the molecular weight of the unknown gas?

Question

A 3.00 gram sample of an unknown gas is found to occupy a volume of 2.43 L at a pressure of 792 mm Hg and a temperature of 58 °C. The molecular weight of the unknown gas is ____ g/mol.

Assume ideal behavior.

Answer 1

To answer this question, you need the ideal gas law: PV=nRT

This question gives us P,V, and T. We always have R since it is a constant, and it also gives us grams.

P = Pressure; V = Volume; n = moles; R = universal gas constant; T = Temperature

P = 792 mm Hg; V = 2.43 L; n = ?; R = 0.082057 L atm mol^-1K^-1

Next think about what you are trying to solve: g/mol. This should be a clue that to find this answer, we will need to solve for n.

Next, notice the units given don't match our R constant. Our units of R are in L atm mol^-1K^-1, while units of Temperature are in Celsius and units of Pressure are in mm Hg. This means that we need to convert our pressure from mm Hg to atm, and convert temperature from Celsius to Kelvin. The conversion for these is pretty straight forward:

K = 273.15 + C; atm = 760 mm Hg

Pressure Conversion:

792 mm Hg x (1 atm/760 mm Hg) = 1.042 atm

Temperature Conversion:

58 C + 273.15 = 331.15 K

Next, let's insert these values into our equation and solve for the only unknown (n):

PV=nRT

(1.042 atm)*(2.43 L) = n (0.082057 L atm mol^-1K^-1)*(331.15K)

[(1.042 atm)*(2.43 L)]/[(0.082057 L atm mol^-1K^-1)*(331.15K)] = n

n = 0.093 moles

So if we have moles and we have grams, we can find molar mass easily since molar mass is just g/mol. The question gives you grams and you just solved for moles.

Molar Mass = 3.00 g / 0.093 mol = 32.258 g/mol

Answer 2

Molecular weight is grams/moles. We are given the grams (3.00 g), so we need to find the moles. For this, we can use PV = nRT and solve for n (moles).

P = 792 mm Hg x 1 atm/760 mm Hg = 1.047 atm

V = 2.43 L

n = ?

R = gas constant = 0.0821 Latm/Kmol

T = 58 + 273 = 331K

n = PV/RT = (1.047 atm)(2.43 L)/(0.0821 Latm/Kmol)(331K)

n = 0.0936 moles

MW = 3.00 g/0.0936 moles = 32.1 g/mole