Question

The empirical formula of a gaseous fluorocarbon is $C{F}_2$. At a certian temperature and pressure, a 1 L volume hold 8.95 g of this fluorocarbon. Under the same conditions, 1 L volume holds only 1.70 g gaseous $F_2$. Thus, fluorocarbon is_______.
(F = 19)

• A. $C{F}_2$
• B. $C_2{F}_4$
• C. $C_3{F}_6$
• D. $C_4{F}_8$

Answer 1

The correct option is D $C_4{F}_8$

Empirical weight of $C{F}_2=50$
Molecular formula = $(C{F}_2{)}_n$
where, $n=\frac{\text{molecular weight}}{\text{empirical weight}}$
Under the condition of same temperature and pressure, all gasses will have equal number of molecules.
Hence,
moles of $F_2$ = moles of the gaseous fluorocarbon

moles of $F_2$ = $\frac{1.7}{38}=\text{moles of fluorocarbon}$
Molar mass of the fluorocarbon$=\frac{\text{mass}}{\text{number of moles}}=\frac{8.95\times 38}{1.7}=200g$

Thus, molecular weight of $(C{F}_2{)}_n=200{\text{g mol}}^{-1}$
$\therefore n=\frac{200}{50}=4$
$\therefore$ Molecular formula = $(C{F}_2{)}_4=C_4{F}_8$