Question #3ef23

As you know, hydrates are salts that have water of hydration associated with their crystal structure.

This means that you need to take into account the added water molecules when trying to figure out how many moles of #"CuSO"""_4# you have in solution.

You can think about hydrates as being impure samples. In this case, each mole of copper(II) sulfate, #"CuSO"""_4#, comes attached to five moles of water.

Let's say that your reaction requires the presene of 0.5 moles of #"CuSO""_4#. For anhydrous copper(II) sulfate, this is equivalent to a mass of

#0.5color(red)(cancel(color(black)("moles"))) * "159.6 g"/(1color(red)(cancel(color(black)("mole")))) = "79.8 g"#

The molar mass of the pentahydrate is #"249.685 g/mol"#. If you use 78.9 g of pentahydrate, you only get

#78.9color(red)(cancel(color(black)("g"))) * ("1 mole CuSO"""_4 * 5"H"_2"O")/(249.7color(red)(cancel(color(black)("g")))) = "0.320 moles CuSO"""_4 * 5"H"_2"O"#

Since each mole of pentahydrate contains one mole of copper(II) sulfate, adding this much pentahydrate will be equivalent to adding 3.20 moles of #"CuSO"""_4#.

In this example, to get 79.8 g of #"CuSO"""_4#, you need to use the percent composition of the hydrate.

#(159.6color(red)(cancel(color(black)("g/mole"))))/(249.7color(red)(cancel(color(black)("g/mole")))) * 100 = 63.9%#

This means that you get 63.9 g of #"CuSO"""_4# for every 100 g of #"CuSO"""_4 * 5"H"_2"O"#. This means that you have

#79.8color(red)(cancel(color(black)("g CuSO"""_4))) * ("100 CuSO"""_4 * 5"H"_2"O")/(63.9color(red)(cancel(color(black)("g CuSO"""_4)))) = "124.9 g CuSO"""_4 * 5"H"_2"O"#

So yes, you need to take into account the water of hydration when doing stoichiometric calculations.

If a problem tells you that 100 g of copper(II) sulfate pentahydrate are dissolved in solution and made to react with a certain mass of sodium carbonate, you need to take into account the water of hydration when you calculate the number of moles of #"CuSO"""_4#.

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