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:

1.00

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98

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0.198

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99

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Solution

The correct option is A $1.00$
The molecular mass of $C_{2} F_{4}$ containing C-12 is $2 (12) + 4 (19) = 100$ g/mol
The molecular mass of $C_{2} F_{4}$ containing one C-12 and one C-13 is $12 + 13 + 4 (19) = 101$ 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 $\frac{2}{200} \times 100 = 1$ %

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Natural abundances of 12C and 13C isotopes of carbon are 99% and 1%, respectively. Assuming they only contribute to the mol. wt. of C2F4, the percentage of C2F4 having a molecular mass of 101 is:

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: