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Zorluk: Çok zorPercentage Composition and Percentage Purity Calculations

A 5.00 g5.00\text{ g} sample of an impure hydrated calcium tetraoxosulfate(VI) salt, CaSO4⋅xH2O\text{CaSO}_4 \cdot x\text{H}_2\text{O}, containing 14%14\% non-volatile, inert impurities by mass, is heated strongly to constant mass to remove all water of crystallization. If the mass of the remaining dry residue is 4.10 g4.10\text{ g}, what is the value of xx? (Relative atomic masses: Ca=40\text{Ca} = 40, S=32\text{S} = 32, O=16\text{O} = 16, H=1\text{H} = 1)

  1. A
    1
  2. 2Cevap
  3. C
    3
  4. D
    5

Cevap

The value of xx is 2.
The original 5.00 g5.00\text{ g} sample contains 14%14\% non-volatile impurity (0.70 g0.70\text{ g}). Upon heating, only water evaporates, leaving 4.10 g4.10\text{ g} of residue. Subtracting the 0.70 g0.70\text{ g} of impurity gives 3.40 g3.40\text{ g} of pure anhydrous CaSO4\text{CaSO}_4 (0.025 mol0.025\text{ mol}). The mass of evaporated water is 5.00 g−4.10 g=0.90 g5.00\text{ g} - 4.10\text{ g} = 0.90\text{ g} (0.050 mol0.050\text{ mol}). Dividing moles of water by moles of anhydrous salt yields x=2x = 2.

Adım Adım Çözüm

1
Calculate the mass of the inert, non-volatile impurity in the original sample.
Mass of impurity=14%×5.00 g=0.70 g\text{Mass of impurity} = 14\% \times 5.00\text{ g} = 0.70\text{ g}.
The sample consists of pure hydrated salt and 14%14\% inert non-volatile impurity.
2
Determine the mass of pure anhydrous CaSO4\text{CaSO}_4 in the dry residue.
Mass of pure CaSO4=4.10 g−0.70 g=3.40 g\text{Mass of pure } \text{CaSO}_4 = 4.10\text{ g} - 0.70\text{ g} = 3.40\text{ g}.
Since the impurity is non-volatile, it remains in the 4.10 g4.10\text{ g} dry residue along with the anhydrous salt.
3
Calculate the mass of water of crystallization driven off.
Mass of H2O=5.00 g−4.10 g=0.90 g\text{Mass of } \text{H}_2\text{O} = 5.00\text{ g} - 4.10\text{ g} = 0.90\text{ g}.
Heating to constant mass removes only the volatile water of crystallization.
4
Find the molar amounts of pure CaSO4\text{CaSO}_4 and H2O\text{H}_2\text{O}.
Moles of CaSO4=3.40 g136 g/mol=0.025 mol\text{Moles of } \text{CaSO}_4 = \frac{3.40\text{ g}}{136\text{ g/mol}} = 0.025\text{ mol}; Moles of H2O=0.90 g18 g/mol=0.050 mol\text{Moles of } \text{H}_2\text{O} = \frac{0.90\text{ g}}{18\text{ g/mol}} = 0.050\text{ mol}.
Molar masses are 136 g/mol136\text{ g/mol} for CaSO4\text{CaSO}_4 and 18 g/mol18\text{ g/mol} for H2O\text{H}_2\text{O}.
5
Determine the mole ratio x=n(H2O)n(CaSO4)x = \frac{n(\text{H}_2\text{O})}{n(\text{CaSO}_4)}.
x=0.0500.025=2x = \frac{0.050}{0.025} = 2.
The formula ratio requires the relative mole ratio of water to anhydrous salt.

Anahtar Kavram

Accounting for non-volatile impurities in hydrated salt analysis to determine water of crystallization.
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Percentage Composition and Percentage Purity Calculations

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