Finding Molar Mass from Gas Mass, Pressure, Volume and Temperature

Two identical 0.500 L bulbs contain different gases. Both are at 100.0 kPa and 25.0°C. One contains 0.888 g of gas, while the other contains 0.444 g.

Predict: do they contain different numbers of moles?

No. If both gases behave ideally, equal pressure, volume, and temperature mean equal amounts of gas. Each bulb contains approximately 0.0202 mol.

The difference is the mass of each mole. The first gas has a molar mass of 44.0 g mol⁻¹, while the second has a molar mass of 22.0 g mol⁻¹.

This is exactly what you can calculate when you know the mass, pressure, volume, and temperature of a gas sample.

The equation you need is:

\[ \boxed{M=\frac{mRT}{PV}} \]

It comes directly from the ideal gas equation. The important part is understanding which measurements belong in the equation and how to correct them before calculating.

01Deriving the molar mass equation

You already know the ideal gas equation:

\[ PV=nRT \]

It connects four properties of a gas:

• \(P\) = absolute pressure of the gas
• \(V\) = volume of the gas
• \(n\) = amount of gas in moles
• \(R\) = ideal gas constant
• \(T\) = absolute temperature in kelvin

The equation tells you how many moles of gas are present if you know its pressure, volume, and temperature.

But we want molar mass, which connects the mass of a sample to its amount in moles.

Molar mass is defined as:

\[ M=\frac{m}{n} \]

where:

• \(M\) = molar mass in g mol⁻¹
• \(m\) = mass of the gas sample in grams
• \(n\) = amount of gas in moles

Rearranging gives:

\[ n=\frac{m}{M} \]

Now substitute this expression into the ideal gas equation:

\[ PV=\frac{m}{M}RT \]

Rearrange to make \(M\) the subject:

\[ PVM=mRT \]

Therefore:

\[ \boxed{M=\frac{mRT}{PV}} \]

That’s the formula you need.

Notice what it actually does: the pressure, volume, and temperature tell you how many moles are present, while the measured mass tells you how much those moles weigh.

You are combining those two pieces of information to find the mass of one mole.

02How the gas measurements are obtained experimentally

Knowing the equation is one thing. Getting reliable measurements is another.

A common experimental approach uses a rigid glass bulb with a known internal volume and a stopcock (a valve that controls gas flow).

The apparatus also needs:

• A vacuum pump to remove the air from the bulb.
• A supply of the unknown, preferably dry, gas.
• A pressure sensor to measure gas pressure.
• A thermometer to measure gas temperature.
• A balance to measure the mass of the bulb.

The experiment follows a particular sequence.

Step 1: Evacuate the bulb.

Connect the bulb to the vacuum pump and remove as much air as possible.

Close the stopcock and weigh the evacuated bulb. Record this as \(m_1\).

Why remove the air first?

If air remains inside, your final sample will contain a mixture of air and the unknown gas. The mass difference will no longer represent just the unknown gas.

Step 2: Introduce the unknown gas.

Connect the bulb to the gas supply and allow the unknown gas to enter.

Close the valve to isolate the sample at the desired pressure.

Step 3: Measure pressure and temperature.

Allow the gas to reach thermal equilibrium with its surroundings.

This means the gas and bulb have had enough time to reach the same temperature.

Record the gas pressure and temperature. Make sure the pressure is absolute pressure, not gauge pressure.

Step 4: Measure the gas mass.

Weigh the filled bulb and record its mass as \(m_2\).

The mass of the gas is the difference:

\[ m=m_2-m_1 \]

The glass bulb itself has not changed mass. The increase on the balance comes from the gas introduced into it.

You now have all four measurements:

One important experimental detail: if gas remains in connecting tubing during the measurement, its volume must also be included. Otherwise, the volume used in the calculation will be too small.

The next step is making sure these measurements use compatible units.

03Getting the units right

For most HSC-style calculations, the easiest combination is:

Using these units means you can substitute directly into:

\[ M=\frac{m(8.314)T}{PV} \]

There are three conversions you should check before reaching for your calculator.

