Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains 80.78%C, 13.56%H, and 5.66%O. A solution of 1.41g of this pheromone in 9.23g of benzene freezes at 2.73 degrees C. what are the molecular formula and molar mass of the compound? the normal freezing point of pure benzene is 5.50 degrees C
delta T = Kf*molality
solve for molality
molality = mols/kg solvent
You know m and kg solvent, solve for moles.
moles = grams/molar mass
solve for molar mass.
To find the molecular formula and molar mass of the compound, we can follow these steps:
Step 1: Calculate the moles of benzene used in the solution.
Molar mass of benzene (C6H6):
C: 12.01 g/mol
H: 1.01 g/mol
Molar mass of benzene = (12.01 g/mol x 6) + (1.01 g/mol x 6) = 78.11 g/mol
Moles of benzene = Mass of benzene used / Molar mass of benzene
Moles of benzene = 9.23 g / 78.11 g/mol = 0.1182 mol
Step 2: Calculate the moles of the pheromone.
Moles of pheromone = Moles of benzene
Since the pheromone and benzene have a 1:1 ratio in the mixture, the moles are the same.
Moles of pheromone = 0.1182 mol
Step 3: Calculate the empirical formula of the pheromone.
The empirical formula gives the simplest whole number ratio of atoms present in a compound.
To find the empirical formula, divide the percentages of each element by their molar masses, then divide by the smallest value obtained.
Element | Mass % | Molar Mass | Moles
----------------------------------------
Carbon | 80.78% | 12.01 g/mol | 6.7310 mol
Hydrogen | 13.56% | 1.01 g/mol | 13.4291 mol
Oxygen | 5.66% | 16.00 g/mol | 0.3538 mol
Dividing by the smallest value (0.3538 mol), we get:
Element | Moles Ratio
-----------------------
Carbon | 6.7310 / 0.3538 = 19.00
Hydrogen | 13.4291 / 0.3538 = 37.94
Oxygen | 0.3538 / 0.3538 = 1.00
Approximating the ratios to the nearest whole number, the empirical formula is C19H38O.
Step 4: Calculate the molar mass of the pheromone.
Molar mass of empirical formula C19H38O:
(C: 12.01 g/mol x 19) + (H: 1.01 g/mol x 38) + (O: 16.00 g/mol x 1) = 270.71 g/mol
So, the molecular formula of the compound is C19H38O and its molar mass is approximately 270.71 g/mol.
To find the molecular formula and molar mass of the compound, we need to determine the empirical formula first. The empirical formula gives the simplest ratio of atoms present in a compound.
Let's start by finding the moles of carbon, hydrogen, and oxygen in the given compound.
1. Calculate the mass of carbon:
Mass of carbon = percentage of carbon * mass of the compound
= 80.78% * (1.41g + 9.23g)
= 10.77g
2. Calculate the moles of carbon:
Moles of carbon = mass of carbon / atomic mass of carbon
= 10.77g / 12.01 g/mol
≈ 0.896 mol
3. Perform similar calculations to find the moles of hydrogen and oxygen:
Moles of hydrogen = mass of hydrogen / atomic mass of hydrogen
= (13.56% * (1.41g + 9.23g)) / 1.008 g/mol
≈ 1.227 mol
Moles of oxygen = mass of oxygen / atomic mass of oxygen
= (5.66% * (1.41g + 9.23g)) / 16.00 g/mol
≈ 0.503 mol
Now, we need to find the simplest whole-number ratio of these moles by dividing them by the smallest value, which is approximately 0.503 mol.
Moles of carbon in simple ratio = 0.896 mol / 0.503 mol ≈ 1.78
Moles of hydrogen in simple ratio = 1.227 mol / 0.503 mol ≈ 2.44
Moles of oxygen in simple ratio = 0.503 mol / 0.503 mol = 1
The empirical formula is C1.78H2.44O1, which can be rounded to C2H3O.
To determine the molecular formula, we need to know the molar mass of the compound. Since we don't have that information yet, we'll assume the empirical formula's molar mass as a placeholder.
1. Calculate the molar mass of the empirical formula (C2H3O):
Molar mass = (atomic mass of carbon * number of carbon atoms) + (atomic mass of hydrogen * number of hydrogen atoms) + (atomic mass of oxygen * number of oxygen atoms)
≈ (12.01 g/mol * 2) + (1.008 g/mol * 3) + (16.00 g/mol * 1)
≈ 24.02 g/mol + 3.024 g/mol + 16.00 g/mol
≈ 43.044 g/mol
Now that we have the molar mass of the empirical formula, we can determine the molecular formula if we know the actual molar mass of the compound.
2. Find the molecular mass of the compound using the cryoscopic constant of benzene:
ΔT = (Kf * molality of solution) / (molar mass of compound)
ΔT = (5.50 - 2.73) °C = 2.77 °C
molality of solution = (1.41g + 9.23g) / (molar mass of benzene in kg)
= 10.64g / 78.11 g/mol
≈ 0.136 mol/kg
Kf = freezing point depression constant of benzene = 5.12 °C/m
Solving for the molar mass of the compound:
2.77 °C = (5.12 °C/m * 0.136 mol/kg) / (molar mass of compound)
molar mass of compound = (5.12 °C/m * 0.136 mol/kg) / 2.77 °C
≈ 0.025018 kg/mol
3. Find the molecular formula:
Number of empirical formula units = molar mass of compound / molar mass of empirical formula
= 0.025018 kg/mol / 0.043044 kg/mol
≈ 0.581
Rounding to the nearest whole number, we get the molecular formula: C2H3O.
Therefore, the molecular formula of the compound is C2H3O, and its molar mass is approximately 43.044 g/mol.