Question

An oxide of aluminum is formed by the reaction of 4.151 g Al with 3.692 g O. How to calculate its empirical formula?

Answer 1

Hint: Empirical formula of a compound is nothing but the simple positive integer of atoms those are present in the given compound. We should know the number of moles of the compound to determine the empirical formula of the compound.
Number of moles of an atom or compound = $\dfrac{\text{weight of the atom or compound}}{\text{molecular weight of the atom or compound}}$

Complete answer:
- In the question it is given, an oxide of aluminum is formed by the reaction of 4.151 g Al with 3.692 g O. we have to calculate the empirical formula of aluminum oxide.
- The given weight of the aluminum is 4.151 g and the molecular weight of the aluminum is 26.98 g.
- Therefore the number of moles of the aluminum can be calculated as follows.
Number of moles of aluminum
\[\begin{align}
  & =\dfrac{\text{weight of the aluminum}}{\text{molecular weight of the aluminum}} \\
 & =\dfrac{4.151}{26.98} \\
 & =0.1539moles \\
\end{align}\]
- The given weight of the oxygen is 3.692 g and the molecular weight of the oxygen is 16 g.
- Therefore the number of moles of the oxygen can be calculated as follows.
Number of moles of oxygen
\[\begin{align}
  & =\dfrac{\text{weight of the oxygen}}{\text{molecular weight of the oxygen}} \\
 & =\dfrac{3.692}{16} \\
 & =0.2308moles \\
\end{align}\]
- Means 0.1539 moles of aluminum reacts with 0.2308 moles of oxygen.
- Now we have to divide the number of moles of aluminum and oxygen to get the mole ratio.
\[\dfrac{\text{aluminum}}{\text{oxygen}}\text{=}\dfrac{\text{0}\text{.1539}}{\text{0}\text{.2308}}\text{=0}\text{.667=}\dfrac{\text{2}}{\text{3}}\]
- Therefore the empirical formula of aluminum oxide is $A{{l}_{2}}{{O}_{3}}$ .

Note:
If we use oxygen in diatomic state we will get the empirical formula as $A{{l}_{4}}{{({{O}_{2}})}_{3}}$. We can write $A{{l}_{4}}{{({{O}_{2}})}_{3}}$ as $A{{l}_{4}}{{O}_{6}}$. The empirical formula of $A{{l}_{4}}{{O}_{6}}$ is $A{{l}_{2}}{{O}_{3}}$ only. Means we can oxygen in a diatomic state or monoatomic state while calculating the empirical formula.