Temperature must be in kelvin

The ideal gas equation uses absolute temperature, not degrees Celsius.

Convert using:

\[ T(\mathrm K)=t(^{\circ}\mathrm C)+273.15 \]

For example:

\[ 25.0^{\circ}\mathrm C=298.15\ \mathrm K \]

Never substitute 25.0 directly into the equation.

Volume must be in litres

If the question gives volume in millilitres, divide by 1000.

For example:

\[ 500\ \mathrm{mL}=0.500\ \mathrm L \]

Pressure must be absolute

This is a particularly important experimental correction.

There are two ways pressure is commonly reported.

• Absolute pressure is measured relative to a perfect vacuum.
• Gauge pressure is measured relative to atmospheric pressure.

The ideal gas equation requires absolute pressure.

The relationship is:

\[ \boxed{P_{\mathrm{abs}}=P_{\mathrm{gauge}}+P_{\mathrm{atm}}} \]

For example, suppose a pressure sensor reads 15.0 kPa gauge pressure, and atmospheric pressure is 100.8 kPa.

Then:

\[ P_{\mathrm{abs}}=15.0+100.8 \]

\[ P_{\mathrm{abs}}=115.8\ \mathrm{kPa} \]

A common mistake is to substitute 15.0 kPa directly into the ideal gas equation. That would make the calculated molar mass much too large because the pressure in the denominator would be far too small.

If you want a deeper explanation of choosing the correct gas constant and matching pressure and volume units, see Choosing R When Pressure Is in kPa and Volume Is in L.

04Worked examples

05What if the gas is collected over water?

So far, both worked examples involved dry gas.

But many chemistry experiments collect gas by displacing water in an inverted measuring cylinder.

This introduces another complication.

The gas inside the cylinder is not just the unknown gas. Some water evaporates, so the collected sample contains both the unknown gas and water vapour.

This means the measured pressure includes the pressure contributed by water vapour.

Dalton’s law of partial pressures tells us that the total pressure of a gas mixture is the sum of its individual partial pressures.

For the collected gas:

\[ P_{\mathrm{total}}=P_{\mathrm{gas}}+P_{\mathrm{H_2O}} \]

Rearranging:

\[ \boxed{P_{\mathrm{gas}}=P_{\mathrm{total}}-P_{\mathrm{H_2O}}} \]

If the water levels inside and outside the measuring cylinder are equal, the total pressure of the gas mixture is equal to atmospheric pressure.

Therefore:

\[ P_{\mathrm{gas}}=P_{\mathrm{atm}}-P_{\mathrm{H_2O}} \]

For example, at 25.0°C:

• Atmospheric pressure = 101.3 kPa
• Water vapour pressure = 3.17 kPa

Therefore:

\[ P_{\mathrm{gas}}=101.3-3.17 \]

\[ P_{\mathrm{gas}}=98.13\ \mathrm{kPa} \]

You must use 98.13 kPa when calculating the amount of the unknown dry gas.

If the water levels are not equal, a hydrostatic pressure correction is also needed.

An important distinction: correcting pressure versus correcting mass

There is one more detail worth understanding.

If the measured mass is already the mass of the dry unknown gas, you only need to correct the pressure for water vapour.

However, if the measured mass includes both the unknown gas and water vapour, correcting pressure alone is not enough.

You must also remove the mass of the water vapour.

First, calculate the amount of water vapour:

\[ n_{\mathrm{H_2O}}= \frac{P_{\mathrm{H_2O}}V}{RT} \]

Then calculate its mass:

\[ m_{\mathrm{H_2O}}=n_{\mathrm{H_2O}}\times18.0 \]

Finally:

\[ m_{\mathrm{dry}}= m_{\mathrm{total}}-m_{\mathrm{H_2O}} \]

You can then use:

\[ \boxed{ M_{\mathrm{dry}}= \frac{m_{\mathrm{dry}}RT} {P_{\mathrm{dry}}V} } \]

The key is consistency: the mass and pressure must refer to the same gas.

A wet sample has water vapour contributing to both its total mass and its total pressure. Removing water vapour from only one of those measurements produces an incorrect molar mass.

06Ideal gas assumptions versus experimental corrections

It is useful to separate two ideas that students sometimes mix up.

The ideal gas equation is a model of gas behaviour. It assumes that gas particles have negligible volume and that intermolecular attractions can be ignored. Their collisions are treated as elastic.

Real gases do not behave perfectly ideally under all conditions.

In particular, deviations become more important at high pressures and low temperatures, especially when a gas approaches condensation.

Experimental corrections are different. They account for what your equipment actually measures.

Correcting gauge pressure or water vapour does not make a real gas ideal. It simply ensures that you are using the correct measurements for the gas you want to study.

There is another useful limitation to remember.

If your unknown sample is actually a mixture of different gases, the equation calculates the mixture’s average molar mass. It does not automatically identify the individual gases present.

For a pure, dry gas behaving ideally, the equation gives the molar mass of that gas.

07Questions and solutions

Question 1

A dry unknown gas has a mass of 0.366 g and occupies a volume of 250 mL.

Its absolute pressure is 100.0 kPa, and its temperature is 27.0°C.

Calculate the molar mass of the gas.

Solution 1

The molar mass is 36.5 g mol⁻¹.

Start by converting the measurements into the required units.

Temperature:

\[ T=27.0+273.15=300.15\ \mathrm K \]

Volume:

\[ V=250/1000=0.250\ \mathrm L \]

Use:

\[ M=\frac{mRT}{PV} \]

Substitute:

\[ M= \frac{0.366\times8.314\times300.15} {100.0\times0.250} \]

\[ M=36.53\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \boxed{M=36.5\ \mathrm{g\,mol^{-1}}} \]

Both pressure and volume were already compatible with the chosen gas constant. The only conversions needed were temperature and volume.

Question 2

An unknown gas is collected over water in an inverted measuring cylinder.

The following measurements are recorded:

• Volume of gas = 0.500 L
• Temperature = 25.0°C
• Atmospheric pressure = 101.3 kPa
• Water vapour pressure at 25.0°C = 3.17 kPa
• Mass of dry unknown gas, measured independently = 0.580 g

The water levels inside and outside the measuring cylinder are equal.

Calculate the molar mass of the unknown gas. Explain why using atmospheric pressure directly would give an incorrect result.

Solution 2

The molar mass of the dry unknown gas is 29.3 g mol⁻¹.

The important step is correcting the pressure before using the molar mass equation.

Because the water levels are equal, the total pressure inside the cylinder equals atmospheric pressure.

However, some of this pressure comes from water vapour.

Calculate the dry gas pressure:

\[ P_{\mathrm{gas}}=101.3-3.17 \]

\[ P_{\mathrm{gas}}=98.13\ \mathrm{kPa} \]

Convert temperature:

\[ T=25.0+273.15=298.15\ \mathrm K \]

Now substitute into:

\[ M=\frac{mRT}{PV} \]

\[ M= \frac{0.580\times8.314\times298.15} {98.13\times0.500} \]

\[ M=29.30\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \boxed{M=29.3\ \mathrm{g\,mol^{-1}}} \]

Using atmospheric pressure directly would give approximately 28.4 g mol⁻¹.

This is an underestimate because atmospheric pressure includes the contribution from water vapour. Using the larger total pressure in the denominator makes the calculated molar mass too small.

Question 3

A chemistry class wants to determine the molar mass of an unknown dry gas.

The available equipment includes:

• A rigid glass bulb with an internal volume of 0.750 L.
• A balance.
• A vacuum pump.
• A cylinder containing the unknown gas.
• A stopcock.
• A pressure sensor.
• A thermometer.

Describe a suitable experimental procedure for determining the molar mass.

