Chapter 3, Problem 3.119P

(a)

Interpretation Introduction

Interpretation:

The number of moles present in $0.588\text{ g}$ of ammonium bromide is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

The formula to calculate mass is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molecular mass}\left(\text{g/mol}\right)\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.119P

The number of moles present in $0.588\text{ g}$ of ammonium bromide is $0.00600\text{ mol}$.

Explanation of Solution

The chemical formula for ammonium bromide is ${\text{NH}}_{\text{4}}\text{Br}$. The molar mass of ammonium bromide is $97.94\text{ g/mol}$.

Substitute $0.588\text{ g}$ for mass and $97.94\text{ g/mol}$ for molar mass in equation (1).

  $\begin{array}{c}{\text{Moles of NH}}_{\text{4}}\text{Br}\left(\text{mol}\right)\text{=}\frac{\text{0}{\text{.588 g of C}}_{\text{7}}{\text{H}}_{\text{8}}}{\text{97}\text{.94 g/mol }\text{}{\text{C}}_{\text{7}}{\text{H}}_{\text{8}}}\\ \text{= 0}\text{.006003675 mol}\\ \approx 0.00600\text{ mol}\text{.}\end{array}$

Conclusion

In $0.588\text{ g}$ of ammonium bromide, the number of moles present is found to be $0.00600\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

The number of potassium ions in $88.5\text{ g}$ of potassium nitrate is to be calculated.

Concept introduction:

The number of moles is calculated by the formula :

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

The formula to calculate mass from moles is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molecular mass}\left(\text{g/mol}\right)\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.119P

The number of potassium ions in $88.5\text{ g}$ of potassium nitrate is $5.27\times {10}^{23}$.

Explanation of Solution

The chemical formula for potassium nitrate is ${\text{KNO}}_{\text{3}}$. The molar mass of potassium nitrate is $\text{101}\text{.11 g/mol}$.

Substitute $\text{88}\text{.5 g}$ for mass and $97.94\text{ g/mol}$ for molar mass in equation (1).

  $\begin{array}{c}{\text{Moles of KNO}}_{\text{3}}\left(\text{mol}\right)\text{=}\frac{\text{88}\text{.5 g }}{\text{101}\text{.11 g/mol }}\\ \text{= 0}\text{.875284 mol}\end{array}$

The formula to evaluate the number of ions is as follows:

  $\text{number of ions}=\left(\text{number of moles}\right)\left(6.022\times {10}^{23}\right)$        (3)

Substitute $\text{ 0}\text{.875284 mol}$ for the number of moles in equation (3).

  $\begin{array}{c}\text{Number of potassium ions}=\left(\text{ 0}\text{.875284 mol}\right)\left(6.022\times {10}^{23}\text{ions/mol}\right)\\ =5.27096\times {10}^{23}\\ \approx 5.27\times {10}^{23}.\end{array}$

Conclusion

In $88.5\text{ g}$ of potassium nitrate, the number of potassium ions is $5.27\times {10}^{23}$.

(c)

Interpretation Introduction

Interpretation:

The mass of $5.85\text{ mol}$ glycerol is to be calculated.

Concept introduction:

The formula to calculate the moles is as follows:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

The formula to calculate mass from moles is as follows:

  $\text{Mass of a given sample}\left(\text{g}\right)=\left(\text{molecular mass}\left(\text{g/mol}\right)\right)\left(\text{number of moles}\right)$        (2)

Answer to Problem 3.119P

The mass of $5.85\text{ mol}$ glycerol is $539\text{ g}$.

Explanation of Solution

The molar mass of glycerol $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}\right)$ is calculated as:

  $\begin{array}{c}\text{Molar mass}=\left(\text{3}\right)\left(\text{12}\text{.01 g/mol}\right)\text{+}\left(\text{8}\right)\left(\text{1}\text{.008 g/mol}\right)\text{+}\left(\text{3}\right)\left(\text{16}\text{.02 g/mol}\right)\\ =\text{92}\text{.09 g/mol}\end{array}$

Substitute $5.85\text{ mol}$ for the number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}$ and $92.09\text{ g/mol}$ for molar mass in equation (2).

