Question

A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid, and the solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. What is the formula of the original halide?

Short answer

Answer

The formula of the original halide is $\mathit{S}\mathit{r}\mathit{C}{\mathit{l}}_{\mathbf{2}}$.

Step-by-step solution

Description of stoichiometry equivalent molar ratios

The term stoichiometry equivalent means that a definite amount of one substance is formed from, produces, or reacts with the definite amount with the other.

In a balanced equation, the number of moles of one substance is stoichiometrically equivalent to the number of moles of any other substance.

Balanced chemical reaction

The balanced chemical equation of this chemical reaction is as follow. Where X stands for halide.

$Sr{X}_2+H_{2}S{O}_4\to SrS{O}_4+2HX$

In a balanced chemical reaction, if you know the number of one substance you can calculate the moles of all the other substances.

Calculating number of moles of SrSO4

The given mass of strontium halide is 0.755 g.

In this reaction,

1 mole of strontium halide will result in 1 mole of strontium sulfate. You have given the mass of SrSO4and can find the mole. The molar mass of SrSO4is 183.68g/mol.

$$\begin{gathered} \mathit{N}\mathit{o}\mathbf{.}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{r}{\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\left(g\right)\mathbf{}\mathit{S}\mathit{r}\mathit{S}{\mathit{O}}_{\mathbf{4}}\mathbf{\times }\frac{\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}\mathbf{S}\mathbf{r}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{S}\mathbf{r}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{755}\mathbf{}\mathit{g}\mathbf{}\mathbf{\times }\mathbf{}\frac{\mathbf{1}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{}}{\mathbf{183}\mathbf{.}\mathbf{68}\mathbf{}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{11}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{3}} \end{gathered}$$

Calculating molar mass of SrX2

Number of moles of SrSO4 is stoichiometrically equivalent to number of moles of SrX2. Number of moles of SrX2 is $\mathbf{4}\mathbf{.}\mathbf{11}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$

$$\begin{gathered} \mathit{N}\mathit{o}\mathbf{.}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{r}{\mathit{X}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\left(g\right)\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{S}\mathbf{r}{\mathbf{X}}_{\mathbf{2}}}{\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{S}\mathbf{r}{\mathbf{X}}_{\mathbf{2}}} \\ \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{r}{\mathit{X}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\left(g\right)\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{S}\mathbf{r}{\mathbf{X}}_{\mathbf{2}}}{\mathbf{n}\mathbf{o}\mathbf{.}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{S}\mathbf{r}{\mathbf{X}}_{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{652}}{\mathbf{4}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{158}\mathbf{.}\mathbf{62}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

Calculating molar mass of X

To calculate the molar mass of $X_2$, you have to subtract molar mass of Sr with the calculated molar mass of $Sr{X}_2$.

Molar mass of Sr is 87.62g/mol.

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathbf{158}\mathbf{.}\mathbf{62}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{-}\mathbf{87}\mathbf{.}\mathbf{62}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{71}\mathbf{.}\mathbf{00}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

To calculate molar mass of X you need to divide molar mass of $X_2$by 2

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{X}\mathbf{=}\frac{\mathbf{71}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{m}\mathbf{o}\mathbf{l}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{5}\mathbf{}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

Calculating the formula of the halide

The calculated molar mass of the halide, X is 35.5 g/mol. This is the atomic mass of the halogen chlorine.

Thus, the formula of the halide is$\mathit{S}\mathit{r}\mathit{C}{\mathit{l}}_{\mathbf{2}}$