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1.8 Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively. Given that the molar mass of the oxide is $159.69 g\ mol^{-1}$ .

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Given that the mass percentage of iron is $69.9\%$ and the mass percentage of oxygen is $30.1\%$

The atomic mass of iron $= 55.85\ u$ .

The atomic mass of oxygen $=16.00\ u$ .

So, the relative moles of iron in iron oxide will be:

$=\frac{mass\ of\ iron\ by\ mass}{Atomic\ mass\ of\ iron}$

$= \frac{69.9}{55.85} = 1.25$

And relative moles of oxygen in iron oxide will be:

$=\frac{mass\ of\ oxygen \ by\ mass}{Atomic\ mass\ of\ oxygen}$

$= \frac{30.01}{16.00} = 1.88$

Hence the simplest molar ratio:

$=\frac{1.25}{1.88}$ or $\Rightarrow 1:1.5 = 2:3$

Therefore, the empirical formula of iron oxide will be $Fe_{2}O_{3}$ .

Now, calculating the molar mass of $Fe_{2}O_{3}$ :

$=(2\times 55.85) + (3\times16.00) = 159.7g\ mol^{-1}$ .

Hence it is matching with the given molar mass of the oxide.

Posted by

Divya Prakash Singh

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