First, we need to determine the formula weight of the hydrate.
Formula weight of Ba(ClO4)2 = 2(Ba) + 2(ClO4) = 2(137.33 g/mol) + 2(99.45 g/mol) = 520.52 g/mol
Next, we can use the given percentage of barium by mass to calculate the mass of barium in one mole of the hydrate.
Mass of Ba in 1 mole of hydrate = 0.352 × formula weight of hydrate
= 0.352 × 520.52 g/mol = 183.45 g/mol
Now, we need to find the mass of water in the hydrate.
Mass of water in 1 mole of hydrate = Formula weight of hydrate − Mass of Ba in 1 mole of hydrate
= 520.52 g/mol − 183.45 g/mol = 337.07 g/mol
Finally, we can use the molar mass of water (18.02 g/mol) to find the value of x.
x = Mass of water in 1 mole of hydrate ÷ Molar mass of water
= 337.07 g/mol ÷ 18.02 g/mol ≈ 18.7
Therefore, the hydrate is Ba(ClO4)2 • 18.7H2O.
A hydrate of barium perchlorate, Ba(ClO4)2 • xH2O, contains 35.2% of barium by mass. Find the value of x.
1 answer