gram/mol, gram/liter

Question

A 3.68 gram sample of an unknown gas is found to occupy a volume of 1.53 L at a pressure of 805 mm Hg and a temperature of 44°C. The molecular weight of the unknown gas is  _____g/mol.

A sample of oxygen gas has a density of  _______g/L at a pressure of 1.11 atm and a temperature of 38°C. Assume ideal behavior.

Answer 1

Molecular weight = gram/mole

You are given the grams (3.68 g), so you need to find the moles. Dead give away to use PV = nRT where

n = moles.

(805 mm Hg/760 mm Hg/atm)(1.53 L) = n(0.0821 Latm/Kmol)(317)

n = (1.06 atm)(1.53 L) / (0.0821 Latm/Kmol)(317K)

n = 0.0622 moles

MW = 3.68 g / 0.0622 moles = 59.2 g/mol

You should be able to do the second part. Density is g/ml. Assume you have 1 L of gas, plug it into PV = nRT (remember to use K not ºC) and solve for n (moles). Multiply moles of O₂ by 32 g/mol to get grams. That will be your answer in g/L.