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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.
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
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
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