Question

Natural abundances of ${}^{12}C$ and ${}^{13}C$ isotopes of carbon are $99\%$ and $1\%$, respectively. Assuming they only contribute to the mol. wt. of $C_2{F}_4$, the percentage of $C_2{F}_4$ having a molecular mass of $101$ is:

• A. $1.00$
• B. $98$
• C. $0.198$
• D. $99$

Answer 1

The correct option is A $1.00$

The molecular mass of $C_2{F}_4$ containing C-12 is $2\left(12\right)+4\left(19\right)=100$ g/mol
The molecular mass of $C_2{F}_4$ containing one C-12 and one C-13 is $12+13+4\left(19\right)=101$ g/mol
Natural abundances of C-12 and C-13 isotopes of carbon are 99% and 1%, respectively.
100 $C_2{F}_4$ molecules will have 200 C atoms out of whic 1% or 2 C atoms will be C-13. Assuming that these two C-13 atoms will be in different $C_2{F}_4$ molecules, the percentage of $C_2{F}_4$ having a molecular mass of 101 is $\frac{2}{200}\times 100=1$ %