Q125CP - Chemistry: Molecular Nature Of Matter And Change

Question

A student forgets to weigh a mixture of sodium bromide dehydrate and magnesium bromide hexahydrate. Upon strong heating, the sample loses 252.1 mg of water. The mixture of anhydrous salts reacts with excess AgNO₃ solution to form 6.00x10^-3 mol of solid AgBr. Find the mass % of each compound in the original mixture.

Step-by-Step Solution

Answer

The mass % of NaBr.2H₂O and MgBr₂.6H₂O are 32.22% and 67.78%, respectively.

Step 1: Mass of water lost

Sodium bromide dehydrate on strong heating loses 2 moles of water

 $\mathbf{NaBr}\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\stackrel{\mathbf{∆}}{\mathbf{\to }}\mathbf{NaBr}\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

 

Molar mass of NaBr.2H₂O = 138.922 g/mol

Molar mass of H₂O = 18.016 g/mol.

So, 138.922 g of NaBr.2H₂O loses 36.032 g of water.

Similarly, Magnesium bromide dehydrate on strong heating loses 6 moles of water

 ${\mathbf{MgBr}}_{\mathbf{2}}\mathbf{.}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\stackrel{\mathbf{∆}}{\mathbf{\to }}{\mathbf{MgBr}}_{\mathbf{2}}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

 

Molar mass of MgBr₂.6H₂O = 292.206 g/mol

Molar mass of H₂O = 18.016 g/mol.

So, 138.922 g of MgBr₂.6H₂O loses 108.096 g of water.

Total mass of the mixture is,

 $\mathbf{138}\mathbf{\times }\mathbf{922}\mathbf{g}\mathbf{+}\mathbf{292}\mathbf{.}\mathbf{206}\mathbf{g}\mathbf{=}\mathbf{431}\mathbf{.}\mathbf{128}\mathbf{g}$

 

Total mass of water lost is,

$\mathbf{36}\mathbf{.}\mathbf{02}\mathbf{g}\mathbf{+}\mathbf{108}\mathbf{.}\mathbf{096}\mathbf{g}\mathbf{=}\mathbf{144}\mathbf{.}\mathbf{128}\mathbf{g}$ 

 

Actual mass of water lost = 252.1 mg = 0.2521 g

 

Mass of the mixture that loses 0.2521 g is,

 $\mathbf{0}\mathbf{.}\mathbf{2521}{\mathbf{gH}}_{\mathbf{2}}\mathbf{O}\mathbf{\times }\frac{\mathbf{431}\mathbf{.}\mathbf{128}\mathbf{g}}{\mathbf{144}\mathbf{.}\mathbf{128}\mathbf{g}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7541}\mathbf{g}$

Step 2: Determination of mass%

Mass of NaBr.2H₂O present in 0.7541 g mixture is,

 $$\begin{gathered} =\mathbf{0}\mathbf{.}\mathbf{7541}\mathbf{g}\mathbf{\times }\frac{\mathbf{138}\mathbf{.}\mathbf{922}\mathbf{g}}{\mathbf{431}\mathbf{.}\mathbf{128}\mathbf{gmixture}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2430}\mathit{g} \end{gathered}$$

 

Mass percent of NaBr.2H₂O is,

 $$\begin{gathered} \mathbf{=}\frac{\mathbf{mass}}{\mathbf{totalmass}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{2430}\mathbf{g}}{\mathbf{0}\mathbf{.}\mathbf{7541}\mathbf{g}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{=}\mathbf{32}\mathbf{.}\mathbf{22}\mathbf{\%} \end{gathered}$$

 

Mass percent of MgBr₂.6H₂O is,

 =100%-32.22%

 =67.78%

Hence, the mass % of NaBr.2H₂O and MgBr₂.6H₂O are 32.22% and 67.78%, respectively.