Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 2, Problem 2.138P
Interpretation Introduction

Interpretation:

The mass percentages of NO in the molecules of nitroglycerin and isoamyl nitrate are to be determined.

Concept introduction:

The molecular mass of a compound is defined as the sum of the atomic masses of atoms of all the elements present in it.

The general formula to calculate the molecular mass of a compound is,

Molecular mass of the compound=[Σ(number of atoms of the element)(atomic mass of the element)] (1)

The mass percent of an element in a compound is defined as the total concentration of the element in the compound.

The general formula to calculate the total mass of a molecule in a compound is,

Total mass of a molecule=[(total number of molecules)(molecular mass )] (2)

The general formula to calculate the mass percent of a molecule in a compound is,

Mass percent=(total mass of the molecule in the compoundmolecular mass of the compound)(100) (3)

Expert Solution & Answer
Check Mark

Answer to Problem 2.138P

The mass percentages of NO in the molecules of nitroglycerin and isoamyl nitrate are 39.64 and 22.54 respectively.

Explanation of Solution

The formula to calculate the molecular mass of NO is,

Molecular mass=[(1)(atomic mass of nitrogen)+(1)(atomic mass of oxygen)] (4)

Substitute 14.01 amu for the atomic mass of nitrogen and 16 amu for the atomic mass of oxygen in equation (4).

Molecular mass=[(1)(14.01 amu)+(1)(16 amu)]=14.01 amu+16 amu=30.01 amu

The molecular formula for nitroglycerin is C3H5N3O9.

The formula to calculate the molecular mass of C3H5N3O9 is,

Molecular mass=[(3)(atomic mass of carbon)+(5)(atomic mass of hydrogen)+(3)(atomic mass of nitrogen)+(9)(atomic mass of oxygen)] (5)

Substitute 12.01 amu for the atomic mass of carbon, 1.008 amu for the atomic mass of hydrogen, 14.01 amu for the atomic mass of nitrogen and 16 amu for the atomic mass of oxygen in equation (5).

Molecular mass=[(3)(12.01 amu)+(5)(1.008 amu)+(3)(14.01 amu)+(9)(16 amu)]=36.03 amu+5.04 amu+42.03 amu+144 amu=227.1 amu

The total number of NO molecules present in one molecule C3H5N3O9 is 3.

The total mass of 3 molecules of NO in a molecule of C3H5N3O9 is calculated as follows:

Total mass=[(total number of molecules)(molecular mass )] (6)

Substitute 3 for the total number of molecules and 30.01 amu for the molecular mass in equation (6).

Total mass=(3)(30.01 amu)=90.03 amu

The formula to calculate the mass percent of NO in a molecule of C3H5N3O9 is,

Mass percent=(total mass of the NO in C3H5N3O9molecular mass of C3H5N3O9)(100) (7)

Substitute 90.03 amu for the total mass of NO in C3H5N3O9 and 227.1 amu for the molecular mass of C3H5N3O9 in equation (7).

Mass percent=(90.03 amu227.1 amu)(100)=(0.3964)(100)=39.64

The molecular formula for isoamyl nitrate is (CH3)2CHCH2CH2ONO2.

The formula to calculate the molecular mass of (CH3)2CHCH2CH2ONO2 is,

Molecular mass=[(5)(atomic mass of carbon)+(11)(atomic mass of hydrogen)+(1)(atomic mass of nitrogen)+(3)(atomic mass of oxygen)] (8)

Substitute 12.01 amu for the atomic mass of carbon, 1.008 amu for the atomic mass of hydrogen, 14.01 amu for the atomic mass of nitrogen and 16 amu for the atomic mass of oxygen in equation (5).

Molecular mass=[(5)(12.01 amu)+(11)(1.008 amu)+(1)(14.01 amu)+(3)(16 amu)]=60.05 amu+11.088 amu+14.01 amu+48 amu=133.15 amu

The total number of NO molecules present in one molecule (CH3)2CHCH2CH2ONO2 is 1.

The total mass of 1 molecule of NO in a molecule of C3H5N3O9 is calculated as follows:

Total mass=[(total number of molecules)(molecular mass )] (9)

Substitute 1 for the total number of molecules and 30.01 amu for the molecular mass in equation (9).

Total mass=(1)(30.01 amu)=30.01 amu

The formula to calculate the mass percent of NO in a molecule of (CH3)2CHCH2CH2ONO2 is,

Mass percent=(total mass of the NO in (CH3)2CHCH2CH2ONO2molecular mass of (CH3)2CHCH2CH2ONO2)(100) (10)

Substitute 30.01 amu90.03 amu for the total mass of NO in (CH3)2CHCH2CH2ONO2 and 133.15 amu for the molecular mass of (CH3)2CHCH2CH2ONO2 in equation (10).

Mass percent=(30.01 amu133.15 amu)(100)=(0.2254)(100)=22.54

Conclusion

The mass percentages of NO in the molecules of nitroglycerin and isoamyl nitrate are 39.64 and 22.54 respectively.

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Chapter 2 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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