Chapter 3, Problem 3.21QP

(a)

Interpretation Introduction

Interpretation: The atoms of titanium in $0.125\text{mole}$ of each of the given compound are to be calculated.

Concept introduction: A mole of a substance is the quantity of a substance which contains the same number of particles as the number of carbon atoms in exactly 12 grams for the carbon-12 isotopes. A mole of particles contains Avogadro’s number $\left(6.022\times {10}^{23}\right)$ of particles. The number of atoms in a substance is equal to the product of moles and Avogadro’s number.

To determine: The atoms of titanium in $0.125\text{mole}$ of ilmenite, ${\text{FeTiO}}_{\text{3}}$.

Answer to Problem 3.21QP

Solution

The number of atoms of titanium in $0.125\text{mole}$ of ${\text{FeTiO}}_{\text{3}}$ is $\underset{\_}{7.53\times 10^{22}atoms}$.

Explanation of Solution

Explanation

Given

One mole ilmenite contains Avogadro’s number of molecules of ilmenite. Therefore, one mole of ilmenite is equivalent to $6.022\times {10}^{23}$ molecules of ilmenite.

So, $1\text{mol}{\text{FeTiO}}_3=6.022\times {10}^{23}\text{molecules}{\text{FeTiO}}_3$.

One molecule of ${\text{FeTiO}}_{\text{3}}$ contains one atom of $\text{Fe}$, one atom of $\text{Ti}$ and three atoms of $\text{O}$.

The number of titanium atoms in $0.125\text{mole}$ of ${\text{FeTiO}}_{\text{3}}$ is calculated by using the formula,

$N=n\times N_A\times \text{number}\text{of}\text{Ti}\text{atoms}\text{in}\text{a}\text{molecule}$

Where,

• $N$ is the number of atoms.
• $N_A$ is the Avogadro’s number of atoms.
• $n$ is the number of moles.

Substitute the values of $n$ and $N_A$ in the above formula to calculate the number of titanium atoms.

$\begin{array}{c}N=0.125\text{mole}\times 6.022\times {10}^{23}\text{molecules/mole}\times \frac{1\text{Ti}\text{atom}}{\text{1}\text{molecule}{\text{FeTiO}}_{\text{3}}}\\ =0.125\times 6.022\times {10}^{23}\times 1\text{atoms}\\ =\underset{\_}{7.53\times 10^{22}atoms}\end{array}$

Hence, the number of atoms of titanium in $0.125\text{mole}$ of ${\text{FeTiO}}_{\text{3}}$ is $\underset{\_}{7.53\times 10^{22}atoms}$.

(b)

Interpretation Introduction

To determine: The atoms of titanium in $0.125\text{mole}$ of titanium (IV) chloride.

Answer to Problem 3.21QP

Solution

The number of atoms of titanium in $0.125\text{mole}$ of ${\text{TiCl}}_{\text{4}}$ is $\underset{\_}{7.53\times 10^{22}atoms}$.

Explanation of Solution

Explanation

Given

One mole titanium (IV) chloride $\left({\text{TiCl}}_{\text{4}}\right)$ contains Avogadro’s number of molecules of ilmenite. Therefore, one mole of ${\text{TiCl}}_{\text{4}}$ is equivalent to $6.022\times {10}^{23}$ molecules of ${\text{TiCl}}_{\text{4}}$.

So, $1\text{mol}{\text{TiCl}}_{\text{4}}=6.022\times {10}^{23}\text{molecules}{\text{TiCl}}_{\text{4}}$.

One molecule of ${\text{TiCl}}_{\text{4}}$ contains one atom of $\text{Ti}$ and four atoms of $\text{Cl}$.

The number of titanium atoms in $0.125\text{mole}$ of ${\text{TiCl}}_{\text{4}}$ is calculated by using the formula,

$N=n\times N_A\times \text{number}\text{of}\text{Ti}\text{atoms}\text{in}\text{a}\text{molecule}$

Where,

• $N$ is the number of atoms.
• $N_A$ is the Avogadro’s number of atoms.
• $n$ is the number of moles.

Substitute the values of $n$ and $N_A$ in the above formula to calculate the number of titanium atoms.

$\begin{array}{c}N=0.125\text{mole}\times 6.022\times {10}^{23}\text{molecules/mole}\times \frac{1\text{Ti}\text{atom}}{\text{1}\text{molecule}{\text{TiCl}}_4}\\ =0.125\times 6.022\times {10}^{23}\times 1\text{atoms}\\ =\underset{\_}{7.53\times 10^{22}atoms}\end{array}$

Hence, the number of atoms of titanium in $0.125\text{mole}$ of ${\text{TiCl}}_{\text{4}}$ is $\underset{\_}{7.53\times 10^{22}atoms}$.

