Chapter 3, Problem 3.18P

(a)

Interpretation Introduction

Interpretation:

The mass $\left(\text{g}\right)$ of $\text{8}\text{.35 mol}$ of $\text{copper}\left(\text{I}\right)\text{carbonate}$ is to be calculated.

Concept introduction:

Molar mass of a substance is defined as the mass of one mole of a chemical entity. It is calculated as the summation of product of number of atoms of each element and the atomic mass of the respective element.

Answer to Problem 3.18P

The mass $\left(\text{g}\right)$ of $\text{8}\text{.35 mol}$ of $\text{copper}\left(\text{I}\right)\text{carbonate}$ is $1.56\times {10}^3\text{ g}$.

Explanation of Solution

$\text{Copper}\left(\text{I}\right)\text{carbonate}$ is an ionic compound with chemical formula ${\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}$.

The formula ot calculate the molar mass of ${\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}$ is,

${\text{Molar Mass of Cu}}_{\text{2}}{\text{CO}}_{\text{3}}=\left[\begin{array}{l}\left(\text{number of Cu atoms}\right)\left(\text{atomic mass}\text{of Cu}\right)\\ +\left(\text{number of}\text{C atoms}\right)\left(\text{atomic}\text{mass}\text{of C}\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]$ (1)

Substitute $2$ for number of $\text{Cu}$ atoms, $\text{63}\text{.55 g/mol}$ for the atomic mass of $\text{Cu}$, $1$ for number of $\text{C}$ atoms, $12.01\text{g/mol}$ for the atomic mass of $\text{C}$, $3$ for number of $\text{O}$ atoms and $16.00\text{g/mol}$ for the atomic mass of $\text{O}$ in equation (1).

$\begin{array}{c}{\text{Molar Mass of Cu}}_{\text{2}}{\text{CO}}_{\text{3}}=\left[\left(2\right)\left(\text{63}\text{.55 g/mol}\right)+\left(1\right)\left(12.01\text{g/mol}\right)+\left(3\right)\left(16.00\text{g/mol}\right)\right]\\ =127.1\text{ g/mol}+12.01\text{g/mol}+48.00\text{g/mol}\\ =187.11\text{g/mol}\end{array}$

The expression to calculate mass $\left(\text{g}\right)$ of ${\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}$ is,

$\text{Mass (g)}=\left({\text{mol of Cu}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\left(\frac{\text{molecular mass}\left(\text{g}\right){\text{ of Cu}}_{\text{2}}{\text{CO}}_{\text{3}}}{{\text{1 mol Cu}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)$ (2)

Substitute $\text{8}\text{.35 mol}$ for mol of ${\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}$ and $187.11\text{g/mol}$ for molecular mass of ${\text{Cu}}_{\text{2}}{\text{CO}}_{\text{3}}$ in equation (2).

$\begin{array}{c}{\text{Mass (g) of Cu}}_{\text{2}}{\text{CO}}_{\text{3}}=\left(\text{8}\text{.35 mol}\right)\left(\frac{\text{187}{\text{.11 g Cu}}_{\text{2}}{\text{CO}}_{\text{3}}}{{\text{1 mol Cu}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\\ =156.24\text{ g}\\ \text{=1}\text{.56}\times {\text{10}}^{\text{3}}{\text{ g Cu}}_{\text{2}}{\text{CO}}_{\text{3}}\end{array}$

Conclusion

The mass $\left(\text{g}\right)$ of $\text{8}\text{.35 mol}$ of $\text{copper}\left(\text{I}\right)\text{carbonate}$ is $1.56\times {10}^3\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

The mass $\left(\text{g}\right)$ of $\text{4}\text{.04}\times {\text{10}}^{\text{20}}\text{ molecules}$ of dinitrogen pentoxide is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in $12\text{ g}$ of $\text{C-12}$. This number is called Avogadro’s number. The value of Avogadro’s number is $\text{6}\text{.022}\times {\text{10}}^{23}\left(\text{entitites/mol}\right)$.

Molar mass is defined as the mass of $1\text{ mol}$ of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

Following are the steps to calculate the mass of a chemical substance when a number of molecules are given.

