An ionic hydrate has the formula Pb3(PO4)4 • xH2O. The total mass of the hydrated ionic compound is 18.3g. After analysis, it was found that the anhydrous compound had a mass of 17.36g. What is the complete formula for this hydrated compound?
It would be great if someone could answer this question as I am not sure what steps to follow, thank you!
To determine the formula of the hydrated compound, we need to find the value of 'x' in the formula Pb3(PO4)4 • xH2O.
First, we find the mass of water in the hydrated compound by subtracting the mass of the anhydrous compound from the total mass of the hydrated compound:
Mass of water = Total mass of hydrated compound - Mass of anhydrous compound
Mass of water = 18.3g - 17.36g
Mass of water = 0.94g
Next, we need to calculate the moles of water using the molar mass of water, which is 18.02 g/mol:
Moles of water = Mass of water / Molar mass of water
Moles of water = 0.94g / 18.02 g/mol
Moles of water = 0.052mol
Since the formula of water is H2O, we know that the number of moles of water is equal to the number of water molecules (x) in the formula. Therefore, x = 0.052.
Now we can write the complete formula for the hydrated compound:
Pb3(PO4)4 • 0.052H2O
Pb3(PO4)4 • xH2O. The total mass of the hydrated ionic compound is 18.3g. After analysis, it was found that the anhydrous compound had a mass of 17.36g.
% Pb3(PO4)4 = 100(17.36/18.3) = 94.86
mass H2O in the 18.3 g sample is 18.3-17.36 = 0.94 g.
% H2O = 100(0.94/18.3) = 5.14%
Take 100 g sample: You have 94.86 g Pb3(PO4)4 and 5.14 g H2O.
molar mass H2O = 18. molar mass Pb3(PO4)4 = 1001.48
mols Pb3(PO4)4 = 94.86/1001.48 = 0.0947
mols H2O = 5.14/18 = 0.286
To find the ratio you want small whole numbers. The easy way to get this to divide everything by the smallest; i.e..
Pb3(PO4)4 = 0.0947/0.0947 = 1.00
H2O = 0.286/0.0947 = 3.02 which rounds to 3.0 so x in the formula is 3.0 and the formula is Pb3(PO4)4*3H2O. I use the * because I can't get the . in the middle.