you heat 4.31 grams of a nickel 2 phosphate hydrate. You heat to a constant mass of 3.09 grams. What is the formula of this hydrate?
Nickel (II)
4.31-3.09 = 1.22 = mass H2O driven off.
mols H2O = 1.22/molar mass = ?
mols Ni3(PO4)2 = 3.09/molar mass?
Now find the molar ratio of the Ni(II) phosphate to H2O. It look like 10 to me but check that.
To determine the formula of the hydrate, you need to find the stoichiometric ratio between the nickel 2 phosphate and the water molecules in the hydrate. Here's how you can do it:
1. Calculate the mass of water lost:
Mass of water lost = Initial mass - Final mass = 4.31 g - 3.09 g = 1.22 g
2. Convert the mass of water lost to moles:
Moles of water = Mass of water / molar mass of water
The molar mass of water (H₂O) is 18.015 g/mol.
Moles of water = 1.22 g / 18.015 g/mol ≈ 0.068 mol
3. Calculate the moles of the nickel 2 phosphate (Ni₃(PO₄)₂):
The problem states that the constant mass obtained after heating is 3.09 grams. So, the remaining mass is due to the nickel 2 phosphate.
Moles of nickel 2 phosphate = Mass of nickel 2 phosphate / molar mass of nickel 2 phosphate
The molar mass of nickel 2 phosphate (Ni₃(PO₄)₂) can be calculated by summing the atomic masses of its constituents: (58.6934 g/mol x 3) + (30.97376 g/mol x 2) + (15.999 g/mol x 8) = 480.042 g/mol
Moles of nickel 2 phosphate = 3.09 g / 480.042 g/mol ≈ 0.00644 mol
4. Determine the mole ratio between nickel 2 phosphate and water:
Divide the moles of water and the moles of nickel 2 phosphate by the smallest value obtained.
Moles ratio Ni₃(PO₄)₂ : H₂O ≈ (0.00644 mol / 0.00644 mol) : (0.068 mol / 0.00644 mol) ≈ 1 : 10.56 ≈ 1 : 10.5 (rounded to one decimal place)
Therefore, the formula of this hydrate is Ni₃(PO₄)₂ · 10.5 H₂O