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:
Hydrates |
Salts |
BaCl_{2} ● 2 H_{2}O |
BaCl_{2} |
CuSO_{4} ● 5 H_{2}O |
CuSO_{4} |
MnSO_{4} ● 1 H_{2}O |
MnSO_{4} |
Na_{2}CO_{3} ● 10 H_{2}O |
Na_{2}CO_{3} |
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 I’m about to give you:
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 CuSO_{4 }● X H_{2}O
The hydrate formula in my example is showing a ratio between the CuSO_{4} (the salt) and the H_{2}O. 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 let’s 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:
Find the grams of the hydrate ( BaCl_{2} ● X H_{2}O )
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 (BaCl_{2} )
We are given the mass after heating (29.87 g) which will be the mass of the crucible + salt ( BaCl_{2} )
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) BaCl_{2} ● X H_{2}O
- 8.57 g (salt) BaCl_{2}
-----------
1.48 g (water)
H_{2}O
SECOND - Find the moles of salt and water
Now we will use the molar masses of the salt ( BaCl_{2} ) and water ( H_{2}O ) to convert the grams of each to moles:
1.48 g H_{2}O 1mol H_{2}O
-------------- x ---------------- = 0.0822 mol H_{2}O
1 18.0 g H_{2}O
8.57 g BaCl_{2} 1 mol BaCl_{2}
----------------
x ------------------- = 0.0411
mol BaCl_{2}
1 208.3 g BaCl_{2}
FINALLY - Divide the moles of water by the moles of salt and you have X
0.0822 mol H_{2}O 2
---------------------- = ------ So 2 = X and the formula of this hydrate is
0.0411 mol BaCl_{2} 1 BaCl_{2}
● 2 H_{2}O
barium
chloride dihydrate