A 13.90\text{ g} sample of hydrated iron(II) tetraoxosulfate(VI), \text{FeSO}_4 \cdot x\te… — JAMB UTME

A $13.90\text{ g}$ sample of hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ , was heated strongly until all water of crystallization was driven off. The mass of the remaining anhydrous salt was $7.60\text{ g}$ . What is the value of $x$ ? $[\text{Fe} = 56, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Answer: 7

Answer

The value of

x

in the hydrated salt formula

\text{FeSO}_4 \cdot x\text{H}_2\text{O}

is 7.

Heating the sample drives off all water of crystallization, leaving only anhydrous

\text{FeSO}_4

. The mass of water is

13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}

. Converting both components to moles gives

0.05\text{ mol}

\text{FeSO}_4

and

0.35\text{ mol}

\text{H}_2\text{O}

. The ratio

\frac{0.35}{0.05} = 7

, yielding

x = 7

Step-by-Step Solution

Find the mass of water lost

Mass of

\text{H}_2\text{O} = 13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}

The loss in mass upon heating represents the driven-off water of crystallization.

Calculate molar masses of anhydrous salt and water

Molar mass of

\text{FeSO}_4 = 152\text{ g/mol}

, Molar mass of

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

Molar masses are required to convert the measured masses into mole quantities.

Calculate the amount in moles of both components

Moles of

\text{FeSO}_4 = \frac{7.60}{152} = 0.05\text{ mol}

; Moles of

\text{H}_2\text{O} = \frac{6.30}{18} = 0.35\text{ mol}

Stoichiometric coefficient

x

represents the mole ratio between water and anhydrous salt.

Determine the mole ratio

x = \frac{0.35\text{ mol}}{0.05\text{ mol}} = 7

Dividing the moles of water of crystallization by the moles of anhydrous salt yields the integer coefficient

x

Key Concept

Determining the formula of a hydrated salt from gravimetric data

Estimated Time:1m 30s

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