Question

170g of a sample of $A{{l}_{2}}{{O}_{3}}$ produces 9g Al.
\[A{{l}_{2}}{{O}_{3}}\to 2Al+\dfrac{3}{2}{{O}_{2}}\]
Find the percentage purity of $A{{l}_{2}}{{O}_{3}}$.

Answer 1

Hint: Percentage purity of a sample indicates the real amount of the compound present in it. The higher the purity of a sample, the more the amount of desired product is formed. Finding the theoretical yield of aluminum (Al) from the pure sample of aluminum oxide $(A{{l}_{2}}{{O}_{3}})$ will help us in calculating the percentage purity of the sample.

Complete step by step solution: From the reaction equation,
\[A{{l}_{2}}{{O}_{3}}\to 2Al+\dfrac{3}{2}{{O}_{2}}\]
We can see that one mole of aluminum oxide $(A{{l}_{2}}{{O}_{3}})$ gives a yield of two moles of aluminum (Al).
We know that molar mass of one mole of aluminum oxide $(A{{l}_{2}}{{O}_{3}})$ is 101.96 g/mol, and the molar mass of one mole of aluminum (Al) is 26.98 g/mol.
So, the mass of two moles of aluminum (Al) will be 53.96 g.
Now, theoretically, 101.96 g of aluminum oxide $(A{{l}_{2}}{{O}_{3}})$ yields 53.98 g of aluminum Al.
Hence using unitary method, we can calculate that 170 g of pure sample of aluminum oxide $(A{{l}_{2}}{{O}_{3}})$ will theoretically yield:
\[\dfrac{53.96}{101.96}\times 170=89.9686\text{ g Aluminum (Al)}\text{.}\]
But we know that only 9 g aluminum (Al) was yielded from a sample of 170 g aluminum oxide $(A{{l}_{2}}{{O}_{3}})$.
So, to calculate percentage purity, we will use the following formula
\[percentage\text{ }purity\text{ }=\text{ }\dfrac{actual\text{ }mass\text{ }of\text{ }compund \text{ }yielded\text{ }from\text{ }sample}{theoretical\text{ }mass\text{ }of\text{ }compound\text{ }yielded\text{ }from\text{ }pure\text{ }sample}\times 100\]
Actual mass of aluminum (Al) yielded from the sample = 9 g
Theoretical mass of aluminum (Al) yielded from pure sample = 89.9686 g.
By substituting the values in the formula, we get
\[percentage\text{ }purity\text{ }=\text{ }\dfrac{9}{89.9686}\times 100\simeq 10%\]
Hence the percentage purity of the given aluminum oxide $(A{{l}_{2}}{{O}_{3}})$ sample is approximately 10%.

Additional information:A sample having 100% purity is virtually impossible. Each sample has some amount of impurity along with the required compound.

Note: It is important to note that to calculate the percentage purity of a sample, we must ensure that the chemical reaction is balanced. A balanced chemical reaction can be used to find out the ratio of number of moles between the reactants and the product, which further can be used to calculate the theoretical yield.