Chapter 2, Problem 2.44E

Write three relationships (equalities) based on the mole concept for each of the following elements:

a. phosphorus

b. aluminum

c. krypton

Interpretation Introduction

(a)

Interpretation:

The three relationships (equalities) based on the mole concept for given elements are to be stated.

Concept introduction:

Molar mass is the sum of the atomic masses of all the atoms present in the chemical formula of any compound. One mole of substance contains $6.02\times {10}^{23}$ atoms or molecules. This number is known as the Avogadro number.

Answer to Problem 2.44E

The three relationships (equalities) based on the mole concept for given element are,

$30.97\text{g}\text{P}=1\text{mol}\text{P}$…(1)

$1\text{mol}\text{P}=6.02\times {10}^{23}\text{P}\text{atoms}$…(2)

$6.02\times {10}^{23}\text{P}\text{atoms}=30.97\text{g}\text{P}$…(3)

Explanation of Solution

The molar mass of phosphorus is $30.97\text{g}/\text{mol}$. This is the mass of one mole of phosphorus. The relationship between the molar mass and moles is as follows:

$30.97\text{g}\text{P}=1\text{mol}\text{P}$…(1)

The conversion factor corresponding to the given relationship is,

$\frac{30.97\text{g}\text{P}}{1\text{mol}}$ and $\frac{1\text{mol}}{30.97\text{g}\text{P}}$

One mole of phosphorus contains $6.02\times {10}^{23}$ atoms. The relationship between the number of atoms and moles is shown below.

$1\text{mol}\text{P}=6.02\times {10}^{23}\text{P}\text{atoms}$…(2)

The conversion factor corresponding to the given relationship is,

$\frac{6.02\times {10}^{23}\text{atoms}}{1\text{mol}}$ and $\frac{1\text{mol}}{6.02\times {10}^{23}\text{atoms}}$

From the equalities (1) and (2), the mass of $6.02\times {10}^{23}$ atoms of phosphorus is $30.97\text{g}$. The relationship between molar mass and number of atoms is shown below.

$6.02\times {10}^{23}\text{P}\text{atoms}=30.97\text{g}\text{P}$…(3)

The conversion factor corresponding to the given relationship is,

$\frac{6.02\times {10}^{23}\text{atoms}}{30.97\text{g}\text{P}}$ and $\frac{30.97\text{g}\text{P}}{6.02\times {10}^{23}\text{atoms}}$

Conclusion

The three relationships (equalities) based on the mole concept for given element are,

$30.97\text{g}\text{P}=1\text{mol}\text{P}$…(1)

$1\text{mol}\text{P}=6.02\times {10}^{23}\text{P}\text{atoms}$…(2)

$6.02\times {10}^{23}\text{P}\text{atoms}=30.97\text{g}\text{P}$…(3)

Interpretation Introduction

(b)

Interpretation:

The three relationships (equalities) based on the mole concept for given element are to be stated.

Concept introduction:

Molar mass is the sum of the atomic masses of all the atoms present in the chemical formula of any compound. One mole of substance contains $6.02\times {10}^{23}$ atoms or molecules. This number is known as the Avogadro number.

Answer to Problem 2.44E

The three relationships (equalities) based on the mole concept for given element are,

$26.98\text{g}\text{Al}=1\text{mol}\text{Al}$…(1)

$1\text{mol}\text{Al}=6.02\times {10}^{23}\text{Al}\text{atoms}$…(2)

$6.02\times {10}^{23}\text{Al}\text{atoms}=26.98\text{g}\text{Al}$…(3)

Explanation of Solution

The molar mass of aluminum is $26.98\text{g}/\text{mol}$. This is the mass of one mole of aluminum. The relationship between the molar mass and moles is as follows:

$26.98\text{g}\text{Al}=1\text{mol}\text{Al}$…(1)

The conversion factor corresponding to the given relationship is,

$\frac{26.98\text{g}\text{Al}}{1\text{mol}}$ and $\frac{1\text{mol}}{26.98\text{g}\text{Al}}$

One mole of aluminum contains $6.02\times {10}^{23}$ atoms. The relationship between the number of atoms and moles is as follows:

$1\text{mol}\text{Al}=6.02\times {10}^{23}\text{Al}\text{atoms}$…(2)

The conversion factor corresponding to the given relationship is,

$\frac{6.02\times {10}^{23}\text{atoms}}{1\text{mol}}$ and $\frac{1\text{mol}}{6.02\times {10}^{23}\text{atoms}}$

From the equalities (1) and (2), the mass of $6.02\times {10}^{23}$ atoms of aluminum is $26.98\text{g}$. The relationship between molar mass and number of atoms is shown below.

