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 2 of 4: Halide identification

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n(\ce{Ag+}) = 0.2 \times 0.010 = 0.0020\ \text{mol}; \quad M(\ce{AgA}) = \dfrac{0.28664}{0.0020} = 143.32\ \text{g mol}^{-1} \Rightarrow \ce{A} = \ce{Cl^-}

Analysis

Precipitating 0.0020 mol gives an adduct with molar mass 143.32 g mol $^{-1}$ , which is exactly silver chloride ( $A_r(\ce{Ag}) + A_r(\ce{Cl}) = 107.87 + 35.45 = 143.32$ ). The 1/5 aliquot (0.23792 g) has 0.002 mol $\ce{Cl^-}$ , so 2.3793 g contains 0.0200 mol $\ce{Cl^-}$ .