elemental analysis indicates the simplest formula of an unknown is C2H4O. when 0.775g of the compound is dissolved in 50g of benzene(C6H6), the solution freezes at 5.06 degree C. what are the molecular weight and the molecular formula of the compound? the freezing point of pure benzene is 5048 degree C, and its Kf is 5.12 degree C/m.
delta Tf= 5.48C-5.06C=0.42C
m=0.42C/(5.12C/m )=0.082m C6H6
(0.082mol/kg) * (50g/1000g) = 4.1E-3 mol compound
0.775g/4.1E-3 = 189g/mol compound
n=(189g/mol) / (44 g/mol) ~ 4
4* (C2H4O) = C8H16O4
** i would like to know is the problem solved correctly.
I don't like to see 4.3 rounded to 4 but I don't see anything wrong with your work. It look fine to me.
Based on your calculations, the molecular weight of the compound is approximately 189 g/mol. However, there seems to be a small error in the determination of the molecular formula. Let's correct it.
You correctly calculated that the compound is present in a concentration of 0.082 mol/kg in the solution. Multiplying this concentration by the mass of benzene (50 g) gives you the number of moles of the compound present: 0.082 mol/kg * (50 g / 1000 g) = 4.1 * 10^-3 mol of the compound.
To calculate the molecular formula, divide the mass of the compound (0.775 g) by the number of moles obtained (4.1 * 10^-3 mol):
0.775 g / (4.1 * 10^-3 mol) ≈ 189 g/mol
The calculated molecular weight matches the previous result.
Now, to determine the molecular formula, divide the molecular weight by the empirical formula weight (44 g/mol for C2H4O):
189 g/mol / 44 g/mol ≈ 4.3
This shows that the molecular formula is approximately 4.3 times the empirical formula, which can be rounded down to 4.
Therefore, the molecular formula of the compound is C8H16O4 (4 * C2H4O), which matches your previous conclusion.
Overall, your calculations and solution are correct. Good job!