Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 3, Problem 3.80QP
Interpretation Introduction

Interpretation: The nitrogen oxides that have more than 50% oxygen by mass are to be predicted. The nitrogen oxides, if any, with same empirical formula are to be stated.

Concept introduction: Molar mass of a compound is sum of the atomic mass of each elements present in the compound.

To determine: The nitrogen that have more than 50% oxygen by mass and nitrogen oxides, if any, with same empirical formula.

Expert Solution & Answer
Check Mark

Answer to Problem 3.80QP

Solution

The compounds that have the same empirical formula are shown in Figure 1. The nitrogen oxides that have more than 50% oxygen by mass are NO , N2O3 , N2O4 , NO2 and N2O2 . The compounds that have the same empirical formula are NO , N2O2 and NO2 and N2O4 .

Explanation of Solution

Explanation

Molecular mass of N2O is calculated by the formula,

Molecular mass of N2O=((NumberofNatoms×MolarmassofN)+(NumberofOatoms×MolarmassofO))

Number of nitrogen atoms is 2 .

Number of oxygen atoms is 1 .

Molar mass of nitrogen is 14.00g/mol .

Molar mass of oxygen is 16.00g/mol .

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

Molecular mass of N2O=(2×14.00g/mol)+(1×16.00)=44g/mol

The percent composition of oxygen (O) in N2O is calculated by the formula,

Percent composition of O=Mass of OMass of N2O×100

Substitute the value of mass of O atoms and mass of N2O in the above equation.

Percent composition of O=16.00g44g×100=36.36% O

Therefore, the percent composition of oxygen in N2O is 36.36% O .

Molecular mass of NO is calculated by the formula,

Molecular mass of NO=((NumberofNatoms×MolarmassofN)+(NumberofOatoms×MolarmassofO))

Number of nitrogen atoms is 1 .

Number of oxygen atoms is 1 .

Molar mass of nitrogen is 14.00g/mol .

Molar mass of oxygen is 16.00g/mol .

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

Molecular mass of N2O=(1×14.00g/mol)+(1×16.00)=30g/mol

The percent composition of oxygen (O) in NO is calculated by the formula,

Percent composition of O=Mass of OMass of NO×100

Substitute the value of mass of O atoms and mass of NO in the above equation.

Percent composition of O=16g30g×100=53.33% O

Therefore, the percent composition of oxygen in NO is 53.33% O .

Molecular mass of N2O3 is calculated by the formula,

Molecular mass of N2O3=(NumberofNatoms×MolarmassofN)+(NumberofOatoms×MolarmassofO)

Number of nitrogen atoms is 2 .

Number of oxygen atoms is 3 .

Molar mass of nitrogen is 14.00g/mol .

Molar mass of oxygen is 16.00g/mol .

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

Molecular mass of N2O3=(2×14.00g/mol)+(3×16.00)=76g/mol

The percent composition of oxygen (O) in N2O3 is calculated by the formula,

Percent composition of O=Mass of OMass of N2O3×100

Mass of 3O atoms is 3×16g/mol=48g/mol .

Substitute the value of mass of O atoms and mass of N2O3 in the above equation.

Percent composition of O=48g76g×100=63.15% O

Therefore, the percent composition of oxygen in N2O3 is 63.15% O .

Molecular mass of N2O2 is calculated by the formula,

Molecular mass of N2O2=(NumberofNatoms×MolarmassofN)+(NumberofOatoms×MolarmassofO)

Number of nitrogen atoms is 2 .

Number of oxygen atoms is 2 .

Molar mass of nitrogen is 14.00g/mol .

Molar mass of oxygen is 16.00g/mol .

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

Molecular mass of N2O2=(2×14.00g/mol)+(2×16.00)=60g/mol

The percent composition of oxygen (O) in N2O2 is calculated by the formula,

Percent composition of O=Mass of OMass of N2O2×100

Mass of 2O atoms is 2×16g/mol=32g/mol .

Substitute the value of mass of O atoms and mass of N2O2 in the above equation.

Percent composition of O=32g60g×100=53.33% O

Therefore, the percent composition of oxygen in N2O2 is 53.33% O .

Molecular mass of NO2 is calculated by the formula,

Molecular mass of NO2=((NumberofNatoms×MolarmassofN)+(NumberofOatoms×MolarmassofO))

Number of nitrogen atoms is 1 .

Number of oxygen atoms is 2 .

Molar mass of nitrogen is 14.00g/mol .

Molar mass of oxygen is 16.00g/mol .

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

Molecular mass of NO2=(1×14.00g/mol)+(2×16.00)=46g/mol

The percent composition of oxygen (O) in NO2 is calculated by the formula,

Percent composition of O=Mass of OMass of N2O2×100

Mass of 2O atoms is 2×16g/mol=32g/mol .

Substitute the value of mass of O atoms and mass of NO2 in the above equation.

Percent composition of O=32g46g×100=69.56% O

Therefore, the percent composition of oxygen in NO2 is 69.56% O .

Molecular mass of N2O4 is calculated by the formula,

Molecular mass ofN2O4=((NumberofNatoms×MolarmassofN)+(NumberofOatoms×MolarmassofO))

Number of nitrogen atoms is 2 .

Number of oxygen atoms is 4 .

Molar mass of nitrogen is 14.00g/mol .

Molar mass of oxygen is 16.00g/mol .

Substitute the value of number of atoms and molar masses of nitrogen and oxygen in the above equation.

Molecular mass of N2O4=(2×14.00g/mol)+(4×16.00)=92g/mol

The percent composition of oxygen (O) in N2O4 is calculated by the formula,

Percent composition of O=Mass of OMass of N2O4×100

Mass of 4O atoms is 4×16g/mol=64g/mol .

Substitute the value of mass of O atoms and mass of N2O4 in the above equation.

Percent composition of O=64g92g×100=69.56% O

Therefore, the percent composition of oxygen in N2O4 is 69.56% O .

The nitrogen oxides that have more than 50% oxygen by mass are NO , N2O3 , N2O4 , NO2 and N2O2 .

The ratio of N:O in NO is 1:1 . Thus the empirical formula of NO is NO . The ratio of N:O in N2O2 is 1:1 . Thus the empirical formula of N2O2 is NO . The compounds that have the same empirical formula are NO and N2O2 .

The ratio of N:O in NO2 is 1:2 . Thus the empirical formula of NO2 is NO2 . The ratio of N:O in N2O4 is 1:2 . Thus the empirical formula of N2O4 is NO2 . The compounds that have the same empirical formula are NO2 and N2O4 .

Conclusion

The nitrogen oxides that have more than 50% oxygen by mass are NO , N2O3 , N2O4 , NO2 and N2O2 . The compounds that have the same empirical formula are NO , N2O2 and NO2 and N2O4

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Chapter 3 Solutions

Chemistry

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