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?
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It would be great if someone could answer this question as I am not sure what steps to follow, thank you!
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.
% 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.
3 answers