Question

Element X forms both a dichloride (XCI₂) and a tetrachloride (XCI₄).Treatment of 10.00 g of XCI₂ with excess chlorine forms 12.55 g of XCI₄.

Calculate the atomic weight (mass) of X and identify X.

Short answer

The atomic weight (mass) of X is $\mathbf{207}\raisebox{1ex}{$\mathbf{g}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{m}\mathbf{o}\mathbf{l}$}\right.$and the identification of X is Pb (lead).

Step-by-step solution

Definition

The molar mass of a materialis defined as the mass of the substancecontained in one moleof the substance, and it is equal to the total massof the component atoms. It may be stated mathematically as follows:
Molar mass$=\frac{Weighting}{numberofmoles}$

Determination of limiting reactant

$XC{L}_2+C{I}_2\to XC{I}_4$

is the balanced equationfor the reaction between XCI₂ and CI₂ that generates XCI₄ as a product.

The excess reactant in the initial mixture is chlorine, whereas the limiting reactant isXCI₂. As a result, the number of moles of XCI₄ created is proportional tothe number of moles of XCI₂ consumed.

Assume that the molar mass of element X is M g/mol.

Determination of atomic mass of X

Calculate the molar mass of the unknown element X by equating the number of moles of XCI₄ with the number of moles of XCI₂.

Moles of XCI₂ = Moles of XCI₄

$\begin{array}{rcl}\frac{10.00g}{\left(M+2\times 35.45{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}\right)}& =& \frac{12.55g}{\left(M+4\times 35.45{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.}\right)}\\ 2.55M& =& 528\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\\ M& =& 207\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\end{array}$

Therefore X's atomic mass is $207\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$.

This atomic mass corresponds to the element Pb (lead), according to the periodic table.