Question

The “alum” used in cooking is potassium aluminum sulfate hydrate, $KAl{(S{O_3})_2}.X{H_2}O$, To find the value of x, a sample of the compound is heated. The mass of the empty crucible is 20.01 g.
The alum hydrate was added to the crucible until the total mass of the crucible and hydrate was 24.75 g. The sample was heated in the crucible until the final mass of the crucible and anhydrous product was 22.5 g.
What is the value of x?

Answer 1

Number of moles $=$ mass in grams/molar mass
The Molar Mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it. Molar mass is calculated in grams.
Mass in grams is the given mass of the substance in the solution.

Complete step by step answer:
To find the mass of alum hydrate,
Given,
Mass of the empty crucible is 20.01 g
The Total mass of crucible and hydrate is 24.75 g
From the above two equations, we can deduct the mass of the alum hydrate
Mass of the alum hydrate $=$ Total mass of the crucible and hydrate $-$ Mass of the empty crucible
$\text{Mass of alum hydrate = }24.75g - 20.01g$
$= 4.74g$
To find the mass of anhydrous alum,
After heating,
The final mass of the crucible and anhydrous alum is 22.5 g
The total mass of the empty crucible is 20.01 g
From the above two equations, we can easily find the mass of anhydrous alum
Mass of anhydrous alum $=$ Total mass of the crucible and anhydrous alum $-$ Total mass of the empty crucible
$= 22.5g - 20.01g$
$= 2.49g$
Mass of water lost on heating Mass of the alum hydrate $-$ Mass of anhydrous alum
$= 4.74g - 2.49g$
$= 2.25g$
Now, to find the value of x in $KAl{(S{O_3})_2}.X{H_2}O$
The Molecular mass of $KAl{(S{O_3})_2} = 39.0983g/mol(K) + 26.9815g/mol(Al) + 2(32.065)g/mol(S) + 6(14.9994)g/mol(O) = 226.2062g/mol(KAl{(S{O_3})_2}$
The Molecular mass of ${H_2}O = 2(1.0079) + 15.9994 = 18.0152g/mol$
The chemical reaction on heating the potassium aluminum sulfate hydrate is as follows:
$KAl{(S{O_3})_2}.X{H_2}O \to KAl{(S{O_3})_2} + x{H_2}O$
Writing the equation in terms of molecular masses,
$(226.2062 + 18.0152x) \to 226.2062g$ –(i)
Writing the equation in terms of given masses,
$4.74g \to 2.49g$ –(ii)
Dividing equation (i) by (ii) will give us the value of x
$\dfrac{{226.2062 + 18.0152x}}{{4.74}} = \dfrac{{226.2062}}{{2.49}}$

On solving, we get, $x = 12$

Note:
One mole of a substance is equal to $6.022 \times {10^{23}}$ units of that substance (such as atoms, molecules or, ions)
The number $6.022 \times {10^{23}}$ is known as Avogadro’s number or Avogadro’s constant.
The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions.
The concentration of a solution is commonly expressed by its molarity, defined as the amount of dissolved substance per unit volume of solution, for which the unit is used in moles per litre.