Calculated how many grams of methane (CH4) are in a sealed 80 mL flask at room temperature (22ºC) and 780 mm Hg of pressure. Show all your work.
To calculate the number of grams of methane in a sealed flask at room temperature and pressure, you need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (mm Hg)
V = volume (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (K)
First, we need to convert the given values to the appropriate units:
Pressure: 780 mm Hg
Volume: 80 mL = 0.08 L (convert from mL to L)
Temperature: 22ºC = 22 + 273.15 = 295.15 K (convert from ºC to K)
Now, we can rearrange the ideal gas law equation to solve for moles (n):
n = (PV) / (RT)
Plugging in the values:
n = (780 mm Hg * 0.08 L) / (0.0821 L·atm/(mol·K) * 295.15 K)
n = (62.4 mm L) / (24.16 L·mm/(mol·K))
Now, we can cancel out the units:
n = 2.58 mol
Since methane (CH4) has a molar mass of 16.04 g/mol, we can now calculate the grams of methane:
grams of CH4 = number of moles * molar mass
grams of CH4 = 2.58 mol * 16.04 g/mol
grams of CH4 = 41.47 g
Therefore, there are approximately 41.47 grams of methane in the 80 mL flask at room temperature and a pressure of 780 mm Hg.