Answer to Question #89622 in Inorganic Chemistry for ghalisaleem

Question

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?

Answer 1

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

n(H₂O) = m(H₂O)/M(H₂O) = [m_sample– 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.