# Question af77e

Apr 22, 2015

The concept is very simple. Hydrates are compounds that contain a certain amount of water. When you heat a hydrate, all the water that's a part of its structure is evaporated, leaving behind the anhydrous salt.

In your case, the hydrated form of iron (II) sulfate weighs 2.78 g. After all the water of hydration is removed, you're left with the anhydrous salt iron (II) sulfate, which weighs 1.52 g.

The difference in weight between the hydrate and the anhydrous salt is the water of hydration.

${m}_{\text{water" = m_"hydrate" - m_"anhydrous salt}}$

${m}_{\text{water" = 2.78 - 1.52 = "1.26 g}}$

To determine the number of water molecules associated with each unit of iron (II) sulfate, you need to use the two compounds' molar masses

1.26cancel("g") * "1 mole water"/(18.015cancel("g")) = "0.0700 moles water"

1.52cancel("g") * ("1 mole "FeSO_4)/(151.91cancel("g")) = "0.0100 moles "# $F e S {O}_{4}$

Divide these numbers by the smallest one to get the mole ratio that exists between water and iron (II) sulfate in the hydrate

${H}_{2} O : \frac{0.0700}{0.0100} = 7$

$F e S {O}_{4} : \frac{0.0100}{0.0100} = 1$

This means that each unit of hydrated iron (II) sulfate contains 7 moles of water

$F e S {O}_{4} \cdot 7 {H}_{2} O$ $\to$ iron (II) sulfate heptahydrate