1. Analysis of Alum Objectives
To determine the percentage of water in alum hydrate and an unknown hydrate
To calculate the water of crystallization of an unknown hydrate
To develop the lab skills for analyzing a hydrate

2. Empirical Formulas of Hydrates
A hydrate is an ionic solid AX that has some water trapped in it
It is written as AX.yH2O
AX is some known ionic compound like say Na2SO4 or BaCl2 etc
Here is an example
In the experiment we know what AX is, but we will not know y this we must find
CoCl2
Anhydrous Colbalt (II) Chloride
CoCl2.6H2O
Colbalt(II)Chloride hexahydrate

3. Hydrates
Prefix
# water of hydration
Hemi
1/2
Mono 1
Sesqui
3/2 Di 2
Tri 3
Tetra 4
Penta 5
Hexa 6
Hepta 7
Octa 8
Nona 9
Deca 10
Undeca 11
Dodeca 12
“A solid compound having a fixed number of water molecules H2O in its composition”
The water in the hydrate is called the water of crystallization
BaCl2.2H2O
2 waters of crystallization
Barium Chloride dihydrate
KAl(SO4)2.12H2O
12 waters of crystallization
Alum hydrate or potassium aluminum sulfate dodecahydrate

4. How much water in my hydrate?
We can measure the mass percent of water in a hydrate quite easily
The mass percent of water in a hydrate is defined as
If we already know the chemical formula for the hydrate we can calculate it theoretically from the molar masses assuming 1 mole of the hydrate

5. Calculate the theoretical % of H2O in alum hydrate (KAl(SO4)2.12H2O)
To do this we figure out how many of each element are present in the formula
1K, 1Al, 2S, ( = 20 O), and 24H
We then calculate the mass of this many moles of each element and add them together
1 x g = g
1 x g = g
2 x g = g
20 x g = g
24 x g = g g
We the apply the formula from the last slide

6. Hydrate(s) + heat  anhydrous salt(s) + water of hydration(g)
Experimental mass % H2O
All hydrates can be heated to remove the water, leaving behind the anhydrous material
Hydrate(s) + heat  anhydrous salt(s) + water of hydration(g)
To get the experimental value we measure the mass before heating mbefore
Heat the crystal to boil off the water
Reweigh mafter
The mass % H2O can be calculated from the formula

7. 1. 150g Alum hydrate when dehydrated leaves a mass of 0
1.150g Alum hydrate when dehydrated leaves a mass of g of the anhydrous product: what is the mass % water?
Given
mafter = g
mbefore = g
Use
𝑚𝑎𝑠𝑠 % 𝐻 2 𝑂= 𝑔 −0.627𝑔 𝑔 ×100%
= 45.5%

8. Experimental Technique
The proficiency of your technique can be inferred by comparing the experimental result (45.5%) with the theoretical value (45.58%)
If your value is too small you haven’t removed all the water so heat some more
If your value is too large you heated it too strongly and as the water came out of the crystal it blew some of the crystal out too! So next time heat it more gently

9. Getting the number of waters of hydration y
We have just calculated % H2O, what if we also knew what the ionic solid was?
Let’s say we knew we had CuSO4 but we didn’t know how many waters were in our hydrate
In other words we want to find y in the formula
CuSO4.yH2O
We can use the following formula to get y
𝑦= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

10. Getting the number of waters of hydration y
2.200g of Copper(II)sulfate hydrate after heating has a mass of g
What is the mass % water?
What is y in the formula of our hydrate if it is written CuSO4.yH2O?
The number of moles of Copper (II) Sulfate
The number of moles of water lost
𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡= −1.407 𝑔 18.0𝑔/𝑚𝑜𝑙 =4.40× 10 −2 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟
What is the formula for the hydrate? CuSO4.5H2O
𝑚𝑎𝑠𝑠 % 𝐻 2 𝑂= 𝑔 −1.407𝑔 𝑔 ×100%
= 36.05%
𝑚𝑜𝑙𝑒𝑠 𝐶𝑢𝑆 𝑂 4 = 𝑚𝑎𝑠𝑠 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = 𝑔 𝑔/𝑚𝑜𝑙 =8.81× 10 −3 𝑚𝑜𝑙𝑒𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
𝑦= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = 4.40× 10 −2 𝑚𝑜𝑙𝑒𝑠 𝑤𝑎𝑡𝑒𝑟 8.81× 10 −3 𝑚𝑜𝑙𝑒𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 =5.00

11. Getting the number of waters of hydration y
You will be given a mystery compound but we will tell the molar mass of the part of the compound that doesn’t have the water. Since this is the formula of what is left one you heat it and remove the water we call the part without the water the ANHYDROUS COMPOUND (AC)
Your formula is therefore AC.yH2O and you must find y
We will tell you the molar mass of AC
Then following what we did for the copper(II) sulfate
𝑦= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑙𝑜𝑠𝑡 𝑚𝑜𝑙𝑒𝑠 𝐴𝐶

12. Procedure
Weigh a clean dry 250-mL beaker covered with a watch glass m1 = mbeaker + mwatchglass
Add about 1g of your unknown to the beaker and reweigh (with the watch glass on) m2 = m1 + mhydrate
Determine the mass of the unknown mbefore = mhydrate = m2 – m1
Support beaker and watchglass on a ring stand with a wire gauze.
Heat gently (burner 3 inches below beaker) – till the moisture has gone (Crystal should turn to powder at this point)
Turn off burner, let it cool for 10 mins then reweigh, the beaker, its contents, and the watchglass together m3
mafter = manhydrous = m3 – m1
mwater= mbefore - mafter
Discard powder in trash, clean beaker and repeat a second trial
Calculate the mass % water for each trial
Calculate y for each trial