Chapter 7, Problem 24E

Questions 23 to 26: Find the number of moles for each mass of substance given.

a) $53.8\text{g}$ beryllium

b) $781\text{g}$ ${\text{C}}_3{\text{H}}_{\text{4}}{\text{Cl}}_4$

c) $0.756\text{g}$ calcium hydroxide

d) $9.94\text{g}$ cobalt $\left(\text{III}\right)$ bromide

e) $8.80\text{g}$ ammonium dichromate (dichromate ion, ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{2-}$)

f) $28.3\text{g}$ magnesium perchlorate

Interpretation Introduction

(a)

Interpretation:

The number of moles of beryllium in $53.8\text{g}$ beryllium is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as that present in $12\text{g}$ of $12\text{C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles of a substance is given as,$n=\frac{m}{M}$

Answer to Problem 24E

The number of moles of beryllium in $53.8\text{g}$ beryllium is $5.97\text{mol}$.

Explanation of Solution

Mass of beryllium is $53.8\text{g}$.

The molar mass of the beryllium atom is $9.0\text{g}/\text{mol}$.

The number of moles of a substance is given as,$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the value of mass and molar mass of beryllium in the equation (1).

$\begin{array}{rll}n& =& \frac{53.8\text{g}}{9.0\text{g}/\text{mol}}\\ & =& 5.97\text{mol}\end{array}$

Therefore, the number of moles of beryllium in $53.8\text{g}$ beryllium is $5.97\text{mol}$.

Conclusion

The number of moles of beryllium in $53.8\text{g}$ beryllium is $5.97\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ in $781\text{g}$ of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as that present in $12\text{g}$ of $\text{12}\text{C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles of a substance is given as,$n=\frac{m}{M}$

Answer to Problem 24E

The number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ in $781\text{g}$ of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is $4.31\text{mol}$.

Explanation of Solution

Mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is $781\text{g}$.

The molar mass of the carbon atom is $12\text{g}/\text{mol}$.

The molar mass of the hydrogen atom is $1\text{g}/\text{mol}$.

The molar mass of the chlorine atom is $35.5\text{g}/\text{mol}$.

The molar mass of compound is the sum of molar mass of individual elements present in it.

Therefore, the molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is

$\begin{array}{l}=3\text{C}+3\text{H}+4\text{Cl}\\ =3\times 12\text{g}/\text{mol}+3\times 1\text{g}/\text{mol}+4\times 35.5\text{g}/\text{mol}\\ =181\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is $181\text{g}/\text{mol}$.The number of moles of a substance is given as,$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the value of mass and molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{781\text{g}}{181\text{g}/\text{mol}}\\ & =& 4.31\text{mol}\end{array}$

Therefore, the number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ in $781\text{g}$ of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is $4.31\text{mol}$.

Conclusion

The number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ in $781\text{g}$ of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{\text{Cl}}_{\text{4}}$ is $4.31\text{mol}$.

Interpretation Introduction

(c)

Interpretation:

The number of moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in $0.756\text{g}$ of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as that present in $12\text{g}$ of $\text{12}\text{C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles of a substance is given as,$n=\frac{m}{M}$

Answer to Problem 24E

The number of moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in $0.756\text{g}$ of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is $0.0102\text{mol}$.

Explanation of Solution

Mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is $0.756\text{g}$.

The molar mass of the calcium is $40\text{g}/\text{mol}$.

The molar mass of the oxygen atom is $16.00\text{g}/\text{mol}$.

The molar mass of the hydrogen atom is $1\text{g}/\text{mol}$.

The molar mass of compound is the sum of molar mass of individual elements present in it.

Therefore, the molar mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$$\begin{array}{rll}& =& 40\text{g}/\text{mol}+2\times 16.00\text{g}/\text{mol}+2\times 1\text{g}/\text{mol}\\ & =& 40\text{g}/\text{mol}+32.00\text{g}/\text{mol}+2\text{g}/\text{mol}\\ & =& 74\text{g}/\text{mol}\end{array}$

Hence, the molar mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is $74\text{g}/\text{mol}$.

The number of moles of a substance is given as,$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the value of mass and molar mass of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{0.756\text{g}}{74\text{g}/\text{mol}}\\ & =& 0.0102\text{mol}\end{array}$

Therefore, the number of moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in $0.756\text{g}$ of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is $0.0102\text{mol}$.

Conclusion

The number of moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in $0.756\text{g}$ of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is $0.0102\text{mol}$.

Interpretation Introduction

(d)

Interpretation:

The number of moles of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ in $9.94\text{g}$ $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as that present in $12\text{g}$ of $\text{12}\text{C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles of a substance is given as,$n=\frac{m}{M}$

Answer to Problem 24E

The number of moles of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ in $9.94\text{g}$ $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is $0.033\text{mol}$.

Explanation of Solution

Mass of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is $9.94\text{g}$.

The chemical formula of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is ${\text{CoBr}}_{\text{3}}$.

The molar mass of the cobalt is $58.93\text{g}/\text{mol}$.

The molar mass of the bromine atom is $79.90\text{g}/\text{mol}$.

The molar mass of compound is the sum of molar mass of individual elements present in it.

