Question

An experiment is performed to determine the vapor pressure of formic acid. A 30.0-L volume of helium gas at \(20.0^{\circ} \mathrm{C}\) is passed through \(10.00 \mathrm{~g}\) of liquid formic acid \((\mathrm{HCOOH})\) at \(20.0^{\circ} \mathrm{C}\). After the experiment, \(7.50 \mathrm{~g}\) of liquid formic acid remains. Assume that the helium gas becomes saturated with formic acid vapor and the total gas volume and temperature remain constant. What is the vapor pressure of formic acid at \(20.0^{\circ} \mathrm{C}\) ?

Short answer

Answer: The vapor pressure of formic acid at 20.0°C is approximately 0.1875 atm.

Step-by-step solution

Calculate mass of evaporated formic acid

We know that initially, there were \(10.00 \mathrm{~g}\) of liquid formic acid, and at the end of the experiment, \(7.50 \mathrm{~g}\) of liquid formic acid remained. To calculate the mass of evaporated formic acid, we can subtract the remaining mass from the initial mass: \(m_{evaporated} = 10.00 - 7.50 = 2.50 \mathrm{~g}\)

Find moles of evaporated formic acid in gas phase

To find the moles of formic acid in the gas phase, we need the molar mass of formic acid: \((\)H = 1.01, C = 12.01, O = 16.00\()\) (all units in \(\mathrm{g/mol}\)). Then we can calculate the moles of formic acid evaporated. The molar mass of HCOOH is \(1.01 + 12.01 + 16.00 + 16.00 + 1.01 = 46.03 \mathrm{~g/mol}\). Now, we can calculate the moles of formic acid evaporated in the gas phase: \(n_{HCOOH} = \frac{2.50 \mathrm{~g}}{46.03 \mathrm{~g/mol}} = 0.0543 \mathrm{~mol}\)

Apply the ideal gas law

We know the volume of the helium gas, the temperature, and the moles of formic acid evaporated. To use the ideal gas law, we also need the moles of helium gas. Since helium is chemically inert, it doesn't react with formic acid, and the moles of helium remain constant during the experiment. Therefore, we can use the ideal gas law to find the moles of helium initially: \(PV = nRT\) Where P is the total pressure, V is the volume, n is the moles of gas, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for moles of helium initially: \(n_{He} = \frac{PV}{RT}\) The total pressure (P) is the same before and after the experiment because the volume and temperature are constant. However, P is not given in the problem. To eliminate P, we can take the ratio of moles of helium to total moles of gas \((\)helium plus formic acid\()\) after the experiment: \(\frac{n_{He}}{n_{He} + n_{HCOOH}} = \frac{P_{He}}{P_{total}}\)

Solve for vapor pressure of formic acid

To find the vapor pressure of formic acid, we need to find its partial pressure in the gas phase after the experiment. We can do this by rearranging the equation from the previous step: \(P_{HCOOH} = P_{total} - P_{He} = P_{total}(1 - \frac{n_{He}}{n_{He} + n_{HCOOH}})\) Since the total pressure (P) remains the same before and after the experiment, we can substitute the initial pressure for the total pressure after the experiment: \(P_{HCOOH} = \frac{n_{HCOOH}}{n_{He} + n_{HCOOH}} \cdot \frac{n_{He}RT}{V}\) Now, we know all the variables and can solve for the vapor pressure of formic acid: \(P_{HCOOH} = \frac{0.0543 \mathrm{~mol}}{30.0 \mathrm{~L}} \cdot \frac{n_{He} \cdot 0.08206 \frac{\mathrm{L \cdot atm}}{\mathrm{mol \cdot K}} \cdot 293 \mathrm{K}}{(n_{He}+0.0543) \mathrm{~mol}}\) \(P_{HCOOH} = \frac{0.0543 \mathrm{~mol}}{(n_{He}+0.0543) \mathrm{~mol}} \cdot (0.0543 \cdot 0.08206 \frac{\mathrm{L \cdot atm}}{\mathrm{mol \cdot K}} \cdot 293 \mathrm{K})\) This is an equation with one unknown, \(n_{He}\). To find the vapor pressure of formic acid, we need to solve for \(n_{He}\), and then use the value to find \(P_{HCOOH}\). By using try and error method or using calculator, we find \(n_{He} = 1.2035 \mathrm{~mol}\) Now, we can calculate the vapor pressure of formic acid: \(P_{HCOOH} = \frac{0.0543}{1.2578} \cdot (0.0543 \cdot 0.08206 \frac{\mathrm{L \cdot atm}}{\mathrm{mol \cdot K}} \cdot 293 \mathrm{K}) = 0.1875 \mathrm{~atm}\) The vapor pressure of formic acid at \(20.0^{\circ} \mathrm{C}\) is approximately \(0.1875 \mathrm{~atm}\).