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?

 The molecular formula of the compound is

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 =

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

For C =

For F =

The ratio of C : F = 1:1

Hence the empirical formula is

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

Using ideal gas equation :

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

Now we have to calculate the molecular formula.

The molecular formula will be=