A hydrate containing aluminium sulphate has the formula #"Al"_2 ("SO"_4)_3 * x"H"_2 "O"# and it contains 11.11% of aluminium by mass. Calculate the value of #x# in the hydrate formula. ?
If possible, could you type out your thought process that went into solving it as well. Thank you
If possible, could you type out your thought process that went into solving it as well. Thank you
1 Answer
Explanation:
Your goal here is to figure out exactly how many moles of water of hydration are present for every mole of hydrate.
As you can see, the chemical formula of the hydrate tells you that you get
You know that the percent composition of aluminium in this unknown aluminium sulfate hydrate is
If you pick a
#11.11 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.4118 moles Al"#
Now, the chemical formula of the anhydrous salt tells you that every mole of aluminium sulfate contains
#"Al"_ color(red)(2)("SO"_4)_3 -> "1 mole Al"_color(red)(2)("SO"_4)_3color(white)(.)"contains"color(white)(.)color(red)(2)color(white)(.)"moles Al"#
This means that the sample contains
#0.4118 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Al"_2("SO"_4)_3)/(2color(red)(cancel(color(black)("moles Al")))) = "0.2059 moles Al"_2("SO"_4)_3#
You can determine the mass of the anhydrous salt present in the hydrate by using the molar mass of aluminium sulfate
#0.2059 color(red)(cancel(color(black)("moles Al"_2("SO"_4)_3))) * "342. 15 g"/(1color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = "70.45 g"#
This implies that the sample contains
#overbrace("100 g")^(color(blue)("mass of hydrate")) - overbrace("70.45 g")^(color(blue)("mass of Al"_2("SO"_4)_3)) = overbrace("29.55 g")^(color(blue)("mass of water"))#
Convert this to moles by using the molar mass of water
#29.55 color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("moles H"_2"O")))) = "1.640 moles H"_2"O"#
So, if you know that you get
#1 color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3))) * ("1.640 moles H"_2"O")/(0.2059color(red)(cancel(color(black)("mole Al"_2("SO"_4)_3)))) = 7.965 ~~ "8 moles H"_2"O"#
Therefore, you can say that for every
#x = 8#
and the unknown hydrate is aluminium sulfate octahydrate,