Step 1: Determine the amount of substance in moles by using Avogadro’s number. The expression to calculate the moles of a chemical substance is as follows:

$\text{Amount }\left(\text{mol }\right)=\left(\text{Given molecules}\right)\left(\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ molecules}}\right)$

Step 2: Multiply the moles with the molar mass of the chemical substance to obtain the mass of chemical substance in grams. The formula to calculate the mass of a substance in grams is as follows:

$\text{Mass }\left(\text{g }\right)=\left(\text{Amount }\left(\text{mol}\right)\right)\left(\frac{\text{No}\text{. of grams}}{\text{1 mol}}\right)$

Answer to Problem 3.18P

The mass $\left(\text{g}\right)$ of $\text{4}\text{.04}\times {\text{10}}^{\text{20}}\text{ molecules}$ of dinitrogen pentoxide is $0.0725\text{ g}$.

Explanation of Solution

Dinitrogen pentoxide is an inorganic compound with chemical formula ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$.

The formula to calculate the molar mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is,

${\text{Molar Mass of N}}_{\text{2}}{\text{O}}_{\text{5}}=\left[\begin{array}{l}\left(\text{number of N atoms}\right)\left(\text{atomic mass}\text{of N}\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]$ (3)

Substitute $2$ for number of $\text{N}$ atoms, $\text{14}\text{.01 g/mol}$ for atomic mass of $\text{N}$, $5$ for number of $\text{O}$ atoms, $16.00\text{g/mol}$ for atomic mass of $\text{O}$ in equation (3).

$\begin{array}{c}{\text{Molar Mass of N}}_{\text{2}}{\text{O}}_{\text{5}}=\left[\left(2\right)\left(\text{14}\text{.01 g/mol}\right)+\left(5\right)\left(16.00\text{g/mol }\right)\right]\\ =28.02\text{ g/mol}+80.00\text{g/mol}\\ =108.02\text{g/mol}\end{array}$

One mole of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ contains $6.022\times {10}^{23}{\text{ N}}_{\text{2}}{\text{O}}_{\text{5}}\text{ molecules}$. Divide $\text{4}\text{.04}\times {\text{10}}^{\text{20}}\text{ molecules}$ of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ by the Avogadro’s number to calculate the moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ as follows:

$\begin{array}{c}{\text{Amount of N}}_{\text{2}}{\text{O}}_{\text{5}}\left(\text{mol}\right)=\left(\text{4}\text{.04}\times {\text{10}}^{\text{20}}{\text{ N}}_{\text{2}}{\text{O}}_{\text{5}}\text{ molecules}\right)\left(\frac{1{\text{ mol N}}_{\text{2}}{\text{O}}_{\text{5}}}{6.022\times {10}^{23}{\text{ N}}_{\text{2}}{\text{O}}_{\text{5}}\text{ molecules}}\right)\\ =6.7087\times {10}^{-4}{\text{ mol N}}_{\text{2}}{\text{O}}_{\text{5}}\end{array}$

The expression to calculate mass $\left(\text{g}\right)$ of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is,

$\text{Mass (g)}=\left({\text{mol of N}}_{\text{2}}{\text{O}}_{\text{5}}\right)\left(\frac{\text{molecular mass}\left(\text{g}\right){\text{ of N}}_{\text{2}}{\text{O}}_{\text{5}}}{{\text{1 mol N}}_{\text{2}}{\text{O}}_{\text{5}}}\right)$ (4)

Substitute $6.7087\times {10}^{-4}\text{ mol}$ for mol of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ and $108.02\text{g/mol}$ for molecular mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ in equation (4).

$\begin{array}{c}{\text{Mass (g) of N}}_{\text{2}}{\text{O}}_{\text{5}}=\left(6.7087\times {10}^{-4}\text{ mol}\right)\left(\frac{\text{ 108}{\text{.02 g N}}_{\text{2}}{\text{O}}_{\text{5}}}{{\text{1 mol N}}_{\text{2}}{\text{O}}_{\text{5}}}\right)\\ =0.072467\text{ g}\\ =\text{0}{\text{.0725 g N}}_{\text{2}}{\text{O}}_{\text{5}}\end{array}$

Conclusion

The mass $\left(\text{g}\right)$ of $\text{4}\text{.04}\times {\text{10}}^{\text{20}}\text{ molecules}$ of dinitrogen pentoxide is $0.0725\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

The number of moles and formula unit in $78.9\text{ g}$ of sodium perchlorate is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in $12\text{ g}$ of $\text{C-12}$. This number is called Avogadro’s number. The value of Avogadro’s number is $\text{6}\text{.022}\times {\text{10}}^{23}\left(\text{entitites/mol}\right)$.

