Question

An ionic compound ${\mathbf{MX}}_{\mathbf{3}}$is prepared according to the following unbalanced chemical equation:

$\mathbf{M}\mathbf{+}{\mathbf{X}}_{\mathbf{2}}\mathbf{\to }\mathit{M}{\mathit{X}}_{\mathbf{3}}$
A 0.105-g sample of ${\mathbf{X}}_{\mathbf{2}}$contains $\mathbf{8}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}$molecules. The compound $\mathit{M}{\mathit{X}}_{\mathbf{3}}$ consists of 54.47% X by mass. What are the identities of M and X, and what is the correct name for $\mathit{M}{\mathit{X}}_{\mathbf{3}}$? Starting with 1.00 g each of M a can be prepared?

Short answer

The identities of M and X are Yttrium and chlorine. The correct name ofcompound${\mathrm{MX}}_3$is Yttrium (III) chloride and the mass${\mathrm{MX}}_3$ of is $1.84\mathrm{g}$.

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:

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

Determination of Atomic mass of chlorine

Let us consider a 0.105-g sample of ${\mathrm{X}}_2$contains $8.92\times {10}^{20}$molecules

Here, 1 mole of ${\mathrm{X}}_2$has a mass

role="math" localid="1655445343246" $$\begin{gathered} =0.105\mathrm{g}\times \left(\frac{6.0233\times {10}^{23}}{8.92\times {10}^{22}}\right) \\ =70.88\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \end{gathered}$$

The Atomic mass of $\mathrm{X}=35.44\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$which is chlorine.

Thus $X_2$denotes ${\mathrm{Cl}}_2$.

Determination of Molar mass of MX2

54.47% ofisX or chlorine

Mass of chlorine in

$\begin{array}{rcl}{\mathrm{MX}}_3& =& 3\left(35.45\right)\\ & =& 106.35\mathrm{g}\end{array}$

$\left(\frac{54.47}{100}\right)\times \mathrm{A}=106.35\mathrm{g}$

Molar mass of${\mathrm{MX}}_3,\mathrm{A}=195.24\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$

Mass of

role="math" localid="1655446108941" $\begin{array}{rcl}\mathrm{M}& =& 195.24\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.-106.35\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

$\begin{array}{rcl}\mathrm{M}& =& 88.9\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$,the element M is yttrium.

Thus the name of compound is Yttrium (III) chloride.

Determination of Moles of chlorine gas and Yttrium

$\begin{array}{rcl}& & 1.00\mathrm{g}\mathrm{M}\mathrm{and}1.00\mathrm{g}\mathrm{of}\mathrm{X}\\ \mathrm{Moles}\mathrm{of}\mathrm{Yttrium}\left(\mathrm{M}\right)& =& \frac{\mathrm{Mass}}{\mathrm{Atomic}\mathrm{mass}}\\ & =& \frac{100\mathrm{g}}{88.9{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 0.01125\mathrm{moles}\end{array}$

$\begin{array}{rcl}\mathrm{Moles}\mathrm{of}\mathrm{Chlorine}\left({\mathrm{X}}_2\right)& =& \frac{100\mathrm{g}}{70.9{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 0.0141\mathrm{moles}\end{array}$

Balanced equation

Thus the balanced equation for the reaction

$2\mathrm{Y}+3{\mathrm{Cl}}_2\to 2{\mathrm{YCl}}_3$

Here, $0.0141$moles of chlorine react with $0.0094$moles of Yttrium to produce $0.0094$moles of ${\mathrm{YCl}}_3$.

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{YCl}}_3& =& \mathrm{Moles}\times \mathrm{Molar}\mathrm{mass}\\ & =& 0.0094\times 195.24\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mole}$}\right.\\ & =& 1.84\mathrm{g}\end{array}$

Therefore the mass of$YC{l}_3$ is $1.84g$.