Q4.121CP - Chemistry: Molecular Nature Of Matter And Change

Question

There are various methods for finding the composition of an alloy (a metal-like mixture). Show that calculating the mass % of Mg in a magnesium-aluminum alloy (d = 2.40 g/cm³) gives the same answer (within rounding) using each of these methods: 

(a) $\mathbf{a}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{263}\mathbf{-}\mathbf{gsample}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{alloy}(\mathbf{d}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Mg}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{74}\mathbf{g}\mathbf{/}{\mathbf{cm}}^3\mathbf{;}\mathbf{dofAl}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{g}\mathbf{/}{\mathbf{cm}}^3)\mathbf{;}$   

(b) an identical sample reacting with excess aqueous HCl forms  $\mathbf{1}\mathbf{.}\mathbf{38}\mathbf{x}{\mathbf{10}}^{-2}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_2\mathbf{;}$ 

(c)  an identical sample reacting with excess O₂  forms 0.483 g of oxide.

Step-by-Step Solution

Answer

The mass percent of Mg calculated by three methods are quite close to each other.

Step 1: ( a ) A 0 . 263 - gsampleofalloy ( dofMg = 1 . 74 g / cm 3 ; dofAl = 2 . 70 g / cm 3 )

Let the mass of Mg be x g.

 

So mass of aluminium will be = (0.263 – x) g

 

Volume of the alloy is,

 $\begin{array}{c}\mathrm{Volume}=\frac{\mathrm{mass}}{\mathrm{density}}\\ =\frac{0.263\mathrm{g}}{2.40\mathrm{g}/{\mathrm{cm}}^3}\\ =0.1096{\mathrm{cm}}^3\end{array}$

Now, volume of Mg and Al is,

 $\begin{array}{c}\mathrm{Mg}=\frac{\mathrm{x}\mathrm{g}}{1.74\mathrm{g}/{\mathrm{cm}}^3}=\frac{\mathrm{x}}{1.74}{\mathrm{cm}}^3\\ \mathrm{Al}=\frac{(0.263-\mathrm{x}\mathrm{g})}{2.70\mathrm{g}/{\mathrm{cm}}^3}=\frac{(0.263-\mathrm{x})}{2.70}{\mathrm{cm}}^3\end{array}$

 

Determine the value of x,

 $\begin{array}{c}\mathrm{Volume}\mathrm{of}\mathrm{alloy}=\mathrm{volume}\mathrm{of}\mathrm{Mg}+\mathrm{volume}\mathrm{of}\mathrm{Al}\\ 0.1096{\mathrm{cm}}^3=\left(\frac{\mathrm{x}}{1.74}\right){\mathrm{cm}}^3+\left(\frac{0.263-\mathrm{x}}{2.70}\right){\mathrm{cm}}^3\\ 0.1096=\frac{0.96\mathrm{x}+0.458}{4.698}\\ \mathrm{x}=0.0592\end{array}$

So, mass of Mg = 0.0592 g

Mass of Al = (0.263 – x) = 0.2037 g

 

Now, the mass percentage is,

 

$\begin{array}{c}\mathrm{Mass}\mathrm{\%}\mathrm{of}\mathrm{Mg}=\frac{\mathrm{Mg}\mathrm{mass}}{\mathrm{total}\mathrm{mass}}\times 100\\ =\frac{0.0593\mathrm{g}}{0.263\mathrm{g}}\times 100\\ =22.5\mathrm{\%}\end{array}$

Step 2: (b) An identical sample reacting with excess aqueous HCl forms 1.38x10 -2 mol of H 2

The reaction is,

 $\begin{array}{c}\mathrm{Mg}+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+{\mathrm{H}}_2\\ 2\mathrm{Al}+6\mathrm{HCl}\to 2{\mathrm{AlCl}}_3+3{\mathrm{H}}_2\end{array}$

 

Moles of H₂ from Mg are,

 $=\mathrm{x}\mathrm{g}\mathrm{Mg}\times \left(\frac{1\mathrm{mol}\mathrm{Mg}}{24.31\mathrm{g}}\right)\times \left(\frac{1\mathrm{mol}{\mathrm{H}}_2}{1\mathrm{mol}\mathrm{Mg}}\right)=\frac{\mathrm{x}}{24.31}\mathrm{mol}$

Moles of H₂ from Al are,

 

 $=(0.263-\mathrm{x}\mathrm{g})\mathrm{Al}\times \left(\frac{1\mathrm{mol}\mathrm{Al}}{26.98\mathrm{g}}\right)\times \left(\frac{3\mathrm{mol}{\mathrm{H}}_2}{2\mathrm{mol}\mathrm{Al}}\right)=\frac{(0.263-\mathrm{x})}{17.98}\mathrm{mol}$

The value of x is,

 

 $\begin{array}{c}1.38\times {10}^{-2}mol=\left(\frac{x}{24.31}\right)+\left(\frac{0.263-x}{17.98}\right)\\ x=0.057g\end{array}$

Mass percent of Mg is,

 $\begin{array}{c}\mathrm{Mass}\mathrm{\%}=\frac{0.057\mathrm{g}}{0.263\mathrm{g}}\times 100\\ =21.7\mathrm{\%}\end{array}$

 

Step 3: (c) an identical sample reacting with excess O 2 forms 0.483 g of oxide

The reaction is,

 $\begin{array}{c}2\mathrm{Mg}+{\mathrm{O}}_2\to 2\mathrm{MgO}\\ 4\mathrm{Al}+3{\mathrm{O}}_2\to 2{\mathrm{Al}}_2{\mathrm{O}}_3\end{array}$

 

Mass of MgO from Mg is,

$=\mathrm{x}\mathrm{g}\mathrm{Mg}\times \left(\frac{1\mathrm{mol}\mathrm{Mg}}{24.31\mathrm{g}}\right)\times \left(\frac{2\mathrm{mol}\mathrm{MgO}}{2\mathrm{mol}\mathrm{Mg}}\right)\times \left(\frac{40.30\mathrm{g}}{1\mathrm{mol}\mathrm{MgO}}\right)=1.66\mathrm{x}\mathrm{g}$

Mass of Al₂O₃ from Al is,

$=(0.263-\mathrm{x}\mathrm{g})\mathrm{Mg}\times \left(\frac{1\mathrm{mol}\mathrm{Al}}{26.98\mathrm{g}}\right)\times \left(\frac{1\mathrm{mol}\mathrm{Al}}{26.98\mathrm{g}}\right)\times \left(\frac{2\mathrm{mol}{\mathrm{Al}}_2{\mathrm{O}}_3}{4\mathrm{mol}\mathrm{Al}}\right)\times \left(\frac{101.96\mathrm{g}}{1\mathrm{mol}{\mathrm{Al}}_2{\mathrm{O}}_3}\right)=1.89(0.263-\mathrm{x})\mathrm{g}$

 

The value of x is,

 $\begin{array}{c}0.483g=1.66x+1.89(0.263-x)g\\ x=0.0611g\end{array}$

 

So, mass of Mg = 0.0611 g

 

Now, mass percent of Mg is,

 $\begin{array}{c}\mathrm{Mass}\mathrm{\%}=\frac{0.0611\mathrm{g}}{0.263\mathrm{g}}\times 100\\ =23.2\mathrm{\%}\end{array}$

Thus, the mass percent of Mg calculated by three methods are quite close to each other.