Molar mass is defined as the mass of $1\text{ mol}$ of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

A formula unit is used for the ionic compound to represent their empirical formula. The steps to determine the formula unit of an ionic compound from the given mass are as follows:

Step 1: Divide the given mass of the ionic compound with the molar mass to calculate the moles. The expression to calculate the moles of an ionic compound when the mass is given is as follows:

$\text{Amount}\left(\text{mol}\right)=\left(\text{Mass }\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Step 2: Multiply the calculated moles with the Avogadro’s number to determine the formula units of an ionic compound. The expression to determine the formula unit is as follows:

$\text{Number of formula units}=\left(\text{Amount }\left(\text{mol}\right)\right)\left(\frac{6.022\times {10}^{23}\text{ formula units}}{1\text{ mol}}\right)$

Answer to Problem 3.18P

The number of moles and formula unit in $78.9\text{ g}$ of sodium perchlorate is $0.644\text{ mol}$ and $3.88\times {10}^{23}\text{ FU}$ respectively.

Explanation of Solution

Sodium perchlorate is an ionic compound with chemical formula ${\text{NaClO}}_{\text{4}}$.

The formula to calculate the molar mass of ${\text{NaClO}}_{\text{4}}$ is,

${\text{Molar Mass of NaClO}}_{\text{4}}=\left[\begin{array}{l}\left(\text{number of Na atoms}\right)\left(\text{atomic mass}\text{of Na}\right)\\ +\left(\text{number of}\text{Cl atoms}\right)\left(\text{atomic}\text{mass}\text{of Cl}\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]$ (5)

Substitute $1$ for number of $\text{Na}$ atoms, $\text{22}\text{.99 g/mol}$ for the atomic mass of $\text{Na}$, 1 for number of $\text{Cl}$ atoms, $35.45\text{g/mol}$ for the atomic mass of $\text{Cl}$, $4$ for number of $\text{O}$ atoms and $16.00\text{g/mol}$ for the atomic mass of $\text{O}$ in equation (5).

$\begin{array}{c}{\text{Molar Mass of NaClO}}_{\text{4}}=\left[\begin{array}{l}\left(1\right)\left(\text{22}\text{.99 g/mol}\right)+\left(1\right)\left(35.45\text{g/mol }\right)\\ +\left(4\right)\left(16.00\text{g/mol }\right)\end{array}\right]\\ =\text{22}\text{.99 g/mol}+35.45\text{g/mol}+64.00\text{g/mol}\\ =122.44\text{g/mol}\end{array}$

The expression to calculate moles of ${\text{NaClO}}_{\text{4}}$ is:

$\text{Moles}\text{of}{\text{NaClO}}_{\text{4}}=\left({\text{given mass of NaClO}}_{\text{4}}\left(\text{g}\right)\right)\left(\frac{1{\text{ mol NaClO}}_{\text{4}}}{\text{molecular mass}\left(\text{g}\right)}\right)$ (6)

Substitute $78.9\text{ g}$ for given mass of ${\text{NaClO}}_{\text{4}}$ and $122.44\text{g/mol}$ for the molecular mass of ${\text{NaClO}}_{\text{4}}$ in equation (6).

$\begin{array}{c}\text{Moles of}\text{}\text{}{\text{NaClO}}_{\text{4}}=\left(\text{78}\text{.9 g}{\text{NaClO}}_{\text{4}}\right)\left(\frac{{\text{1mol NaClO}}_{\text{4}}}{122.44{\text{ g NaClO}}_{\text{4}}}\right)\\ =0.644397\text{mol}\\ =0.644{\text{ mol NaClO}}_{\text{4}}\end{array}$

The expression to calculate formula units (FU) of ${\text{NaClO}}_{\text{4}}$ is:

${\text{FU of NaClO}}_{\text{4}}=\left({\text{mol of NaClO}}_{\text{4}}\right)\left(\frac{6.022\times {10}^{23}{\text{ FU NaClO}}_{\text{4}}}{{\text{1mol NaClO}}_{\text{4}}}\right)$ (7)

Substitute $0.644397\text{mol}$ for mol of ${\text{NaClO}}_{\text{4}}$ in equation (7).

