Chapter 4, Problem 3CP

(a)

Interpretation Introduction

Interpretation:

The number of molecules that are found in $1.0\text{ lb}$ of sucrose has to be calculated.

Concept Introduction:

Steps to find number of molecules from mass:

  $\left(\text{Gram}\right)\underset{\text{g}\times \frac{\text{1 mol}}{\text{g}}=\text{mol}}{\overset{\text{Divide by molar mass}}{\to }}\left(\text{moles}\right)\underset{\text{mol}\times \frac{6.022\times {10}^{23}\text{ particles}}{\text{1 mol}}=\text{particles}}{\overset{\text{multiply by Avogadro's number }}{\to }}\left(\begin{array}{l}\text{Particles (or)}\\ \text{molecules}\end{array}\right)$

Explanation of Solution

Given:

$1.0\text{ lb}$ of sucrose, ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}$

The molar mass of sucrose is $\text{342}\text{.3 g/mol}$

Determination of number of molecules:

Convert the given amount of sucrose into grams.

  $\text{1}\text{.0 lb sucrose = 453}\text{.592 gram sucrose}$

Determine the moles of sucrose present in given amount as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{453}\text{.592 gram}}{\text{342}\text{.3 g/mol}}\\ =1.325\text{ mol}\end{array}$

The number of molecules is found as follows,

  $1.325\text{ mol}\times \frac{6.022\times {10}^{23}\text{ molecules}}{\text{ 1 mol}}=7.979\times {10}^{23}\text{ molecules}$

Therefore, the number of molecules of sucrose is $7.979\times {10}^{23}\text{ molecules}$.

(b)

Interpretation Introduction

Interpretation:

The number of molecules that are found in $\text{1}\text{.57 kg}$ of ${\text{N}}_{\text{2}}\text{O}$ has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given:

$\text{1}\text{.57 kg}$ of anesthetic, ${\text{N}}_{\text{2}}\text{O}$

The molar mass of ${\text{N}}_{\text{2}}\text{O}$ is $\text{44}\text{.013 g/mol}$

Determination of number of molecules:

Convert the given amount of ${\text{N}}_{\text{2}}\text{O}$ into grams.

  $\text{1}\text{.57 kg = 1570 gram}$

Determine the moles of ${\text{N}}_{\text{2}}\text{O}$ present in given amount as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{1570 gram}}{\text{44}\text{.013 g/mol}}\\ =35.671\text{ mol}\end{array}$

The number of molecules is found as follows,

  $35.671\text{ mol}\times \frac{6.022\times {10}^{23}\text{ molecules}}{\text{ 1 mol}}=2.148\times {10}^{25}\text{ molecules}$

Therefore, the number of molecules of ${\text{N}}_{\text{2}}\text{O}$ is $2.148\times {10}^{25}\text{ molecules}$.

(c)

Interpretation Introduction

Interpretation:

The number of molecules that are found in $4\times {10}^5\text{ tons}$ of ${\text{SO}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given:

$4\times {10}^5\text{ tons}$ of ${\text{SO}}_{\text{2}}$

The molar mass of ${\text{SO}}_{\text{2}}$ is $\text{64}\text{.066 g/mol}$

Determination of number of molecules:

Convert the given amount of ${\text{SO}}_{\text{2}}$ into grams.

  $\text{4}\times {\text{10}}^5\text{ tons = 3}\text{.6287}\times {\text{10}}^{11}\text{ gram}$

Determine the moles of ${\text{SO}}_{\text{2}}$ present in given amount as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{3}\text{.6287}\times {\text{10}}^{11}\text{ gram }}{\text{64}\text{.066 g/mol}}\\ =5.66\times {10}^9\text{ mol}\end{array}$

The number of molecules is found as follows,

  $5.66\times {10}^9\text{ mol}\times \frac{6.022\times {10}^{23}\text{ molecules}}{\text{ 1 mol}}=3.408\times {10}^{33}\text{ molecules}$

Therefore, the number of molecules of ${\text{SO}}_{\text{2}}$ is $3.408\times {10}^{33}\text{ molecules}$.

(d)

Interpretation Introduction

Interpretation:

The number of molecules that are found in $25.0\text{ lb}$ of ${\text{SiO}}_{\text{2}}$  has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given:

$25.0\text{ lb}$ of ${\text{SiO}}_{\text{2}}$

The molar mass of ${\text{SiO}}_{\text{2}}$ is $\text{60}\text{.08 g/mol}$

Determination of number of molecules:

Convert the given amount of ${\text{SiO}}_{\text{2}}$ into grams.

  $\text{25}{\text{.0 lb SiO}}_{\text{2}}\text{ = 11339}{\text{.8 gram SiO}}_{\text{2}}$

Determine the moles of ${\text{SiO}}_{\text{2}}$ present in given amount as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{11339}\text{.8 gram}}{\text{60}\text{.08 g/mol}}\\ =188.745\text{ mol}\end{array}$

The number of molecules is found as follows,

  $188.745\text{ mol}\times \frac{6.022\times {10}^{23}\text{ molecules}}{\text{ 1 mol}}=1.1366\times {10}^{26}\text{ molecules}$

Therefore, the number of molecules of ${\text{SiO}}_{\text{2}}$ is $1.1366\times {10}^{26}\text{ molecules}$.