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 1 of 4: Water determined via thionyl chloride

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n(\ce{BaSO4}) = \dfrac{14.004}{233.4} = 0.0600\ \text{mol} \Rightarrow n(\ce{H2O}) = 0.0600\ \text{mol}

Analysis

Each mole of crystal water reacts as $\ce{H2O + SOCl2 -> SO2 + 2HCl}$ . Oxidation of $\ce{SO2}$ by $\ce{H2O2}$ yields $\ce{SO4^{2-}}$ , precipitating 14.004 g of $\ce{BaSO4}$ (contains 13.74 % S, matching $\ce{BaSO4}$ exactly). Thus the 2.3793 g sample contains 0.0600 mol of $\ce{H2O}$ .