A hydrate of Na2CO3 contains 276.0g. It was heated strongly to drive off the water. The mass of the anhydrous Na2CO3 was found to be 102.0g.Find the Empirial formula. And also name it. Thank you

Expert Answers
sanjeetmanna eNotes educator| Certified Educator

Mass of hydrated Na2CO3 = 276.0 g

Mass of Anydrous Na2CO3 = 102.0 g

Difference in mass = 276 - 102 = 174g

Amount of water dehydrated = 174 g

One mole of water = 18 g

So for 174 g there would be 9.66 ≈ 10 moles of water.

Therefore the Emprical formula = Na2CO3.10H2O

Name = sodium carbonate decahydrate or Hydrated sodium carbonate.

qweem | Student

First determine the number of moles of Na2CO3 after heating:

Formula weight of Na2CO3=

2 Na - 2x22.99

1 C  - 12.01

3O  - 3x15.99

Na2CO3=105.96

 

Moles Na2CO3 = mass after heating / formula weight

= 102.0g/105.96 = 0.96 moles Na2CO3

 

The weight change from heating is a result of water loss, so the weight of water in the hydrate was initial weight minus final weight:

276.0g - 102.0g = 174.0g

 

The molecular weight of water is 18.01g/mole, so the number of moles of water is weight in hydrate from water divided by its molecular weight:

174.0g/18.01g = 9.67 moles water

 

Divide this by the number of moles of Na2CO3 to find the number of moles of water per mole of Na2CO3.

9.67/9.6 = 10

 

So the empirical formula is Na2CO3 x 10H2O