Respuesta :
Answer: The molecular formula of the compound is [tex]C_2F_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 38.734 g
Mass of F = 61.266 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of F =[tex]\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{61.266g}{19g/mole}=3.2245moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.2278}{3.2245}=1[/tex]
For F =[tex]\frac{3.2245}{3.2245}=1[/tex]
The ratio of C : F = 1:1
Hence the empirical formula is [tex]CF[/tex]
The empirical weight of [tex]CF[/tex] = 1(12)+1(19)= 31 g.
Using ideal gas equation :
[tex]PV=nRT[/tex]
where,
n = number of moles of gas = ?
P = pressure of the gas = 1.10 atm
T = temperature of the gas = 288.0 K
R = gas constant = 0.0821 L.atm/mole.K
V = volume of gas = 10.0 mL =0.01 L
[tex]1.10\times 0.01=n\times 0.0821\times 288.0[/tex]
[tex]n=4.65\times 10^{-4}[/tex]
[tex]n=\frac{\text {given mass}}{\text {Molar mass}}[/tex]
[tex]4.65\times 10^{-4}=\frac{0.02887}{\text {Molar mass}}[/tex]
[tex]{\text {Molar mass}}=62.0[/tex]
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{62.0}{31.0}=2[/tex]
The molecular formula will be=[tex]2\times CF=C_2F_2[/tex]
