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Difficulty: HardWater of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

A 10.0 g10.0\text{ g} sample of hydrated magnesium tetraoxosulfate(VI), MgSO4⋅xH2O\text{MgSO}_4 \cdot x\text{H}_2\text{O}, is heated to constant mass in a crucible, leaving a residue of 4.88 g4.88\text{ g} of anhydrous salt. When the anhydrous residue is exposed to moist ambient air at room temperature, it absorbs water vapor until its mass increases back to 10.0 g10.0\text{ g} without forming a liquid solution. What is the integer value of xx, and what atmospheric behavior does the anhydrous salt display during moisture absorption? [Relative atomic masses: Mg=24,S=32,O=16,H=1][\text{Relative atomic masses: } \text{Mg} = 24, \text{S} = 32, \text{O} = 16, \text{H} = 1]

  1. x=7x = 7, and the salt displays hygroscopyAnswer
  2. B
    x=7x = 7, and the salt displays deliquescence
  3. C
    x=5x = 5, and the salt displays hygroscopy
  4. D
    x=5x = 5, and the salt displays efflorescence

Answer

x=7x = 7, and the salt displays hygroscopy
The mass of water lost from 10.0 g10.0\text{ g} of MgSO4⋅xH2O\text{MgSO}_4 \cdot x\text{H}_2\text{O} is 5.12 g5.12\text{ g}. Converting the mass of anhydrous MgSO4\text{MgSO}_4 (4.88 g4.88\text{ g}) and water (5.12 g5.12\text{ g}) to moles gives 0.04067 mol0.04067\text{ mol} and 0.2844 mol0.2844\text{ mol} respectively, yielding x=7x = 7. Because the anhydrous salt absorbs atmospheric moisture without dissolving into a liquid solution, it demonstrates hygroscopy.

Step-by-Step Solution

1
Calculate the mass of water lost upon heating.
Mass of H2O=10.0 g−4.88 g=5.12 g\text{Mass of } \text{H}_2\text{O} = 10.0\text{ g} - 4.88\text{ g} = 5.12\text{ g}
Heating to constant mass drives off all water of crystallization from the hydrated crystal structure.
2
Determine the molar masses of anhydrous MgSO4\text{MgSO}_4 and H2O\text{H}_2\text{O}.
Molar mass of MgSO4=24+32+(4×16)=120 g/mol\text{MgSO}_4 = 24 + 32 + (4 \times 16) = 120\text{ g/mol}; Molar mass of H2O=(2×1)+16=18 g/mol\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}.
Required to convert sample masses into chemical mole quantities.
3
Calculate the mole ratio of water of crystallization to anhydrous salt to find xx.
Moles of MgSO4=4.88120=0.04067 mol\text{Moles of } \text{MgSO}_4 = \frac{4.88}{120} = 0.04067\text{ mol}; Moles of H2O=5.1218=0.2844 mol\text{Moles of } \text{H}_2\text{O} = \frac{5.12}{18} = 0.2844\text{ mol}; x=0.28440.04067=7x = \frac{0.2844}{0.04067} = 7.
The coefficient xx represents the integer ratio of moles of water to moles of anhydrous salt.
4
Identify the atmospheric behavior of the anhydrous salt upon absorbing moisture without forming a solution.
The behavior is termed hygroscopy (or hygroscopic nature).
Hygroscopic substances absorb moisture from air without forming a liquid solution, whereas deliquescent substances absorb enough moisture to dissolve into a solution.

Key Concept

Stoichiometric determination of water of crystallization and conceptual differentiation between hygroscopy and deliquescence.
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Topics

Water of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

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