The given table is to be completed. Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, Moles = Given mass / Molar mass . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

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Answer 1

Question

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Conclusion

The complete table for the given amount of substances is given above.

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Answer 2

Question

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Conclusion

The complete table for the given amount of substances is given above.

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Answer 3

Question

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Conclusion

The complete table for the given amount of substances is given above.

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Answer 4

Question

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Conclusion

The complete table for the given amount of substances is given above.

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Answer 5

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Conclusion

The complete table for the given amount of substances is given above.

Answer 6

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Conclusion

The complete table for the given amount of substances is given above.

Answer 7

Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $Moles=Given mass/Molar mass$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Answer 8

Expert Solution & Answer

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C 6 H 6$	$3.27× 10 22$ molecules $C 6 H 6$	$3.92× 10 23$ total atoms in $C 6 H 6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The given mass of benzene $(C6H6)$ is $4.24 g$ .

The number of moles of $C6H6$ can be calculated as follows.

$Moles=Given massMolar mass$ (1)

The molar mass of $C6H6$ is $78.11 g/mol$ .

Substitute the values in equation (1).

$Moles of C6H6=4.24 g78.11 g/mol=0.0543 mol$

The number of molecules can be calculated as follows:

$Number of molecules=Moles×Avogadro's number$ (2)

The value of Avogadro’s number is $6.023×1023 mol−1$ .

Substitute the values in equation (2).

$Number of molecules=Moles×6.023×1023 mol−1$ (3)

Substitute the respective values in equation (3).

$Number of C6H6 molecules=0.0543 mol×6.023×1023 mol−1=3.27×1022$

The number of atoms can be calculated as follows.

$Total atoms=Atoms per molecule×Total molecules$ (4)

The total number of atoms in $C6H6$ is $12$ .

Substitute the respective values in equation (4).

$Total atoms in C6H6=12×3.27×1022=3.92×1023$

The given moles of water $(H2O)$ are $0.224 mol$ .

The molar mass of water is $18.015 g/mol$ .

The mass of water can be calculated using equation (1) as follows.

$0.224 mol=Mass of H2O18.015 g/molMass of H2O=0.224 mol×18.015 g/molMass of H2O=4.035 g$

The number of molecules of $H2O$ can be calculated using equation (3) as follows.

$Number of H2O molecules=0.224 mol×6.023×1023 mol−1=1.35×1023$

The number of atoms in $H2O$ molecule are $3$ .

The total number of atoms in $H2O$ can be calculated using equation (4) as follows.

$Total atoms in H2O=3×1.35×1023=4.05×1023$

The given number of molecules of $CO2$ are $2.71×1022$ .

The number of moles of $CO2$ can be calculated using equation (3) as follows.

$2.71×1022=Moles of CO2×6.023×1023 mol−1Moles of CO2=2.71×10226.023×1023 mol−1Moles of CO2=0.045 mol$

The molar mass of $CO2$ is $44.01 g/mol$ .

The mass of $CO2$ can be calculated using equation (1) as follows.

$0.045 mol=Mass of CO244.01 g/molMass of CO2=0.045 mol×44.01 g/molMass of CO2=1.98 g$

The number of atoms in $CO2$ molecules are $3$ .

The total atoms in $CO2$ can be calculated using equation (4) as follows.

$Total atoms in CO2=3×2.71×1022=8.13×1022$

The given total number of atoms in $CH3OH$ is $3.35×1022$ atoms.

The number of atoms in $CH3OH$ molecule are $6$ .

The total number of molecules in $CH3OH$ can be calculated using equation (4) as follows.

$3.35×1022 atoms=6 atoms×Total molecules in CH3OHTotal molecules in CH3OH=3.35×1022 atoms6 atomsTotal molecules in CH3OH=5.58×1021$

The number of moles of $CH3OH$ can be calculated using equation (3) as follows.

$5.58×1021=Moles of CH3OH×6.023×1023 mol−1Moles of CH3OH=5.58×10216.023×1023 mol−1Moles of CH3OH=9.264×10−3 mol$

The molar mass of $CH3OH$ is $32.04 g/mol$ .

The mass of $CH3OH$ can be calculated using equation (1) as follows.

$9.264×10−3 mol=Mass of CH3OH32.04 g/molMass of CH3OH=9.264×10−3 mol×32.04 g/molMass of CH3OH=0.297 g$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24 g$ $C6H6$	$0.0543 mol$ $C6H6$	$3.27×1022$ molecules $C6H6$	$3.92×1023$ total atoms in $C6H6$ sample
$4.035 g$ $H2O$	$0.224 mol$ $H2O$	$1.35×1023$ molecules $H2O$	$4.05×1023$ total atoms in $H2O$ sample
$1.98 g$ $CO2$	$0.045 mol$ $CO2$	$2.71×1022$ molecules $CO2$	$8.13×1022$ total atoms in $CO2$ sample
$0.297 g$ $CH3OH$	$9.264×10−3 mol$ $CH3OH$	$5.58×1021$ molecules $CH3OH$	$3.35×1022$ total atoms in $CH3OH$ sample