$\begin{array}{c}{\text{FU of NaClO}}_{\text{4}}=\left(\text{0}{\text{.644397mol NaClO}}_{\text{4}}\right)\left(\frac{6.022\times {10}^{23}{\text{ FU NaClO}}_{\text{4}}}{{\text{1mol NaClO}}_{\text{4}}}\right)\\ =3.88056\times {10}^{23}\text{ FU}\\ =\text{3}\text{.88}\times {10}^{23}{\text{ FU NaClO}}_{\text{4}}\end{array}$

Conclusion

The number of moles and formula unit in $78.9\text{ g}$ of sodium perchlorate is $0.644\text{ mol}$ and $3.88\times {10}^{23}\text{ FU}$ respectively.

(d)

Interpretation Introduction

Interpretation:

The number of sodium ions, perchlorate ions, chlorine atoms and oxygen atoms in the mass of compound ${\text{NaClO}}_{\text{4}}$ is to be calculated.

Concept introduction:

A formula unit is used for the ionic compound to represent their empirical formula. The molecular formula of a compound tells the number of atoms/ions of each element present in the compound.

Number of ions in a chemical compound is directly linked to the formula unit of the compound.

Answer to Problem 3.18P

The number of sodium ions, perchlorate ions, chlorine atoms are $3.88\times {10}^{23}$ and oxygen atoms are $1.55\times {10}^{24}$ in the mass of compound ${\text{NaClO}}_{\text{4}}$.

Explanation of Solution

From the formula of ${\text{NaClO}}_{\text{4}}$, it is concluded that 1 sodium ion, one perchlorate ion, one chlorine atom and four oxygen atoms are present in one formula unit of ${\text{NaClO}}_{\text{4}}$.

The expression to calculate the number of ions/atoms in ${\text{NaClO}}_{\text{4}}$ is:

$\text{Number of ions}=\left({\text{FU of NaClO}}_{\text{4}}\right)\left(\frac{\text{Number of ions}/\text{atoms}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)$ (8)

Substitute $\text{3}\text{.88056}\times {10}^{23}\text{ FU}$ for ${\text{NaClO}}_{\text{4}}$ and $1$ for number of ${\text{Na}}^{+}$ ion in equation (8).

$\begin{array}{c}{\text{Number of Na}}^{\text{+}}\text{ ions}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{1{\text{ Na}}^{\text{+}}\text{ ion}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =3.88\times {10}^{23}{\text{ Na}}^{+}\text{ ions}\end{array}$

Substitute $\text{3}\text{.88056}\times {10}^{23}\text{ FU}$ for ${\text{NaClO}}_{\text{4}}$ and $1$ for number of ${\text{ClO}}_{\text{4}}{}^{-}$ ion in equation (8).

$\begin{array}{c}{\text{Number of ClO}}_{\text{4}}{}^{-}\text{ ions}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{1{\text{ ClO}}_{\text{4}}{}^{-}\text{ ion}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =3.88\times {10}^{23}{\text{ ClO}}_{\text{4}}{}^{-}\text{ ions}\end{array}$

Substitute $\text{3}\text{.88056}\times {10}^{23}\text{ FU}$ for ${\text{NaClO}}_{\text{4}}$ and $1$ for number of $\text{Cl}$ atom in equation (8).

$\begin{array}{c}\text{Number of Cl atoms}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{1\text{ Cl atom}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =3.88\times {10}^{23}\text{ Cl atoms}\end{array}$

Substitute $\text{3}\text{.88056}\times {10}^{23}\text{ FU}$ for ${\text{NaClO}}_{\text{4}}$ and $4$ for number of $\text{O}$ atom in equation (8).

$\begin{array}{c}\text{Number of O atoms}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{\text{ 4 O atom}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =1.55\times {10}^{24}\text{ O atoms}\end{array}$

Conclusion

The number of sodium ions, perchlorate ions, chlorine atoms are $3.88\times {10}^{23}$ and oxygen atoms are $1.55\times {10}^{24}$ in the mass of compound ${\text{NaClO}}_{\text{4}}$.