A 17.2\text{ g} sample of hydrated calcium tetraoxosulfate(VI), \text{CaSO}_4 \cdot x\text… — JAMB UTME

A $17.2\text{ g}$ sample of hydrated calcium tetraoxosulfate(VI), $\text{CaSO}_4 \cdot x\text{H}_2\text{O}$ , is heated at $120^\circ\text{C}$ until it partially dehydrates, losing $2.70\text{ g}$ of water vapor to form plaster of Paris, $\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}$ . What is the value of $x$ , the number of molecules of water of crystallization per formula unit in the original hydrated salt? $[\text{Ca} = 40, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Cevap: 2

Cevap

The value of

x

in the hydrated salt formula is 2.

Applying mass conservation, the mass of plaster of Paris (

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

) remaining is

17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

, corresponding to

0.10\text{ mol}

. The mass of evolved water is

2.70\text{ g}

, which equals

0.15\text{ mol}

. The mole ratio of evolved water to salt formula units is

\frac{0.15}{0.10} = 1.5

. Since the reaction is

\text{CaSO}_4 \cdot x\text{H}_2\text{O} \rightarrow \text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} + (x - 0.5)\text{H}_2\text{O}

x - 0.5 = 1.5 \implies x = 2

Adım Adım Çözüm

Calculate the molar masses of

\text{CaSO}_4

\text{H}_2\text{O}

, and the residue

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

Molar mass of

\text{CaSO}_4 = 40 + 32 + (4 \times 16) = 136\text{ g/mol}

;

\text{H}_2\text{O} = 18\text{ g/mol}

;

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = 136 + (0.5 \times 18) = 145\text{ g/mol}

Molar masses are necessary to perform mole calculations from mass measurements.

Determine the mass and amount in moles of plaster of Paris formed.

Mass of residue

= 17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

. Moles of

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = \frac{14.50\text{ g}}{145\text{ g/mol}} = 0.10\text{ mol}

Subtracting the mass of lost water gives the mass of solid product remaining.

Calculate the moles of water vapor driven off.

Moles of

\text{H}_2\text{O} = \frac{2.70\text{ g}}{18\text{ g/mol}} = 0.15\text{ mol}

Determining the quantity of lost water allows finding the mole ratio of lost water to salt units.

Relate the moles of lost water to the stoichiometry of partial dehydration to solve for

x

Moles of water lost per mole of salt

= \frac{0.15\text{ mol}}{0.10\text{ mol}} = 1.5\text{ mol}

. Since partial dehydration yields

(x - 0.5)

moles of lost water,

x - 0.5 = 1.5 \implies x = 2

Connecting empirical mole ratios to chemical formula coefficients yields the integer hydration number

x

Anahtar Kavram

Stoichiometric determination of water of crystallization from mass loss during partial dehydration.

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