Chapter 7, Problem 23E

Interpretation Introduction

(a)

Interpretation:

The number of moles of oxygen in $6.79\text{g}$ oxygen 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 $\text{12C}$. The number of particles present in one mole of a substance is $6.02\times {10}^{23}$ particles. The number of moles a substance is given as,$n=\frac{m}{M}$

Answer to Problem 23E

The number of moles of oxygen in $6.79\text{g}$ oxygen is $0.212\text{mol}$.

Explanation of Solution

The given mass of oxygen is $6.79\text{g}$.

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.

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& 2\text{O}\\ & =& 2\times 16.00\text{g}/\text{mol}\\ & =& 32.00\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\text{O}}_2$ is $32.00\text{g}/\text{mol}$.

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 oxygen in the equation (1).

$\begin{array}{rll}n& =& \frac{6.79\text{g}}{32.00\text{g}/\text{mol}}\\ & =& 0.212\text{mol}\end{array}$

Therefore, the number of moles of oxygen in $6.79\text{g}$ oxygen is $0.212\text{mol}$].

Conclusion

The number of moles of oxygen in $6.79\text{g}$ oxygen is $0.212\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The number of moles of magnesium nitrate in $9.05\text{g}$ magnesium nitrate is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{12C}$. 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 23E

The number of moles of magnesium nitrate in $9.05\text{g}$ magnesium nitrate is $0.0610\text{mol}$.

Explanation of Solution

The given mass of magnesium nitrate is $9.05\text{g}$.

The molar mass of the magnesium atom is $24.31\text{g}/\text{mol}$.

The molar mass of the nitrogen atom is $14.01\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& \text{Mg}+\text{2N}+\text{6O}\\ & =& 24.31\text{g}/\text{mol}+2\times 14.01\text{g}/\text{mol}+6\times 16.00\text{g}/\text{mol}\\ & =& 148.33\text{g}/\text{mol}\end{array}$

Hence, the molar mass of $\text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $148.33\text{g}/\text{mol}$.

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 magnesium nitrate in the equation (1).

$\begin{array}{rll}n& =& \frac{9.05\text{g}}{148.33\text{g}/\text{mol}}\\ & =& 0.0610\text{mol}\end{array}$

Therefore, the number of moles of magnesium nitrate in $9.05\text{g}$ magnesium nitrate is $0.0610\text{mol}$.

Conclusion

The number of moles of magnesium nitrate in $9.05\text{g}$ magnesium nitrate is $0.0610\text{mol}$.

Interpretation Introduction

(c)

Interpretation:

The number of moles of aluminum oxide in $0.770\text{g}$ aluminum oxide is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{12C}$. 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 23E

The number of moles of aluminum oxide in $0.770\text{g}$ aluminum oxide is $0.00755\text{mol}$.

Explanation of Solution

The given mass of aluminum oxide is $0.770\text{g}$.

The molar mass of the aluminum atom is $26.98\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& 2\text{Al}+3\text{O}\\ & =& 2\times 26.98\text{g}/\text{mol}+3\times 16.00\text{g}/\text{mol}\\ & =& 101.96\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ is $101.96\text{g}/\text{mol}$.

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 aluminum oxide in the equation (1).

$\begin{array}{rll}n& =& \frac{0.770\text{g}}{101.96\text{g}/\text{mol}}\\ & =& 0.00755\text{mol}\end{array}$

Therefore, the number of moles of aluminum oxide in $0.770\text{g}$ aluminum oxide is $0.00755\text{mol}$.

Conclusion

The number of moles of aluminum oxide in $0.770\text{g}$ aluminum oxide is $0.00755\text{mol}$.

Interpretation Introduction

(d)

Interpretation:

The number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in $659\text{g}$${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{12C}$. 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 23E

The number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in $659\text{g}$${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $14.3\text{mol}$.

Explanation of Solution

The given mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $659\text{g}$.

The molar mass of the carbon atom is $12.01\text{g}/\text{mol}$.

The molar mass of the hydrogen atom is $1.008\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& 2\text{C}+\text{O}+\text{6H}\\ & =& 2\times 12.01\text{g}/\text{mol}+16.00\text{g}/\text{mol}+6\times 1.008\text{g}/\text{mol}\\ & \approx & 46.07\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $46.07\text{g}/\text{mol}$.

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 ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{659\text{g}}{46.07\text{g}/\text{mol}}\\ & =& 14.3\text{mol}\end{array}$

Therefore, the number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in $659\text{g}$${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $14.3\text{mol}$.

Conclusion

The number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ in $659\text{g}$${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $14.3\text{mol}$.

Interpretation Introduction

(e)

Interpretation:

The number of moles of ammonium carbonate in $0.394\text{g}$ ammonium carbonate is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{12C}$. 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 23E

The number of moles of ammonium carbonate in $0.394\text{g}$ ammonium carbonate is $0.00410\text{mol}$.

Explanation of Solution

The given mass of ammonium carbonate is $0.394\text{g}$.

The molar mass of the carbon atom is $12.01\text{g}/\text{mol}$.

The molar mass of the hydrogen atom is $1.008\text{g}/\text{mol}$.

The molar mass of the nitrogen atom is $14.01\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& 2\text{N}+8\text{H}+\text{C}+3\text{O}\\ & =& 2\times 14.01\text{g}/\text{mol}+8\times 1.008\text{g}/\text{mol}+12.01\text{g}/\text{mol}+3\times 16.00\text{g}/\text{mol}\\ & \approx & 96.09\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $96.09\text{g}/\text{mol}$.

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 carbonate in the equation (1).

$\begin{array}{rll}n& =& \frac{0.394\text{g}}{96.09\text{g}/\text{mol}}\\ & =& 0.00410\text{mol}\end{array}$

Therefore, the number of moles of ammonium carbonate in $0.394\text{g}$ ammonium carbonate is $0.00410\text{mol}$.

Conclusion

The number of moles of ammonium carbonate in $0.394\text{g}$ ammonium carbonate is $0.00410\text{mol}$.

Interpretation Introduction

(f)

Interpretation:

The number of moles of lithium sulfide in $34.0\text{g}$ lithium sulfide is to be calculated.

Concept introduction:

A mole of a substance is defined as the same number of particles of the substance as present in $12\text{g}$ of $\text{12C}$. 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 23E

The number of moles of lithium sulfide in $34.0\text{g}$ lithium sulfide is $0.740\text{mol}$.

Explanation of Solution

The given mass of lithium sulfide is $34.0\text{g}$.

The molar mass of the sulfur atom is $32.06\text{g}/\text{mol}$.

The molar mass of the lithium atom is $6.94\text{g}/\text{mol}$.

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

$\begin{array}{rll}\text{Therefore, the molar mass of is}& =& 2\text{Li}+\text{S}\\ & =& 2\times 6.94\text{g}/\text{mol}+32.06\text{g}/\text{mol}\\ & \approx & 45.94\text{g}/\text{mol}\end{array}$

Hence, the molar mass of ${\text{Li}}_{\text{2}}\text{S}$ is $45.94\text{g}/\text{mol}$.

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 lithium sulfide in the equation (1).

$\begin{array}{rll}n& =& \frac{34.0\text{g}}{45.94\text{g}/\text{mol}}\\ & =& 0.740\text{mol}\end{array}$

Therefore, the number of moles of lithium sulfide in $34.0\text{g}$ lithium sulfide is $0.740\text{mol}$.

Conclusion

The number of moles of lithium sulfide in $34.0\text{g}$ lithium sulfide is $0.740\text{mol}$.