Chapter 2, Problem 2.116AP

(a)

Interpretation Introduction

To determine: The percentage of aluminum present in $12.04\text{carats}$ of ruby.

Interpretation: The percentage of aluminum present in $12.04\text{carats}$ of ruby is to be calculated.

Concept introduction: The mass of aluminum present in ruby is calculated by the formula,

$m_{\text{Al}}=\frac{M_{\text{Al}}}{M_R}\times m_{\text{R}}$

Answer to Problem 2.116AP

The percentage of aluminum present in ruby is $\underset{\_}{52.90\%}$.

Explanation of Solution

Given

The chemical formula of ruby is ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$.

The mass of ruby is $12.04\text{carats}$.

The mass of aluminum present in ruby is calculated by the formula,

$m_{\text{Al}}=\frac{M_{\text{Al}}}{M_R}\times m_{\text{R}}$ (1)

Where,

• $M_{\text{Al}}$ is the molar mass of aluminum.
• $M_R$ is the molar mass of ruby.
• $m_{\text{R}}$ is the given mass of ruby.

The relationship between $\text{carat}$ and $\text{mg}$ is,

$1\text{carat}=200.0\text{mg}$

Convert $12.04\text{carats}$ to $\text{mg}$ using the above relation.

$\begin{array}{c}12.04\text{carats}=12.04\times 200.0\text{mg}\\ =2408\text{mg}\end{array}$

The relationship between $\text{mg}$ and $\text{g}$ is,

$1\text{g}=1000\text{mg}$

Convert $2408\text{mg}$ to $\text{g}$ using above relation.

$\begin{array}{c}2408\text{mg}=\frac{2408}{1000}\text{g}\\ =2.408\text{g}\end{array}$

The molar mass of aluminum is $27\text{g}/\text{mol}$.

The molar mass of ruby $\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\right)$ is $\begin{array}{c}=2\text{Al}+3\text{O}\\ =2\times 27\text{g}/\text{mol}+3\times 16\text{g}/\text{mol}\\ =102\text{g}/\text{mol}\end{array}$

The molar mass of two aluminum atoms is $\begin{array}{c}=2\text{Al}\\ =2\times 27\text{g}/\text{mol}\\ =54\text{g}/\text{mol}\end{array}$

Substitute the molar mass of ruby, molar mass of aluminum and given mass of ruby in equation (1).

$\begin{array}{c}{m}_{\text{Al}}=\frac{54\text{g}/\text{mol}}{102\text{g}/\text{mol}}\times 2.408\text{g}\\ =0.5294\times 2.408\text{g}\\ =1.274\text{g}\end{array}$

The percentage of aluminum is calculated by the formula,

$P_{\%}=\frac{m_{\text{Al}}}{m_{\text{R}}}\times 100$

Substitute the mass of aluminum present in ruby and mass of ruby in the above formula.

$\begin{array}{c}{P}_{\%}=\frac{1.274\text{g}}{2.408\text{g}}\times 100\\ =52.90\%\end{array}$

The percentage of aluminum present in ruby is $\underset{\_}{52.90\%}$.

(b)

Interpretation Introduction

To determine: The volume of ruby.

Interpretation: The density of ruby is given to be $4.02{\text{g/cm}}^3$. The volume of ruby is to be calculated.

Concept introduction: The volume of the ruby is calculated by the formula,

$V=\frac{m_{\text{R}}}{d}$

Answer to Problem 2.116AP

The volume of ruby is $\underset{\_}{0.5990ml}$.

Explanation of Solution

The volume of the ruby is calculated by the formula,

$V=\frac{m_{\text{R}}}{d}$

Where,

• $m_{\text{R}}$ is the mass of ruby.
• $d$ is the density of ruby.

The mass of ruby is $2.408\text{g}$.

Substitute the mass and density of ruby in the above formula.

$\begin{array}{c}V=\frac{2.408\text{g}}{4.02\text{g}/{\text{cm}}^3}\\ =0.5990{\text{cm}}^3\end{array}$

The relationship between ${\text{cm}}^{\text{3}}$ and $\text{ml}$ is,

$1{\text{cm}}^3=1\text{ml}$

Convert $0.5990{\text{cm}}^3$ to $\text{ml}$.

$0.5990{\text{cm}}^3=0.5990\text{ml}$

Therefore, the volume of ruby is $\underset{\_}{0.5990ml}$.

Conclusion

1. a. The percentage of aluminum present in ruby is $\underset{\_}{52.90\%}$.
2. b. The volume of ruby is $\underset{\_}{0.5990ml}$.