$6.02\times {10}^{23}\text{Al}\text{atoms}=26.98\text{g}\text{Al}$…(3)

The conversion factor corresponding to the given relationship is,

$\frac{6.02\times {10}^{23}\text{atoms}}{26.98\text{g}\text{Al}}$ and $\frac{26.98\text{g}\text{Al}}{6.02\times {10}^{23}\text{atoms}}$

Conclusion

The three relationships (equalities) based on the mole concept for given element are,

$26.98\text{g}\text{Al}=1\text{mol}\text{Al}$…(1)

$1\text{mol}\text{Al}=6.02\times {10}^{23}\text{Al}\text{atoms}$…(2)

$6.02\times {10}^{23}\text{Al}\text{atoms}=26.98\text{g}\text{Al}$…(3)

Interpretation Introduction

(c)

Interpretation:

The three relationships (equalities) based on the mole concept for given element are to be stated.

Concept introduction:

Molar mass is the sum of the atomic masses of all the atoms present in the chemical formula of any compound. One mole of substance contains $6.02\times {10}^{23}$ atoms or molecules. This number is known as the Avogadro number.

Answer to Problem 2.44E

The three relationships (equalities) based on the mole concept for given element are,

$83.80\text{g}\text{Kr}=1\text{mol}\text{Kr}$…(1)

$1\text{mol}\text{Kr}=6.02\times {10}^{23}\text{Kr}\text{atoms}$…(2)

$6.02\times {10}^{23}\text{Kr}\text{atoms}=83.80\text{g}\text{Kr}$…(3)

Explanation of Solution

The molar mass of krypton is $83.80\text{g}/\text{mol}$. This is the mass of one mole of krypton. The relationship between the molar mass and moles is shown below.

$83.80\text{g}\text{Kr}=1\text{mol}\text{Kr}$…(1)

The conversion factor corresponding to the given relationship is,

$\frac{83.80\text{g}\text{Kr}}{1\text{mol}}$ and $\frac{1\text{mol}}{83.80\text{g}\text{Kr}}$

One mole of krypton contains $6.02\times {10}^{23}$ atoms. The relationship between the number of atoms and moles is shown below.

$1\text{mol}\text{Kr}=6.02\times {10}^{23}\text{Kr}\text{atoms}$…(2)

The conversion factor corresponding to the given relationship is,

$\frac{6.02\times {10}^{23}\text{atoms}}{1\text{mol}}$ and $\frac{1\text{mol}}{6.02\times {10}^{23}\text{atoms}}$

From the equalities (1) and (2), the mass of $6.02\times {10}^{23}$ atoms of krypton is $83.80\text{g}$. The relationship between the molar mass and number of atoms is as follows:

$6.02\times {10}^{23}\text{Kr}\text{atoms}=83.80\text{g}\text{Kr}$…(3)

The conversion factor corresponding to the given relationship is,

$\frac{6.02\times {10}^{23}\text{atoms}}{83.80\text{g}\text{Kr}}$ and $\frac{83.80\text{g}\text{Kr}}{6.02\times {10}^{23}\text{atoms}}$

Conclusion

The three relationships (equalities) based on the mole concept for given element are,

$83.80\text{g}\text{Kr}=1\text{mol}\text{Kr}$…(1)

$1\text{mol}\text{Kr}=6.02\times {10}^{23}\text{Kr}\text{atoms}$…(2)

$6.02\times {10}^{23}\text{Kr}\text{atoms}=83.80\text{g}\text{Kr}$…(3)