Consider compound Na₂CO₃⦁10H₂O. This compound is a hydrate.What is the percent of water in this compound?

1 answer

To find the percent of water in the hydrate Na₂CO₃·10H₂O (sodium carbonate decahydrate), we need to calculate the molar masses of both the anhydrous sodium carbonate (Na₂CO₃) and the water (H₂O) in the hydrate.

  1. Calculate the molar mass of Na₂CO₃ (sodium carbonate):

    • Sodium (Na): 22.99 g/mol × 2 = 45.98 g/mol
    • Carbon (C): 12.01 g/mol × 1 = 12.01 g/mol
    • Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol

    Total for Na₂CO₃ = 45.98 + 12.01 + 48.00 = 105.99 g/mol

  2. Calculate the molar mass of water (H₂O):

    • Hydrogen (H): 1.01 g/mol × 2 = 2.02 g/mol
    • Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol

    Total for H₂O = 2.02 + 16.00 = 18.02 g/mol

  3. Calculate the molar mass of the hydrate Na₂CO₃·10H₂O:

    • Since there are 10 water molecules in the hydrate:
    • Total for 10 H₂O = 10 × 18.02 g/mol = 180.20 g/mol

    Therefore, the molar mass of Na₂CO₃·10H₂O = 105.99 g/mol + 180.20 g/mol = 286.19 g/mol

  4. Calculate the percent of water in the hydrate: \[ \text{Percent of water} = \left(\frac{\text{mass of water}}{\text{mass of the hydrate}}\right) \times 100 \] \[ \text{Percent of water} = \left(\frac{180.20}{286.19}\right) \times 100 \approx 62.91% \]

Therefore, the percent of water in Na₂CO₃·10H₂O is approximately 62.91%.