Answer to Question #154395 in General Chemistry for tarik

Question

A 2.78 g sample of hydrated iron (II) sulfate was heated to remove all water of hydration. The mass of the anhydrous compound was 1.52 g. Determine the molecular formula of the original hydrated compound 

Answer 1

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_water = m_hydrate - m_{anhydrous salt}

m_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

1.26g ($\frac{1molewater}{18.015g}$) = 0.0700mole water

1.52g ($\frac{1moleFeSO_4}{151.91g}$) = 0.0100mole FeSO₄

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

H₂O = 0.0700/0.0100 = 7 mole

FeSO₄ = 0.0100/0.0100 = 1 mole

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

FeSO₄ • 7 H₂O iron (II) sulfate heptahydrate