Natural abundances of ^{12}C and ^{13}C isoto | vedclass

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(A) In the molecular formula $C_2F_4$,there are $4$ fluorine atoms. The atomic mass of $F$ is $19$,so $4$ fluorine atoms contribute $76$ to the molar

Vedclass · Accepted Answer

$1.98$

Vedclass · Answer

(A) In the molecular formula $C_2F_4$,there are $4$ fluorine atoms. The atomic mass of $F$ is $19$,so $4$ fluorine atoms contribute $76$ to the molar mass. Possible molar masses of $C_2F_4$ depend on the carbon isotopes: $100$ $(76 + 12 + 12)$,$101$ $(76 + 12 + 13)$,and $102$ $(76 + 13 + 13)$. Given the abundances: $^{12}C = 99\%$ $(0.99)$ and $^{13}C = 1\%$ $(0.01)$. The probability of having a molar mass of $101$ corresponds to the combination of one $^{12}C$ and one $^{13}C$ atom. Since there are two carbon positions,the combinations are $(^{12}C, ^{13}C)$ or $(^{13}C, ^{12}C)$. Percentage of $C_2F_4$ with molar mass $101 = 2 	imes (0.99 	imes 0.01) 	imes 100 = 1.98\%$.

Natural abundances of $^{12}C$ and $^{13}C$ isotopes of carbon are $99\,\%$ and $1\,\%$,respectively. Assuming they only contribute to the molecular weight of $C_2F_4$,the percentage of $C_2F_4$ having a molecular mass of $101$ is $.....$

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