Examkin

Question

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

A 14.3 g14.3\text{ g} sample of hydrated sodium trioxocarbonate(IV), Na2CO3⋅xH2O\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}, was heated strongly in a crucible to constant mass. After complete heating, the mass of the remaining anhydrous residue was found to be 5.3 g5.3\text{ g}. Given the relative atomic masses (Na=23,C=12,O=16,H=1)(\text{Na} = 23, \text{C} = 12, \text{O} = 16, \text{H} = 1), what is the value of xx in the formula of the hydrated salt?

  1. 1010Answer
  2. B
    77
  3. C
    55
  4. D
    11

Answer

The value of xx is 1010, giving the formula Na2CO3⋅10H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}.
Heating to constant mass completely removes the water of crystallization (9.0 g9.0\text{ g} of H2O\text{H}_2\text{O}, equal to 0.50 mol0.50\text{ mol}). The remaining 5.3 g5.3\text{ g} of anhydrous Na2CO3\text{Na}_2\text{CO}_3 equals 0.05 mol0.05\text{ mol}. Dividing 0.50 mol0.50\text{ mol} by 0.05 mol0.05\text{ mol} yields a mole ratio of 1010, which means x=10x = 10.

Step-by-Step Solution

1
Calculate the mass of water lost during heating
Mass of H2O=14.3 g−5.3 g=9.0 g\text{Mass of H}_2\text{O} = 14.3\text{ g} - 5.3\text{ g} = 9.0\text{ g}
The loss in mass upon heating to constant mass corresponds directly to the driven-off water of crystallization.
2
Determine the molar masses of anhydrous Na2CO3\text{Na}_2\text{CO}_3 and H2O\text{H}_2\text{O}
Molar mass of Na2CO3=(2×23)+12+(3×16)=106 g/mol\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\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 required to convert the given masses into mole quantities.
3
Calculate the number of moles of anhydrous salt and water
Moles of Na2CO3=5.3 g106 g/mol=0.05 mol\text{Moles of Na}_2\text{CO}_3 = \frac{5.3\text{ g}}{106\text{ g/mol}} = 0.05\text{ mol}; Moles of H2O=9.0 g18 g/mol=0.50 mol\text{Moles of H}_2\text{O} = \frac{9.0\text{ g}}{18\text{ g/mol}} = 0.50\text{ mol}
Moles equal mass divided by molar mass.
4
Calculate the mole ratio to find xx
x=Moles of H2OMoles of Na2CO3=0.500.05=10x = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of Na}_2\text{CO}_3} = \frac{0.50}{0.05} = 10
The coefficient xx represents the stoichiometric ratio of moles of water of crystallization per mole of anhydrous salt.

Key Concept

Determination of Water of Crystallization in Hydrated Salts
Rate this question

Topics

Water of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

More questions from this topic