Examkin

Question

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

A 4.99 g4.99\text{ g} sample of hydrated copper(II) tetraoxosulfate(VI), CuSO4⋅xH2O\text{CuSO}_4 \cdot x\text{H}_2\text{O}, is heated to constant mass, yielding 3.19 g3.19\text{ g} of anhydrous CuSO4\text{CuSO}_4. Separately, a salt whose saturated solution vapor pressure is greater than the partial pressure of water vapor in the surrounding atmosphere loses its water of crystallization to become an anhydrous powder upon exposure to air. [Cu=63.5,S=32,O=16,H=1][\text{Cu} = 63.5, \text{S} = 32, \text{O} = 16, \text{H} = 1]. What is the value of xx in the hydrated salt, and what term describes the atmospheric behavior of the second salt?

  1. x=5x = 5, and the second salt undergoes efflorescenceAnswer
  2. B
    x=5x = 5, and the second salt undergoes deliquescence
  3. C
    x=7x = 7, and the second salt undergoes efflorescence
  4. D
    x=7x = 7, and the second salt undergoes hygroscopy

Answer

The value of xx is 5, and the atmospheric behavior described is efflorescence.
Calculating the moles of anhydrous CuSO4\text{CuSO}_4 (3.19 g/159.5 g/mol=0.020 mol3.19\text{ g} / 159.5\text{ g/mol} = 0.020\text{ mol}) and water (1.80 g/18 g/mol=0.100 mol1.80\text{ g} / 18\text{ g/mol} = 0.100\text{ mol}) yields a mole ratio x=0.100/0.020=5x = 0.100 / 0.020 = 5. Additionally, when a hydrate's vapor pressure is higher than the atmospheric partial pressure of water vapor, it loses water of crystallization to the surrounding air, which is defined as efflorescence.

Step-by-Step Solution

1
Calculate the molar masses of anhydrous CuSO4\text{CuSO}_4 and H2O\text{H}_2\text{O}.
Molar mass of CuSO4=63.5+32+(4×16)=159.5 g/mol\text{Molar mass of CuSO}_4 = 63.5 + 32 + (4 \times 16) = 159.5\text{ g/mol}; Molar mass of H2O=(2×1)+16=18 g/mol\text{Molar mass of H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}.
Molar masses are needed to convert the experimental masses into mole quantities.
2
Determine the mass and moles of anhydrous CuSO4\text{CuSO}_4 and water of crystallization lost.
Mass of CuSO4=3.19 g⇒n(CuSO4)=3.19159.5=0.020 mol\text{Mass of CuSO}_4 = 3.19\text{ g} \Rightarrow n(\text{CuSO}_4) = \frac{3.19}{159.5} = 0.020\text{ mol}. Mass of H2O=4.99 g−3.19 g=1.80 g⇒n(H2O)=1.8018=0.100 mol\text{Mass of H}_2\text{O} = 4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g} \Rightarrow n(\text{H}_2\text{O}) = \frac{1.80}{18} = 0.100\text{ mol}.
Subtracting the anhydrous mass from the total hydrated mass gives the mass of lost water.
3
Calculate the stoichiometric integer ratio xx.
x=n(H2O)n(CuSO4)=0.1000.020=5x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)} = \frac{0.100}{0.020} = 5.
The value of xx is the mole ratio of water to anhydrous salt.
4
Identify the atmospheric phenomenon based on vapor pressure conditions.
The phenomenon is efflorescence.
When the vapor pressure of a hydrated salt's water of crystallization exceeds the atmospheric vapor pressure, the salt spontaneously loses water to the atmosphere, becoming anhydrous or lower-hydrated powder.

Key Concept

Water of crystallization stoichiometry and vapor pressure behavior in efflorescence vs deliquescence
Rate this question

Topics

Water of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

More questions from this topic