theoretical percent sulfate in AlK(SO4)2*H2O
(2xmolarmass SO4/molar mass AlK(SO4)2.H2O)*100.
Let me point out that alum is AlK(SO4)2.12H2O. I don't know if your problem is real or a typo. Just look at it to see.
so it would be like 96.06g/258.33g?
no it would be 192.14/258.33?
The 192.14 is ok but I think you omitted the water in the 258. I think that is supposed to be something like 276 or so. Check it out.
To determine the theoretical percent sulfate in AlK(SO4)2•H2O, we need to calculate the molar mass of sulfate (SO4).
The molar mass of sulfur (S) is 32.06 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Since there are four oxygen atoms in sulfate, the total molar mass of sulfate is:
(32.06 g/mol) + (4 * 16.00 g/mol) = 96.06 g/mol.
Next, we need to calculate the molar mass of AlK(SO4)2•H2O. The molar mass of aluminum (Al) is 26.98 g/mol, the molar mass of potassium (K) is 39.10 g/mol, the molar mass of water (H2O) is 18.02 g/mol, and the molar mass of sulfate is 96.06 g/mol.
So, the molar mass of AlK(SO4)2•H2O is:
(26.98 g/mol) + (39.10 g/mol) + (2 * 96.06 g/mol) + (18.02 g/mol) = 276.22 g/mol.
Now, let's find the molar mass of sulfate in AlK(SO4)2•H2O:
(2 * 96.06 g/mol) = 192.12 g/mol.
Finally, we can determine the theoretical percent sulfate in AlK(SO4)2•H2O by dividing the molar mass of sulfate by the molar mass of AlK(SO4)2•H2O and multiplying by 100:
(192.12 g/mol) / (276.22 g/mol) * 100 = 69.59%
Therefore, the theoretical percent sulfate in AlK(SO4)2•H2O is 69.59%.
Find the total molecular mass of it
one Al, one K, two S, eight O, and one H2O
divide that sum (the mole mass) into the mass of two SO4 ions.