Question

A compound of carbon, hydrogen, and oxygen was burned in oxygen, and 2.00 g of the compound produced 2.868 g CO2 and 1.567 g H2O. In another experiment 0.1107g of the compound was dissolved in 25.0 g of water. This solution had a freezing point of -0.0894 degrees C. What is the molecular formula of the compound?

Please show work

Answer 1

Convert 2.868g CO2 to g C.
2.868 x (atomic mass C/molar mass CO2)
Convert 1.567g H2O to g H (not H2).
1.567 x (2*atomic mass H/molar mass H2O)
mass O = 2.00 - mass C - mass H.

Convert g C, H, O to mols.
mols C, H, O, = grams/atomic mass.

Now find the ratio of the element to each other for the empirical formula.

The other part of the problem is to determine the molar mass.
delta T = Kf*m
Solve for m

m = mols/kg solvent
Solve for mol.

mols = grams/molar mass\
Solve for molar mass

Then empirical mass x n = molar mass
Solve for n, round to a whole number, and the molecular formula is
(empirical formula)_n\
Post your work if you get stuck.

Answer 2

Here's what I did so far

CO2 2.868 * (12.01/44.01)=.7826
H2O 1.567 * (2*1.008/18.016)=.1753

mass O =2.00-.7826-.1753= 1.0421

C= .7826/12.01= .0651
H= .1753/1.008= .1739
O= 1.0421/16.00= .0651
then i divided each number by .0651 and got 1:2.6:1

from here on i'm stuck finding the molar mass.
not sure what the constants are in
delta T=Kf*m

Answer 3

So far so good except you didn't finish the empirical formula. There are only whole numbers in empirical formulas. Your is 1:2.6:1(actually it is 1:2.67:1. The easiest way to get whole numbers is to try multiples; i.e., 2 x each gives 2:5.34:2. That doesn't do it so try 3. That gives 3:8.01:3. That looks good so we could round off to C3H8O3. The empirical mass for that is about 92.

delta T you have in the problem as 0.0894. Kf is 1.86. Solve for m = molality. I'll leave it with you to finish but I'll check back periodically if you need more assistance.

Answer 4

Thanks DrBob222!!
the help is much appreciated!