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A 5.000 gram sample of hydrated nickel (II) cyanide, Ni(CN)2·xH2O is heated according to the procedure...

A 5.000 gram sample of hydrated nickel (II) cyanide, Ni(CN)2·xH2O is heated according to the procedure described in this laboratory. After heating, the sample is found to weigh 3.029 grams. How many moles of H2O were present in the sample? How many moles of Ni(CN)2 were present?

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Answer #1

Due to heating the water of crystallisation is lost .

Loss of water = (5.000 - 3.029) = 1.971 g.

Moles of water present = (1.971/18)

= 0.1095.

Mass of Ni(CN)2 = 3.029 grams

Molar mass of Ni(CN)2 = 110.73 g/mol.

Then moles of Ni(CN)2 = ( 3.029/110.73) = 0.0273

Now,

Ni(CN)2 : H2O

= 0.0273 : 0.1095

1:4

Hence, formula is Ni(CN)2, 4 H2O.

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