Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 3, Problem 3.77QP

(a)

Interpretation Introduction

Interpretation: The percent composition of the given compounds is 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 percent composition of Na2O .

(a)

Expert Solution
Check Mark

Answer to Problem 3.77QP

Solution

Percent composition of Na2O is 74.19% Na_ and 25.81% O_ .

Explanation of Solution

Explanation

The molar mass of Na2O is calculated as,

Molar mass=(Number of Na atoms×Molar mass of Na+Number of O atoms×Molar mass of O)

In Na2O , number of Na atoms =2 , number of O atoms =1 .

Molar mass of Na =22.98 g/mol and molar mass of O =16 g/mol .

Substitute the number of sodium and oxygen atoms and their molar masses in the above equation.

Molar mass=2×22.98+1×16=61.98 g/mol

The percent composition of sodium ( Na ) is calculated by the formula,

Percent composition of Na=Mass of NaMass of Na2O×100

Mass of 2 atoms of Na atoms =2×22.99 g=45.98 g .

Substitute the value of mass of Na and mass of Na2O in the above equation.

Percent composition of Na=45.98 g61.98 g×100=74.19% Na_

Therefore, the percent composition of Na is 74.19% Na_ .

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

Percent composition of O=Mass of OMass of Na2O×100

Mass of 1 atom of O =1×16 g=16 g .

Substitute the value of mass of O and mass of Na2O in the above equation.

Percent composition of O=16 g61.98 g×100=25.81% O_

Therefore, the percent composition of O is 25.81% O_ .

Percent composition of Na2O is 74.19% Na_ and 25.81% O_ .

(b)

Interpretation Introduction

To determine: The percent composition of NaOH .

(b)

Expert Solution
Check Mark

Answer to Problem 3.77QP

Solution

Percent composition of NaOH is 54.48% Na_ and 40.00% O_ and 2.52% H_ .

Explanation of Solution

Explanation

The molar mass of NaOH is calculated as,

Molar mass=(Number of Na atoms×Molar mass of Na+Number of O atoms×Molar mass of O+Number of H atoms×Molar mass of H)

In NaOH , number of Na atoms =1 , number of O atoms =1 and number of H atoms =1 .

Molar mass of Na =22.98 g/mol and molar mass of O =16 g/mol and molar mass of H is 1.008 g/mol

Substitute the number of sodium, oxygen and hydrogen atoms and their molar masses in the above equation.

Molar mass=1×22.98+1×16+1×1.008=40.00 g/mol

The percent composition of sodium ( Na ) is calculated by the formula,

Percent composition of Na=Mass of NaMass of NaOH×100

Substitute the value of mass of Na and mass of NaOH in the above equation.

Percent composition of Na=22.99 g40.00 g×100=57.48% Na_

Therefore, the percent composition of Na is 57.48% Na_ .

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

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

Mass of 1 atom of O =1×16 g=16 g .

Substitute the value of mass of O and mass of NaOH in the above equation.

Percent composition of O=16 g40.00 g×100=40.00% O_

Therefore, the percent composition of O is 40.00% O_ .

The percent composition of hydrogen ( H ) is calculated by the formula,

Percent composition of H=Mass of HMass of NaOH×100

Mass of 1 atom of H =1×1.008 g=1.008 g .

Substitute the value of mass of H and mass of NaOH in the above equation.

Percent composition of H=1.008 g40.00 g×100=2.52% H_

Therefore, the percent composition of H is 2.52% H_ .

Percent composition of NaOH is 54.48% Na_ and 40.00% O_ and 2.52% H_ .

(c)

Interpretation Introduction

To determine: The percent composition of NaHCO3 .

(c)

Expert Solution
Check Mark

Answer to Problem 3.77QP

Solution

Percent composition of NaHCO3 is 27.37% Na_ , 1.20% H_ , 14.30% C_ and 57.14% O_

Explanation of Solution

Explanation

The molar mass of NaHCO3 is calculated as,

Molar mass=(Number of Na atoms×Molar mass of Na+Number of O atoms×Molar mass of O+Number of H atoms×Molar mass of H+Number of C atoms×Molar mass of C)

In NaHCO3 , number of Na atoms =1 , number of O atoms =3 , number of C atoms =1 and number of H atoms =1 .

