What's the significance of finding the formula of a Hydrate?

Click For Summary
SUMMARY

The discussion centers on the significance of determining the formula for Copper(II) Sulfate xHydrate, specifically identifying it as Pentahydrate (CuSO4·5H2O). Understanding the formula of hydrates is crucial for accurate stoichiometric calculations in chemical reactions, particularly when using compounds with water of crystallization. For example, when using sucrose (C22H22O11·5H2O) in reactions, knowing the anhydrous mass is essential for precise measurements and yield predictions.

PREREQUISITES
  • Basic understanding of chemical formulas and stoichiometry
  • Familiarity with hydrates and their properties
  • Knowledge of heat application in chemical reactions
  • Experience with laboratory measurements and calculations
NEXT STEPS
  • Research the process of determining the formula of hydrates through laboratory experiments
  • Learn about stoichiometric calculations involving hydrates and anhydrous compounds
  • Explore the implications of water of crystallization in chemical reactions
  • Study the thermal decomposition of hydrates and its applications in chemistry
USEFUL FOR

Chemistry students, laboratory technicians, and educators interested in the practical applications of hydrates in chemical reactions and stoichiometry.

Hazelhaze
Messages
1
Reaction score
0
Well, we basically did a lab where we found the formula of Copper(II) Sulfate xHydrate (Found it was Penta after the calculations), and I've come to the final question of my lab. "What is the significance of finding the formula of a Hydrate." I really have no clue to be honest, and was hoping for someone to enlighten me on what exactly it is.

Thanks for your time!

~ Adam
 
Physics news on Phys.org
If I had an experiment planned that used sucrose as a reactant, and I needed 200 grams of it for the reaction, I would need to consider that the sucrose I have contains a crystalline structure with H20 (Might be something like C22H22O11*5H20). Simply measuring out 200 grams of the hydrous compound would not suffice because only the anhydrous part is what I need for the reaction. If I knew the formula for the compound I have, I could figure out how much I need for the reaction (and there would also be water as a product).

For instance, C22H22O11*5H20 -> C22H22O11 + 5H20 with heat applied.
(I don't know how to type the yield symbol with the delta over it)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
109K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
18K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 25 ·
Replies
25
Views
5K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
13K