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
(empirical formula)n\
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
delta T=Kf*m
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.
Thanks DrBob222!!
the help is much appreciated!
To determine the molecular formula of the compound, we need to calculate the empirical formula first and then use additional information to find the molecular formula.
1. Determine the number of moles of CO2 and H2O produced:
- CO2: mass of CO2 / molar mass of CO2 = 2.868 g / 44.01 g/mol = 0.065 mol
- H2O: mass of H2O / molar mass of H2O = 1.567 g / 18.015 g/mol = 0.087 mol
2. Find the ratio of the moles of carbon to hydrogen in the compound:
- CO2 contains 1 carbon atom and 2 oxygen atoms, so for the compound: 0.065 mol CO2 * (1 mol C / 1 mol CO2) = 0.065 mol C
- H2O contains 2 hydrogen atoms and 1 oxygen atom, so for the compound: 0.087 mol H2O * (2 mol H / 1 mol H2O) = 0.174 mol H
The ratio of carbon to hydrogen is 0.065 mol C : 0.174 mol H, which can be simplified to 1:2.5.
3. Calculate the empirical formula:
- Divide the subscripts (moles) by the smallest value to get the simplest whole-number ratio: C1:H2.5
- Multiply by 2 to get whole-number ratios: C2:H5
Now we have determined the empirical formula is C2H5.
4. Calculate the molar mass of the empirical formula:
- C2: 2 carbon atoms * 12.01 g/mol = 24.02 g/mol
- H5: 5 hydrogen atoms * 1.01 g/mol = 5.05 g/mol
The molar mass of the empirical formula C2H5 is 29.07 g/mol.
5. Calculate the molar mass of the compound:
To calculate the molar mass of the compound (MM), use the formula:
MM = (mass of solute / moles of solute) * (grams of solvent / mass of solvent)
MM = (0.1107 g / 0.025 mol) * (25.0 g / 0.1107 g) = 250 g/mol
6. Determine the molecular formula:
Divide the molar mass of the compound by the molar mass of the empirical formula to find the ratio: 250 g/mol / 29.07 g/mol ≈ 8.61
The ratio of the molar mass of the compound to the molar mass of the empirical formula is approximately 8.61.
Multiply the subscripts of the empirical formula by the ratio:
C2H5 * 8.61 = C17.22H43.05
Since we can't have fractional subscripts in a molecular formula, we round to the nearest whole number:
C17H43
Therefore, the molecular formula of the compound is C17H43.