What mass of CH4 is contained in a 443 mL sample of CH4 at −32◦C and 282 torr?
Answer in units of g.
I tried so many times and I don't get it!
Use PV = nRT and solve for n.
P = (282/760) atm
V = 0.443 L
R = 0.08206 L*atm/mol*K
T = -32+273 = ?K
Solve for n = number of mols. Then
mols = grams/molar mass You know mols and molar mass, solve for grams.
To find the mass of CH4 in the given sample, we need to use the ideal gas law and the molar mass of CH4.
The formula for the ideal gas law is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(K·mol)) (Note: Make sure to convert the pressure to atm if it's given in torr)
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
So, T = -32 + 273.15 = 241.15 K
Next, convert the pressure from torr to atm:
1 atm = 760 torr
So, 282 torr = 282/760 atm = 0.371 atm
Now, let's rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the given values:
n = (0.371 atm) * (0.443 L) / [(0.0821 L·atm/(K·mol)) * (241.15 K)]
Now, calculating:
n = 0.00699 mol
The molar mass of CH4 can be found from its chemical formula, which is 16.04 g/mol.
Finally, to find the mass (m) in grams, we can use the formula:
m = n * M
Substituting the known values:
m = (0.00699 mol) * (16.04 g/mol) = 0.111 g
Therefore, the mass of CH4 in the 443 mL sample at -32°C and 282 torr is 0.111 g.