### Hydrate Lab Calculations

Example #1: A hydrate of aluminum chloride, AlCl3 nH2O was used. The following data were collected:

 empty crucible 15.807 g crucible+contentsafter first heating 16.503 g crucible+hydrate 17.061 g crucible+contentsafter second heating 16.499g

Using the above information, answer the following questions:

(a) What was the mass of the anhydrous salt?
(b) How many moles of anhydrous salt is this?
(c) What was the mass of the water driven off?
(d) How many moles of water is this?
(e) Determine the mole ratio of water : anhydrous form.
(f) What is the fomula of this hydrate?

Not sure what a crucible looks like? Here is a photo. It's held by a clay triangle and is being heated. Note the gap between crucible and lid. This is to allow steam to escape.

Solution:

(a) What was the mass of the anhydrous salt?

16.499 g − 15.807 g = 0.692 g

(b) How many moles of anhydrous salt is this?

0.692 g / 133.341 g/mol = 0.0051897 mol (carry some extra digits)

(c) What was the mass of the water driven off?

17.061 g − 16.499 g = 0.562 g

(d) How many moles of water is this?

0.562 g / 18.015 gmol = 0.0311962 mol (more extra digits)

(e) Determine the mole ratio of water : anhydrous form.

0.0311962 mol / 0.0051897 mol

In small whole numbers, this is 6 to 1

(f) What is the fomula of this hydrate?

AlCl3 6H2O

Example #2: Solid copper(II) chloride forms a hydrate of formula CuCl2 xH2O. A student heated a sample of hydrated copper(II) chloride, in order to determine the value of x. The following results were obtained:

mass of crucible = 16.221 g
mass of crucible and hydrated copper(II) chloride = 18.360 g
mass of crucible and anhydrous copper(II) chloride = 17.917 g

From these data, determine the value of x and write the complete formula for hydrated copper(II) chloride.

Solution:

1) Determine the mass of the anhydrate and of the water that was lost:

CuCl2 ---> 17.917 − 16.221 = 1.696 g
H2O ---> 18.360 − 17.917 = 0.443 g

2) Determine moles of each:

CuCl2 ---> 1.696 g / 134.452 g/mol = 0.012614 mol
H2O ---> 0.443 g / 18.015 g/mol = 0.02459 mol

3) Divide by smallest:

CuCl2 ---> 0.012614 mol / 0.012614 mol = 1
H2O ---> 0.02459 mol / 0.012614 mol = 1.95

x = 2

CuCl2 2H2O

Example #3: A hydrate of magnesium chloride is present and the following data is collected:

mass of crucible = 22.130 grams
mass of crucible + hydrate = 25.290 grams
mass of crucible and contents after heating = 23.491 grams

What is the complete formula of this hydrate?

Solution:

1) Mass of hydrate:

25.290 g − 22.130 g = 3.160 g

2) Mass of anhydrate:

23.491 g − 22.130 g = 1.181 g

3) Water lost:

3.160 g − 1.181 g = 1.979 g

4) Moles MgCl2:

1.181 g / 95.211 g/mol = 1.24 mol

5) Moles water lost:

1.979 g / 18.015 g/mol = 0.109853 mol

6) Molar ratio of MgCl2 to water is:

1 : 11.3

7) Within fairly reasonable experimental error, the formula of the hydrate is:

MgCl2 · 12H2O

The Wiki page for magnesium chloride shows hydrates with 12, 8, 6, 4 and 2 waters of hydration exist. No example of 11 waters of hydration has ever been found. The one that exists at room temperature is the one with 6. The one with 12 loses 4 waters of hydration above −16.4 °C. So, while the problem above does not occur near room temperature, it theoretically could occur.

Example #4: Two documents:

(b) Here's a worksheet with a solved example using lab data followed by eight other hydrate problems, all of which have (hand-written) solutions. The original location of this document is here.

Example #5: I started the embedded video at the point where the discussion of the calculations start.