Chapter 2.9, Problem 34AAP

To determine

The chemical formula of the inter-metallic compound.

Answer to Problem 34AAP

The chemical formula of the inter-metallic compound is ${\text{MgAl}}_{\text{5}}$.

Explanation of Solution

Write the expression to calculate number of gram-moles of manganese $\left(n_{\text{Mg}}\right)$.

 $n_{\text{Mg}}=\frac{m_{\text{Mg}}}{M_{\text{Mg}}}$                                                                                                             (I)

Here, molar mass of manganese is $M_{\text{Mg}}$ and mass of manganese is $m_{\text{Mg}}$.

Write the expression to calculate number of gram-moles of aluminum $\left(n_{\text{Al}}\right)$.

 $n_{\text{Al}}=\frac{m_{\text{Al}}}{M_{\text{Al}}}$                                                                                                            (II)

Here, molar mass of aluminum is $M_{\text{Al}}$ and mass of aluminum is $m_{\text{Al}}$.

Write the expression to calculate total number of gram-moles of compound $\left(n\right)$.

 $n=n_{\text{Mg}}+n_{\text{Al}}$                                                                                                        (III)

Write the expression to calculate gram-mole fraction of manganese $\left(x\right)$.

 $x=\frac{n_{\text{Mg}}}{n}$                                                                                               (IV)

Write the expression to calculate gram-mole fraction of aluminum $\left(y\right)$.

 $y=\frac{n_{\text{Al}}}{n}$                                                                                                  (V)

Conclusion:

It is given that the inter-metallic compound has $\text{84}\text{.32wt\%}$ aluminum and $\text{15}\text{.68wt\%}$ manganese. Then, the masses of aluminum and manganese are $\text{84}\text{.32}\text{g}$ and $\text{15}\text{.68}\text{g}$ respectively.

The molar masses of manganese and aluminum are $24.31\text{g}/\text{mol}$ and $26.98\text{g}/\text{mol}$ respectively.

Substitute $24.31\text{g}/\text{mol}$ for $M_{\text{Mg}}$ and $\text{15}\text{.68}\text{g}$ for $m_{\text{Mg}}$ in Equation (I).

 $\begin{array}{c}{n}_{\text{Mg}}=\frac{\text{15}\text{.68}\text{g}}{24.31\text{g}/\text{mol}}\\ =0.645\text{mol}\end{array}$

Substitute $26.98\text{g}/\text{mol}$ for $M_{\text{Al}}$ and $84.32\text{g}$ for $m_{\text{Al}}$ in Equation (II).

 $\begin{array}{c}{n}_{\text{Al}}=\frac{84.32\text{g}}{26.96\text{g}/\text{mol}}\\ =3.125\text{mol}\end{array}$

Substitute $0.645\text{mol}$ for $n_{\text{Mg}}$ and $3.125\text{mol}$ for $n_{\text{Al}}$ in Equation (III).

 $\begin{array}{c}n=0.645\text{mol}+3.125\text{mol}\\ =3.770\text{mol}\end{array}$

Substitute $0.645\text{mol}$ for $n_{\text{Mg}}$ and $3.770\text{mol}$ for $n$ in Equation (IV).

 $\begin{array}{c}x=\frac{0.645\text{mol}}{3.770\text{mol}}\\ =0.17\end{array}$

Substitute $3.125\text{mol}$ for $n_{\text{Al}}$ and $3.770\text{mol}$ for $n$ in Equation (V).

 $\begin{array}{c}y=\frac{3.125\text{mol}}{3.770\text{mol}}\\ =0.83\end{array}$

The general expression for chemical formula of inter-metallic compound of magnesium and aluminium.

  ${\text{Mg}}_x{\text{Al}}_y$

The gram-mole fractions of manganese and aluminum present in the compound are 0.17 and 0.83 respectively.

Then, the chemical formula of the inter-metallic compound is,

  ${\text{Mg}}_{0.17}{\text{Al}}_{\text{0}\text{.83}}$

To make the mole fractions into whole number, multiply the $x,y$ values by 6.

  $\begin{array}{c}x=0.17\times 6\\ =1.02\\ \simeq 1\end{array}$

  $\begin{array}{c}y=0.83\times 6\\ =4.98\\ \simeq 5\end{array}$

Thus, the chemical formula of the inter-metallic compound is ${\text{MgAl}}_{\text{5}}$.