A 39.09 gram sample of a hydrate of CoSO4 was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, 23.03 grams of the dehydrated compound remained. What is the formula of the hydrate?
let the formula be CoSO4.xH2O
mass of = H2O = mass of hydrated salt - mass of anhydrous salt
mass of = H2O = 39.09 g - 23.03 g
mass of = H2O = 16.06 g
Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
mass(H2O)= 16.06 g
number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(16.06 g)/(18.016 g/mol)
= 0.8914 mol
Molar mass of CoSO4,
MM = 1*MM(Co) + 1*MM(S) + 4*MM(O)
= 1*58.93 + 1*32.07 + 4*16.0
= 155 g/mol
mass(CoSO4)= 23.03 g
number of mol of CoSO4,
n = mass of CoSO4/molar mass of CoSO4
=(23.03 g)/(155 g/mol)
= 0.1486 mol
X = mol (H2O)/mol (CoSO4)
X = 0.8914 / 0.1486
X = 6
so,
the formula be CoSO4.6H2O
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