Question

1. there are two stable isotopes of chlorine, $^{35}cl$ (34.9688 u) and $^{37}cl$ (36.9659 u), with % natural abundances of 75.8% and 24.2% respectively. carbon also has two stable isotopes, $^{12}c$ (12.00 u, 98.9%) and $^{13}c$ (13.00 u, 1.1%).

a) what is the atomic number of cl? (z = ?)
b) how many neutrons are present in $^{37}cl$?
c) assuming there is only one isotope for h (1h, 1.000 u, 100%), how many signals to you expect to see in a full mass spectrum of $ch_3cl$? do not include any fragments! (hint: determine all possible combinations of isotopes)

Explanation:

Step1: Recall atomic number concept

The atomic number of chlorine (Cl) is 17, not 35 (35 is the mass - number of one of its isotopes).

Step2: Calculate neutrons in $^{37}$Cl

The atomic number of Cl is $Z = 17$. For an isotope $^{A}X$, the number of neutrons $n=A - Z$. For $^{37}$Cl, $A = 37$ and $Z = 17$, so $n=37 - 17=20$.

Step3: Determine mass - spectrum signals of $CH_3Cl$

Carbon has 2 isotopes ($^{12}C$ and $^{13}C$) and chlorine has 2 isotopes ($^{35}Cl$ and $^{37}Cl$), and hydrogen has 1 isotope ($^{1}H$).
The possible combinations for $CH_3Cl$ are:

1. $^{12}C - ^{1}H_3-^{35}Cl$
2. $^{12}C - ^{1}H_3-^{37}Cl$
3. $^{13}C - ^{1}H_3-^{35}Cl$
4. $^{13}C - ^{1}H_3-^{37}Cl$

So there are 4 possible signals.

Answer:

a) 17
b) 20
c) 4