Question

a 15.79 g sample of a hydrate of an iron(ii) sulfate compound was heated, without decomposing the sulfate to drive off the water. the mass was reduced to 8.63 g . what is the formula of the hydrate?

Answer 1

The formula of the hydrated salt  would be $FeSO_4.7H_2O$

Stoichiometric problem

The formula of the hydrated salt of iron (II) sulfate would be in the form, $FeSO_4.xH_2O$. The number of moles of water of hydration can be determined using the empirical formula approach.

Mass of hydrated salt = 15.79 g

Mass of anhydrous salt = 8.63 g

Mass of water of hydration = 15.79 - 8.63

                                             = 7.16 g

Now, let's find the equivalent mole of each component of the hydrated salt.

Molar mass of iron (II) sulfate = 151.908 g/mol

Molar mass of water of hydration = 18.01 g/mol

Mole of iron (II) sulfate = 8.63/151.908

                                      = 0.0568

Mole of water of hydration = 7.16/18.01

                                            = 0.3976

Divide by the smallest mole:

Iron (II) sulfate = 0.0568/0.0568 = 1

Water of hydration = 0.3976/0.0568 = 7

In other words, x, which represents the number of moles of water of hydration, is 7.

Hence, the formula of the hydrated salt would be  $FeSO_4.7H_2O$.

More on the formula of hydrated salts can be found here: https://brainly.com/question/18264811

#SPJ1