Question

The composition of a sample of iron oxide is $ F{e_{0.93}}{O_{1.0}} $ . What is the percentage of iron present in the form of $ Fe\left( {III} \right) $ .
 $ (i){\text{ 5}}{\text{.053}} $
 $ (ii){\text{ 10}}{\text{.053}} $
 $ (iii){\text{ 15}}{\text{.053}} $
 $ (iv){\text{ }} $ None of the above

Answer 1

Hint: For a neutral compound the amount of negative charge is equal to the amount of positive. The sum of both polarity of charges in a neutral compound is always equal to zero. We have to find the amount of both $ Fe\left( {III} \right) $ and $ Fe\left( {II} \right) $ for the given molecule. Then we can calculate the percentage of $ Fe\left( {III} \right) $ in the given compound.
Percentage of $ Fe\left( {III} \right){\text{ = }}\dfrac{{{\text{amount of F}}{{\text{e}}^{ + 3}}{\text{ present in sample}}}}{{{\text{ amount of Fe in sample}}}}{\text{ }} \times {\text{ 100}} $

Complete answer:
We know that for the neutral compound the amount of negative charge and positive is equal. We can also say that when the overall charge of a compound is zero then the sum of charge of each atom of compound is equal to zero. Here in $ F{e_{0.93}}{O_{1.0}} $ we have iron and oxygen atoms. But we know that iron forms two ions which are $ F{e^{ + 2}} $ and $ F{e^{ + 3}} $ . Thus both of these ions are present in the given compound. Also we know that oxygen have $ - 2 $ charge.
Now if we assume the amount of $ F{e^{ + 2}} $ present in $ F{e_{0.93}}{O_{1.0}} $ be $ = {\text{ }}x{\text{ mole}} $
Then amount of $ F{e^{ + 3}} $ present in $ F{e_{0.93}}{O_{1.0}} $ will be $ = {\text{ }}\left( {{\text{0}}{\text{.93 - x }}} \right){\text{ mole}} $
Since the net charge of the compound is zero we can write as,
 $ 2x{\text{ + 3}}\left( {0.93{\text{ - x}}} \right){\text{ + 1}}\left( { - 2} \right){\text{ = 0}} $
 $ 2x{\text{ + 2}}{\text{.73 - 3x - 2 = 0}} $
 $ {\text{x = 0}}{\text{.79}} $
Thus the moles of $ F{e^{ + 2}} $ present in $ F{e_{0.93}}{O_{1.0}} $ is $ 0.79{\text{ mole}} $ .
Moles of $ F{e^{ + 3}} $ present in $ F{e_{0.93}}{O_{1.0}} $ will be $ = {\text{ }}\left( {{\text{0}}{\text{.93 - x }}} \right){\text{ mole}} $
Moles of $ F{e^{ + 3}} $ present in $ F{e_{0.93}}{O_{1.0}} $ will be $ = {\text{ }}\left( {{\text{0}}{\text{.93 - 0}}{\text{.79 }}} \right){\text{ mole}} $
Moles of $ F{e^{ + 3}} $ present in $ F{e_{0.93}}{O_{1.0}} $ will be $ = {\text{ 0}}{\text{.14 mole}} $
Now the percentage of $ F{e^{ + 3}} $ can be calculated as,
Percentage of $ Fe\left( {III} \right){\text{ = }}\dfrac{{{\text{amount of F}}{{\text{e}}^{ + 3}}{\text{ present in sample}}}}{{{\text{ amount of Fe in sample}}}}{\text{ }} \times {\text{ 100}} $
Percentage of $ Fe\left( {III} \right){\text{ = }}\dfrac{{0.14{\text{ }} \times {\text{ 56}}}}{{0.93{\text{ }} \times {\text{ 56}}}}{\text{ }} \times {\text{ 100}} $
Percentage of $ Fe\left( {III} \right){\text{ = 15}}{\text{.05 % }} $
Thus the percentage of $ Fe\left( {III} \right) $ present in the given sample is $ {\text{ 15}}{\text{.05 % }} $ .

Note:
The amount of iron present in a compound is calculated by multiplying its moles with the atomic mass of iron. If the compound has some overall charge then the sum of charge of all atoms is equal to that overall charge. We can also find the percentage of $ F{e^{ + 2}} $ in the similar way.