The key difference between CuSO4 and CuSO4 5H2O is that CuSO4 is amorphous, whereas CuSO4 5H2O is crystalline.
CuSO4 is the chemical formula of copper(II) sulfate, while CuSO4 5H2O is the hydrated form of copper(II) sulfate. The term hydrated indicates that this compound has one or more water molecules in association with it. Therefore, CuSO4 is the usual name for the anhydrous form.
CONTENTS
1. Overview and Key Difference
2. What is CuSO4
3. What is CuSO4 5H2O
4. Side by Side Comparison – CuSO4 vs CuSO4 5H2O in Tabular Form
5. Summary
What is CuSO4?
CuSO4 is copper(II) sulfate that has the copper metal in +2 oxidation state. It is an inorganic compound that has no water molecules associated with it. Therefore, we call it the anhydrous form of copper sulfate. Moreover, this anhydrous compound occurs as a white powder.
The industrial production of copper sulfate involves treating the copper metal with sulfuric acid in hot and concentrated form. Also, it is possible to produce this compound using oxides of copper as well. Here, it is done by treating copper oxide with diluted sulfuric acid. Besides, slowly leaching low-grade copper ore in the air is another method of production. It is possible to use bacteria to catalyze this process.
Figure 02: Copper Sulfate Anhydrous
When considering the chemical properties of this compound, the molar mass is 159.6 g/mol. It appears in grey-white colour. The density is 3.60 g/cm3. When considering the melting point of copper sulfate, it is 110 °C, and upon further heating, the compound decomposes.
What is CuSO4 5H2O?
CuSO4 5H2O is copper(II) sulfate pentahydrate. It has five water molecules associated with the copper sulfate molecule. It appears as a bright blue colour solid. Besides, it is the most common hydrated form of copper sulfate. Further, some common names for this compound are blue vitriol, bluestone, vitriol of copper, Roman vitriol, etc.
Figure 02: Appearance of Copper Sulfate Pentahydrate
Moreover, this compound exothermally dissolves in water. Then, it forms an aqua complex containing one CuSO4 molecule in association with six water molecules, and this complex has an octahedral molecular geometry. The molar mass of it is 249.65 g/mol. When considering the melting point, upon heating above 560 °C, the compound decomposes. That mean; the compound decomposes before melting. There, this compound removes two water molecules at 63 °C and two more at 109 °C. Moreover, the final water molecule is released at 200 °C.
What is the Difference Between CuSO4 and CuSO4 5H2O?
CuSO4 is copper(II) sulfate that has the copper metal in +2 oxidation state. CuSO4 5H2O is copper(II) sulfate pentahydrate. The key difference between CuSO4 and CuSO4 5H2O is that CuSO4 is amorphous, whereas CuSO4 5H2O is crystalline. Furthermore, copper sulfate is anhydrous whereas copper sulfate pentahydrate is a hydrated form.
Moreover, a further difference between CuSO4 and CuSO4 5H2O is their melting point; melting point of CuSO4 is 110 °C, and upon further heating, the compound decomposes, while CuSO4 5H2O compound decomposes before melting.
Below infographic shows more information regarding the difference between CuSO4 and CuSO4 5H2O.
Summary – CuSO4 vs CuSO4 5H2O
CuSO4 is copper(II) sulfate that has the copper metal in +2 oxidation state. CuSO4 5H2O is copper(II) sulfate pentahydrate. In short, the key difference between CuSO4 and CuSO4 5H2O is that CuSO4 is amorphous, whereas CuSO4 5H2O is crystalline.
Reference:
1. “Copper Sulfate Pentahydrate.” National Center for Biotechnology Information. PubChem Compound Database, U.S. National Library of Medicine, Available here.
Image Courtesy:
1. “Copper sulfate” By Stephanb – Own work, (CC BY-SA 3.0) via Commons Wikimedia
2. “Copper sulfate anhydrous” By W. Oelen – (CC BY-SA 3.0) via Commons Wikimedia