Question

170g of a sample of $A{l_2}{O_3}$ produces 9g of$Al$.
$A{l_2}{O_3} \to 2Al\, + \,\dfrac{3}{2}{O_2}$
Find % purity of $A{l_2}{O_3}$?

Answer 1

Purity indicates the actual amount of the compound present in the sample. This is because impurities are present along with the desired compound. Hence, in many chemical reactions, the percent purity of the final product is determined. Higher is the purity more will be the amount of the desired compound in the formed product.

Complete solution:
To determine the percent purity it is important to determine the theoretical yield of the given reaction.
Here, the reaction given is as follows:
$A{l_2}{O_3} \to 2Al\, + \,\dfrac{3}{2}{O_2}$
As the reaction has given is the balanced chemical reaction indicates 1 mole of aluminium oxide gives 2 moles of aluminium.
The molar mass of the aluminium oxide $101.96\,gmo{l^{ - 1}}$ is and the molar mass of aluminium is $26.98\,gmo{l^{ - 1}}$.
$1 mole \, of \,A{l_2}{O_3} = 2\,moles\, of \,Al$
$101.96\,g\,A{l_2}{O_3} = 2 \times\, 26.98\,g\,Al$
$101.96\,g\,A{l_2}{O_3} = 53.96\,g\,Al$
Now,
$101.96\,g\,A{l_2}{O_3} = 53.96\,g\,Al$
$170\,g\,A{l_2}{O_3} = 53.96\,g\,Al \times \dfrac{{170\,g\,A{l_2}{O_3}}}{{101.96\,g\,A{l_2}{O_3}}} = 89.9686\,g Al$
Thus, the theoretical yield is $89.9686\,g\, Al$.
But we know that the actual yield of the aluminium obtained from the 170g of aluminium oxide is 9g.
Now, determine the percent purity of the aluminium oxide as follows:

$percentage purity = \dfrac{{mass\,of\,the\,pure\,substance}}{{mass\,of\,sample}} \times 100\%$
Here, in the formula mass of the pure substance is taken as the theoretical yield obtained and the mass of the sample is nothing but the mass of the sample given.
$percentage \, purity = \dfrac{{9\,g}}{{89.9686\,g }} \times 100\%$
$percentage\, purity = 10.00\%$

Thus, the percent purity of the aluminium oxide is 10%.

Note: To determine the percent purity of the material it is important to know the balanced chemical reaction from which we can determine the mole ratio between the reactants and products. From the mole ratio using the molar masses of the species, the theoretical yield can be determined. The theoretical yield is nothing but the mass of the pure sample.