Question

A crystalline salt on being rendered anhydrous loses 45.6% of it's weight. The composition of anhydrous salt is Al - 10.5%, K - 15.1%, S - 24.8%, O - 49.6%, Calculate the molecular formula of anhydrous and crystalline salt ​

Answer 1

Answer:

An empirical formula of hydrated salt will be $KAlS_{2}I. 12 H_{2}O$

Explanation:

The molecular masses of Aluminium, potassium, Sulphur and Iodide are 27 g/mol, 39 g/mol, 32 g/mol and 127 g/mol respectively.

The %age composition of an anhydrous saltwill be :  

Aluminium = 10.5%

Potassium = 15.1%

Sulphur = 24.8%  

Iodide = 49.6%.

Hence, 100 g of an anhydrous salt includes :

10.5 g = Aluminium

15.1 g = Potassium

24.8 g = Sulphur

49.6 g = Iodide.

10.5 g Aluminium will correspond to 10.527 = 0.39 moles

15.1 g potassium will correspond to 15.139 = 0.39 moles

24.8 g Sulphur will correspond to 24.832 = 0.775 moles

49.6 g Iodide will correspond to 49.6127 = 0.39 moles

Hence, the ratio of the no. of moles of Al: K :S: I will be:

0.39 : 0.39 : 0.775 : 0.39

Then, we divide this by 0.39

Then,

Al:K:S:I  

1:1:2:1.

The empirical formula of an anhydrous salt will be KAlS2I.

Formula mass= 39+27+32+32+127= 257g/mol.

Formula wt. of hydrated salt will be= 100100−45.6×257=472.

Mass of water of crystalline salt= 472−257=215g.

21518=12 water molecules.

An empirical formula of hydrated salt will be KAlS2I.12H2O.

Answer 2

Answer:

Explanation:

On heating the crystalline salt, it loses its water and become anhydrous. The weight lost in this manner is that of water which according to question forms 45.6% of total weight of crystalline salt. Hence rest of  salt 54.4% is that of anhydrous salt.

Let us consider 100 grams  (54.4%) of anhydrous salt.

Total mass of crystalline salt = 100/0.544 = 183.82 grams

mass of water = 183.82 - 100 = 83.82 grams

The composition of the salt is;

Aluminum = 10.5 grams = 10.5/27 = 0.38 moles

Potassium = 15.1 grams = 15.1/39 = 0.38 moles

Sulphur = 24.8 grams = 24.8/32 = 0.775 moles

Oxygen = 49.6 grams = 49.6/16 = 3.1 moles

Water = 83.82 grams = 83.82/18 = 4.65 moles

Hence the empirical formula of anhydrous and crystalline salt:

Al₀.₃₈K₀.₃₈S₀.₇₇₅O₃.₁ and Al₀.₃₈K₀.₄₈S₀.₇₇₅O₁.₅₅4.65H₂O respectively.

Dividing whole terms by 0.38 and rounding off to nearest digit,we get.

AlKS₂O₈ and AlKS₂O₈ 11.92 H₂O

Formula of anhydrous salt = KAlS₂O₈

Formula of crystalline salt = KAlS₂O₈ 12 H₂O