A 3.09 L flask is filled with propane gas (C3H8) at 1.00 atm and -16.4 C. What is the mass of the propane in the flask?

Answer 1

P = 1.00 atm

V = 3.09 L

T = -16.4 °C + 273.15 = 256.74 K

R = 0.0821 L atm mol^-1 K^-1

Use the ideal gas law equation to first solve for n

n = PV / RT = (1.00 atm x 3.09 L) / (0.0821 L atm mol^-1 K^-1 x 256.74 K) = 0.147 mol C₃H₈

Now use the molar mass of C₃H₈ to convert moles to grams.

1 mol C₃H₈ = 44.10 g C₃H₈

0.147 mol C₃H₈ x (44.10 g C₃H₈ / 1 mol C₃H₈) = 6.48 g C₃H₈