Question

When copper(II) sulfate pentahydrate, \(\mathrm{CuSO}_{4}+\mathrm{SH}_{2} \mathrm{O}\), is heated, it decomposes to the dehydrated form. The waters of hydration are released from the solid crystal and form water vapor. The hydrated form is medium blue, and the dehydrated solid is light blue. The balanced equation is $$ \operatorname{CuSO}_{4} \cdot \mathrm{SH}_{2} \mathrm{O}(s) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) What is the molar mass of \(\mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O}\) ? (b) What is the molar mass of \(\mathrm{CuSO}_{4}\) ? (c) If \(1.00 \mathrm{~g} \mathrm{CuSO}_{4} 5 \mathrm{H}_{2} \mathrm{O}\) is decomposed to \(\mathrm{CuSO}_{4}\) predict the mass of the remaining light blue solid.

Short answer

(a) The molar mass of \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\) is approximately 249.68 g/mol. (b) The molar mass of \(\mathrm{CuSO}_{4}\) is approximately 159.62 g/mol. (c) The predicted mass of the remaining light blue solid after decomposing 1.00 g of \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\) is approximately 0.64 g.

Step-by-step solution

Molar mass of \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\)

First calculate the molar mass of \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\). This comprises of copper (Cu), sulfur (S), oxygen (O) and Hydrogen (H). From the Periodic table: the atomic masses are Cu = 63.5g/mol, S = 32.07g/mol, O = 16.00 g/mol, and H = 1.008 g/mol. By summing the product of each element's atomic mass and the number of its atoms in \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\), we can calculate the molar mass: molar mass = 1(63.5 g/mol) + 1(32.07 g/mol) + 9(16.00 g/mol) + 10(1.008 g/mol).

Molar mass of \(\mathrm{CuSO}_{4}\)

Calculate the molar mass of \(\mathrm{CuSO}_{4}\). This comprises of copper (Cu), sulfur (S), and oxygen (O). By summing up the product of each element's atomic mass and the number of its atoms in \(\mathrm{CuSO}_{4}\), the molar mass can be calculated: molar mass = 1(63.5 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol).

Mass of the remaining solid (\(\mathrm{CuSO}_{4}\))

Finally, predict the mass of the remaining light blue solid (\(\mathrm{CuSO}_{4}\)) when \(1.00 g\) of \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\) is decomposed. This can be found by simply taking the ratio of the molar mass of \(\mathrm{CuSO}_{4}\) (from Step 2) and \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\) (from Step 1) and multiplying it by the original mass (1.00 g) of hydrated salt: mass = (molar mass of \(\mathrm{CuSO}_{4}\) / molar mass of \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\)) * original mass of hydrated salt.

Key concepts

Molar Mass

Molar mass is an essential concept in chemistry as it allows you to convert between the mass of a chemical substance and the amount in moles. It is defined as the mass of one mole of a substance, expressed in grams per mole (g/mol). To calculate the molar mass, sum the atomic masses of all atoms present in the molecular formula. For example, to find the molar mass of copper(II) sulfate pentahydrate, we add up the atomic masses of copper (Cu), sulfur (S), oxygen (O), and hydrogen (H).
By looking at a periodic table, we find the atomic masses:

• Copper (Cu) = 63.5 g/mol
• Sulfur (S) = 32.07 g/mol
• Oxygen (O) = 16.00 g/mol
• Hydrogen (H) = 1.008 g/mol

Next, multiply the atomic mass of each element by the number of times that element appears in the compound and add them together. This provides the compound's molar mass, crucial for calculating reactions and conversions.

Copper(II) Sulfate Pentahydrate

Copper(II) sulfate pentahydrate, commonly denoted as \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}\), is a blue crystalline solid used in various industrial applications, including as a fungicide and herbicide. This compound contains copper sulfate and five water molecules attached as waters of crystallization. These water molecules are included in the compound's crystal lattice structure, affecting its mass and color.
When heated, the pentahydrate undergoes chemical decomposition, losing its water molecules. The process changes the compound from a vibrant blue to a light blue, dehydrated form. The chemical equation representing this decomposition is:\[\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\text{heat}}{\longrightarrow} \mathrm{CuSO}_{4}(s) + 5 \mathrm{H}_{2} \mathrm{O}(g)\]Understanding the role of the water molecules is vital for calculations involving mass changes during heating or dehydration.

Water Vapor

Water vapor is the gaseous phase of water, formed when water molecules gain enough energy to overcome their intermolecular attractions. In chemical reactions, particularly decomposition reactions like with copper(II) sulfate pentahydrate, water vapor is released as a byproduct when water molecules separate from the crystalline structure due to heat.
When copper(II) sulfate pentahydrate is heated, the water molecules in the crystal lattice escape as water vapor. The release of water vapor marks the transformation of the compound to its anhydrous form. Identifying this process is key for understanding how changes in temperature can induce phase changes in a substance and influence both its physical appearance and weight.
Recognizing water vapor's role in reactions and natural processes emphasizes the connection between physical and chemical changes in substances.