Q. All the oxygen in a $0.5434 \,g$ sample of a pure oxide of iron is removed by reduction in a stream of $H _2$. The loss in weight is $0.1210\, g$. hence, formula of the iron oxide is $( Fe =56)$
Some Basic Concepts of Chemistry
Solution:
Iron oxide $=0.5434\, g$
Oxygen lost as $H _2 O =0.1210\, g$
Iron $=0.5434-0.1210=0.4224 \,g$
Element
Amount
%
Mole
Ratio
Iron (56)
0
1224 g
77.73
1.39
1
Oxygen(16)
0.1210 g
22.26
1.39
1
Thus $FeO$
| Element | Amount | % | Mole | Ratio |
|---|---|---|---|---|
| Iron (56) | 0 1224 g | 77.73 | 1.39 | 1 |
| Oxygen(16) | 0.1210 g | 22.26 | 1.39 | 1 |