Question

A hydrate containing aluminium sulphate has the formula $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$ and it contains $11.11\% $ of aluminium by mass. Calculate the value of $x$ in the hydrate formula?

Answer 1

Hint: Aluminium sulfate exists as off-white solid substance which is stable in nature in its hydrated form. Hydrates are formed when a water molecule is entrapped with the chemical compound.

Complete answer:
Chemical formula of aluminium sulphate is $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$, in the given chemical formula we see that water molecules are attached with chemical compound therefore, aluminium sulfate comes under the category of hydrates.
Hydrate molecules have better solubility than solvates.
In order to determine the value $x$ in the given compound, we have to follow some step:
Firstly, we have to calculate the molecular mass of aluminium sulphate in anhydrous form.
This can be calculated by adding atomic mass of aluminium, sulphur and oxygen.
Now we have to calculate the molecular mass of water.
Then we have to create a mathematical equation which includes the mass of both anhydrous aluminium sulphate as well as water molecules.
In question it is mention that $11.11\% $ of aluminium is present in $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$ molecules. It means that for every $100$ g of $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$, $11.11$ g of aluminium is present.
Now we calculate the number of moles of aluminium present in $11.11$ g.
$mole = \dfrac{m}{M}$
Atomic mass of aluminium is $27$.
So moles will be- $mole = \dfrac{{11.11}}{{27}}$
After calculation, we find that
$mole = 0.411$
It means that $0.411$ moles of aluminium is present in $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$.
From the chemical formula of $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$, we see that one mole of aluminium sulfate contain two moles of aluminium. Therefore, total amount of aluminium in $0.411$ moles is
Total moles of aluminium present in one mole of anhydrous aluminium sulfate is $ = \dfrac{{0.411}}{2}$
After solving, we get that-
Total moles of aluminium present in one mole of anhydrous aluminium sulfate is $0.205$ moles of aluminium.
Molar mass of aluminium is $27$g, it means that weight of one mole of aluminium is $27$g, but we have $0.205$ moles of aluminium, so total mass of aluminium in aluminium sulfate is:
Mass of aluminium$ = $$0.205$$ \times 27$
After solving, we get
Mass of aluminium$ = $$70.45$g
It means that $100$g of $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$ contain $70.45$g of aluminium. Hence, the mass of water in compound will be
$100 - $$70.45$$ = 29.55$g of water.
As we know mass of one mole of water is $18$g, so the total moles of water present in compound will be:
$mole = \dfrac{m}{M}$
$mole = \dfrac{{29.55}}{{18}}$
After solving we get,
$mole = 1.641$
Hence, $1$ mole of $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$ contains $0.2059$ moles of aluminium and $1.641$ moles of water.
Now we can calculate moles of water per mole of the $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$:
$x = \dfrac{{1.641}}{{0.2059}}$
After calculating we get
$x = 7.9 \simeq 8$
$ \Rightarrow $ Hence, value of $x$ in molecule $A{l_2}{\left( {S{O_4}} \right)_3}.x{H_2}O$ is $8$. It shows that $8$ molecules of water are attached with hydrated aluminium sulfate.

Note:
In addition to water and aluminium sulfate, impurities presence in the compound aggregates and forms a large particle which is easier to remove. Aluminum sulfate was earlier used in production of paper for commercial use. It is also used to coagulate blood while cut or in an accident.