In your response, explain:

a) Why the bulb should be evacuated before introducing the unknown gas.

b) Why the gas should be allowed to reach thermal equilibrium before recording its temperature.

c) How a gauge pressure measurement should be converted before using the ideal gas equation.

d) Two possible systematic errors and how each could affect the calculated molar mass.

Solution 3

A suitable experiment involves measuring the mass of a known volume of gas at a known absolute pressure and temperature.

The molar mass can then be calculated using:

\[ M=\frac{mRT}{PV} \]

a) Evacuating the bulb

First, connect the bulb to the vacuum pump and remove as much air as possible.

Close the stopcock and weigh the evacuated bulb.

This initial weighing provides a reference mass.

If air remains inside, the final sample will contain air mixed with the unknown gas. The measured mass difference will not represent the mass of a pure unknown gas sample.

b) Allowing thermal equilibrium

Introduce the unknown gas and isolate it using the stopcock.

Allow the bulb to reach thermal equilibrium with its surroundings before measuring the temperature.

This is important because the ideal gas equation requires the actual temperature of the gas sample.

If the bulb has just been filled and the gas is warmer or cooler than the surroundings, using the surrounding room temperature would introduce an error.

c) Correcting gauge pressure

If the pressure sensor records gauge pressure, add atmospheric pressure:

\[ P_{\mathrm{abs}}=P_{\mathrm{gauge}}+P_{\mathrm{atm}} \]

Absolute pressure must be used in the ideal gas equation because pressure is measured relative to a perfect vacuum.

d) Systematic errors

Two possible errors are:

• Using gauge pressure directly instead of absolute pressure. This makes the pressure used in the denominator too small, causing the calculated molar mass to be too large.
• Ignoring the volume of gas in connecting tubing. If the actual gas volume is larger than the volume entered into the calculation, the calculated molar mass will be too large.

Other possible errors include gas leakage, incomplete evacuation, moisture contamination, and inaccurate temperature measurements.

The experiment is only reliable when the mass, pressure, volume, and temperature all describe the same isolated gas sample.

Question 4

A sealed rigid bulb contains a dry unknown gas.

Initially, the gas has:

• Mass = 0.420 g
• Volume = 0.300 L
• Absolute pressure = 98.0 kPa
• Temperature = 20.0°C

The bulb is heated while its valve remains closed.

At the final temperature of 80.0°C, the measured pressure is 118 kPa.

A student attempts to calculate the molar mass using the initial pressure and final temperature.

Explain why this approach is incorrect.

Then calculate the molar mass using both the initial and final measurements. Determine whether the two results are consistent.

Solution 4

The molar mass is 34.8 g mol⁻¹, and both sets of measurements give the same result within the stated precision.

The student’s mistake is combining measurements from different states of the gas.

When a fixed amount of gas is heated in a rigid sealed container, its volume and amount remain constant. Its pressure increases as its absolute temperature increases.

You must use pressure and temperature measurements from the same state.

Step 1: Calculate the molar mass using the initial conditions.

Convert the initial temperature:

\[ T_1=20.0+273.15 \]

\[ T_1=293.15\ \mathrm K \]

Substitute:

\[ M_1= \frac{0.420\times8.314\times293.15} {98.0\times0.300} \]

\[ M_1=34.82\ \mathrm{g\,mol^{-1}} \]

Step 2: Calculate the molar mass using the final conditions.

Convert the final temperature:

\[ T_2=80.0+273.15 \]

\[ T_2=353.15\ \mathrm K \]

Use the measured final pressure of 118 kPa:

\[ M_2= \frac{0.420\times8.314\times353.15} {118\times0.300} \]

\[ M_2=34.83\ \mathrm{g\,mol^{-1}} \]

Both calculations give:

\[ \boxed{M=34.8\ \mathrm{g\,mol^{-1}}} \]

Step 3: Check whether the pressure change makes sense.

For a fixed amount of gas at constant volume:

\[ \frac{P_1}{T_1}=\frac{P_2}{T_2} \]

Rearrange:

\[ P_2=\frac{P_1T_2}{T_1} \]

Substitute:

\[ P_2=\frac{98.0\times353.15}{293.15} \]

\[ P_2=118.06\ \mathrm{kPa} \]

This agrees with the measured pressure of 118 kPa.

