Chapter 13, Problem 13.27P

Propose two molecular formulas for each molecular ion: (a) 102; (b) 98; (c) 119; (d) 74

Interpretation Introduction

(a)

Interpretation: Two molecular formulas for the given molecule are to be proposed.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The compound that contain $\text{C, H}$ and $\text{O}$ atoms always have a molecular ion with even mass.

Answer to Problem 13.27P

The possible molecular formulas are ${\text{C}}_8{\text{H}}_6$, ${\text{C}}_7{\text{H}}_{18}$ and ${\text{C}}_6{\text{H}}_{14}\text{O}$.

Explanation of Solution

The molecular ion peak of a given compound is observed at $m\text{/}z=102$.

Possible hydrocarbons are calculated as,

• Divide $102$ by $12$ (mass of one carbon atom). This gives the maximum number of carbon atoms.

$\frac{102}{12}=8$ Carbon atoms (remainder is $6$).

Possible molecular formula is ${\text{C}}_8{\text{H}}_6$.

• Replace one carbon by $12$ hydrogen atoms for another possible molecular formula.

${\text{C}}_8{\text{H}}_{\text{6}}\underset{{\text{+12H}}^{\text{'}}\text{s}}{\overset{-\text{1C}}{\to }}{\text{C}}_7{\text{H}}_{\text{18}}$

The compound that contain $\text{C, H}$ and $\text{O}$ atoms always have a molecular ion with even mass.

Possible hydrocarbons with $\text{C, H}$ and $\text{O}$ atoms are calculated as,

• Substitute one $\text{O}$ for ${\text{CH}}_4$.

${\text{C}}_7{\text{H}}_{18}\underset{\text{+1 O}}{\overset{-{\text{CH}}_4}{\to }}{\text{C}}_6{\text{H}}_{14}\text{O}$.

Hence, the possible molecular formulas are ${\text{C}}_8{\text{H}}_6$, ${\text{C}}_7{\text{H}}_{18}$ and ${\text{C}}_6{\text{H}}_{14}\text{O}$.

Conclusion

The possible molecular formulas are ${\text{C}}_8{\text{H}}_6$, ${\text{C}}_7{\text{H}}_{18}$ and ${\text{C}}_6{\text{H}}_{14}\text{O}$.

Interpretation Introduction

(b)

Interpretation: Two molecular formulas for the given molecule are to be proposed.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The compound that contain $\text{C, H}$ and $\text{O}$ atoms always have a molecular ion with even mass.

Answer to Problem 13.27P

The possible molecular formulas are ${\text{C}}_8{\text{H}}_2$, ${\text{C}}_7{\text{H}}_{14}$ and ${\text{C}}_6{\text{H}}_{10}\text{O}$.

Explanation of Solution

The molecular ion peak of a given compound is observed at $m\text{/}z=98$.

Possible hydrocarbons are calculated as,

• Divide $98$ by $12$ (mass of one carbon atom). This gives the maximum number of carbon atoms.

$\frac{98}{12}=8$ Carbon atoms (remainder =$2$) $\to {\text{C}}_8{\text{H}}_2$

• Replace one carbon by $12$ hydrogen atoms for another possible molecular formula.

${\text{C}}_8{\text{H}}_2\underset{{\text{+12H}}^{\text{'}}\text{s}}{\overset{-\text{1C}}{\to }}{\text{C}}_7{\text{H}}_{14}$

The compound that contain $\text{C, H}$ and $\text{O}$ atoms always have a molecular ion with even mass.

Possible hydrocarbons with $\text{C, H}$ and $\text{O}$ atoms are calculated as,

• Substitute one $\text{O}$ for ${\text{CH}}_4$.

${\text{C}}_7{\text{H}}_{14}\underset{\text{+1 O}}{\overset{-{\text{CH}}_4}{\to }}{\text{C}}_6{\text{H}}_{10}\text{O}$.

