# Chemistry

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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 geometry shape around each carbon atom?

Polar or Nonpolar?

• Chemistry - ,

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 - ,

For n I got 26.18 mols. For the mass I got 31 because it was CF.

• Chemistry - ,

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 - ,

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 - ,

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 5E-4 but that is just a close estimate.
Then n = grams/molar mass and
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 - ,

F2O3H