Calculate the percentage of water in each of the following hydrates.
Aluminum bromide hexahydrate
I just need help setting this up. I can work it from there
Would it be:
AlBr * 6H2O
Did I write it right?
Al = 26.981
3Br = 239.727
12H = 12.096
6O = 95.994
Total = 374.798
Total H2O = 108.090
108.090/374.798 x 100% = 28.840%
Correct? are my sig figs messed up in anyway?
AlBr3 * 6H20
we will be happy to critique your thinking or work.
It is fine. The sig figures are ok, IF those figs in the individual elements are correct.
Oh dear, it seems like you may have forgotten to use subscripts! The correct formula for aluminum bromide hexahydrate is actually AlBr3 · 6H2O. Remember, the subscript outside the parentheses applies to everything inside. Now you're all set to calculate the percentage of water in this hydrate. Go ahead and solve it from there!
To calculate the percentage of water in aluminum bromide hexahydrate, you need to determine the molar mass of water and the molar mass of the entire compound.
First, let's determine the molar mass of water (H2O). The molar mass of hydrogen (H) is approximately 1.01 g/mol, and oxygen (O) has a molar mass of about 16.00 g/mol. Therefore, the molar mass of water is:
2(1.01 g/mol H) + 1(16.00 g/mol O) = 18.02 g/mol
Now let's calculate the molar mass of the entire compound, aluminum bromide hexahydrate (AlBr3·6H2O).
The molar mass of aluminum (Al) is approximately 26.98 g/mol, bromine (Br) has a molar mass of about 79.90 g/mol, and we already calculated that water (H2O) has a molar mass of 18.02 g/mol.
Therefore, the molar mass of aluminum bromide hexahydrate is:
1(26.98 g/mol Al) + 3(79.90 g/mol Br) + 6(18.02 g/mol H2O) = total molar mass
Now that you have the molar mass of water and the molar mass of the entire compound, you can calculate the percentage of water in aluminum bromide hexahydrate by using the formula:
(6 * molar mass of water) / molar mass of compound x 100%