Molar mass of Na =22.98 g/mol , molar mass of C =12 g/mol , molar mass of O =16 g/mol and molar mass of H is 1.008 g/mol

Substitute the number of sodium, carbon, oxygen and hydrogen atoms and their molar masses in the above equation.

Molar mass=1×22.98+3×16+1×1.008+1×12=84.01 g/mol

The percent composition of sodium ( Na ) is calculated by the formula,

Percent composition of Na=Mass of NaMass of NaHCO3×100

Substitute the value of mass of Na and mass of NaHCO3 in the above equation.

Percent composition of Na=22.99 g84.01 g×100=24.37% Na_

Therefore, the percent composition of Na is 24.37% Na_ .

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

Percent composition of O=Mass of OMass of NaHCO3×100

Mass of 3 atom of O =3×16 g=48 g .

Substitute the value of mass of O and mass of NaHCO3 in the above equation.

Percent composition of O=48 g84.01 g×100=57.14% O_

Therefore, the percent composition of O is 57.14% O_ .

The percent composition of hydrogen ( H ) is calculated by the formula,

Percent composition of H=Mass of HMass of NaHCO3×100

Mass of 1 atom of H =1×1.008 g=1.008 g .

Substitute the value of mass of H and mass of NaHCO3 in the above equation.

Percent composition of H=1.008 g84.01 g×100=1.20% H_

Therefore, the percent composition of H is 1.20% H_ .

The percent composition of carbon (C) is calculated by the formula,

Percent composition of C=Mass of CMass of NaHCO3×100

Mass of 1 atom of C =1×12 g=12 g .

Substitute the value of mass of C and mass of NaHCO3 in the above equation.

Percent composition of C=12.01 g84.01 g×100=14.30% C_

Therefore, the percent composition of C is 14.30% C_ .

Percent composition of NaHCO3 is 27.37% Na_ , 1.20% H_ , 14.30% C_ and 57.14% O_

(d)

Interpretation Introduction

To determine: The percent composition of Na2CO3 .

(d)

Expert Solution
Check Mark

Answer to Problem 3.77QP

Solution

Percent composition of Na2CO3 is 43.38% Na_ , 11.33% C_ and 45.28% O_ .

Explanation of Solution

Explanation

The molar mass of Na2CO3 is calculated as,

Molar mass=(Number of Na atoms×Molar mass of Na+Number of O atoms×Molar mass of O+Number of C atoms×Molar mass of C)

In Na2CO3 , number of Na atoms =2 , number of O atoms =3 , number of C atoms =1 .

Molar mass of Na =22.98 g/mol , molar mass of C =12 g/mol , molar mass of O =16 g/mol .

Substitute the number of sodium, carbon, oxygen atoms and their molar masses in the above equation.

Molar mass=2×22.98+3×16+1×12=106 g/mol

The percent composition of sodium ( Na ) is calculated by the formula,

Percent composition of Na=Mass of NaMass of Na2CO3×100

Mass of 2 Na atoms 2×22.98g=45.98 g .

Substitute the value of mass of Na and mass of Na2CO3 in the above equation.

Percent composition of Na=45.98 g106 g×100=43.38% Na_

Therefore, the percent composition of Na is 43.38% Na_ .

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

Percent composition of O=Mass of OMass of Na2CO3×100

Mass of 3 atom of O =3×16 g=48 g .

Substitute the value of mass of O and mass of Na2CO3 in the above equation.

Percent composition of O=48 g106 g×100=45.28% O_

Therefore, the percent composition of O is 45.28% O_ .

The percent composition of carbon ( C ) is calculated by the formula,

Percent composition of C=Mass of CMass of Na2CO3×100

Mass of 1 atom of C =1×12 g=12 g .

Substitute the value of mass of C and mass of Na2CO3 in the above equation.

Percent composition of C=12.01 g106 g×100=11.33% C_

Therefore, the percent composition of C is 11.33% C_ .

Conclusion

  1. a. Percent composition of Na2O is 74.19% Na_ and 25.81% O_ .
  2. b. Percent composition of NaOH is 54.48% Na_ and 40.00% O_ and 2.52% H_ .
  3. c. Percent composition of NaHCO3 is 27.37% Na_ , 1.20% H_ , 14.30% C_ and 57.14% O_ .
  4. d. Percent composition of Na2CO3 is 43.38% Na_ , 11.33% C_ and 45.28% O_

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Chapter 3 Solutions

Chemistry

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