Question

A salt contains only barium and one of the halide ions. A 0.158-9 sample of the salt wasdissolved in water, and an excess of sulfuric acid was added to form barium sulphate,$\left(BaS{O}_4\right)$, which was filtered, dried, and weighed. Its mass was found to be 0.124 g. What is the formula of the barium halide?

Short answer

The formula ofthe barium halide is${\mathrm{BaBr}}_2$.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one moleof 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 mass of BaSO4

Here the mass of${\mathrm{BaSO}}_4\mathrm{=}0.124\mathrm{g}$

We can calculate the mass of $\mathrm{Ba}$ in 0.158 g of ${\mathrm{BaX}}_2$salt (X-halogen, $\mathrm{Ba}$is charged 2+, X are charged 1-, so formula must be ${\mathrm{BaX}}_2$) using the mass of${\mathrm{BaSO}}_4$

$\begin{array}{rcl}\mathrm{M}\left({\mathrm{BaSO}}_4\right)& =& 37.3+32.07+4\times 15.999\\ & =& 233.266\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

$\begin{array}{rcl}\mathrm{M}\left(\mathrm{Ba}\right)& =& 0.124\mathrm{g}{\mathrm{BaSO}}_4·\left(\frac{137.3\mathrm{Ba}}{233.366\mathrm{g}{\mathrm{BaSO}}_4}\right)\\ & =& 0.073\mathrm{g}\end{array}$

Determination of mass of X

$\begin{array}{rcl}MassofXin0.158gofBa{X}_2& =& 0.158g-0.073g\\ Ba{X}_2& =& 0.085g\\ n\left(Ba\right)& =& 0.073gBa\left(\frac{1mol}{137.3gBa}\right)\\ & =& 5.32\times {10}^{-4}\\ n\left(X\right)& =& 5.32\times {10}^{-4}molBa\left(\frac{2molX}{1molBa}\right)\\ & =& 1.06\times {10}^{-4}molX\end{array}$

Determination the molar mass of X

Thus the calculation of molar mass of X,

$\begin{array}{rcl}\mathrm{M}\left(\mathrm{X}\right)& =& 1\mathrm{mol}·\left(\frac{0.085}{1.06\times {10}^{-3}\mathrm{mol}}\right)\\ & =& 79.93\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

Therefore X is ${\mathrm{B}}_{\mathrm{r}}$and ${\mathrm{BaBr}}_2$the formula isfrom theperiodic table.