A 1.065-g sample of an unknown substance is dissolved in 30.00 g of benzene; the freezing point of the solution is 4.25 degrees C. The compound is 50.69% C, 4.23% H, and 45.08% O by mass.
Info you might need on benzene:
Normal Freezing point,degrees C: 5.53
Kf, degrees m^1: 5.12
Normal boiling point, degrees C: 80.10
Kb, degrees C m^-1: 2.53
Answer: C6H6O4
Follow the problems you've already done above on freezing points.
To determine the identity of the unknown substance, we can use the concept of freezing point depression. Freezing point depression is a colligative property, meaning it depends on the number of solute particles present in a solution, rather than the nature of the solute itself.
First, we need to calculate the molality (moles of solute per kilogram of solvent) of the solution. We know the mass of the solvent (benzene) is 30.00 g, and we can calculate the moles of benzene using its molar mass. The molar mass of benzene (C6H6) is 78.11 g/mol.
moles of benzene = mass of benzene / molar mass of benzene
moles of benzene = 30.00 g / 78.11 g/mol
Now, let's calculate the moles of the unknown substance using the freezing point depression formula:
ΔTf = Kf * molality
where ΔTf is the freezing point depression, Kf is the cryoscopic constant of benzene (5.12 °C m^-1), and molality is the molality of the solution.
We know that the freezing point of the solution is 4.25 °C (lower than the normal freezing point of benzene, 5.53 °C). Therefore, ΔTf = 5.53 °C - 4.25 °C = 1.28 °C.
Now, we can solve for molality:
1.28 °C = 5.12 °C m^-1 * molality
molality = 1.28 °C / 5.12 °C m^-1
Next, let's calculate the moles of the unknown substance:
moles of unknown substance = molality * kg of benzene
We know the mass of benzene is 30.00 g, which is equivalent to 0.03000 kg.
moles of unknown substance = molality * 0.03000 kg
Finally, let's calculate the moles of each element (C, H, and O) present in the unknown substance using their mass percentages.
mass of C = 1.065 g * 50.69% = 0.540 g
mass of H = 1.065 g * 4.23% = 0.045 g
mass of O = 1.065 g * 45.08% = 0.480 g
moles of C = mass of C / molar mass of C
moles of H = mass of H / molar mass of H
moles of O = mass of O / molar mass of O
The molar masses of C, H, and O are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively.
Now, divide each of the moles by the smallest number of moles to obtain the empirical formula:
Empirical formula = C:H:O
Finally, round off the empirical formula to obtain the molecular formula.
So, the unknown compound is C6H6O4.