Chem--PLEASE HELP 911
posted by MSJO on .
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
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
Post your work if you get stuck.
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
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.
the help is much appreciated!