Analysis of a compound indicates that it is 49.02% carbon, 2.743% hydrogen, and 48.23% chlorine by mass. A solution is prepared by dissolving 3.150 grams of the compound in 25.00 grams of benzene, C6H6. Benzene has a normal freezing point of 5.50degreeC and the solution freezes at 1.12degreeC. The molal freezing point constant,kf, for benzene is 5.12C/molal.

1.Find the empirical formula of this compound.
2.Using the freezing point data calculate the molar mass of the compound.
3. Calculate the mole fraction of benzene in the solution.
4. The vapor pressure of benzene at 35degreeC is 150.0mmHg. Calculate the vapor pressure of benzene over the solution described in this problem at 35degreeC.

Part 1.

Take a 100 g sample. That will give you
49.02 g C, 2.743 g H, and 48.23 g Cl.
Convert grams to moles.
49.02/12.01 = ??
2.743/1.008 = ??
48.23/35.45 = ??
Find the mole ratio of the elements to each other. The easy way to do this is to divide the smallest number by itself (thus assuring that will be 1.00), then divide the other two numbers by the same small number. Round to whole numbers. That will give you the empirical formula.

part 2.
delta T = Kf*molality
solve for molality.
m = moles/kg solvent.
Solve for moles.
moles = grams/molar mass
solve for molar mass.

part 3.
mole fraction benzene = moles benzene/total moles. You have grams benzene which can be convert to moles and you have moles of the unknown compound.

part 4.
Pbenzene = Xbenzene*Pnormal benzene.
post your work if you get stuck.

1) C=3 ;H=2 and Cl=1

so the empirical formula of the compound is C3H2Cl.

I seem not to get the others clear.

C=3 ;H=2 and Cl=1. Therefore the empirical formula of the compound is C3H2Cl

The others don't seem clear to me

To answer these questions, we will need to use various concepts from chemistry such as empirical formula, molal freezing point constant, molar mass, and mole fraction. Let's go through each question step by step:

1. Finding the empirical formula of the compound:
The empirical formula represents the simplest whole number ratio of atoms in a compound. To find the empirical formula, we first need to determine the number of moles of each element present in the compound.

Given:
- Compound contains 49.02% carbon, 2.743% hydrogen, and 48.23% chlorine by mass.

Assume we have 100 grams of the compound:
- Mass of carbon = 49.02 grams
- Mass of hydrogen = 2.743 grams
- Mass of chlorine = 48.23 grams

Now, we need to convert the masses to moles using the atomic masses:
- Atomic mass of carbon (C) = 12.01 g/mol
- Atomic mass of hydrogen (H) = 1.008 g/mol
- Atomic mass of chlorine (Cl) = 35.45 g/mol

Number of moles of each element:
- Moles of carbon = 49.02 g / 12.01 g/mol
- Moles of hydrogen = 2.743 g / 1.008 g/mol
- Moles of chlorine = 48.23 g / 35.45 g/mol

Now, divide each number of moles by the smallest number of moles to get the simplest ratio. In this case, it is the number of moles of chlorine.
- Carbon: 49.02 / 12.01 ≈ 4.08
- Hydrogen: 2.743 / 1.008 ≈ 2.72
- Chlorine: 48.23 / 35.45 ≈ 1.36

To get whole numbers, we multiply these values by 4 to eliminate the decimal places:
- Carbon: 4.08 * 4 = 16.32
- Hydrogen: 2.72 * 4 = 10.88
- Chlorine: 1.36 * 4 = 5.44

The ratio of atoms in the empirical formula is approximately C16H11Cl5.

2. Calculating the molar mass of the compound using freezing point data:
The change in freezing point of the solution can be used to calculate the molar mass of the compound. The formula to calculate molar mass is:

Molar mass = (ΔT / Kf) * m

Given:
- Freezing point depression (ΔT) = 5.50°C - 1.12°C = 4.38°C
- Molal freezing point constant (Kf) for benzene = 5.12°C/molal
- Mass of benzene (solvent) = 25.00 g

We convert the change in temperature to Kelvin:
- ΔT = 4.38°C = 4.38 K

The molality (m) of the solution is the number of moles of solute per kilogram of solvent.
- Moles of solute = 3.150 g / molar mass of the empirical formula
- Moles of benzene = 25.00 g / molar mass of benzene (C6H6)

Substituting the values in the formula for molar mass, we have:
molar mass = (4.38 K / 5.12°C/molal) * (3.150g / moles of solute)

Calculate the moles of solute using the empirical formula:
- Moles of solute = 3.150 g / molar mass of empirical formula = 3.150 g / (16 x 12.01 g/mol + 11 x 1.008 g/mol + 5 x 35.45 g/mol)

Substitute this value back into the formula and solve for the molar mass.

3. Calculating the mole fraction of benzene in the solution:
The mole fraction (X) is the ratio of moles of one component (benzene) to the total moles in the solution.

Moles of benzene = mass of benzene / molar mass of benzene
Moles of solute = mass of solute / molar mass of the empirical formula

Total moles in the solution = moles of benzene + moles of solute

Mole fraction (X) of benzene = moles of benzene / total moles in the solution

4. Calculating the vapor pressure of benzene over the solution at 35°C:
We can use Raoult's Law to calculate the vapor pressure of benzene over the solution.

P(total) = P(benzene) * (mole fraction of benzene)

Given:
- Vapor pressure of benzene at 35°C = 150.0 mmHg
- Mole fraction of benzene (calculated in the previous step)

Substitute the values into the formula and calculate the vapor pressure of benzene over the solution.

By following these steps and performing the necessary calculations, you should be able to find the answers to all the questions.