EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 21, Problem 21.25P

(a)

Interpretation Introduction

Interpretation:

Plausible formula for CnHxOyNz has to be suggested when nominal mass is 79 and ratio of M++1/M+ peak is 5.9 %.

Concept Introduction:

Nominal mass may be defined as mass of most abundant isotope of a particular ion or molecule. For instance, carbon exist as two isotopes C-12 and C-13. C-12 is abundant as much as 98.9 % while abundance of C-13 is only 1.1 %. So clearly nominal mass of carbon is 12 g.

(a)

Expert Solution
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Explanation of Solution

Odd nominal mass of 79 suggests there are odd numbers of N atoms. So it can be 1, 3 or any odd number.

Natural abundance of C-13 is 1.1 %. Hence number of C atoms is calculated as follows:

  M++1M+=n(1.1 %)n=5.9 %1.1 %=5.35

Hence number of C atom is 5.

Since each C has mass of 12, so five C atoms contribute a mass of 60 to overall formula CnHxOyNz. This leaves 19 mass that indicates possibility of one N atom and remainare five H atoms. So with no O atom, the plausible formula of compound is C5H5N.

(b)

Interpretation Introduction

Interpretation:

Plausible formula for CnHxOyNz has to be suggested when nominal mass is 123 and ratio of M++1/M+ peak is 6.1 %.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
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Explanation of Solution

Odd nominal mass of 123 suggests there are odd numbers of N atoms. So it can be 1, 3 or any odd number.

Natural abundance of C-13 is 1.1 %. Hence number of C atoms is calculated as follows:

  M++1M+=n(1.1 %)n=6.1 %1.1 %=5.55 or 6

Hence number of C atom is 5 or 6.

Since each C has mass of 12 so six C atoms are assumed in formula then they contribute a mass of 72 to overall formula CnHxOyNz. This leaves 51 as mass that indicates possibility of only one N atom. If molecule has 3 N atoms then it would have contributed 41 and then no possibility of O atom exists.

With 1N, and 2 O and remainder 5 H atoms, contribution of 51 gets satisfied and thus one plausible formula is C6H5O2N.

When five C atoms are assumed in formula, they contribute a mass of 60 to overall formula CnHxOyNz. This leaves 63 as mass that indicates possibility of 3 N atoms and this would have contributed 42 and then one possibility of 1 O atom and remainder five H atoms. Thus, another possible formula is C5H5ON3.

(c)

Interpretation Introduction

Interpretation:

Plausible formula for CnHxOyNz has to be suggested when nominal mass is 148 and ratio of M++1/M+ peak is 7.4 %.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
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Explanation of Solution

Even nominal mass of 148 suggests there are even numbers of N atoms. So it can be 2, 4 or any even number.

Natural abundance of C-13 is 1.1 %. Hence number of C atoms is calculated as follows:

  M++1M+=n(1.1 %)n=7.4 %1.1 %=6.77 or 6

Hence number of C atom is 7 or 6.

Plausible formulas with 6 C atoms that satisfy nominal mass of 148 include C6H12O4 and C6H16O2N2.

Plausible formulas with 7 C atoms that satisfy nominal mass of 148 include C7H4O2N2 and C7O4.

(d)

Interpretation Introduction

Interpretation:

Plausible formula for CnHxOyNz has to be suggested when nominal mass is 168 and ratio of M++1/M+ peak is 12.5 %.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Even nominal mass of 168 suggests there are even numbers of N atoms. So it can be 2, 4 or any even number.

Natural abundance of C-13 is 1.1 %. Hence number of C atoms is calculated as follows:

  M++1M+=n(1.1 %)n=12.5 %1.1 %=11.3611 or 12

Hence number of C atom is 11 or 12.

Plausible formulas with 11 C atoms that satisfy nominal mass of 168 includes C11H20O, C11H4O2 or C11H8N2.

Plausible formulas with 12 C atoms that satisfy nominal mass of 168 include C12H24 and C12H8O.

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