Hence, the possible molecular formulas are ${\text{C}}_8{\text{H}}_2$, ${\text{C}}_7{\text{H}}_{14}$ and ${\text{C}}_6{\text{H}}_{10}\text{O}$.

Conclusion

The possible molecular formulas are ${\text{C}}_8{\text{H}}_2$, ${\text{C}}_7{\text{H}}_{14}$ and ${\text{C}}_6{\text{H}}_{10}\text{O}$.

Interpretation Introduction

(c)

Interpretation: Two molecular formulas for the given molecule are to be proposed.

Concept introduction: The molecular ion has odd mass, the compound may contain $\text{C, H}$ and $\text{N}$ atoms. The compound that contains an odd number of $\text{N}$ atoms gives an odd molecular ion.

Answer to Problem 13.27P

The possible molecular formulas are ${\text{C}}_9{\text{H}}_{11}$ and ${\text{C}}_8{\text{H}}_9\text{N}$.

Explanation of Solution

The molecular ion peak of a given compound is observed at $m\text{/}z=119$.

Possible hydrocarbons are calculated as,

• Divide $73$ by $12$ (mass of one carbon atom). This gives the maximum number of carbon atoms.

$\frac{119}{12}=9$ Carbon atoms (remainder = $11$) $\to {\text{C}}_9{\text{H}}_{11}$.

Possible compounds with $\text{C, H}$ and $\text{N}$.

• Substitute one $\text{N}$ for ${\text{CH}}_2$.

${\text{C}}_9{\text{H}}_{11}\underset{\text{+1 N}}{\overset{-{\text{CH}}_2}{\to }}{\text{C}}_8{\text{H}}_9\text{N}$.

Hence, the possible molecular formulas are ${\text{C}}_9{\text{H}}_{11}$ and ${\text{C}}_8{\text{H}}_9\text{N}$.

Conclusion

The possible molecular formulas are ${\text{C}}_9{\text{H}}_{11}$ and ${\text{C}}_8{\text{H}}_9\text{N}$.

Interpretation Introduction

(d)

Interpretation: Two molecular formulas for the given molecule are to be proposed.

Concept introduction: Molecular mass is the sum of the atomic weights of each constituent element multiplied by the number of atoms of that element. The compound that contain $\text{C, H}$ and $\text{O}$ atoms always have a molecular ion with even mass.

Answer to Problem 13.27P

The possible molecular formulas are ${\text{C}}_6{\text{H}}_2$ and ${\text{C}}_4{\text{H}}_{10}\text{O}$.

Explanation of Solution

The molecular ion peak of a given compound is observed at $m\text{/}z=74$.

Possible hydrocarbons is calculated as,

• Divide $74$ by $12$ (mass of one carbon atom). This gives the maximum number of carbon atoms.

$\frac{74}{12}=6$ Carbon atoms (remainder =$2$) $\to {\text{C}}_6{\text{H}}_2$

• Replace one carbon by $12$ hydrogen atoms for another possible molecular formula.

${\text{C}}_6{\text{H}}_2\underset{{\text{+12H}}^{\text{'}}\text{s}}{\overset{-\text{1C}}{\to }}{\text{C}}_5{\text{H}}_{14}$

The compound that contain $\text{C, H}$ and $\text{O}$ atoms always have a molecular ion with even mass.

Possible hydrocarbons with $\text{C, H}$ and $\text{O}$ atoms are calculated as,

• Substitute one $\text{O}$ for ${\text{CH}}_4$.

${\text{C}}_5{\text{H}}_{14}\underset{\text{+1 O}}{\overset{-{\text{CH}}_4}{\to }}{\text{C}}_4{\text{H}}_{10}\text{O}$.

Hence, the possible molecular formulas are ${\text{C}}_6{\text{H}}_2$ and ${\text{C}}_4{\text{H}}_{10}\text{O}$.

Conclusion

The possible molecular formulas are ${\text{C}}_6{\text{H}}_2$ and ${\text{C}}_4{\text{H}}_{10}\text{O}$.