4 вариант 1 FeO 2 Fe(OH)2 3 FeS 4 FeCl2 5 FeSO3 6 FeSO4 7 FeCO3 8 Fe(NO3)2 9 Fe 3(PO4)2

Определим массовые доли элементов в соединениях:

1. FeO
   • Mr(FeO) = 56 + 16 = 72
   • w(Fe) = (56/72) * 100% = 77.78%
   • w(O) = (16/72) * 100% = 22.22%
2. Fe(OH)2
   • Mr(Fe(OH)2) = 56 + 2(16 + 1) = 56 + 34 = 90
   • w(Fe) = (56/90) * 100% = 62.22%
   • w(O) = (32/90) * 100% = 35.56%
   • w(H) = (2/90) * 100% = 2.22%
3. FeS
   • Mr(FeS) = 56 + 32 = 88
   • w(Fe) = (56/88) * 100% = 63.64%
   • w(S) = (32/88) * 100% = 36.36%
4. FeCl2
   • Mr(FeCl2) = 56 + 2*35.5 = 56 + 71 = 127
   • w(Fe) = (56/127) * 100% = 44.09%
   • w(Cl) = (71/127) * 100% = 55.91%
5. FeSO3
   • Mr(FeSO3) = 56 + 32 + 3*16 = 56 + 32 + 48 = 136
   • w(Fe) = (56/136) * 100% = 41.18%
   • w(S) = (32/136) * 100% = 23.53%
   • w(O) = (48/136) * 100% = 35.29%
6. FeSO4
   • Mr(FeSO4) = 56 + 32 + 4*16 = 56 + 32 + 64 = 152
   • w(Fe) = (56/152) * 100% = 36.84%
   • w(S) = (32/152) * 100% = 21.05%
   • w(O) = (64/152) * 100% = 42.11%
7. FeCO3
   • Mr(FeCO3) = 56 + 12 + 3*16 = 56 + 12 + 48 = 116
   • w(Fe) = (56/116) * 100% = 48.28%
   • w(C) = (12/116) * 100% = 10.34%
   • w(O) = (48/116) * 100% = 41.38%
8. Fe(NO3)2
   • Mr(Fe(NO3)2) = 56 + 2*(14 + 3*16) = 56 + 2*(14 + 48) = 56 + 2*62 = 56 + 124 = 180
   • w(Fe) = (56/180) * 100% = 31.11%
   • w(N) = (28/180) * 100% = 15.56%
   • w(O) = (96/180) * 100% = 53.33%
9. Fe3(PO4)2
   • Mr(Fe3(PO4)2) = 3*56 + 2*(31 + 4*16) = 168 + 2*(31 + 64) = 168 + 2*95 = 168 + 190 = 358
   • w(Fe) = (168/358) * 100% = 46.93%
   • w(P) = (62/358) * 100% = 17.32%
   • w(O) = (128/358) * 100% = 35.75%

Ответ: Массовые доли элементов определены.