Chapter 4, Problem 53QP

How many formula units are in 250.0 g of the following substances?

$\begin{array}{l}\left(\text{a}\right){\text{ Br}}_{\text{2}}\\ \left(\text{b}\right){\text{ MgCl}}_{\text{2}}\\ \left(\text{c}\right){\text{ H}}_{\text{2}}\text{O}\\ \left(\text{d}\right)\text{ Fe}\end{array}$

Interpretation Introduction

Interpretation:

The number of formula units in $250\text{ g}$ of ${\text{Br}}_2$ .

Explanation of Solution

For calculating the number of formula units present, the conversion map is as follows:

$\text{Mass in grams}\stackrel{\text{MM}}{\to }\text{Moles}\stackrel{{\text{N}}_{\text{A}}}{\to }\text{Number of formula units}$

The units of molar mass is $\text{g/mol}$ . It connects moles and mass. The units of Avogadro’s number is $\text{molecule/mole}$ . It connects moles and molecules.

The number of formula units present in one sand grain is determined as follows:

$\text{Number of formula units}=\frac{\text{Given mass}}{\text{Molar mass}}\times {\text{N}}_{\text{A}}$ …..(1)

Here, ${\text{N}}_{\text{A}}$ is Avogadro’s number having a value $6.022\times {10}^{23}{\text{mol}}^{-1}$ .

The molar mass of ${\text{Br}}_2$ is $159.80\text{g/mol}$ . The weight of ${\text{Br}}_2$ is $250\text{g}$ . Substitute these values in above equation as follows:

$\begin{array}{c}\text{Formula units}{\text{Br}}_2=\frac{250\cancel{\text{g}}}{\text{159}\text{.80}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =9.421\times {10}^{23}\text{ formula units}{\text{Br}}_2\end{array}$

Therefore, $9.421\times {10}^{23}$ formula units are present in $250\text{ g}$ of ${\text{Br}}_2$ .

Interpretation Introduction

Interpretation:

The number of formula units in $250\text{ g}$ of ${\text{MgCl}}_2$ is to be determined.

Explanation of Solution

The molar mass of ${\text{MgCl}}_2$ is $95.21\text{g/mol}$ . The weight of ${\text{MgCl}}_2$ is $250\text{g}$ . Substitute these values in equation (1) to determine the number of formula units present in $250\text{ g}$ of ${\text{MgCl}}_2$ as follows:

$\begin{array}{c}\text{Formula units}{\text{MgCl}}_2=\frac{250\cancel{\text{g}}}{\text{95}\text{.21}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =1.581\times {10}^{24}\text{ formula units}{\text{MgCl}}_2\end{array}$

Therefore, $1.581\times {10}^{24}$ formula units are present in $250\text{ g}$ of ${\text{MgCl}}_2$ .

Interpretation Introduction

Interpretation:

The number of formula units in $250\text{ g}$ of ${\text{H}}_2\text{O}$ is to be determined.

Explanation of Solution

The molar mass of ${\text{H}}_2\text{O}$ is $18.02\text{g/mol}$ . The weight of ${\text{H}}_2\text{O}$ is $250\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $250\text{ g}$ of ${\text{H}}_2\text{O}$ as follows:

$\begin{array}{c}\text{Formula units}{\text{H}}_2\text{O}=\frac{\text{250 }\cancel{\text{g}}}{\text{18}\text{.02}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =8.355\times {10}^{24}{\text{ formula units H}}_2\text{O}\end{array}$

Therefore, $8.355\times {10}^{24}$ formula units are present in $250\text{ g}$ of ${\text{H}}_2\text{O}$ .

Interpretation Introduction

Interpretation:

The number of formula units in $250\text{ g}$ of Fe is to be determined.

Explanation of Solution

The molar mass of Fe is $55.85\text{g/mol}$ . The weight of Fe is $250\text{ g}$ . Substitute these values in equation (1) to determine the number of formula units present in $250\text{ g}$ of Fe as follows:

$\begin{array}{c}\text{Formula units}\text{Fe}=\frac{\text{250 }\cancel{\text{g}}}{\text{55}\text{.85}\cancel{\text{g}}\text{/}\cancel{\text{mol}}}\times 6.022\times {10}^{23}\cancel{{\text{mol}}^{-1}}\\ =2.696\times {10}^{24}\text{ formula units Fe}\end{array}$

Therefore, $2.696\times {10}^{24}$ formula units are present in $250\text{ g}$ of Fe.