 53.33 %.Determine the formula of this compound.

given:
ω(Fе) = 46,67  %
ω(S) = 53,33  %
find :
Fех Sу  -?
Solution:
m(Fе)= (m(〖Fе〗_х S_у )•ω(Fе))/(100 %)= (100 г•46,67 %)/(100 %)=46,67 g;
m(S)= (m(〖Fе〗_х S_у )•ω(S))/(100 %)= (100 г•53,33 %)/(100 %)=53,33 g;
М(Fе) = 56 g/mol,
ν(Fе)= (46,67g)/(56 g/mol)   =0,83 mol ;
М(S) = 32 g/mol,
ν(S)= (53,33 g )/(32 g/mol)   =1,66 mol ;
ν(Fе)  :  ν(S)  =  х  :  у =  0,83 : 1,66
х∶у= (0,83)/(0,83) ∶ (1,66)/(0,83)=1∶2 .
formula iron compound FеS2.
Answer: FеS2.