Posted by Anon on .
A 0.02882 g sample of gas occupies 10.0mL 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 geometry shape around each carbon atom?
Polar or Nonpolar?

Chemistry 
DrBob222,
Use PV = nRT and solve for n = numbre of mols. Substitute into n = grams/molar mass. You know mols and grams, solve fo rmolar mass. Use that number later.
Take a 100 g sample which gives you
38.734 g C and 61.266 g F. Convert to mols.
38.734/12 = ?
61.266/19 = ?
Find the ratio of C to F with the smallest number being 1.00. The easy way to do that is to divide the smaller number by itself; then divide the other number by the same small number. That gives you the empirical formula.
Calculate the mass of the empirical formula and substitute into
empirical formula x ?number = molar mass from above and round ?number to a whole number.
The molecular formula is (CxHy)_{?}
Post your work if you get stuck. 
Chemistry 
Anon,
For n I got 26.18 mols. For the mass I got 31 because it was CF.

Chemistry 
DrBob222,
26.18 is not right. CF is the empirical formula. The empirical mass is not 31. You need the molar mass from mols and grams.

Chemistry 
Anon,
The moles of C is 3.227. The moles of F is 3.224. Therefore the empirical formula is CF. How would I find the empirical mass? Your explanation is confusing.

Chemistry 
DrBob222,
Yes, I said the empirical formula of CF is correct. My error for empirical mass; it is 12 + 19 = 31 as you noted.
Here is how you do the molar mass.
PV = nRT
P = 1.10 atm
V = 10.0 mL = 0.0100 L
n = solve for this
R = 0.08206 L*atm/mol*K
T = 288.5 K.
Solve for n = about 5E4 but that is just a close estimate.
Then n = grams/molar mass and
about 5E4 = 0.02882/molar mass.
Solve for molar mass which is approximately 60.
The molecular formula then is
empirical formula x ? number = molar mass
31 x ? = 60
? = about 2 when rounded to a whole number; theefore, the molecular formula is
(empirical formula)_{2} or
C2F2
You should go through and confirm those numbers since they are just estimates on my part. 
Chemistry 
Anonymous,
F2O3H