Question

A sample of Substances with empirical formula ${\mathbf{XBr}}_{\mathbf{2}}$ weighs 0.5000g. When it is dissolved in water and all its bromine is converted to insoluble$\mathbf{AgBr}$ by addition of an excess of silver nitrate, the mass of the resulting $\mathbf{AgBr}$is found to be 1.0198g. The chemical reaction is

$\mathit{X}\mathit{B}{\mathit{n}}_{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\to }\mathbf{2}\mathit{A}\mathit{g}\mathit{B}\mathit{r}\mathbf{+}\mathit{X}{\left(N{O}_3\right)}_{\mathbf{2}}$

1. Calculate molecular mass of${\mathbf{XBr}}_{\mathbf{2}}$
2. Calculate atomic mass of X and give its nature and symbol.

Short answer

1. Molecular mass of $XB{r}_2$is 184.12g

b. Atomic mass of X is 24.32g

Name and symbol of X is magnesium and ‘Mg’ respectively

The dibromide of unknown element ‘X’ reacts in excess silver nitrate and forms silver bromide and nitrate

Equation of reaction is

${XBr}_2+2AgN{O}_2\to 2AgBr+X{\left(N{O}_3\right)}_2$

We assume molar mass of X be$mgmo{l}^{-1}$

Hence

Molar mass of${XBr}_2$=$\left(m+2\times 79.90\right)gmo{l}^{-1}$

Molar mass of$X{\left(N{O}_3\right)}_2=\left[m+2\left(14.01+3\times 16.00\right)\right]gmo{l}^{-1}$

Step-by-step solution

Mass of AgBr

Mass of AgBr Produced

According to balanced equation

produces 2 moles of AgBr

So, mass of AgBr produced from 0.5000g ${XBr}_2$is

$$\begin{gathered} 0.5000gXB{r}_2\times \frac{1molXB{r}_2}{\left(m+2\times 79.90g\right)XB{r}_2}\times \frac{2mol.AgBr}{1molXB{r}_2}\times \frac{187.77gAgBr}{1molAgBr} \\ =\frac{187.77}{\left(m+2\times 79.90\right)}gAgBr \end{gathered}$$

Molecular mass of XBr2

Molecular mass of$\mathit{X}\mathit{B}{\mathit{r}}_{\mathbf{2}}$

0.5000g $XB{r}_2$produces 1.0198g AgBr, So their amounts can be equated as follows:

$$\begin{gathered} \frac{187.77}{\left(m+2\times 79.90\right)}g=1.0198g \\ m+159.8=\frac{187.77}{1.0198}=184.12 \\ m=184.12-159.8=24.32gmol \end{gathered}$$

Hence atomic mass of element x is 24.82 and molar mass of $XB{r}_2$is 184.12g

Atomic mass, Name and Symbol of x

• From the previous step we find, Atomic mass of x is 24.32g
• Magnesium has the atomic mass 24 which is very close to atomic mass of x i.e. 24.32

Hence Name and symbol of element probably isMagnesium andMg respectively