A gas has a density of #3.18*g*L^-1# at a temperature of #347*K#, and a pressure of #1.2*atm#. What is its molecular mass?

Question

A gas has a density of #3.18*g*L^-1# at a temperature of #347*K#, and a pressure of #1.2*atm#.  What is its molecular mass?

Peyton Adams · Accepted Answer

We use the Ideal Gas equation to get #"molecular mass"=75.5*g*mol^-1#

We assume ideality, and so #PV=nRT#, and thus,

#PV=("mass"/"molar mass")RT# because #n="mass"/"molar mass"#

On rearrangement, #"molar mass"="mass"/Vxx(RT)/P#

But #"mass"/V=rho, "density"#, and thus...........

#"molar mass"=(rhoRT)/P=#

#(3.18*g*L^-1xx0.0821*(L*atm)/(K*mol)xx347*K)/(1.2*atm)#

I am not immune to errors, so let's just cancel out the units to make sure we got this right.

#"Molar mass"=(3.18*g*cancel(L^-1)xx0.0821*cancel(L*atm)/(cancelK*mol)xx347*cancelK)/(1.2*cancel"atm")#

And this provides, I believe, an answer of...

#75.5*g*mol^-1#. Because we have got consistent units, I think our order of operations is correct.

Xavier Avery · Answer

undefined To find the molecular mass of the gas, you can use the ideal gas law equation:

$$ PV = nRT $$

Rearrange the equation to solve for the number of moles ($ n $):

$$ n = \frac{PV}{RT} $$

Then, use the formula for density to find the number of moles per liter ($ n/V $):

$$ \frac{n}{V} = \frac{\text{density}}{molar\ volume} $$

Finally, rearrange the equation to solve for the molecular mass ($ M $):

$$ M = \frac{molar\ mass}{molar\ volume} = \frac{molar\ mass \times RT}{PV} $$

Plug in the given values:

$$ M = \frac{molar\ mass \times 0.0821 \times 347}{1.2} $$

Given that the density is $ 3.18 \, g/L $, and the molar mass is in grams per mole, we can directly use the density as the molar mass:

$$ M = \frac{3.18 \times 0.0821 \times 347}{1.2} $$

$$ M = 74.45 \, g/mol $$

So, the molecular mass of the gas is approximately $ 74.45 \, g/mol $.