Question

You want to determine the value of x in hydrated iron II sulfate, FeSO4 xH2O. In the laboratory you weigh out 1.983 g of the hydrated salt. After thorough heating, 1.084 g of the anhydrous salt remains. What is the value of x?

Answer 1

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:

$1mol(FeSO_{4} ).\frac{152gFeSO_{4}}{1mol(FeSO_{4} )} .\frac{0.899gH_{2}O }{1.084gFeSO_{4}} =126gH_{2}O$

As a result 1 mol of FeSO₄ is accompanied by 126 g of water. If we pass that mass into moles:

$126gH_{2}O.\frac{1molH_{2}O}{18gH_{2}O} =7molH_{2}O$

Finally, the number of moles of water in the molecular formula is x = 7.