Molar Mass of Gas

 

Learning Objective

  • Apply knowledge of molar mass to the Ideal Gas Law


Key Points

    • The molecular weight (molar mass) of any gas is the mass of one particle of that gas multiplied by Avogadro's number (6.02 x 1023).
    • Knowing the molar mass of an element or compound can help us stoichiometrically balance a reaction equation.
    • The average molar mass of a mixture of gases is equal to the sum of the mole fractions of each gas (xi) multiplied by the molar mass (Mi) of that particular gas:
      Mˉ=∑ixiMi\bar { M} =\sum _{ i }^{ }{ { x }_{ i }{ M }_{ i } }
      .




Terms

  • stoichiometrythe study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions (chemical equations)
  • molar massthe mass of one mole of an element or compound
  • ideal gasa hypothetical gas whose molecules exhibit no interaction and undergo elastic collision with each other and with the walls of the container


Molar Mass of Gases and Gas Mixtures

Molar mass (M) is equal to the mass of one mole of a particular element or compound; as such, molar masses are expressed in units of grams per mole (g mol–1) and are often referred to as molecular weights. The molar mass of a particular gas is therefore equal to the mass of a single particle of that gas multiplied by Avogadro's number (6.02 x 1023 ). To find the molar mass of a mixture of gases, you need to take into account the molar mass of each gas in the mixture, as well as their relative proportion.

The average molar mass of a mixture of gases is equal to the sum of the mole fractions of each gas, multiplied by their respective molar masses:

Mˉ=∑ixiMi\bar { M} =\sum _{ i }^{ }{ { x }_{ i }{ M }_{ i } }


The molar volumes of all gases are the same when measured at the same temperature and pressure (22.4 L at STP), but the molar masses of different gases will almost always vary.

Calculating Molar Mass using the Ideal Gas Equation

The molar mass of an ideal gas can be determined using yet another derivation of the Ideal Gas Law:
PV=nRTPV=nRT
.

We can write n, number of moles, as follows:

n=mMn=\frac{m}{M}


where m is the mass of the gas, and M is the molar mass. We can plug this into the Ideal Gas Equation:

PV=(mM)RTPV=\left(\frac{m}{M}\right)RT


Rearranging, we get:

PVRT=mM\frac{PV}{RT}=\frac{m}{M}


Finally, putting the equation in terms of molar mass, we have:

M=mRTPVM=\frac{mRT}{PV}


This derivation of the Ideal Gas Equation is useful in determining the molar mass of an unknown gas.

Example

  • An unknown gas with a mass of 205 g occupies a volume of 20.0 L at standard temperature and pressure. What is the molar mass of the gas?
  • M=mRTPVM=\frac{mRT}{PV}


M=(205)(0.0821)(273)(1.0)(20.0)=230gmolM=\frac{(205)(0.0821)(273)}{(1.0)(20.0)}=230\frac{g}{mol}


Ideal Gas Law Practice Problems with Molar Mass - YouTubeHow to set up and solve ideal gas law problems that involve molar mass and converting between grams and moles.

 





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