Question

When aluminum metal is heated with an element from Group 6A of the periodic table, an ioniccompound forms. When the experiment is performed with an unknown Group 6A element, the product is 18.56% Al by mass. What is the formula of the compound?

Short answer

The formula of the compoundis${\mathrm{Al}}_2{\mathrm{Se}}_3$and the mass ${\mathrm{Al}}_2{\mathrm{Se}}_3$of the compound is$79.029\mathrm{amu}$ .

Step-by-step solution

Definition

Themolar mass of a material is defined asthe mass of the substance contained inone mole of the substance, and it isequal to the total massof the component atoms. It may be stated mathematically as follows:
$\mathrm{Molar}\mathrm{Mass}=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{Number}\mathrm{of}\mathrm{moles}}$

Determination of mole of Aluminum

Because the combination generated by Al and VI Group is an ionic compound, we may predict a ${\mathrm{Al}}_2{\mathrm{X}}_3$ type compound, as 6A Group elements have a -2 oxidation state, whilst all other elements have a +3 state.

Average of ${\mathrm{Al}}_2{\mathrm{X}}_3$, Number of Al atoms$=1$

‘X’ atoms role="math" localid="1655210029440" $$\begin{gathered} =\frac{2}{3} \\ =1.5 \end{gathered}$$

Here, Mass of ${\mathrm{Al}}_2{\mathrm{X}}_3=100\mathrm{g}$

Thus the Mass of $\mathrm{Al}=18.56\mathrm{g}$

role="math" localid="1655210345973" $\begin{array}{rcl}\mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{X}& =& 100-18.56\mathrm{g}\\ \mathrm{X}& =& 81.44\mathrm{g}\end{array}$role="math" localid="1655210311472" $\begin{array}{rcl}\mathrm{Moles}\mathrm{of}\mathrm{Aluminium}& =& \frac{18.56\mathrm{g}}{26.981}\\ & =& 0.687\mathrm{moles}\end{array}$

Determination of Number of moles

$\frac{81.44}{X}$

Where X is the atomic mass of element ‘X’

To determine the least number of atoms in an ionic compound, divide each number of moles by the smallest number. The number of atoms in Al equals one.

As a result, divide each number of moles by 0.687 moles of AI.

$\begin{array}{rcl}\frac{0.687}{0.687}& =& 1\mathrm{Atom}\mathrm{of}\mathrm{Al}\\ \frac{81.44}{{\displaystyle \frac{\mathrm{X}}{0.687}}}& =& 1.5\mathrm{Atom}\mathrm{of}\mathrm{X}\\ \frac{81.44}{\left(\mathrm{X}\right)\left(0.687\right)}& =& 1.5\mathrm{Atom}\mathrm{of}\mathrm{X}\\ \mathrm{X}& =& 79.029\mathrm{amu}\end{array}$

The atomic mass of

$79.029\mathrm{amu}$is similar to that of 'Se'.

${\mathrm{AlSe}}_3$, should be the ionic compound.