Hydrate Tutorial & Example Problem Std 3


First of all look at this list of hydrates and salts and see if you can determine the difference:




BaCl2 ● 2 H2O


CuSO4 ● 5 H2O


MnSO4 ● 1 H2O


Na2CO3 ● 10 H2O


Did you see and understand what the differences are? (see answer at bottom of this page)


Now that you know the difference between a salt and a hydrate, you will understand the plan for solving all hydrate problems Im about to give you:


Solving Hydrate Problems


You will always be given enough information to find the grams of the salt and the grams of water present. From that you will be able to:

find moles of salt

find moles of water

find X in the formula of the hydrate ex CuSO4 X H2O


The hydrate formula in my example is showing a ratio between the CuSO4 (the salt) and the H2O. To find this ratio, we simply set up a fraction like this:


mol water X (x mol water)

------------- = -------------------------

mol salt 1 (1 mol salt)


After doing this calculation we will round X to be a whole number.



Now lets work a homework problem from p. 349, problem 154, which uses data from a hydrate experiment given in Table 11-3. Try working this problem following the steps given above. After working it on your own, check with the detailed solution on the next page.







Answer to difference between salt and hydrate - hydrates have water and salt, salt has no water attached to it.


SOLUTION p. 349 #154:

We will follow the steps given above to solve this problem:

FIRST - Find the grams of the salt and the water

Find the grams of the hydrate ( BaCl2 ● X H2O )

We are given mass of crucible empty (21.30 g) and crucible + hydrate (31.35).

So subtract the empty from the crucible + hydrate to get mass of the hydrate

31.35 g (crucible + hydrate)

- 21.30 g (crucible)


10.05 g = g of hydrate


Find the grams of salt (BaCl2 )

We are given the mass after heating (29.87 g) which will be the mass of the crucible + salt ( BaCl2 )

So subtract the mass of the crucible from the crucible + salt

29.87 g (crucible + salt)

- 21.30 g (crucible)


8.57 g = grams of salt


Now that we know the salt and hydrate masses, the difference will be the mass of the water.

10.05 g (hydrate) BaCl2 ● X H2O

- 8.57 g (salt) BaCl2


1.48 g (water) H2O

SECOND - Find the moles of salt and water

Now we will use the molar masses of the salt ( BaCl2 ) and water ( H2O ) to convert the grams of each to moles:


1.48 g H2O 1mol H2O

-------------- x ---------------- = 0.0822 mol H2O

1 18.0 g H2O


8.57 g BaCl2 1 mol BaCl2

---------------- x ------------------- = 0.0411 mol BaCl2

1      208.3 g BaCl2

FINALLY - Divide the moles of water by the moles of salt and you have X


0.0822 mol H2O 2

---------------------- = ------ So 2 = X and the formula of this hydrate is

0.0411 mol BaCl2 1 BaCl2 ● 2 H2O

barium chloride dihydrate