An unknown compound contains only the three elements C,H, and O. A pure sample of the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.
a.)Determine the empirical formula of the compound.
b.)A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze at a temperature 15.2 Celsius below the normal freezing point of pure camphor. Determine the molar mass and apparent molecular formula of the compound. (The molal freezing-point depression constant, kf, for camphor is 40.0 kg-K-mol-1.)
c.)When 1.570 grams of the compound is vaporized at 300 °C and 1.00 atmosphere, the gas occupies a volume of 577 milliliters. What is the molar mass of the compound based on this result?
d.)Briefly describe what occurs in solution that accounts for the difference between the results obtained in parts (b) and (c).
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A.) C7H12O2
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To solve this problem, we'll go step by step for each part:
a.) Determine the empirical formula of the compound.
To find the empirical formula, we need to determine the ratio of elements in the compound. We're given the percentages of carbon and hydrogen, so we assume the remaining percentage is oxygen (since it contains only C, H, and O).
Let's assume we have 100 grams of the compound. Therefore, 65.60 grams are carbon and 9.44 grams are hydrogen. To find the amount of oxygen, we subtract the mass of carbon and hydrogen from the total mass (100 grams):
Mass of oxygen = 100 grams - (65.60 grams + 9.44 grams) = 24.96 grams
Now we need to convert these masses into moles. We'll use the molar mass of each element to do that. The molar masses are approximately:
C = 12 g/mol
H = 1 g/mol
O = 16 g/mol
Number of moles of carbon = 65.60 g / 12 g/mol = 5.47 mol
Number of moles of hydrogen = 9.44 g / 1 g/mol = 9.44 mol
Number of moles of oxygen = 24.96 g / 16 g/mol = 1.56 mol
The empirical formula gives the simplest whole-number ratio of the elements. Therefore, to simplify the ratio, we divide each number of moles by the smallest number (1.56 mol in this case):
Number of moles of carbon (simplified) = 5.47 mol / 1.56 mol ≈ 3.51 mol
Number of moles of hydrogen (simplified) = 9.44 mol / 1.56 mol ≈ 6.05 mol
Number of moles of oxygen (simplified) = 1.56 mol / 1.56 mol = 1 mol
The empirical formula of the compound is therefore C3.51H6.05O.
b.) Determine the molar mass and apparent molecular formula of the compound.
Now, let's calculate the molar mass and the molecular formula using the given freezing point depression.
Mass of the compound = 1.570 grams
Mass of camphor = 16.08 grams
Freezing point depression = 15.2 Celsius
First, we need to convert the freezing point depression to Kelvin:
Freezing point depression (K) = 15.2 Celsius + 273.15 = 288.35 K
The molal freezing-point depression constant (kf) for camphor is given as 40.0 kg-K-mol-1. To find the moles of the compound, we need to calculate the number of moles of camphor:
Number of moles of camphor = Mass of camphor / Molar mass of camphor
Molar mass of camphor = (Mass of camphor) / (Number of moles of camphor)
Number of moles of camphor = 16.08 g / (40.0 kg-K-mol-1 * 0.28835 kg) = 0.111 mol
Now we can find the molar mass of the compound:
Molar mass of the compound = (Mass of the compound) / (Number of moles of the compound)
Molar mass of the compound = 1.570 g / (0.111 mol) = 14.14 g/mol
Finally, to find the apparent molecular formula, we divide the molar mass of the compound by the empirical formula's molar mass. The empirical formula's molar mass can be calculated by summing up the molar masses of each element in the empirical formula:
Empirical formula molar mass = (3.51 mol * 12 g/mol) + (6.05 mol * 1 g/mol) + (1 mol * 16 g/mol) ≈ 70 g/mol
Apparent Molecular formula = (Molar mass of the compound) / (Empirical formula molar mass) ≈ 14.14 g/mol / 70 g/mol = 0.202
Since the apparent molecular formula should be a whole number, we can multiply all the subscripts by 5 to get the simplest whole-number ratio:
Apparent Molecular Formula = C17H30O5
c.) Determine the molar mass of the compound based on vaporization.
Given:
Mass of the compound = 1.570 grams
Volume of the gas = 577 milliliters
Temperature = 300 °C = 573.15 K
Pressure = 1.00 atmosphere
Using the ideal gas law, we can find the number of moles of the gas:
n = PV / RT
P = Pressure = 1.00 atm
V = Volume = 577 mL = 0.577 L
R = Ideal gas constant = 0.0821 L*atm/(mol*K)
T = Temperature = 573.15 K
n = (1.00 atm * 0.577 L) / (0.0821 L*atm/(mol*K) * 573.15 K)
≈ 0.1074 mol
Now, we can find the molar mass of the compound:
Molar mass of the compound = Mass of the compound / Number of moles of the compound
Molar mass of the compound = 1.570 g / 0.1074 mol ≈ 14.60 g/mol
d.) Briefly describe what occurs in solution that accounts for the difference between parts (b) and (c).
In part (b), the compound is dissolved in camphor, which leads to the formation of a solution. The freezing point of the solution is lower than the freezing point of the pure camphor due to the presence of the solute particles. This phenomenon is known as freezing point depression.
In part (c), the compound is vaporized, which involves the compound transitioning from the liquid phase to the gas phase. The molar mass determined in this part is based on the behavior of the compound as a gas.
The difference between the results in parts (b) and (c) arises from the different environmental conditions and the state of the compound (solid/liquid in (b) and gas in (c)). In solution, the compound interacts with the solvent molecules, affecting the freezing point, while in the gas phase, the compound behaves independently, allowing for different measurements and calculations to be made.