the molar mass of a newly synthesized jluorocarbon was measured in a gas microbalance, This device consists of a gkass bulb forming one end of a beam, the whole surrouned by a closed container. The beam is pivoted and the balance point is attained by raising the pressure of gais in the container, so increasing the buoyancy of the enclosed bulb. In one experiment, the balance point was reached when the fluorocarbon lpressure was 327.22 torr; for the same setting of the pivot a balance was reached when CHF3 (MM = 70.014g) was introduced at 423.22 torr. A repeat exlperiment with a different setting of the pivot required a pressure of 293.22 torr of the fluorocarbon and 427.22 torr of the CHF3. What is the molar mass of the fluorocarbon? Suggest a molecular formula.

Question

the molar mass of a newly synthesized jluorocarbon was measured in a gas microbalance, This device consists of a gkass bulb forming one end of a beam, the whole surrouned by a closed container. The beam is pivoted and the balance point is attained by raising the pressure of gais in the container, so increasing the buoyancy of the enclosed bulb. In one experiment, the balance point was reached when the fluorocarbon lpressure was 327.22 torr; for the same setting of the pivot a balance was reached when CHF3 (MM = 70.014g) was introduced at 423.22 torr. A repeat exlperiment with a different setting of the pivot required a pressure of 293.22 torr of the fluorocarbon and 427.22 torr of the CHF3. What is the molar mass of the fluorocarbon? Suggest a molecular formula.

Please help me solve this problem. Thank you so much

Answer 1

To solve this problem, we can use the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

Let's assume the volume and temperature are constant in this experiment, and the number of moles of the CHF3 is equal in both experiments since the pressure of CHF3 is the same. Therefore, we can write the following equation for the first experiment:

(Pressure of fluorocarbon) * V = (Number of moles of fluorocarbon) * R * T

Since V, R, and T are constant, we can rewrite the equation as:

(Pressure of fluorocarbon) = (Number of moles of fluorocarbon) * C1 (where C1 is a constant)

Similarly, for the second experiment:

(Pressure of fluorocarbon) = (Number of moles of fluorocarbon) * C2 (where C2 is a constant)

Now, let's solve the equations. Subtracting the second equation from the first equation, we get:

(Pressure of fluorocarbon) - (Pressure of fluorocarbon) = (Number of moles of fluorocarbon) * (C1 - C2)

(327.22 - 293.22) torr = (Number of moles of fluorocarbon) * (C1 - C2)

34 torr = (Number of moles of fluorocarbon) * (C1 - C2)

Next, divide both sides of the equation by (C1 - C2) to solve for the number of moles of fluorocarbon:

(Number of moles of fluorocarbon) = 34 torr / (C1 - C2)

Since we know the molar mass of CHF3 is 70.014 g/mol, we can calculate the number of moles of CHF3 using the equation:

(Number of moles of CHF3) = (Pressure of CHF3) * V / (R * T)

Substituting the given values, we get:

(Number of moles of CHF3) = (423.22 torr) * V / (R * T)

Now, we can set up the following proportion:

(Number of moles of CHF3) / (Number of moles of fluorocarbon) = (molar mass of CHF3) / (molar mass of fluorocarbon)

(423.22 torr * V) / (R * T) / (34 torr / (C1 - C2)) = 70.014 g/mol / (molar mass of fluorocarbon)

Simplifying the equation, we get:

(molar mass of fluorocarbon) = (70.014 g/mol) * (34 torr * V) / (423.22 torr * (C1 - C2))

Now, substitute the given values and solve for the molar mass of the fluorocarbon. Finally, based on the molar mass, suggest a molecular formula that fits the molar mass of the fluorocarbon.