Mass spectrometry — practice questions

To calculate the relative atomic mass, we use the formula: $A_r = \frac{\sum (m/z \times \text{abundance})}{\text{total abundance}}$

1. Calculate the sum of (mass × abundance) for each isotope:
   • $(90 \times 51.5) = 4635$
   • $(91 \times 11.2) = 1019.2$
   • $(92 \times 17.1) = 1573.2$
   • $(94 \times 17.4) = 1635.6$
   • $(96 \times 2.8) = 268.8$
2. Sum these products: $4635 + 1019.2 + 1573.2 + 1635.6 + 268.8 = 9131.8$
3. Calculate the total abundance: $51.5 + 11.2 + 17.1 + 17.4 + 2.8 = 100.0$
4. Divide the sum of products by the total abundance: $A_r = \frac{9131.8}{100.0} = 91.318$
5. Round to one decimal place as requested: $A_r = 91.3$

First, confirm the $M_r$ of propan-1-ol: $(3 \times 12.0) + (8 \times 1.0) + (1 \times 16.0) = 36.0 + 8.0 + 16.0 = 60.0$. This matches the molecular ion peak.

• Peak at $m/z = 59$: This corresponds to a mass loss of 1 from the molecular ion ($60 - 59 = 1$). This is characteristic of the loss of a single hydrogen atom (a hydrogen radical, $H\cdot$). The fragment is $[CH_3CH_2CHOH]^+$.

• Peak at $m/z = 45$: This corresponds to a mass loss of 15 from the molecular ion ($60 - 45 = 15$). A mass of 15 corresponds to a methyl group ($CH_3$). The bond between the first and second carbon has broken, losing a $CH_3\cdot$ radical. The fragment is $[CH_2CH_2OH]^+$.

• Peak at $m/z = 31$: This corresponds to a mass loss of 29 from the molecular ion ($60 - 31 = 29$). A mass of 29 corresponds to an ethyl group ($CH_2CH_3$). The bond between the first carbon and the oxygen has broken, losing a $CH_3CH_2\cdot$ radical. The fragment is $[CH_2OH]^+$. This is a very common and stable fragment for primary alcohols.