Question

Iron oxide has formula fe0 94o what fraction of fe exists as fe 3

Answer 1

Fe2+)x (Fe3+)yO.

On looking at the given formula of the compound

x + y = 0.93 ... (1)

Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen. Therefore,

2x + 3y = 2 ... (2)

⇒ x + 3/2y = 1... (3)

On subtracting equation (1) from equation (3) we have

3/2y − y = 1 − 0. 93

⇒ 1/2y = 0.07

⇒ y = 0.14

On putting the value of y in equation (1) we get,

x + 0.14 = 0.93

⇒ x = 0.93 – 0.14

x = 0.79

Fraction of Fe2+ ions present in the sample = 0.79/0.93 = 0.81

Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.

Answer 2

Given:

Iron oxide with the formula $\[F{e_{0.94}}O\]$

To Find: Fraction of $Fe$ exists as $\[F{e^{3 + }}\]$

Solution:

Suppose that the total number of Oxygen atoms are $100$.

Therefore,

Total number of $Fe$ atoms $\[ = 94\]$

If the number of  $Fe$ atoms  that exist as $\[F{e^{2 + }}\]$ is $x$, then

Number of  $Fe$ atoms  that exist as $\[F{e^{3 + }}\]$ $=94-x$

As the compound is neutral, therefore, we have,

$\[\begin{gathered} x( + 2) + (94 - x)( + 3) = 100 \times 2 \hfill \\ 2x + 282 - 3x = 200 \hfill \\ x = 82 \hfill \\ \end{gathered} \]$

Therefore, the fraction of $Fe$ atoms that exist as $\[F{e^{3 + }}\]$ can be given as:

$\[\begin{gathered} Fractio{n_{(F{e^{3 + }})}} = \frac{{94 - x}}{{100}} \hfill \\ Fractio{n_{(F{e^{3 + }})}} = \frac{{94 - 82}}{{100}} \hfill \\ Fractio{n_{(F{e^{3 + }})}} = \frac{{12}}{{100}} = 0.12 \hfill \\ \end{gathered} \]$

Hence, the fraction of $Fe$ exists as $\[F{e^{3 + }}\]$ is $\[0.12\]$.

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