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Zorluk: OrtaWater of Crystallization, Efflorescence, Deliquescence, and Hygroscopy

A 5.00 g5.00\text{ g} sample of hydrated copper(II) tetraoxosulfate(VI), CuSO4⋅xH2O\text{CuSO}_4 \cdot x\text{H}_2\text{O}, was heated strongly in a crucible until all water of crystallization was driven off, leaving an anhydrous residue of mass 3.20 g3.20\text{ g}. What is the value of xx in the formula of the hydrated salt? [Cu=64,S=32,O=16,H=1][\text{Cu} = 64, \text{S} = 32, \text{O} = 16, \text{H} = 1]

  1. A
    22
  2. B
    33
  3. 55Cevap
  4. D
    1010

Cevap

The value of xx in the hydrated salt formula is 55.
Heating the hydrated salt drives off all water of crystallization. The mass of water lost is 5.00 g−3.20 g=1.80 g5.00\text{ g} - 3.20\text{ g} = 1.80\text{ g}. Converting both the anhydrous salt (3.20 g3.20\text{ g}) and water (1.80 g1.80\text{ g}) to moles using their respective molar masses (160 g/mol160\text{ g/mol} and 18 g/mol18\text{ g/mol}) yields 0.020 mol0.020\text{ mol} of CuSO4\text{CuSO}_4 and 0.100 mol0.100\text{ mol} of H2O\text{H}_2\text{O}. The mole ratio 0.100/0.0200.100 / 0.020 simplifies to 55, making 55 the correct value for xx.

Adım Adım Çözüm

1
Calculate the mass of water of crystallization driven off.
Mass of H2O=5.00 g−3.20 g=1.80 g\text{Mass of H}_2\text{O} = 5.00\text{ g} - 3.20\text{ g} = 1.80\text{ g}
The loss in mass upon heating equals the mass of water lost from the hydrated salt.
2
Determine the molar masses of anhydrous CuSO4\text{CuSO}_4 and H2O\text{H}_2\text{O}.
M(CuSO4)=64+32+(4×16)=160 g/molM(\text{CuSO}_4) = 64 + 32 + (4 \times 16) = 160\text{ g/mol} and M(H2O)=(2×1)+16=18 g/molM(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18\text{ g/mol}
Molar masses are required to convert the masses of salt and water into mole amounts.
3
Calculate the number of moles of anhydrous CuSO4\text{CuSO}_4 and H2O\text{H}_2\text{O}.
n(CuSO4)=3.20160=0.020 moln(\text{CuSO}_4) = \frac{3.20}{160} = 0.020\text{ mol} and n(H2O)=1.8018=0.100 moln(\text{H}_2\text{O}) = \frac{1.80}{18} = 0.100\text{ mol}
Moles are obtained by dividing mass by molar mass (n=m/Mn = m / M).
4
Determine the stoichiometric ratio x=n(H2O)n(CuSO4)x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)}.
x=0.100 mol0.020 mol=5x = \frac{0.100\text{ mol}}{0.020\text{ mol}} = 5
The mole ratio gives the number of water molecules of crystallization bound per mole of anhydrous salt.

Anahtar Kavram

Determination of Water of Crystallization by Gravimetric Analysis
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Konular

Water of Crystallization, Efflorescence, Deliquescence, and Hygroscopy