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.
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1. To find the empirical formula of the compound, we need to determine the ratio of the elements present in the compound. We can assume a 100g sample of the compound, which means that it contains 49.02g of carbon, 2.743g of hydrogen, and 48.23g of chlorine.
To find the mole ratio, we need to convert the masses of each element into moles. The molar mass of carbon (C) is 12.01 g/mol, the molar mass of hydrogen (H) is 1.01 g/mol, and the molar mass of chlorine (Cl) is 35.45 g/mol.
For carbon:
moles of carbon = mass of carbon / molar mass of carbon
moles of carbon = 49.02g / 12.01g/mol = 4.08 mol
For hydrogen:
moles of hydrogen = mass of hydrogen / molar mass of hydrogen
moles of hydrogen = 2.743g / 1.01g/mol = 2.71 mol
For chlorine:
moles of chlorine = mass of chlorine / molar mass of chlorine
moles of chlorine = 48.23g / 35.45g/mol = 1.36 mol
Now, we need to find the simplest whole number ratio of these elements. Divide each mole value by the smallest mole value (1.36 mol) to get:
Carbon: 4.08 mol / 1.36 mol = 3
Hydrogen: 2.71 mol / 1.36 mol = 2
Chlorine: 1.36 mol / 1.36 mol = 1
So, the empirical formula of the compound is C3H2Cl.
2. To calculate the molar mass of the compound, we need to find the mass of one mole of the compound. We already know the empirical formula is C3H2Cl.
The molar mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, and chlorine (Cl) is 35.45 g/mol.
Total molar mass = (3 * molar mass of carbon) + (2 * molar mass of hydrogen) + (1 * molar mass of chlorine)
Total molar mass = (3 * 12.01 g/mol) + (2 * 1.01 g/mol) + (1 * 35.45 g/mol)
Total molar mass = 36.03 g/mol + 2.02 g/mol + 35.45 g/mol
Total molar mass = 73.5 g/mol
The molar mass of the compound is 73.5 g/mol.
3. To calculate the mole fraction of benzene (C6H6), we need to know the moles of benzene and the moles of the compound.
The moles of benzene can be calculated using its molar mass (78.11 g/mol) and its mass in the solution (25.00 g).
moles of benzene = mass of benzene / molar mass of benzene
moles of benzene = 25.00 g / 78.11 g/mol = 0.32 mol
The moles of the compound were not given directly, but since we know its mass (3.150 g) and its molar mass (73.5 g/mol), we can calculate its moles.
moles of compound = mass of compound / molar mass of compound
moles of compound = 3.150 g / 73.5 g/mol = 0.043 mol
The mole fraction of benzene (Xbenzene) can be calculated as:
Xbenzene = moles of benzene / (moles of benzene + moles of compound)
Xbenzene = 0.32 mol / (0.32 mol + 0.043 mol)
Xbenzene = 0.32 mol / 0.363 mol
Xbenzene = 0.88
The mole fraction of benzene in the solution is 0.88.
4. To calculate the vapor pressure of benzene over the solution described, we can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction.
The vapor pressure of pure benzene at 35°C is given as 150.0 mmHg.
vapor pressure of benzene over solution = Xbenzene * vapor pressure of pure benzene
vapor pressure of benzene over solution = 0.88 * 150.0 mmHg
vapor pressure of benzene over solution = 132.0 mmHg
The vapor pressure of benzene over the solution described in this problem at 35°C is 132.0 mmHg.