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

1 Answer
Sep 15, 2017

Answer:

#x = 8#

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 #x# moles of water of hydration for every #1# mole of hydrate. Keep this in mind.

You know that the percent composition of aluminium in this unknown aluminium sulfate hydrate is #11.11%#, which means that for every #"100 g"# of hydrate you get #"11.11 g"# of aluminium.

If you pick a #"100-g"# sample of this hydrate, you can calculate the number of moles of aluminium it contains by using the molar mass of this element.

#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 #2# moles of aluminium.

#"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 #1.640# moles of water for #0.2059# moles of anhydrous aluminium sulfate, you can say that #1# mole of aluminium sulfate will 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 #1# mole of anhydrous aluminium sulfate you get #8# moles of water, so

#x = 8#

and the unknown hydrate is aluminium sulfate octahydrate, #"Al"_2("SO"_4)_3 * 8"H"_2"O"#.