Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 8, Problem 85AP

Complete the following table.

l>

Mass of Sample Moles of Sample Molecules in Sample Atoms in Sample
4.24 g

msp;  C 6 H 6



0.224 mol

msp;  H 2 O



msp;  2 .71 × 10 22 molecules

msp;  CO 2

1.26 mol HCl

td>



msp;  2 .21 × 10 24 molecules

msp;  H 2 O
0.297 g

msp;  CH 3 OH

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

The given table should be completed.

Mass of sample Moles of sample Molecules in sample Atoms in sample
4.24 gC6H6
0.224 molH2O
2.71×1022 moleculesCO2
1.26 molHCl
4.21×1024 moleculesH2O
0.297 gCH3OH

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

n=mM

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbolNA.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

(6.023×1023 molecules1 mol).

Answer to Problem 85AP

Mass of sample Moles of sample Molecules in sample Atoms in sample
4.24 gC6H6 0.0543 molC6H6 3.27×1022 molecules C6H6 3.924×1023 atoms C6H6
4.032 gH2O 0.224 molH2O 3.27×1022 molecules H2O 3.924×1023 atoms H2O
1.98 gCO2 4.5×102 molCO2 2.71×1022 moleculesCO2 8.13×1022 atomsCO2
45.94 gHCl 1.26 molHCl 7.58×1023 moleculesHCl 15.18×1023 atomsHCl
125.8 gH2O 6.98 molH2O 4.21×1024 moleculesH2O 12.63×1024 atomsH2O
0.297 gCH3OH 0.000927 molCH3OH 5.58×1021 moleculesCH3OH 3.35×1022 atomsCH3OH

Explanation of Solution

Mass ofC6H6 is 4.24 g and molar mass ofC6H6 is 78.11 g/mol thus, number of moles can be calculated as follows:

n=4.24 g78.11 g/mol=0.0543 mol

Since, 1 mol of a substance contains6.023×1023 molecules thus; number of molecules in 0.0543 mol can be calculated as follows:

Nmolecules=0.0543 mol(6.023×1023 molecules1 mol)=3.27×1022 molecules

Since, 1 molecule ofC6H6 contains 12 atoms thus, number of atoms in3.27×1022 molecules ofC6H6 will be:

Natoms=12(3.27×1022)=3.924×1023 atoms

Number of moles ofH2O is 0.244 mol and molar mass ofH2O is 18 g/mol thus, mass can be calculated as follows:

m=n×M

Putting the values,

m=0.224 mol×18 g/mol=4.032 g

Since, 1 mol of a substance contains6.023×1023 molecules thus; number of molecules in 0.224 mol can be calculated as follows:

Nmolecules=0.224 mol(6.023×1023 molecules1 mol)=1.35×1023 molecules

Since, 1 mole of water molecule contains 3 atoms thus, number of atoms will be:

Natoms=3(1.35×1023)=4.05×1023 atoms

Number of molecules ofCO2 is2.71×1022 molecules

Since, 1 mol of a substance contains6.023×1023 molecules thus; number of moles can be calculated as follows:

n=2.71×1022 molecules(1 mol6.023×1023 molecules)=4.5×102 mol

Molar mass of carbon dioxide is 44.01 g/mol thus, mass of carbon dioxide is:

m=n×M=4.5×102 mol×44.01 g/mol=1.98 g

Since, 1 molecule of carbon dioxide contains 3 atoms thus, number of atoms can be calculated as follows:

Natoms=3×2.71×1022=8.13×1022 atoms

Number of moles of HCl is 1.26 mol and molar mass of HCl is 36.46 g/mol thus, mass can be calculated as follows:

m=n×M

Putting the values,

m=1.26 mol×36.46 g/mol=45.94 g

Since, 1 mol of a substance contains6.023×1023 molecules thus; number of molecules in 1.26 mol can be calculated as follows:

Nmolecules=1.26 mol(6.023×1023 molecules1 mol)=7.58×1023 molecules

Since, 1 mol of HCl contains 2 atoms thus, number of atoms can be calculated as follows:

Natoms=2×7.58×1023 =15.18×1023 atoms

Number of molecules ofH2O is4.21×1024 atoms

Since, 1 mol of a substance contains6.023×1023 molecules thus; number of moles can be calculated as follows:

n=4.21×1024 molecules(1 mol6.023×1023 molecules)=6.98 mol

Molar mass of water is 18 g/mol thus, mass of water is:

m=n×M=6.98 mol×18 g/mol=125.8 g

Since, 1 molecule of water contains 3 atoms thus, number of atoms can be calculated as follows;

Natoms=3×4.21×1024 atoms=12.63×1024 atoms

Mass ofCH3OH is 0.297 g and molar mass is 32.04 g/mol thus, number of moles can be calculated as follows:

n=mM=0.297 g32.04 g/mol=0.000927 mol

Since, 1 mol of a substance contains6.023×1023 molecules thus, number of molecules can be calculated as follows:

Nmolecules=0.000927 mol(6.023×1023 molecules1 mol)=5.58×1021 molecules

Since, 1 mol ofCH3OH contains 6 atoms thus, number of atoms can be calculated as follows:

Natoms=6×5.58×1021=3.35×1022 atoms.

Conclusion

The complete table is as follows:

Mass of sample Moles of sample Molecules in sample Atoms in sample
4.24 gC6H6 0.0543 molC6H6 3.27×1022 molecules C6H6 3.924×1023 atoms C6H6
4.032 gH2O 0.224 molH2O 3.27×1022 molecules H2O 3.924×1023 atoms H2O
1.98 gCO2 4.5×102 molCO2 2.71×1022 moleculesCO2 8.13×1022 atomsCO2
45.94 gHCl 1.26 molHCl 7.58×1023 moleculesHCl 15.18×1023 atomsHCl
125.8 gH2O 6.98 molH2O 4.21×1024 moleculesH2O 12.63×1024 atomsH2O
0.297 gCH3OH 0.000927 molCH3OH 5.58×1021 moleculesCH3OH 3.35×1022 atomsCH3OH

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Chapter 8 Solutions

Introductory Chemistry: A Foundation

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