  $\begin{array}{c}{\text{Mass of C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}\left(\text{g}\right)=\left(92.09\text{ g/mol}\right)\left(\text{5}\text{.85 mol}\right)\\ =538.7265\text{ g}\\ \approx 539\text{ g}\text{.}\end{array}$

Conclusion

The mass of $5.85\text{ mol}$ glycerol is found to be $539\text{ g}$.

(d)

Interpretation Introduction

Interpretation:

The volume of $2.85\text{ mol}$ chloroform is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

The relation between density, mass and volume is as follows:

  $\text{Density}\left(\text{g/mL}\right)\text{ =}\frac{\text{Mass}\left(\text{g}\right)}{\text{Volume}\left(\text{mL}\right)}$        (2)

Answer to Problem 3.119P

The volume of $2.85\text{ mol}$ chloroform is $230\text{ mL}$.

Explanation of Solution

The chemical formula for chloroform is ${\text{CHCl}}_{\text{3}}$. The molar mass of chloroform is $\text{119}\text{.37 g/mol}$.

Substitute $\text{2}\text{.85 mol}$ for the number of moles of chloroform and $\text{119}\text{.37 g/mol}$ for molar mass of chloroform in equation (2).

  $\begin{array}{c}{\text{Mass of C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{O}}_{\text{3}}\left(\text{g}\right)=\left(\text{ 2}\text{.85 g/mol}\right)\left(\text{119}\text{.37 mol}\right)\\ =\text{340}\text{.2045 g}\end{array}$

The density of chloroform is $1.48\text{ g/mL}$.

Rearrange equation (2) for volume as follows:

  $\text{Volume}\left(\text{mL}\right)\text{=}\frac{\text{Mass}\left(\text{g}\right)}{\text{Density}\left(\text{g/mL}\right)}$

Substitute $1.48\text{ g/mL}$ for density and $\text{340}\text{.2045 g}$ for mass in the above formula.

  $\begin{array}{c}\text{Volume}\left(\text{mL}\right)\text{=}\frac{340.2045\left(\text{g}\right)}{1.48\left(\text{g/mL}\right)}\\ =229.868\text{ mL}\\ \approx 230\text{ mL}\text{.}\end{array}$

Conclusion

Hence the volume of $2.85\text{ mol}$ chloroform is found to be $230\text{ mL}$.

(e)

Interpretation Introduction

Interpretation:

The number of sodium ions present in $\text{2}\text{.11 mol}$ sodium carbonate is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

Answer to Problem 3.119P

The number of sodium ions present in $\text{2}\text{.11 mol}$ sodium carbonate is $2.54\times {10}^{24}$.

Explanation of Solution

One mole of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ gives $\text{2 mol}$ of ${\text{Na}}^{+1}$ ions hence $\text{2}\text{.11 mol}$${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ would yield:

  $\begin{array}{c}{\text{Moles of Na}}^{+1}=\left(2.11\text{ mol}\right)\left(\frac{2\text{ mol}}{1\text{ mol}}\right)\\ =4.22\text{ mol}\end{array}$

The formula to evaluate the number of ions is as follows:

  $\text{number of ions}=\left(\text{number of moles}\right)\left(6.022\times {10}^{23}\right)$        (3)

Substitute $\text{4}\text{.22 mol}$ number of moles in equation (3).

  $\begin{array}{c}{\text{Number of Na}}^{+1}\text{ ions}=\left(\text{ 4}\text{.22 mol}\right)\left(6.022\times {10}^{23}\text{ions/mol}\right)\\ =25.412\times {10}^{23}\text{ ions}\\ \approx 2.54\times {10}^{24}\text{ ions}\text{.}\end{array}$

Conclusion

The number of sodium ions present in $\text{2}\text{.11 mol}$ sodium carbonate is found to be $2.54\times {10}^{24}$.

(f)

Interpretation Introduction

Interpretation:

The number of atoms in $\text{25}\text{.0 μg}$ cadmium is to be calculated.

