Calculate the mass % water in beryllium phosphite heptahydrate

Calculate the mass % water in beryllium phosphite heptahydrate.

The correct answer and explanation is:

To calculate the mass percent of water in beryllium phosphite heptahydrate (BeHPO₃·7H₂O), follow these steps:

Step 1: Determine the molar mass of the compound.

First, calculate the molar masses of the components:

  1. Beryllium phosphite (BeHPO₃):
    • Be = 9.012 g/mol
    • H = 1.008 g/mol (x1 for 1 hydrogen atom)
    • P = 30.974 g/mol
    • O = 16.00 g/mol (x3 for 3 oxygen atoms)
    So, the molar mass of BeHPO₃ = 9.012 + 1.008 + 30.974 + (3 × 16.00) = 9.012 + 1.008 + 30.974 + 48.00 = 89.994 g/mol.
  2. Water molecules (H₂O):
    • H = 1.008 g/mol (x2 for 2 hydrogen atoms)
    • O = 16.00 g/mol
    So, the molar mass of H₂O = (2 × 1.008) + 16.00 = 18.016 g/mol.
  3. Water in beryllium phosphite heptahydrate (BeHPO₃·7H₂O):
    There are 7 water molecules, so:
    • Molar mass of water in heptahydrate = 7 × 18.016 = 126.112 g/mol.

Step 2: Calculate the total molar mass of BeHPO₃·7H₂O.

The total molar mass of beryllium phosphite heptahydrate is the sum of the molar masses of BeHPO₃ and the water molecules:

Total molar mass = 89.994 g/mol (BeHPO₃) + 126.112 g/mol (7H₂O) = 216.106 g/mol.

Step 3: Calculate the mass percent of water.

The mass percent of water is calculated as: Mass percent of water=(Mass of waterTotal mass)×100\text{Mass percent of water} = \left( \frac{\text{Mass of water}}{\text{Total mass}} \right) \times 100 Mass percent of water=(126.112216.106)×100≈58.4%\text{Mass percent of water} = \left( \frac{126.112}{216.106} \right) \times 100 \approx 58.4\%

Explanation:

In this compound, beryllium phosphite heptahydrate contains seven molecules of water for every formula unit of BeHPO₃. By calculating the molar masses of both the anhydrous part (BeHPO₃) and the water molecules, and determining the ratio of the mass of water to the total mass, we find that approximately 58.4% of the mass of beryllium phosphite heptahydrate is due to water. This calculation is crucial in understanding how much water is present in the hydrated form compared to the anhydrous form.

Scroll to Top