Question

A 0.02882 g sample of gas occupies 10.0-mL at 288.5 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?

What is the molecular formula of the compound?

Draw the Lewis structure for this compound

Is this bond polar?

Identify the geometry around each carbon atom.

Answer 1

Given that;

Sample = 0.02882 g

Volume =10.0-mL, 0.010 L

Temperature = 288.5 K

Pressure = 1.10 atm.

38.734% C and 61.266% F

Calculate moles

PV = nRT

n = PV/(RT)

n = (1.10)(0.01)/(0.082*288.5)

= 0.000464978 mol

MW = mass / mol = 0.02882 /0.000464978 = 61.98 g /mol

Now calculate the moles C and F in the compound as follows:

Assume that compound is 100 g then

Amount of C =38.734 g

Amount of F = 61.266 g

Number of moles of C = 38.734 g /12.01 g/ mole = 3.23 mole

Number of moles of F = 61.266 g /18.99 g/ mole = 3.23 mole

Now calculate the molar ratio as follows:

C: 3.23 /3.23 =1

F= 3.23/3.23=1

Empirical formula = CF

Empirical mass = Mass of C+ mass of F = 12.01 + 18.99 = 31 g/ mole

Molar mass = 61.98 g /mol

Then calculate the number of empirical units in molecular formula as follows:

n   = 61.98 g /mol / 31 g/ mole

=2

So, the molecular formula is CF*2 = C2F2