Concept introduction:

The symbol and molar mass of cadmium is $\text{Cd}$ $112.4\text{ g/mol}$ respectively.

The number of moles is calculated by the formula:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

The formula to evaluate the number of atoms is as follows:

  $\text{Number of atoms}=\left(\text{number of moles}\right)\left(6.022\times {10}^{23}\right)$        (3)

The conversion factor for converting $\text{μg}$ to $\text{g}$ is as follows:

$1\text{ μg}={10}^{-6}\text{g}$

Answer to Problem 3.119P

The number of atoms in $\text{25}\text{.0 μg}$ of cadmium is $1.34\times {10}^{17}$.

Explanation of Solution

Since $1\text{ μg}={10}^{-6}\text{ g}$ so $\text{25}\text{.0 μg}$ is equal to $\text{25}\text{.0}\times {10}^{-6}\text{ g}$.

Substitute $\text{25}\text{.0}\times {10}^{-6}\text{g}$ for mass and $112.4\text{ g/mol}$ for molar mass in equation (1).

  $\begin{array}{c}\text{Moles of Cd}\left(\text{mol}\right)\text{=}\frac{\text{25}\text{.0}\times {10}^{-6}\text{ g}}{\text{112}\text{.4 g/mol }}\\ \text{= 2}\text{.54128}\times {\text{10}}^{24}\text{ mol}\\ \approx \text{2}\text{.54}\times {\text{10}}^{24}\text{ mol}\end{array}$

In order to calculate the number of atoms from the number of moles substitute $\text{2}\text{.54}\times {\text{10}}^{24}\text{mol}$ for moles of cadmium in equation (3).

  $\begin{array}{c}\text{Number of atoms}=\left(\text{2}\text{.224199}\times {\text{10}}^{-7}\text{ mol}\right)\left(6.022\times {10}^{23}\text{}\text{ atoms/mol}\right)\\ =1.3394126\times {10}^{17}\text{ atoms}\\ \approx 1.34\times {10}^{17}\text{ atoms}\text{.}\end{array}$

Conclusion

The number of atoms in $\text{25}\text{.0 μg}$ of cadmium is found to be $1.34\times {10}^{17}$.

(g)

Interpretation Introduction

Interpretation:

The number of atoms in $\text{0}\text{.0015 mol}$ of fluorine gas is to be calculated.

Concept introduction:

The number of moles is calculated by the formula:

  $\text{Number of moles}=\frac{\text{given mass of sample}\left(\text{g}\right)}{\text{molecular mass (g/mol)}}$        (1)

Answer to Problem 3.119P

The number of atoms in $\text{0}\text{.0015 mol}$ of fluorine gas is $1.8\times {10}^{21}$.

Explanation of Solution

The chemical formula of fluorine gas is ${\text{F}}_{\text{2}}$. $\text{1 mol}$ fluorine gas has $\text{2 mol}$ $\text{F}$ atoms.

Therefore, the number of fluorine atoms in $\text{0}\text{.0015 mol}$ of fluorine gas is calculated as,

  $\begin{array}{c}\text{Moles of F atoms}=\left(\text{0}\text{.0015 mol}\right)\left(\frac{2\text{ mol}}{1\text{ mol}}\right)\\ =3\times {10}^{-3}\text{ mol}\end{array}$

In order to calculate the number of atoms from the number of moles substitute $\text{2}\text{.54}\times {\text{10}}^{24}\text{mol}$ for moles in equation (3).

  $\begin{array}{c}\text{Number of atoms}=\left(3\times {10}^{-3}\text{ mol}\right)\left(6.022\times {10}^{23}\text{ atoms/mol}\right)\\ =1.8066\times {10}^{21}\text{ atoms}\\ \approx 1.8\times {10}^{21}\text{ atoms}\text{.}\end{array}$

Conclusion

The number of fluorine atoms present in $\text{0}\text{.0015 mol}$ of fluorine gas is found to be $1.8\times {10}^{21}$.