Question

A 0.02887 g sample of gas occupies 10.0 mL at 288.0 K and 1.10 atm. Upon further analysis, the compound is found to be 38.734 % C and 61.266 % F . What is the molecular formula of the compound?

Answer 1

Answer:  The molecular formula of the compound is $C_2F_2$

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 =$\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{61.266g}{19g/mole}=3.2245moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.2278}{3.2245}=1$

For F =$\frac{3.2245}{3.2245}=1$

The ratio of C : F = 1:1

Hence the empirical formula is $CF$

The empirical weight of $CF$ = 1(12)+1(19)= 31 g.

Using ideal gas equation :

$PV=nRT$

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

$1.10\times 0.01=n\times 0.0821\times 288.0$

$n=4.65\times 10^{-4}$

$n=\frac{\text {given mass}}{\text {Molar mass}}$

$4.65\times 10^{-4}=\frac{0.02887}{\text {Molar mass}}$

${\text {Molar mass}}=62.0$

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{62.0}{31.0}=2$

The molecular formula will be=$2\times CF=C_2F_2$