Question

Vanadic ion,${\mathbf{V}}^{\mathbf{3}\mathbf{+}}$ , forms green salts and is a good reacting agent, being itself changed in neutral solutions to the nearly colorless ion$\mathbf{V}{\left(\mathbf{OH}\right)}_{\mathbf{4}}^{\mathbf{+}}$ . Suppose that 15.0 mL of a$\mathbf{0}.\mathbf{200}\mathbf{M}$ solution of vanadic sulfate,${\mathbf{V}}_{\mathbf{2}}{\left({\mathbf{SO}}_{\mathbf{4}}\right)}_{\mathbf{3}}$ , was needed to reduce completely a $0.540g$sample of an unknown substance$X$ . If each molecule of $X$accepted just one electron, what is the molecular weight of$X$ ? Suppose that each molecule of $X$accepted three electrons; what would be the molecular weight of $X$then?

Short answer

Molecular weight of$X$is$45.0\text{g}/\text{mol}$ .

Molecular weight of $X$ if it accept three electron is data-custom-editor="chemistry" $135\text{g}/\text{mol}$.

Step-by-step solution

Given data

$15.0\text{mL}$ of a 0.200-M solution of vanadic sulfate, was needed to reduce completely a $0.540g$ sample of an unknown substance $X$.

The molecular weight of a given molecule is the “sum of the atomic weights” of all the atoms present in a given molecule. It was measured in role="math" localid="1663407368487" $\mathbf{g}/\mathbf{mol}$.

Calculate the molecular weight of  X

In neutral solutions ${\text{V}}^{3+}$, changes to$\text{V}(\text{OH}{)}_4^{+}$.

The redox reaction between${\text{V}}_2{\left({\text{SO}}_4\right)}_3$and$X$.

${\text{V}}^{3+}\to {\text{V}}^{5+}+2{\text{e}}^{-}$

$2{\text{V}}^{3+}+4\text{X}\to 2{\text{V}}^{5+}+4{\text{X}}^{-}$

$15.0\text{mL}$of $0.200{\text{MV}}_2{\left({\text{SO}}_4\right)}_3$required to reduce$\text{0.540 g}$of X.

$$\begin{gathered} {\text{molV}}_2{\left({\text{SO}}_4\right)}_3=15.0\text{mL}\times \frac{0.200\text{mol}}{1000\text{mL}} \\ =3.00\times {10}^{-3}\text{mol} \end{gathered}$$

$$\begin{gathered} \text{mol}\text{X}=3.00\times {10}^{-3}\text{molV}{V}_2{\left(S{O}_4\right)}_3\times \frac{2{\text{molV}}^{3+}}{lmol{V}_2{\left(S{O}_4\right)}_3}\times \frac{2\text{molX}}{lmol{V}^{3+}} \\ =12.0\times {10}^{-3}\text{molX} \end{gathered}$$

The molecular weight is calculated as:

Calculate the molecular weight of X  (for three electrons)

If the molecule$X$accept three electrons, the equation becomes:

$3{\text{V}}^{3+}+2\text{X}\to 3{\text{V}}^{5+}+2{\text{X}}^{3-}$

$$\begin{gathered} \text{mol}\text{X}=3.00\times {10}^{-3}{\text{molV}}_2{\left(S{O}_4\right)}_3\times \frac{2{\text{molV}}^{3+}}{1{\text{molV}}_2{\left(S{O}_4\right)}_3}\times \frac{2\text{molX}}{3{\text{molV}}^{3+}} \\ =4.00\times {10}^{-3}\text{molX} \end{gathered}$$

The molecular weight is calculated as:

data-custom-editor="chemistry" $$\begin{gathered} X=\frac{0.540\text{g}}{4.0\times {10}^{-3}\text{mol}} \\ =135\text{g}/\text{mol} \end{gathered}$$