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Difficulty: Very hardWater of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

A 6.95 g6.95\text{ g} sample of hydrated iron(II) tetraoxosulfate(VI), FeSO4⋅xH2O\text{FeSO}_4 \cdot x\text{H}_2\text{O}, is dissolved in dilute tetraoxosulfate(VI) acid and made up to 250 cm3250\text{ cm}^3 in a volumetric flask. A 25.0 cm325.0\text{ cm}^3 portion of this solution requires 25.0 cm325.0\text{ cm}^3 of 0.020 mol dm−30.020\text{ mol dm}^{-3} acidified potassium tetraoxomanganate(VII), KMnO4\text{KMnO}_4, for complete titration. Given the relative atomic masses Fe=56\text{Fe} = 56, S=32\text{S} = 32, O=16\text{O} = 16, and H=1\text{H} = 1, what is the integer value of xx, the number of molecules of water of crystallization per formula unit?

Answer: 7

Answer

The integer value of x is 7.
By using the titration volume and concentration of acidified potassium tetraoxomanganate(VII), the amount of iron(II) ions in the sample is calculated. Knowing the total mole amount of anhydrous iron(II) tetraoxosulfate(VI) allows determination of the mass of the anhydrous salt component (3.80 g). Subtracting this from the initial hydrated sample mass (6.95 g) gives the mass of water of crystallization (3.15 g). Dividing the moles of water (0.175 mol) by the moles of anhydrous salt (0.025 mol) yields exactly 7 water molecules of crystallization per formula unit.

Step-by-Step Solution

1
Calculate the moles of KMnO4\text{KMnO}_4 consumed in the titration
n(KMnO4)=0.020 mol dm−3×0.0250 dm3=0.00050 moln(\text{KMnO}_4) = 0.020 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.00050 \text{ mol}
Concentration and volume of titrant are provided.
2
Determine moles of Fe2+\text{Fe}^{2+} present in the 25.0 cm325.0\text{ cm}^3 aliquot using the redox reaction stoichiometry
n(Fe2+)25cm3=5×0.00050 mol=0.0025 moln(\text{Fe}^{2+})_{25\text{cm}^3} = 5 \times 0.00050 \text{ mol} = 0.0025 \text{ mol}
The mole ratio of MnO4−\text{MnO}_4^- to Fe2+\text{Fe}^{2+} in acidic redox titration is 1:51:5 according to MnO4−+5Fe2++8H+→Mn2++5Fe3++4H2O\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}.
3
Calculate the total moles of FeSO4\text{FeSO}_4 in the original 250 cm3250\text{ cm}^3 volumetric flask
n(FeSO4)total=0.0025 mol×(250 cm325.0 cm3)=0.025 moln(\text{FeSO}_4)_{\text{total}} = 0.0025 \text{ mol} \times \left(\frac{250\text{ cm}^3}{25.0\text{ cm}^3}\right) = 0.025 \text{ mol}
The aliquot represents one-tenth of the total solution volume.
4
Calculate the mass of anhydrous FeSO4\text{FeSO}_4 in the sample
Molar mass of FeSO4=56+32+(4×16)=152 g mol−1\text{FeSO}_4 = 56 + 32 + (4 \times 16) = 152 \text{ g mol}^{-1}. Mass =0.025 mol×152 g mol−1=3.80 g= 0.025 \text{ mol} \times 152 \text{ g mol}^{-1} = 3.80 \text{ g}.
Converting moles of anhydrous salt to mass using molar mass.
5
Determine the mass and moles of water of crystallization
Mass of H2O=6.95 g−3.80 g=3.15 g\text{H}_2\text{O} = 6.95 \text{ g} - 3.80 \text{ g} = 3.15 \text{ g}. Moles of H2O=3.15 g18 g mol−1=0.175 mol\text{H}_2\text{O} = \frac{3.15 \text{ g}}{18 \text{ g mol}^{-1}} = 0.175 \text{ mol}.
Subtracting anhydrous mass from initial mass gives water of crystallization mass, converted to moles using molar mass of H2O=18 g mol−1\text{H}_2\text{O} = 18 \text{ g mol}^{-1}.
6
Compute the hydration coefficient x=n(H2O)n(FeSO4)x = \frac{n(\text{H}_2\text{O})}{n(\text{FeSO}_4)}
x=0.175 mol0.025 mol=7x = \frac{0.175 \text{ mol}}{0.025 \text{ mol}} = 7
The coefficient xx represents the mole ratio of water of crystallization to anhydrous salt.

Key Concept

Quantitative determination of water of crystallization in hydrated salts via redox volumetric analysis
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Topics

Water of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

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