Answer:
The value of x is 7.
Explanation:
When a hydrated salt is heated, the water evaporates because it has a lower boiling point (373 K) and what remains is the anhydrous salt (without water) The mass of the hydrated salt is equal to the sum of the anhydrous salt and the water. Since we have 2 of these 3 data, we can look for the mass of water.
mass of FeSO₄.xH₂O = mass of FeSO₄ + mass of H₂O
mass of H₂O = mass of FeSO₄.xH₂O - mass of FeSO₄
mass of H₂O = 1.983g - 1.084g = 0.899g
This means that 0.899g of water accompany 1.084 g of the salt. We want to know how many grams of water accompany 1 mol of the salt (in the molecular formula we have 1 mol of FeSO₄). Since the molar mass of is 152g/mol, we can establish:
[tex]1mol(FeSO_{4} ).\frac{152gFeSO_{4}}{1mol(FeSO_{4} )} .\frac{0.899gH_{2}O }{1.084gFeSO_{4}} =126gH_{2}O[/tex]
As a result 1 mol of FeSO₄ is accompanied by 126 g of water. If we pass that mass into moles:
[tex]126gH_{2}O.\frac{1molH_{2}O}{18gH_{2}O} =7molH_{2}O[/tex]
Finally, the number of moles of water in the molecular formula is x = 7.
The formula of the hydrated salt is FeSO4 . 7H2O
We know that;
Number of moles of hydrated salt = Number of moles of anhydrous salt
Number of moles of hydrated salt = mass of hydrated salt/molar mass of hydrated salt
Number of moles of anhydrous salt = mass of anhydrous salt/molar mass of anhydrous salt
Mass of hydrated salt = 1.983 g
Mass of anhydrous salt = 1.084 g
Molar mass of hydrated salt = 152 + 18x g/mol
Molar mass of anhydrous salt = 152 g/mol
Therefore;
[tex]1.983 / 152 + 18x = 1.084/ 152\\ 152( 1.983) = 1.084(152 + 18x)\\301.4 = 164.768 + 19.512x\\301.4 - 164.768 = 19.512x\\x = 301.4 - 164.768/ 19.512\\x = 7[/tex]
The formula of the hydrated salt is FeSO4 . 7H2O
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