Question

An oxide of iron contains 72.2%of metal by mass determine the formula of this iron oxide

Answer 1

Answer & Explanation:

According to the question,

There is an oxide of iron which must be either Fe₂O₃ or, Fe₃O₄.

Now,

It says that 72.2% is metal of that compound (Iron is a metal).

So,

Mass of Fe₂O₃ = (56*2)+(16*3) g

= 150 g

Mass of Fe₃O₄ = (56*3)+(16*4) g

= 232 g

Now, we've to find that between these 2 compounds, which one has 72.2 % Iron,

= Percentage of Iron in Fe₂O₃ = (112/150*100)%

= 74.6 %

= Percentage of Iron in Fe₃O₄ = (168/232*100)%

= 72.2%

Thus, it clearly shows that the metal is Fe₃O₄.

Hope this helps.