A hydrate of iron $(I I I)$ thiocyanate $F e (S C N)_{3}$ , was found to contain $19 % H_{2} O$ . What is the formula of the hydrate.
(Given: Molecular weight of $F e = 56 u, S = 32 u$ )

F e (S C N)_{3} .3 H_{2} O

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F e (S C N)_{3} .6 H_{2} O

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F e (S C N)_{2} .2 H_{2} O

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F e (S C N)_{3} . H_{2} O

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Solution

The correct option is A $F e (S C N)_{3} .3 H_{2} O$
Let the hydrate be $F e (S C N)_{3} . m H_{2} O$
$Molecular weight of hydrate = 56 + 3 \times (32 + 12 + 14) + 18 m$
$= 230 + 18 m$
$∴ % of H_{2} O = \frac{18 m \times 100}{230 + 18 m} = 19$
$\Rightarrow m = 3$
$∴ Formula is F e (S C N)_{3} .3 H_{2} O$

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