A 16.1\text{ g} sample of hydrated sodium tetraoxosulfate(VI), \text{Na}_2\text{SO}_4 \cdo… — JAMB UTME

A $16.1\text{ g}$ sample of hydrated sodium tetraoxosulfate(VI), $\text{Na}_2\text{SO}_4 \cdot x\text{H}_2\text{O}$ , is heated in a crucible until all the water of crystallization is driven off. The mass of the remaining anhydrous salt is $7.1\text{ g}$ . Calculate the value of $x$ . [Relative atomic masses: $\text{Na} = 23$ , $\text{S} = 32$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Answer: 10

Answer

The integer value of

x

is 10.

Heating

16.1\text{ g}

of hydrated sodium tetraoxosulfate(VI) yields

7.1\text{ g}

of anhydrous

\text{Na}_2\text{SO}_4

(

0.05\text{ mol}

) and releases

9.0\text{ g}

of water (

0.5\text{ mol}

). The mole ratio of

\text{H}_2\text{O}

\text{Na}_2\text{SO}_4

0.5 / 0.05 = 10

, giving

x = 10

Step-by-Step Solution

Calculate the mass of water lost upon heating

Mass of

\text{H}_2\text{O} = 16.1\text{ g} - 7.1\text{ g} = 9.0\text{ g}

The difference between the initial hydrated mass and final anhydrous mass represents the driven-off water of crystallization.

Calculate molar masses of anhydrous salt and water

Molar mass of

\text{Na}_2\text{SO}_4 = 142\text{ g/mol}

, Molar mass of

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

Required to convert sample masses to mole quantities.

Calculate moles of anhydrous salt and water, and find the mole ratio

Moles of

\text{Na}_2\text{SO}_4 = 0.05\text{ mol}

, Moles of

\text{H}_2\text{O} = 0.5\text{ mol}

x = \frac{0.5}{0.05} = 10

The subscript

x

gives the ratio of moles of water to moles of anhydrous salt per mole of compound.

Key Concept

Stoichiometric determination of water of crystallization in hydrated salts

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