I had a 1.094g of mystery Alum. "AB(SO4)c. dH2O" (I assume that that the Alum is KAl(SO4)2-12H2O but we don't know A B c or d, we do know that mass percent of water is 51.9)
After processing the alum (Ba(NO3)2 and HNO3 through a gooch crucible) we end up with 1.739g of BaSO4.
We know that SO4 is 42.15% of BASO4 thus
SO4 has a mass of .7330 g
assuming we captured all of the SO4 of the original Alum.
so if we subtract .7330g from 1.094g = .361 g
therefor .361g is the the AB and the H2O of the Alum
but if water is 51 of the mass of the Alum. I did something wrong any ideas.
You're right. Something is wrong but it must have been in the experimental details. Your math is not quite right but it won't make that much difference.
I think the 42.15% SO4 in BaSO4 is not quite right. I think that should be (96/233.39)*100 = 41.13%. That won't make that much difference. Mass SO4 = 1.739 x (96/233.39) = 0.7153g
I don't think the H2O calculation is correct either. 1.094 x 0.519 = 0.5678 and 1.094-0.5678 = 0.5262g for AB(SO4)c.
So you are faced with the situation of sulfate weighing more than the anhydrous salt.
To find the correct values for A, B, c, and d in the formula AB(SO4)c*dH2O, we need to use the given information and perform some calculations.
First, let's calculate the mass of water in the original Alum. We know that the mass percent of water is 51.9%. So, we can calculate the mass of water as follows:
Mass of water = 51.9% * 1.094g = 0.567786g ≈ 0.568g
Now, let's calculate the mass of AB in the original Alum. Subtracting the mass of water from the total mass of the original Alum gives us:
Mass of AB = Mass of original Alum - Mass of water = 1.094g - 0.568g = 0.526g
Next, we need to calculate the molar mass of AB(SO4)c*dH2O. We can determine the molar mass of each component separately and then add them together. Let's assume the molar mass of AB is x g/mol:
Molar mass of AB(SO4)c*dH2O = x + (32.06 * 4) + (16 * c) + (1.01 * 2 * d)
We know that the molar mass of water (H2O) is 18.02 g/mol, so we can substitute this value into the equation:
x + (32.06 * 4) + (16 * c) + (1.01 * 2 * d) = x + 128.24 + 16c + 2.02d
Since we don't know the specific values for A, B, c, and d, we cannot determine the exact molar mass at this point. However, we can use the information you provided about the mass of BaSO4 formed to make further calculations.
You mentioned that 1.739g of BaSO4 is formed from the reaction between the Alum and Ba(NO3)2 and HNO3. From this, we can calculate the moles of SO4 present in BaSO4:
Moles of SO4 = Mass of BaSO4 / Molar mass of BaSO4
Moles of SO4 = 1.739g / 233.39 g/mol (assuming the molar mass of BaSO4) = 0.007455 mol
Since the original Alum gave us all the SO4, we can now calculate the moles of AB:
Moles of AB = Moles of SO4
Moles of AB = 0.007455 mol
Now, we can calculate the molar mass of AB (the unknown AB compound) by dividing the mass of AB by the moles of AB:
Molar mass of AB = Mass of AB / Moles of AB
Molar mass of AB = 0.526g / 0.007455 mol
We obtain the molar mass of AB.
By performing these calculations, you should be able to determine the values for A, B, c, and d in the formula AB(SO4)c*dH2O.