Chapter 16.13, Problem 88AAP

To determine

The weight percentage of iron in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$.

The weight percentage of boron in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$.

The weight percentage of silicon in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$.

Answer to Problem 88AAP

The weight percentage of iron in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$ is $91.7\%$.

The weight percentage of boron in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$ is $2.97\%$.

The weight percentage of silicon in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$ is $5.33\%$.

Explanation of Solution

Write the expression to calculate mass of iron in the metallic glass composition $\left(m_{\text{Fe}}\right)$.

$m_{\text{Fe}}=n_{\text{Fe}}{M}_{\text{Fe}}$                                                                                                             (I)

Here, number of moles of iron is $n_{\text{Fe}}$ and molar mass of iron is $M_{\text{Fe}}$.

Write the expression to calculate mass of boron in the metallic glass composition $\left(m_{\text{B}}\right)$.

$m_{\text{B}}=n_{\text{B}}{M}_{\text{B}}$                                                                                                              (II)

Here, number of moles of boron is $n_{\text{B}}$ and molar mass of boron is $M_{\text{B}}$.

Write the expression to calculate mass of silicon in the metallic glass composition $\left(m_{\text{Si}}\right)$.

$m_{\text{Si}}=n_{\text{Si}}{M}_{\text{Si}}$                                                                                                            (III)

Here, number of moles of boron is $n_{\text{Si}}$ and molar mass of boron is $M_{\text{Si}}$.

Write the expression to calculate total mass of the metallic glass composition $\left(m\right)$.

$m=m_{\text{Fe}}+m_{\text{B}}+m_{\text{Si}}$                                                                                                (IV)

Write the expression to calculate weight percentage of iron in the metallic glass composition $\left(\%W_{\text{Fe}}\right)$.

$\%W_{\text{Fe}}=\frac{m_{\text{Fe}}}{m}\times 100\%$                                                                                               (V)

Write the expression to calculate weight percentage of boron in the metallic glass composition $\left(\%W_{\text{B}}\right)$.

$\%W_{\text{B}}=\frac{m_{\text{B}}}{m}\times 100\%$                                                                                              (VI)

Write the expression to calculate weight percentage of silicon in the metallic glass composition $\left(\%W_{\text{Si}}\right)$.

$\%W_{\text{Si}}=\frac{m_{\text{Si}}}{m}\times 100\%$                                                                                              (VII)

Conclusion:

Substitute $0.78\text{mol}$ for $n_{\text{Fe}}$ and $55.85\text{g}/\text{mol}$ for $M_{\text{Fe}}$ in Equation (I).

$\begin{array}{c}{m}_{\text{Fe}}=\left(0.78\text{mol}\right)\left(55.85\text{g}/\text{mol}\right)\\ =43.56\text{g}\end{array}$

Substitute $0.13\text{mol}$ for $n_{\text{B}}$ and $10.81\text{g}/\text{mol}$ for $M_{\text{B}}$ in Equation (II).

$\begin{array}{c}{m}_{\text{B}}=\left(0.13\text{mol}\right)\left(10.81\text{g}/\text{mol}\right)\\ =1.41\text{g}\end{array}$

Substitute $0.09\text{mol}$ for $n_{\text{Si}}$ and $28.09\text{g}/\text{mol}$ for $M_{\text{Si}}$ in Equation (III).

$\begin{array}{c}{m}_{\text{Si}}=\left(0.09\text{mol}\right)\left(28.09\text{g}/\text{mol}\right)\\ =2.53\text{g}\end{array}$

Substitute $43.56\text{g}$ for $m_{\text{Fe}}$, $1.41\text{g}$ for $m_{\text{B}}$ and $2.53\text{g}$ for $m_{\text{Si}}$ in Equation (IV).

$\begin{array}{c}m=43.56\text{g}+1.41\text{g}+2.53\text{g}\\ =47.50\text{g}\end{array}$

Substitute $43.56\text{g}$ for $m_{\text{Fe}}$ and $47.50\text{g}$ for $m$ in Equation (V).

$\begin{array}{c}\%W_{\text{Fe}}=\frac{43.56\text{g}}{47.50\text{g}}\times 100\%\\ =91.7\%\end{array}$

Thus, the weight percentage of iron in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$ is $91.7\%$.

Substitute $1.41\text{g}$ for $m_{\text{B}}$ and $47.50\text{g}$ for $m$ in Equation (VI).

$\begin{array}{c}\%W_{\text{B}}=\frac{1.41\text{g}}{47.50\text{g}}\times 100\%\\ =2.97\%\end{array}$

Thus, the weight percentage of boron in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$ is $2.97\%$.

Substitute $2.53\text{g}$ for $m_{\text{Si}}$ and $47.50\text{g}$ for $m$ in Equation (VII).

$\begin{array}{c}\%W_{\text{Si}}=\frac{2.53\text{g}}{47.50\text{g}}\times 100\%\\ =5.33\%\end{array}$

Thus, the weight percentage of silicon in the metallic glass composition of ${\text{Fe}}_{\text{78}}{\text{B}}_{\text{13}}{\text{Si}}_{\text{9}}$ is $5.33\%$.