An organic compound with the empirical formula C2H50 is tested for use as an antifreeze. When a 0.365 gram sample of this compound is dissolved in 25.0 grams of water, the freezing point of the solution is -0.201°C. If the freezing point depression constant of water is 1.86°C/m, find the molecular formula of this compound.
Use the formula ΔTf = (i) (kf) (m)
Thank you in advance!
Use the formula ΔTf = (i) (kf) (m)
Thank you in advance!
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The molar mass of C2H5O is 45.0 g / mol.
0.365 g C2H50 x (1 mol / 45.0 g) = 0.00810 mol
molarity = moles of solute / mass of solvent (in kg)
= 0.00810 mol C2H5O / 0.0250 kg H2O
= 0.324 mol / kg
The Van Hoff's factor is i = 1 because C2H5O is not soluble.
ΔTf = (i) (kf) (m)
0.201°C = (1) (1.86°C/m) m
If we solve for m, we get m = 0.108 mol / kg
Therefore (0.324 mol / kg) / (0.108 mol / kg) = 3.00
That means that the experimental molarity is three times the theoretical molarity.
Therefore 3 x C2H50 = ''C6H15O3'' is the molecular formula.
0.365 g C2H50 x (1 mol / 45.0 g) = 0.00810 mol
molarity = moles of solute / mass of solvent (in kg)
= 0.00810 mol C2H5O / 0.0250 kg H2O
= 0.324 mol / kg
The Van Hoff's factor is i = 1 because C2H5O is not soluble.
ΔTf = (i) (kf) (m)
0.201°C = (1) (1.86°C/m) m
If we solve for m, we get m = 0.108 mol / kg
Therefore (0.324 mol / kg) / (0.108 mol / kg) = 3.00
That means that the experimental molarity is three times the theoretical molarity.
Therefore 3 x C2H50 = ''C6H15O3'' is the molecular formula.