Find the molecular weight and density of gas in lighter knowing pressure, volume, and mass in an experiment?

Question

Experiment design:
a long tube is filled with water and is placed upside down (without losing any of its water) in a beaker of water bath. a lighter is then placed below the opening of the tube in the water bath and its switch is pressed to allow the gas in the lighter to be released directly into the tube. the water in the tube bubbles and is gradually replaced by the gas in the lighter.

initial mass of the lighter=12g
final mass of lighter (after releasing gas)=11.95g
initial mass of water= 18g
final amount of water in the tube=14g
after the gas fills a certain amount of space around the top of the tube, the distance between the line for meniscus of the water remaining near the bottom of the upside down tube and the surface of water in the beaker of water containing the tube is 50 (mm or ml, not sure about the unit)

How would you calculate the molecular weight and the density of the gas in the lighter?

Answer 1

OK, let's start with what we know.
The tube is full of water at the start, and at the end, there is 50ml of gas in the tube (since that's where the meniscus is).

V = 50ml

The mass of the gas is found by subtracting the initial and final weight of the lighter:

m = 12 - 11.95 = .05g

Density = m/V = .05g / 50 ml = 0.001 g/ml or g/cm^3

To find the molecular weight given pressure and volume and temperature, use the Ideal Gas Law:
PV = nRT

Since R is a constant, you find n (# of moles) using the pressure, volume, and temperature of the gas. The molecular weight is found by dividing the mass that we found above by the number of moles.

Answer 2

Or use a million mol of gasoline occupies 22.4dm3 at STP. a million) so vol of 80 g SO3 at 20C an 0.9 atm = 22.4 x 293/273 x a million/0.9 dm3. and density = mass/vol. 2) vol at STP = 10 x 273/303 x 0.seventy 9/a million dm3 n = 22.4/ vol M = m/n