Question

A 0.755-g sample of hydrated copper(II) sulfate$\left(C_u{\mathrm{SO}}_4·{\mathrm{XH}}_{2}O\right)$ was heated carefully until it had changed completely to anhydrous copper(II) sulfate$\left(C_U{\mathrm{SO}}_4\right)$ with a mass of 0.483 g. Determine the value of x. [This number is called the “number of waters of hydration" of copper(II) sulfate $\mathbf{\left(}{\mathbf{C}}_{\mathbf{U}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\right)}$. It specifies the number of water molecules per formula unit of in the hydrated crystal.]

Short answer

The value of xin the given compound formula is5and the formula is ${\mathrm{C}}_{\mathrm{u}}{\mathrm{SO}}_4·5{\mathrm{H}}_2\mathrm{O}$.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

$\mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{Number}\mathrm{of}\mathrm{moles}}$

Determine the moles of CuSO4

Here,

The mass of

$\begin{array}{rcl}{\mathrm{H}}_2\mathrm{O}& =& 0.755\mathrm{g}-0.483\mathrm{g}\\ & =& 0.272\mathrm{g}\end{array}$

Let’s calculate the moles,

$\begin{array}{rcl}{\mathrm{C}}_{\mathrm{u}}{\mathrm{SO}}_4:\mathrm{n}\left({\mathrm{C}}_{\mathrm{u}}{\mathrm{SO}}_4\right)& =& \frac{\mathrm{m}\left({\mathrm{C}}_{\mathrm{u}}{\mathrm{SO}}_4\right)}{\mathrm{Mr}\left({\mathrm{C}}_{\mathrm{u}}{\mathrm{SO}}_4\right)}\\ & =& \frac{0.483\mathrm{g}}{159.62{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 0.00303\mathrm{mol}\end{array}$

Determine the moles of H2O

Let’s calculate the moles,

$\begin{array}{rcl}{\mathrm{H}}_2\mathrm{O}:\mathrm{n}\left({\mathrm{H}}_2\mathrm{O}\right)& =& \frac{\mathrm{m}\left({\mathrm{H}}_2\mathrm{O}\right)}{\mathrm{Mr}\left({\mathrm{H}}_2\mathrm{O}\right)}\\ & =& \frac{0.272\mathrm{g}}{18.02{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 0.0151\mathrm{mol}\end{array}$

Determine the Compound formula

$4.98\mathrm{mol}\left({\mathrm{H}}_2\mathrm{O}\right):1\mathrm{mol}\left({\mathrm{CuSO}}_4\right)$

Therefore the compound formula is ${\mathrm{CuSO}}_4·{\mathrm{H}}_2\mathrm{O}$.