Chapter 4, Problem 29P

4-45 Calculate the number of moles in:

(a) 32 g of methane, CH₄

(b) 345.6 g of nitric oxide, NO

(c) 184.4 g of chlorine dioxide, ClO₂

(d) 720. g of glycerin, C₃H₈O₃

Interpretation Introduction

(a)

Interpretation:

To calculate the number of moles in 32 g of Methane.

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, and molar mass comes out in grams.

Hence molar mass is the sum of 1 mole of atoms present in a given compound.

The formula to calculate number of moles present is given by-

$\text{Number of moles = }\frac{\text{Given mass}}{\text{Molar Mass}}$.

Answer to Problem 29P

The number of moles in 32 g of Methane is 2.

Explanation of Solution

In case of Methane, ${\text{CH}}_{\text{4}}$

Molar mass is calculated as sum of 4 Hydrogen atoms and 1 Carbon atom.

$\begin{array}{l}M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{C}{\text{H}}_4=Mass\text{ 4 }\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 \text{C}\text{a}rb\text{O}n\text{a}t\text{O}m.\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{C}{\text{H}}_4=\left(4\times 1\right)+\left(1\times 12\right)\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{C}{\text{H}}_4=16\text{ g}\hfill \end{array}$

Given mass = 32 g.

$\begin{array}{l}\text{Number of moles = }\frac{\text{Given mass}}{\text{Molar Mass}}\\ \text{Number of moles = }\frac{\text{32}}{\text{16}}\\ \text{Number of moles =2}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

To calculate the number of moles in 345.6 f of Nitric Oxide $\text{NO}$

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

Hence molar mass is the sum of 1 mole of atoms present in a given compound.

The formula to calculate number of moles present is given by-

$\text{Number of moles = }\frac{\text{Given mass}}{\text{Molar Mass}}$.

Answer to Problem 29P

The number of moles in 345.6 of Nitric Oxide $\text{NO}$ is 11.52.

Explanation of Solution

In case of Nitric Oxide $\text{NO}$

Molar mass is calculated as sum of 1 Nitrogen atoms and 1 Oxygen atom.

$\begin{array}{l}M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{N}\text{O}=Mass1\text{N}\text{i}tr\text{O}g\text{e}n\text{a}t\text{O}m+1oxyg\text{e}n\text{a}t\text{O}m\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{N}\text{O}=\left(1\times 14\right)+\left(1\times 16\right)\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{N}\text{O}=\text{ 30 g}\hfill \end{array}$

Given mass = 345.6 g.

$\begin{array}{l}Numb\text{e}rofm\text{O}\text{l}\text{e}s = \frac{G\text{i}v\text{e}nmass}{M\text{O}\text{l}\text{a}rMass} \\ Numb\text{e}rofm\text{O}\text{l}\text{e}s = \frac{345.6}{30} \\ Numb\text{e}rofm\text{O}\text{l}\text{e}s =11.52\end{array}$.

Interpretation Introduction

(c)

Interpretation:

To calculate the number of moles in 184.4 g of Chlorine dioxide ${\text{ClO}}_{\text{2}}$

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

Hence molar mass is the sum of 1 mole of atoms present in a given compound.

The formula to calculate number of moles present is given by-

$\text{Number of moles = }\frac{\text{Given mass}}{\text{Molar Mass}}$.

Answer to Problem 29P

The number of moles in 184.4 g of Chlorine dioxide ${\text{ClO}}_{\text{2}}$ is 2.73.

Explanation of Solution

In case of Chlorine dioxide ${\text{ClO}}_{\text{2}}$

Molar mass is calculated as sum of 1 Chlorine atom and 2 Oxygen atoms.

$\begin{array}{l}M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{C}\text{l}{\text{O}}_2=Mass1\text{C}\text{H}\text{l}\text{O}rin\text{e}\text{a}t\text{O}m\text{ + }2oxyg\text{e}n\text{a}t\text{O}ms\mathrm{..}\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{C}\text{l}{\text{O}}_2=\left(1\times 35.5\right)+\left(2\times 16\right)\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof\text{C}\text{l}{\text{O}}_2=\text{ 67}\text{.5 g}\hfill \end{array}$

Given mass = 184.4 g.

$\begin{array}{l}\text{Number of moles = }\frac{\text{Given mass}}{\text{Molar Mass}}\\ \text{Number of moles = }\frac{\text{184}\text{.4}}{\text{67}\text{.5}}\\ \text{Number of moles =2}\text{.73}\end{array}$.

Interpretation Introduction

(d)

Interpretation:

To calculate the number of moles in 720 g of glycerin ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}$

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

Hence molar mass is the sum of 1 mole of atoms present in a given compound.

The formula to calculate number of moles present is given by-

$\mathrm{Number}ofm\text{O}\text{l}\text{e}s = \frac{G\text{i}v\text{e}nmass}{M\text{O}\text{l}\text{a}rMass}$.

Answer to Problem 29P

The number of moles in 720 g of glycerin ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}$ is 7.82.

Explanation of Solution

In case of Methane, ${\text{CH}}_{\text{4}}$

Molar mass is calculated as sum of 8 Hydrogen atoms, 3 Oxygen atoms and 3 Carbon atoms.

$\begin{array}{l}M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_3{\text{H}}_8{\text{O}}_3=Mass8\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+3oxyg\text{e}n\text{a}t\text{O}ms+3 \text{C}\text{a}rb\text{O}n\text{a}t\text{O}ms\mathrm{..}\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_3{\text{H}}_8{\text{O}}_3=\left(8\times 1\right)+\left(3\times 16\right)+\left(3\times 12\right)\hfill \\ M\text{O}\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_3{\text{H}}_8{\text{O}}_3=92\text{ g}\hfill \end{array}$

Given mass = 720 g.

$\begin{array}{l}\text{Number of moles = }\frac{\text{Given mass}}{\text{Molar Mass}}\\ \text{Number of moles = }\frac{\text{720}}{\text{92}}\\ \text{Number of moles =7}\text{.82}\end{array}$.