Chapter 12, Problem 99A

Interpretation Introduction

Interpretation: The beryllium mass in $147\text{ g}$ of the mineral beryl $\left({\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}\right)$ is to be calculated.

Concept Introduction: To get the mass of a compound, multiply the molar mass of each component by the number of atoms of that component.

Answer to Problem 99A

The mass of $\text{Be}$ in ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is $7.3943\text{ g}$ .

Explanation of Solution

The given compound is the mineral beryl $\left({\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}\right)$ .

The given mass of ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is $147\text{ g}$ .

The molar mass of $\text{Be}$ is $9.012{\text{ g mol}}^{-1}$

The molar mass of $\text{Al}$ is $26.98{\text{ g mol}}^{-1}$

The molar mass of $\text{Si}$ is $28.09{\text{ g mol}}^{-1}$

The molar mass of $\text{O}$ is $16.00{\text{ g mol}}^{-1}$

The molar mass of ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is as follows:

  $\begin{array}{c}{\text{Molar mass Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}=\left(\begin{array}{l}3\times \text{Molar mass Be+2}\times \text{Molar mass Al}\\ \text{+6}\times \text{Molar mass Si+18}\times \text{Molar mass O}\end{array}\right)\\ =\left(\begin{array}{c}3\times 9.012{\text{ g mol}}^{-1}+2\times 26.98{\text{ g mol}}^{-1}+6\times 28.09{\text{ g mol}}^{-1}\\ +18\times 16.00{\text{ g mol}}^{-1}\end{array}\right)\\ =27.036{\text{ g mol}}^{-1}+53.96{\text{ g mol}}^{-1}+168.54{\text{ g mol}}^{-1}+288.00{\text{ g mol}}^{-1}\\ =537.536{\text{ g mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is $537.536{\text{ g mol}}^{-1}$ .

The formula to calculate the number of moles is as follows:

  $n=\frac{w}{M}$

Where,

• $M$ is the molar mass.
• $n$ is the number of moles.
• $w$ is the mass.

Substitute the values of $w$ and $M$ in the above formula.

  $\begin{array}{c}n=\frac{w}{M}\\ =\frac{147\bcancel{\text{g}}}{537.536\bcancel{\text{g}}{\text{ mol}}^{-1}}\\ =0.2735\text{ mol}\end{array}$

Therefore, the number of moles is $0.2735\text{ mol}$ .

The one mole of ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ contains three moles of $\text{Be}$ .

The formula to calculate the moles of $\text{Be}$ in ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is as follows:

  $n_{\text{Be}}=\frac{3\text{ mol}\times n}{1\text{ mol}}$

Where,

• $n_{\text{Be}}$ is the moles of $\text{Be}$ .

Substitute the value of $n$ in the above formula.

  $\begin{array}{c}{n}_{\text{Be}}=\frac{3\text{ mol}\times n}{1\text{ mol}}\\ =\frac{3\bcancel{\text{mol}}\times 0.2735\text{ mol}}{1\bcancel{\text{mol}}}\\ =0.8205\text{ mol}\end{array}$

Therefore, the moles of $\text{Be}$ in ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is $0.8205\text{ mol}$ .

The formula to calculate the mass of $\text{Be}$ in ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is as follows:

  $w_{\text{Be}}=n_{\text{Be}}{M}_{\text{Be}}$

Where,

• $M_{\text{Be}}$ is the molar mass of $\text{Be}$ .
• $w_{\text{Be}}$ is the mass of $\text{Be}$ .

Substitute the values of $M_{\text{Be}}$ and $n_{\text{Be}}$ in the above formula.

  $\begin{array}{c}{w}_{\text{Be}}=n_{\text{Be}}{M}_{\text{Be}}\\ =0.8205\bcancel{\text{mol}}\times 9.012\text{ g }\bcancel{{\text{mol}}^{-1}}\\ =7.3943\text{ g}\end{array}$

Therefore, the mass of $\text{Be}$ in ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is $7.3943\text{ g}$ .

Conclusion

The mass of $\text{Be}$ in ${\text{Be}}_{\text{3}}{\text{Al}}_{\text{2}}{\text{Si}}_{\text{6}}{\text{O}}_{\text{18}}$ is $7.3943\text{ g}$ .