Question

A compound of aluminum and chlorine is composed of 9.0 g Al for every 35.5 g of chlorine. The empirical formula of compound is:
A- $AlCl$
B- $AlCl_{ 2 }$
C- $AlCl_{ 3 }$
D- $AlCl_{ 4 }$

Answer 1

Hint: Find out the number of gram-atoms of aluminum and chlorine in the given compound. We will obtain the formula of the given compound. Simplify the ratio of both atoms such that they cannot be simplified further. The compound with the simplest ratio is called the empirical formula of the compound.


Complete step by step solution:
We first find the number of gram atoms of aluminum and chlorine is present.
A given weight of Al is 9 g. The gram atomic weight of Al is 27 g.
Number of gram atoms = $\dfrac { 9 }{ 27 } \quad =\quad \dfrac { 1 }{ 3 } $
A given weight of chlorine is 35.5g. The gram atomic weight of Cl is 35.5g.
Number of gram atoms = $\dfrac { 35.5 }{ 35.5 } \quad =\quad 1$
Number of atoms of chlorine per one atom of aluminum = $\dfrac { 1 }{ \left( \dfrac { 1 }{ 3 } \right) } \times 1\quad =\quad 3\times 1\quad =\quad 3$
If the number atoms of one element are 1 then that will be the simplest formula of the compound. The simplest formula is nothing but the empirical formula of the compound. In the empirical formula, number atoms will be an integer.
An empirical formula for each atom of aluminum there is 3 atoms of chlorine.
The empirical formula is $AlCl_{ 3 }$.
So, the correct answer is “Option C”.

Note: We need to take the number of atoms of each element in the empirical formula as integers even though the number of gram atoms was fractional. We can’t determine the compound formula from the empirical formula of the compound.