Answer to Question #89622 in Inorganic Chemistry for ghalisaleem

Question #89622
A 32.86g sample of hydrate cocl2 was heated thoroughly in a crucible,until it weight remain constant. after heating,17.98g of the dehydrated compound remain in the crucible.what is the formula of the hydrate?
1
Expert's answer
2019-05-14T03:40:56-0400

n(CoCl2) = m(CoCl2)/M(CoCl2) = 17.98g / 130g mol-1 = 0.1383 mol.

n(H2O) = m(H2O)/M(H2O) = [msample– m(CoCl2)]/ M(H2O) = (32.86g–17.98g)/18g mol-1 = 0.8267 mol.

n(CoCl2) : n(H2O) = 0.1383 : 0.8267 = 0.1383/0.1383 : 0.8267/0.1383 = 1 : 5.98 ≈ 1 : 6.

Therefore, the formula of the hydrate is CoCl2×6H2O.


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