Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# I don't understand the concept of hydrates; A 2.78g sample of hydrated FeSO4 was heated to remove all the water of the hydration. The mass of the anhydrous FeSO4 was 1.52g. Calculate the number of water molecules associated with each unit of FeSO4?

The concept is very simple. Hydrates are 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

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

1.52cancel("g") * ("1 mole "FeSO_4)/(151.91cancel("g")) = "0.0100 moles " FeSO_4

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

H_2O: 0.0700/0.0100 = 7

FeSO_4: 0.0100/0.0100 = 1

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

FeSO_4 * 7H_2O -> iron (II) sulfate heptahydrate