Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
5th Edition
ISBN: 9780393615296
Author: Rein V. Kirss (Author), Natalie Foster (Author), Geoffrey Davies (Author) Thomas R. Gilbert (Author)
Publisher: W. W. Norton
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Chapter 3, Problem 3.82QP
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 3.82QP

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

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)

Ch. 3.4 - Prob. 11PECh. 3.5 - Prob. 12PECh. 3.5 - Prob. 13PECh. 3.6 - Prob. 14PECh. 3.6 - Prob. 15PECh. 3.6 - Prob. 16PECh. 3.7 - Prob. 17PECh. 3.8 - Prob. 18PECh. 3.8 - Prob. 19PECh. 3.9 - Prob. 20PECh. 3.9 - Prob. 21PECh. 3.9 - Prob. 22PECh. 3.9 - Prob. 23PECh. 3 - Prob. 3.1VPCh. 3 - Prob. 3.2VPCh. 3 - Prob. 3.3VPCh. 3 - Prob. 3.4VPCh. 3 - Prob. 3.5VPCh. 3 - Prob. 3.6VPCh. 3 - Prob. 3.7VPCh. 3 - Prob. 3.8VPCh. 3 - Prob. 3.9VPCh. 3 - Prob. 3.10VPCh. 3 - Prob. 3.11QPCh. 3 - Prob. 3.12QPCh. 3 - Prob. 3.13QPCh. 3 - Prob. 3.14QPCh. 3 - Prob. 3.15QPCh. 3 - Prob. 3.16QPCh. 3 - Prob. 3.17QPCh. 3 - Prob. 3.18QPCh. 3 - Prob. 3.19QPCh. 3 - Prob. 3.20QPCh. 3 - Prob. 3.21QPCh. 3 - Prob. 3.22QPCh. 3 - Prob. 3.23QPCh. 3 - Prob. 3.24QPCh. 3 - Prob. 3.25QPCh. 3 - Prob. 3.26QPCh. 3 - Prob. 3.27QPCh. 3 - Prob. 3.28QPCh. 3 - Prob. 3.29QPCh. 3 - Prob. 3.30QPCh. 3 - Prob. 3.31QPCh. 3 - Prob. 3.32QPCh. 3 - Prob. 3.33QPCh. 3 - Prob. 3.34QPCh. 3 - Prob. 3.35QPCh. 3 - Prob. 3.36QPCh. 3 - Prob. 3.37QPCh. 3 - Prob. 3.38QPCh. 3 - Prob. 3.39QPCh. 3 - Prob. 3.40QPCh. 3 - Prob. 3.41QPCh. 3 - Prob. 3.42QPCh. 3 - Prob. 3.43QPCh. 3 - Prob. 3.44QPCh. 3 - Prob. 3.45QPCh. 3 - Prob. 3.46QPCh. 3 - Prob. 3.47QPCh. 3 - Prob. 3.48QPCh. 3 - Prob. 3.49QPCh. 3 - Prob. 3.50QPCh. 3 - Prob. 3.51QPCh. 3 - Prob. 3.52QPCh. 3 - Prob. 3.53QPCh. 3 - Prob. 3.54QPCh. 3 - Prob. 3.55QPCh. 3 - Prob. 3.56QPCh. 3 - Prob. 3.57QPCh. 3 - Prob. 3.58QPCh. 3 - Prob. 3.59QPCh. 3 - Prob. 3.60QPCh. 3 - Prob. 3.61QPCh. 3 - Prob. 3.62QPCh. 3 - Prob. 3.63QPCh. 3 - Prob. 3.64QPCh. 3 - Prob. 3.65QPCh. 3 - Prob. 3.66QPCh. 3 - Prob. 3.67QPCh. 3 - Prob. 3.68QPCh. 3 - Prob. 3.69QPCh. 3 - Prob. 3.70QPCh. 3 - Prob. 3.71QPCh. 3 - Prob. 3.72QPCh. 3 - Prob. 3.73QPCh. 3 - Prob. 3.74QPCh. 3 - Prob. 3.75QPCh. 3 - Prob. 3.76QPCh. 3 - Prob. 3.77QPCh. 3 - Prob. 3.78QPCh. 3 - Prob. 3.79QPCh. 3 - Prob. 3.80QPCh. 3 - Prob. 3.81QPCh. 3 - Prob. 3.82QPCh. 3 - Prob. 3.83QPCh. 3 - Prob. 3.84QPCh. 3 - Prob. 3.85QPCh. 3 - Prob. 3.86QPCh. 3 - Prob. 3.87QPCh. 3 - Prob. 3.88QPCh. 3 - Prob. 3.89QPCh. 3 - Prob. 3.90QPCh. 3 - Prob. 3.91QPCh. 3 - Prob. 3.92QPCh. 3 - Prob. 3.93QPCh. 3 - Prob. 3.94QPCh. 3 - Prob. 3.95QPCh. 3 - Prob. 3.96QPCh. 3 - Prob. 3.97QPCh. 3 - Prob. 3.98QPCh. 3 - Prob. 3.99QPCh. 3 - Prob. 3.100QPCh. 3 - Prob. 3.101QPCh. 3 - Prob. 3.102QPCh. 3 - Prob. 3.103QPCh. 3 - Prob. 3.104QPCh. 3 - Prob. 3.105QPCh. 3 - Prob. 3.106QPCh. 3 - Prob. 3.107QPCh. 3 - Prob. 3.108QPCh. 3 - Prob. 3.109QPCh. 3 - Prob. 3.110QPCh. 3 - Prob. 3.111QPCh. 3 - Prob. 3.112QPCh. 3 - Prob. 3.113QPCh. 3 - Prob. 3.114QPCh. 3 - Prob. 3.115QPCh. 3 - Prob. 3.116QPCh. 3 - Prob. 3.117QPCh. 3 - Prob. 3.118QPCh. 3 - Prob. 3.119QPCh. 3 - Prob. 3.120QPCh. 3 - Prob. 3.121APCh. 3 - Prob. 3.122APCh. 3 - Prob. 3.123APCh. 3 - Prob. 3.124APCh. 3 - Prob. 3.125APCh. 3 - Prob. 3.126APCh. 3 - Prob. 3.127APCh. 3 - Prob. 3.128APCh. 3 - Prob. 3.129APCh. 3 - Prob. 3.130APCh. 3 - Prob. 3.131APCh. 3 - Prob. 3.132APCh. 3 - Prob. 3.133APCh. 3 - Prob. 3.134APCh. 3 - Prob. 3.135APCh. 3 - Prob. 3.136APCh. 3 - Prob. 3.137APCh. 3 - Prob. 3.138APCh. 3 - Prob. 3.139APCh. 3 - Prob. 3.140APCh. 3 - Prob. 3.141APCh. 3 - Prob. 3.142APCh. 3 - Prob. 3.143APCh. 3 - Prob. 3.144APCh. 3 - Prob. 3.145APCh. 3 - Prob. 3.146APCh. 3 - Prob. 3.147APCh. 3 - Prob. 3.148AP
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