Problem 2

A sample of 2.3793 g of a crystallohydrate of the type

\ce{M_xA_y . z H2O}

(where M is a metal) reacted with an excess of

\ce{SOCl2}

. The gaseous products were absorbed in an aqueous solution containing barium chloride, hydrochloric acid and hydrogen peroxide. Small carried-over amounts of

\ce{SOCl2}

had been frozen out. The mass of the deposited precipitate was 14.004 g and contained 13.74 mass % of sulphur. In another experiment, 1.1896 g of the initial hydrate was dissolved in water and made up to 100 cm

^3

. One fifth of this solution required 10 cm

^3

of 0.2 M

\ce{AgNO3}

solution for complete precipitation (yielding 0.28664 g of precipitate). (a) Calculate the formula of the crystallohydrate. (b) Given that the hydrate can contain at most 7 moles of water per mole of hydrate, name another hypothetical hydrate that is ruled out by this limit.

Step 3 of 4: Molar ratio and valence search

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n(\ce{Cl^-}) : n(\ce{H2O}) = 0.0200 : 0.0600 = 1 : 3

Analysis

The ratio $\ce{Cl^-} : \ce{H2O}$ is $1 : 3$ . For $v=1$ : $\ce{MCl.3H2O} \Rightarrow M(\ce{M}) = 29.5$ (no metal fits). For $v=2$ : $\ce{MCl2.6H2O} \Rightarrow M(\text{hydrate}) = 237.93$ , $M(\ce{M}) = 237.93 - 70.91 - 108.09 = 58.93$ g mol $^{-1}$ , corresponding to cobalt ( $\ce{Co}$ ). Thus the hydrate is $\ce{CoCl2 . 6H2O}$ .