Question

When 5.60g of an iron oxide is heated with carbon, 3.92g of iron is produced. Calculate the empirical formula of the iron oxide

Answer 1

Answer:

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Answer 2

Given:

Mass of iron oxide$\[ = 5.60g\]$

Mass of iron$\[ = 3.92g\]$

To Find: Empirical formula of iron oxide

Solution:

According to law of conservation of mass,

Mass of iron oxide $=$ Mass of iron $+$ Mass of oxygen

Therefore,

Mass of oxygen $=$ Mass of iron oxide $-$ Mass of iron

Mass of oxygen $\[ = 5.60g - 3.92g\]$

Mass of oxygen $\[ = 1.68g\]$

Now, the number of moles of iron and oxygen can be given as

Number of moles of oxygen $=\frac{Mass of oxygen}{Molar mass of oxygen}$

Number of moles of oxygen $\[ = \frac{{1.68g}}{{16}}\]$

Number of moles of oxygen $=0.105$

Similarly,

Number of moles of iron $=\frac{Mass of iron}{Molar mass of iron}$

Number of moles of iron $\[ = \frac{{3.92g}}{{55.845}}\]$

Number of moles of iron $=0.07$

The smallest whole number ratio of the number of moles of iron and oxygen in iron oxide can be given as

Oxygen: $\[\frac{{0.105}}{{0.07}} = 1.5\]$

Iron: $\[\frac{{0.07}}{{0.07}} = 1\]$

To change the fractional value into a whole number, multiply both values by 2.

Therefore, the empirical formula of iron oxide can be given as

$\[{(F{e_1}{O_{1.5}})_2} = F{e_2}{O_3}\]$

Hence, the empirical formula of iron oxide is $\[F{e_2}{O_3}\]$.