Therefore, the molar mass of $\text{cobalt}\left(\text{III}\right)\text{bromide}$

$\begin{array}{rll}& =& \text{Co}+3\text{Br}\\ & =& 58.93\text{g}/\text{mol}+3\times 79.90\text{g}/\text{mol}\\ & =& 298.63\text{g}/\text{mol}\end{array}$

Hence, the molar mass of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is $298.63\text{g}/\text{mol}$.

The number of moles of a substance is given as,$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the value of mass and molar mass of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{9.94\text{g}}{298.63\text{g}/\text{mol}}\\ & =& 0.033\text{mol}\end{array}$

Therefore, the number of moles of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ in $9.94\text{g}$ $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is $0.033\text{mol}$.

Conclusion

The number of moles of $\text{cobalt}\left(\text{III}\right)\text{bromide}$ in $9.94\text{g}$ $\text{cobalt}\left(\text{III}\right)\text{bromide}$ is $0.033\text{mol}$.

Interpretation Introduction

(e)

Interpretation:

The number of moles of ammonium dichromate in $8.80\text{g}$ ammonium dichromate is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as that present in $12\text{g}$ of $\text{12}\text{C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles of a substance is given as,$n=\frac{m}{M}$

Answer to Problem 24E

The number of moles of ammonium dichromate in $8.80\text{g}$ ammonium dichromate is $0.034\text{mol}$.

Explanation of Solution

Mass of ammonium dichromate is $8.80\text{g}$.

The chemical formula of ammonium dichromate is ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$.

The molar mass of the nitrogen atom is $14\text{g}/\text{mol}$.

The molar mass of the hydrogen atom is $1\text{g}/\text{mol}$.

The molar mass of the chromium atom is $51.99\text{g}/\text{mol}$.

The molar mass of the oxygen atom is $16.00\text{g}/\text{mol}$.

The molar mass of compound is the sum of molar mass of individual elements present in it.

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& 2\text{N}+8\text{H}+2\text{Cr}+7\text{O}\\ & =& 2\times 14\text{g}/\text{mol}+8\times 1\text{g}/\text{mol}+2\times 51.99\text{g}/\text{mol}+7\times 16\text{g}/\text{mol}\\ & =& 28\text{g}/\text{mol}+8\text{g}/\text{mol}+103.98\text{g}/\text{mol}+112\text{g}/\text{mol}\\ & =& 251.98\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is $251.98\text{g}/\text{mol}$.

The number of moles of a substance is given as,$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the value of mass and molar mass of ammonium dichromate in the equation (1).

$\begin{array}{rll}n& =& \frac{8.80\text{g}}{251.98\text{g}/\text{mol}}\\ & =& 0.034\text{mol}\end{array}$

Therefore, the number of moles of ammonium dichromate in $8.80\text{g}$ ammonium dichromate is $0.034\text{mol}$.

Conclusion

The number of moles of ammonium dichromate in $8.80\text{g}$ ammonium dichromate is $0.034\text{mol}$.

Interpretation Introduction

(f)

Interpretation:

The number of moles of magnesium perchlorate in $28.3\text{g}$ magnesium perchlorate is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as that present in $12\text{g}$ of $\text{12}\text{C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles of a substance is given as,$n=\frac{m}{M}$

Answer to Problem 24E

The number of moles of magnesium perchlorate in $28.3\text{g}$ magnesium perchlorate is $0.126\text{mol}$.

Explanation of Solution

The mass of magnesium perchlorate is $28.3\text{g}$.

The chemical formula of magnesium perchlorate is $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_2$.

The molar mass of the magnesium atom is $24.30\text{g}/\text{mol}$.

The molar mass of the oxygen atom is $16\text{g}/\text{mol}$.

The molar mass of the chlorine atom is $35.5\text{g}/\text{mol}$.

The molar mass of compound is the sum of molar mass of individual elements present in it.

Therefore, the molar mass of $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_2$

$\begin{array}{rll}& =& \text{Mg}+\text{2Cl}+\text{8O}\\ & =& 24.30\text{g}/\text{mol}+2\times 35.5\text{g}/\text{mol}+8\times 16\text{g}/\text{mol}\\ & =& 24.30\text{g}/\text{mol}+71\text{g}/\text{mol}+128\text{g}/\text{mol}\\ & =& 223.3\text{g}/\text{mol}\end{array}$

Hence, the molar mass of $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_2$ is $223.3\text{g}/\text{mol}$.

The number of moles of a substance is given as,$n=\frac{m}{M}$ …(1)

Where,

• $m$ represents the mass of the substance.

• $M$ represents the molar mass of the substance.

Substitute the value of mass and molar mass of $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_2$ in the equation (1).

$\begin{array}{rll}n& =& \frac{28.3\text{g}}{223.3\text{g}/\text{mol}}\\ & =& 0.126\text{mol}\end{array}$

Therefore, the number of moles of magnesium perchlorate in $28.3\text{g}$ magnesium perchlorate is $0.126\text{mol}$.

Conclusion

The number of moles of magnesium perchlorate in $28.3\text{g}$ magnesium perchlorate is $0.126\text{mol}$.