Question

Calculate the ratio of the ${(M+2)}^{+}$to $M^{+}$and the ${(M+4)}^{+}$to $M^{+}$peak heights for

(a)$C_{10}{\mathrm{H}}_6{\mathrm{Br}}_2$, (b)$C_3{H}_{7}ClBr$and (c)$C_6{H}_{4}C{l}_2$.

Short answer

The ratios obtained for ${(M+2)}^{+}$to $M^{+}$and ${(M+4)}^{+}$to $M^{+}$are:

(a) $1.96$and $0.96$.

(b) $1.30$and $0.32$

(c)$0.65$and$0.106$

Step-by-step solution

Molecular formula

The number of atoms of each element in one molecule of a compound is expressed by the molecular formula.

Calculation of ratio for C10H6Br2 is as follows :

The molecular formula given is $C_{10}{H}_{6}B{r}_2$.

For every $100$${}^{79}Br$atoms, but there are role="math" localid="1645520143733" $98{}^{81}Br$atoms. This molecular formula has two atoms of bromine. So, the ratio can be calculated as :

$$\begin{gathered} \frac{{(M+2)}^{+}}{M^{+}}=\frac{2\times 98}{100} \\ =1.96 \\ \frac{{(M+4)}^{+}}{M^{+}}={\left(\frac{98}{100}\right)}^2 \\ =0.96 \end{gathered}$$

Hence, the ratios are $1.96$and $0.96$.

Calculation of ratio for C3H7ClBr is as follows :

The molecular formula given is $C_3{H}_{7}ClBr$.

For every$100$${}^{35}Cl$atoms, but there are$32.5$${}^{37}Cl$atoms. So the ratio can be calculated as :

$$\begin{gathered} \frac{{(M+2)}^{+}}{M^{+}}=\left(\frac{1\times 98}{100}\right)+\left(\frac{1\times 32.5}{100}\right) \\ =1.30 \\ {\frac{(M+4)}{M^{+}}}^{+}=\left(\frac{1\times 98}{100}\right)\times \left(\frac{1\times 32.5}{100}\right) \\ =0.32 \end{gathered}$$

Hence, the ratios are $1.30$and $0.32$.

Calculation of ratio for C6H4Cl2 is as follows :

The molecular formula given is $C_6{H}_{4}C{L}_2$.

For every $100$${}^{35}Cl$atoms, but there are $32.5$${}^{37}Cl$atoms. The molecular formula has two atoms of chlorine. So the ratio can be calculated as :

$$\begin{gathered} \frac{{(M+2)}^{+}}{M^{+}}=\frac{2\times 32.5}{100} \\ =0.65 \\ \frac{{(M+4)}^{+}}{M^{+}}={\left(\frac{32.5}{100}\right)}^2 \\ =0.106 \end{gathered}$$

Hence, the ratios are$0.65$and$0.106$.