Q12-12-15 - Organic Chemistry (Mcmurry)

Question

Write molecular formulas for compounds that show the following molecular ions in their high-resolution mass spectra, assuming that C, H, N, and O might be present. The exact atomic masses are: 1.00783 \(\left( {{}^1{\bf{H}}} \right)\), 12.00000 \(\left( {{}^{12}C} \right)\), 14.00307 \(\left( {{}^{14}N} \right)\), 15.99491 \(\left( {{}^{16}O} \right)\).

(a) \({{\bf{M}}^ + }{\rm{ =  98}}{\rm{.0844}}\)  

(b \({{\bf{M}}^ + }{\rm{ =  123}}{\rm{.0320}}\)

 

 

Step-by-Step Solution

Answer

Molecular formula of molecule:

(a) \({C_5}H{}_{10}{N_2}\)

(b) \({C_6}H{}_5N{O_2}\)

Step 1: Molecular mass

Molecular mass of a substance is the sum of atomic masses of all its atoms in a molecule.

Step 2: Calculation of molecular formula

• Let’s assume these molecules only carry hydrogen and carbon. 
• Molecular mass of carbon is 12 and molecular mass of hydrogen is 1.
• Sum of molecular mass of C and H is \(12 + 1 = 13\)
• Now divide the given molecular mass by 13.
• The quotient indicates the number of carbon and number of hydrogen is equal to the quotient plus the remainder.

 

Step 3: Molecular formula of given molecules possible only with C and H

(a) Molecule with molecular weight 98 : 

• Given by: \(\frac{{98}}{{13}} = 7\frac{7}{{13}}\)
• This indicates that there are 7 carbon and 14 hydrogen.
• So, the formula is \({C_7}H{}_{14}\), maximum can have 7 carbon. 

 

(b) Molecule with molecular weight 123 : 

• Given by: \(\frac{{123}}{{13}} = 9\frac{6}{{13}}\) 
• This indicates that there are 9 carbon and 17 hydrogen.
• So, the formula is  \({C_9}H{}_{17}\)

Step 4: Actual molecules possible

Molecular mass of oxygen is 16 and of nitrogen is 14.

As oxygen or nitrogen is also part of molecule so all possible combinations are obtained by adjusting number of atoms to fit given molecular mass.

(a) Molecule with molecular weight 98 has two possible structures: \({C_6}H{}_{10}O\) and \({C_5}H{}_{10}{N_2}\).

\(\) 

(b) Molecule with molecular weight 123 is:

\({C_6}H{}_5N{O_2}\) as molecular mass is odd number here so hydrogen must be in odd number to get precise calculation.

Step 5: Mass of each predicted molecule

• Precise molecular mass of carbon is 12.0g/mol, mass of hydrogen is 1.00783g/mol, mass of nitrogen is 14.00307 and mass of oxygen is 15.99491g/mol.
• For (a) with molecular weight 98.0844:

\[\begin{array}{c}{C_6}H{}_{10}O = \left( {6 \times molar{\rm{ }}mass{\rm{ of C}}} \right) + \left( {1 \times molar{\rm{ }}mass{\rm{ of O}}} \right) + \left( {10 \times molar{\rm{ }}mass{\rm{ of H}}} \right)\\ = \left( {6 \times 12} \right) + \left( {1 \times 15.99491} \right) + \left( {10 \times 1.00783} \right)\\ = \left( {72 + 15.99491 + 10.00783} \right)\\ = 98.00274g\end{array}\]

 

\[\begin{array}{c}{C_5}H{}_{10}{N_2} = \left( {5 \times molar{\rm{ }}mass{\rm{ of C}}} \right) + \left( {2 \times molar{\rm{ }}mass{\rm{ of N}}} \right) + \left( {10 \times molar{\rm{ }}mass{\rm{ of H}}} \right)\\ = \left( {5 \times 12} \right) + \left( {2 \times 14.00307} \right) + \left( {10 \times 1.00783} \right)\\ = \left( {60 + 28.00614 + 10.00783} \right)\\ = 98.01397g\end{array}\]

 

      It can be seen latter molecule is more precise and hence this is the correct structure.

 

• For (b) with molecular weight 23.0320:

\[\begin{array}{c}{C_6}H{}_5N{O_2} = \left( {6 \times mol.{\rm{ mass of C}}} \right) + \left( {1 \times mol.{\rm{ }}mass{\rm{ of N}}} \right) + \left( {5 \times mol.{\rm{ }}mass{\rm{ of H}}} \right) + \left( {2 \times mol.{\rm{ }}mass{\rm{ of O}}} \right)\\ = \left( {6 \times 12} \right) + \left( {1 \times 14.00307} \right) + \left( {5 \times 1.00783} \right) + \left( {2 \times 15.99491} \right)\\ = \left( {72 + 14.00307 + 5.03915 + 31.98982} \right)\\ = 123.03204g\end{array}\]       It can be seen this molecule is precise and hence is correct structure.