Problem 121
Question
Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathrm{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 0.984 atm and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)
Step-by-Step Solution
Verified Answer
The molecular formula of cyclopropane is C3H6. Compared to Ar, cyclopropane is more likely to deviate from ideal-gas behavior at moderately high pressures and room temperature. Cyclopropane effuses more slowly than methane.
1Step 1: Determine the empirical formula
First, we can assume that we have 100 g of cyclopropane. This means that we have 85.7 g of carbon (C) and 14.3 g of hydrogen (H). To find the molar ratio, we need to divide the mass of each element by its molar mass:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol
Moles of C = (85.7 g) / (12.01 g/mol) = 7.14 mol
Moles of H = (14.3 g) / (1.008 g/mol) = 14.2 mol
Now, we need to find the simplest whole-number ratio between moles of C and H. We can divide both values by the smallest number, 7.14.
C : H = (7.14 / 7.14) : (14.2 / 7.14) ≈ 1 : 2
Thus, the empirical formula is C1H2.
2Step 2: Find the molar mass of the sample
We'll use the ideal gas law to find the molar mass of the sample:
PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature in Kelvin.
First, convert the temperature to Kelvin:
T = 50.0°C + 273.15 = 323.15 K
Rearrange the ideal gas law:
n = (PV) / (RT)
Plug in the given values and convert the units:
P = 0.984 atm
V = 1.00 L
R = 0.0821 (L · atm)/(K · mol)
T = 323.15 K
n = ((0.984 atm)(1.00 L)) / (0.0821 (L · atm)/(K · mol)(323.15 K)) ≈ 0.0381 mol
Now, we can calculate the molar mass of the sample:
Molar mass = (mass of sample) / (number of moles) = (1.56 g) / (0.0381 mol) ≈ 41 g/mol
3Step 3: Determine the molecular formula
Divide the molar mass of the sample by the molar mass of the empirical formula:
(Molar mass of the sample) / (Molar mass of empirical formula) = (41 g/mol) / (12.01 g/mol + 2(1.008 g/mol)) ≈ 3
Since the ratio is close to a whole number (3), we can multiply the empirical formula (C1H2) by 3 to get the molecular formula:
Cyclopropane molecular formula: C3H6
4Step 4: Analyze deviations from ideal-gas behavior and compare with Ar
Cyclopropane's molecular formula (C3H6) is larger and more complex than Ar. Due to its larger size and possible intermolecular forces, cyclopropane is more likely to deviate from ideal-gas behavior at moderately high pressures and room temperature compared to Ar.
5Step 5: Compare effusion rates of cyclopropane and methane
Using Graham's Law of Effusion, we know that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:
Rate_A/Rate_B = sqrt(Molar mass_B/Molar mass_A)
Methane (CH4) has a molar mass of 16 g/mol, while cyclopropane (C3H6) has a molar mass of about 42 g/mol (found in step 3).
Rate_cyclopropane/Rate_methane = sqrt(16 g/mol / 42 g/mol) ≈ 0.61
Since the ratio is less than 1, cyclopropane effuses more slowly than methane.
Key Concepts
Understanding the Empirical FormulaApplying the Ideal Gas LawExploring Graham's Law of Effusion
Understanding the Empirical Formula
The empirical formula represents the simplest whole-number ratio of elements in a compound. It is a crucial first step in determining the molecular formula, which shows the actual number of atoms in a molecule. To find the empirical formula of cyclopropane, we begin by assuming a 100g sample, which simplifies the calculation of percentages to actual masses of carbon (C) and hydrogen (H).
By dividing the mass of each element by their respective molar masses, we get the number of moles, which we then use to find the simplest ratio. In the case of cyclopropane, the empirical formula turns out to be C1H2. This process reveals not only the types of atoms in the molecule but also their basic ratio, setting the stage for further analysis to determine the molecular formula.
By dividing the mass of each element by their respective molar masses, we get the number of moles, which we then use to find the simplest ratio. In the case of cyclopropane, the empirical formula turns out to be C1H2. This process reveals not only the types of atoms in the molecule but also their basic ratio, setting the stage for further analysis to determine the molecular formula.
Applying the Ideal Gas Law
The ideal gas law is a fundamental relationship between pressure (P), volume (V), number of moles (n), the ideal gas constant (R), and temperature (T). In this context, it helps to find the molar mass of a gaseous sample. For cyclopropane, the given conditions of pressure, volume, and temperature allow us to calculate the number of moles using the formula PV = nRT.
After finding the moles, the molar mass is determined by dividing the mass of the sample by the calculated number of moles. This step is not only vital for finding the molecular formula but also for understanding gas behavior in different conditions, which can be essential in various scientific and industrial applications.
After finding the moles, the molar mass is determined by dividing the mass of the sample by the calculated number of moles. This step is not only vital for finding the molecular formula but also for understanding gas behavior in different conditions, which can be essential in various scientific and industrial applications.
Exploring Graham's Law of Effusion
Graham's Law of Effusion states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. Effusion is the process by which gas escapes through a tiny opening, and this law helps us understand how quickly different gases will effuse under the same conditions.
In comparing cyclopropane to methane, we use their molar masses in the law's formula to find that cyclopropane, being the heavier gas, will effuse more slowly than methane. This principle not only aids us in predicting the behavior of gases but also has practical applications, such as in separating isotopes or in identifying unknown gases.
In comparing cyclopropane to methane, we use their molar masses in the law's formula to find that cyclopropane, being the heavier gas, will effuse more slowly than methane. This principle not only aids us in predicting the behavior of gases but also has practical applications, such as in separating isotopes or in identifying unknown gases.
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