Ideal Gas Law and Molar Mass of a Gas

Ideal Gas Law and Molar Mass of a Gas

The ideal gas law can be used to determine the molar mass of a gas. Recall that molar mass has the units grams per mole. Given the volume, pressure and temperature of a gas, we can use the ideal gas law to determine the number of moles of that gas. If we are also given the mass of that gas, we can divide its mass by the number of moles to determine the grams per mole, which is equal to molar mass.

(1)molar-mass-gas

 

Rewriting this equation in terms of moles:

 

(2)  molesn

 

This confirms what we already know, that to convert mass to moles, we divide mass by its molar mass.

Recall the ideal gas law:

(3) ideal-gas-law-small

 

If we substitute equation (2) for moles (n) into equation (3), we get:

(4) PV

 

Rewriting:

(5) PV2

 

To solve for molar mass, we first cross-multiply:

(6) PV3

 

And divide each side by PV:

(7) ideal-gas-law-mm

 

Equation (7) can be used to solve for the molar mass of any gas, for which you know its mass, temperature, pressure and volume.

Sample Problem

A 0.276 g sample of gas occupies a volume of 0.270 L at 739 mm Hg and 98 °C. Calculate the molecular weight of this gas.

Molecular weight is the same thing as molar mass. We want to find the molecular weight of this gas:

Wanted: ? Molar mass

Given:

mass = 0.276 g

V = 0.270 L

T = 98 °C

P = 739 mm Hg

As with all gas law problems, we need to convert degrees Celsius to Kelvin:

371k

 

And all units must match the units of Rydberg’s constant L-atm-K-1-mol-1. So we must convert mm Hg to atm pressure.

972atm

 

Given:

mass = 0.276 g

V = 0.270 L

T = 98 °C = 371.15 K

P = 739 mm Hg = 0.972 atm

Plugging our values into Equation (7):

(7)ideal-gas-law-mm

 

32gmol

 

The molecular weight of this gas is 32.0 g/mol.

Sample Problem

Calculate the mass, in grams, of 3.50 L of NO gas measured at 35 °C and 835 mm Hg.

In this problem, we’re given the molar mass, or at least the identity of the gas, NO for which we can find its molar mass. And, we’re asked to determine the mass in grams.

To rewrite equation (7) to solve for grams, we multiply each side by PV/RT:

(8)PV4

 

The resulting equation is:

(9) ideal-gas-law-g

 

We will use this equation (9) to solve for mass in grams.

We need molar mass, P, V and T:

Given:

Molar mass of NO = 30.01 g/mol

V = 3.50 L

T = 35 °C

P = 835 mm Hg

Once again, we need to convert temperature to Kelvin, and pressure to atm.

308k

 

110a

 

Our updated given values are:

Given:

Molar mass of NO = 30.01 g/mol

V = 3.50 L

T = 35 °C = 308.15 K

P = 835 mm Hg = 1.10 atm

Plugging these values into equation (9)

457g

 

The mass of this sample of NO gas is 4.57 g.