(c)

Interpretation Introduction

To determine: The atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$.

Answer to Problem 3.21QP

Solution

The number of atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{1.51\times 10^{23}atoms}$.

Explanation of Solution

Explanation

Given

One mole ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ contains Avogadro’s number of molecules of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$. Therefore, one mole of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ is equivalent to $6.022\times {10}^{23}$ molecules of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$.

So, $1\text{mol}{\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}=6.022\times {10}^{23}\text{molecules}{\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$.

One molecule of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ contains two atoms of $\text{Ti}$ and three atoms of $\text{O}$.

The number of titanium atoms in $0.125\text{mole}$ of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated by using the formula,

$N=n\times N_A\times \text{number}\text{of}\text{Ti}\text{atoms}\text{in}\text{a}\text{molecule}$

Where,

• $N$ is the number of atoms.
• $N_A$ is the Avogadro’s number of atoms.
• $n$ is the number of moles.

Substitute the values of $n$ and $N_A$ in the above formula to calculate the number of titanium atoms.

$\begin{array}{c}N=0.125\text{mole}\times 6.022\times {10}^{23}\text{molecules/mole}\times \frac{2\text{Ti}\text{atom}}{\text{1}\text{molecule}{\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}}\\ =0.125\times 6.022\times {10}^{23}\times 2\text{atoms}\\ =\underset{\_}{1.51\times 10^{23}atoms}\end{array}$

Hence, the number of atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{1.51\times 10^{23}atoms}$.

(d)

Interpretation Introduction

To determine: The atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_3{\text{O}}_5$.

Answer to Problem 3.21QP

Solution

The number of atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_3{\text{O}}_5$ is $\underset{\_}{2.26\times 10^{23}atoms}$.

Explanation of Solution

Explanation

Given

One mole ${\text{Ti}}_3{\text{O}}_5$ contains Avogadro’s number of molecules of ${\text{Ti}}_3{\text{O}}_5$. Therefore, one mole of ${\text{Ti}}_3{\text{O}}_5$ is equivalent to $6.022\times {10}^{23}$ molecules of ${\text{Ti}}_3{\text{O}}_5$.

So, $1\text{mol}{\text{Ti}}_3{\text{O}}_5=6.022\times {10}^{23}\text{molecules}{\text{Ti}}_3{\text{O}}_5$.

One molecule of ${\text{Ti}}_3{\text{O}}_5$ contains three atoms of $\text{Ti}$ and five atoms of $\text{O}$.

The number of titanium atoms in $0.125\text{mole}$ of ${\text{Ti}}_3{\text{O}}_5$ is calculated by using the formula,

$N=n\times N_A\times \text{number}\text{of}\text{Ti}\text{atoms}\text{in}\text{a}\text{molecule}$

Where,

• $N$ is the number of atoms.
• $N_A$ is the Avogadro’s number of atoms.
• $n$ is the number of moles.

Substitute the values of $n$ and $N_A$ in the above formula to calculate the number of titanium atoms.

$\begin{array}{c}N=0.125\text{mole}\times 6.022\times {10}^{23}\text{molecules/mole}\times \frac{3\text{Ti}\text{atom}}{\text{1}\text{molecule}{\text{Ti}}_3{\text{O}}_5}\\ =0.125\times 6.022\times {10}^{23}\times 3\text{atoms}\\ =\underset{\_}{2.26\times 10^{23}atoms}\end{array}$

Hence, the number of atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_3{\text{O}}_5$ is $\underset{\_}{2.26\times 10^{23}atoms}$.

Conclusion

1. a. The number of atoms of titanium in $0.125\text{mole}$ of ${\text{FeTiO}}_{\text{3}}$ is $\underset{\_}{7.53\times 10^{22}atoms}$.
2. b. The number of atoms of titanium in $0.125\text{mole}$ of ${\text{TiCl}}_{\text{4}}$ is $\underset{\_}{7.53\times 10^{22}atoms}$.
3. c. The number of atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{1.51\times 10^{23}atoms}$.
4. d. The number of atoms of titanium in $0.125\text{mole}$ of ${\text{Ti}}_3{\text{O}}_5$ is $\underset{\_}{2.26\times 10^{23}atoms}$