The student’s tempting approach would have used the initial pressure with the final temperature:

\[ M=\frac{0.420\times8.314\times353.15} {98.0\times0.300} \]

This gives approximately 41.9 g mol⁻¹, which is incorrect.

The pressure increased during heating. Using the initial pressure with the final temperature ignores that change and incorrectly suggests that the gas has a different molar mass.

The molar mass does not change when you heat the sample. The pressure and temperature change together, leaving the calculated value of \(M\) unchanged.

Question 5

An unknown gas is collected over water at 25.0°C.

The gas sample is in equilibrium with liquid water, and the water levels inside and outside the collecting cylinder are equal.

The following measurements are recorded:

• Total gas volume = 0.500 L
• Total pressure = 101.3 kPa
• Total measured gas mass = 0.620 g
• Water vapour pressure at 25.0°C = 3.17 kPa
• Temperature = 25.0°C

The measured mass of 0.620 g includes both the unknown gas and the water vapour. Assume there are no liquid water droplets in the measured sample.

Use 18.0 g mol⁻¹ as the molar mass of water.

Calculate the molar mass of the dry unknown gas.

Two students suggest different approaches:

• Student A subtracts the water vapour pressure but uses the full measured mass of 0.620 g.
• Student B uses the total pressure and total mass without making any corrections, arguing that both measurements describe the same mixture.

Evaluate both approaches and determine which gives the molar mass of the dry unknown gas.

Solution 5

The molar mass of the dry unknown gas is 30.7 g mol⁻¹.

Neither student’s approach gives the correct molar mass of the dry gas.

The important detail is that water vapour contributes to both the measured pressure and the measured mass. We must remove its contribution from both.

Step 1: Calculate the partial pressure of the unknown gas.

Using Dalton’s law:

\[ P_{\mathrm{dry}}=P_{\mathrm{total}}-P_{\mathrm{H_2O}} \]

\[ P_{\mathrm{dry}}=101.3-3.17 \]

\[ P_{\mathrm{dry}}=98.13\ \mathrm{kPa} \]

Step 2: Calculate the amount of water vapour.

Convert the temperature:

\[ T=25.0+273.15=298.15\ \mathrm K \]

Use the ideal gas equation for the water vapour:

\[ n_{\mathrm{H_2O}}= \frac{P_{\mathrm{H_2O}}V}{RT} \]

Substitute:

\[ n_{\mathrm{H_2O}}= \frac{3.17\times0.500} {8.314\times298.15} \]

\[ n_{\mathrm{H_2O}}=0.0006394\ \mathrm{mol} \]

Step 3: Calculate the mass of water vapour.

Using:

\[ m=nM \]

\[ m_{\mathrm{H_2O}}= 0.0006394\times18.0 \]

\[ m_{\mathrm{H_2O}}=0.0115\ \mathrm g \]

Step 4: Calculate the mass of the dry unknown gas.

Subtract the water vapour mass from the total measured mass:

\[ m_{\mathrm{dry}}=0.620-0.0115 \]

\[ m_{\mathrm{dry}}=0.6085\ \mathrm g \]

Step 5: Calculate the molar mass of the dry unknown gas.

Now use the corrected pressure and corrected mass:

\[ M_{\mathrm{dry}}= \frac{m_{\mathrm{dry}}RT} {P_{\mathrm{dry}}V} \]

Substitute:

\[ M_{\mathrm{dry}}= \frac{0.6085\times8.314\times298.15} {98.13\times0.500} \]

\[ M_{\mathrm{dry}}=30.74\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \boxed{M_{\mathrm{dry}}=30.7\ \mathrm{g\,mol^{-1}}} \]

Why Student A is incorrect

Student A correctly removes the water vapour contribution from the pressure but leaves its mass in the numerator.

This produces approximately:

\[ M_{\mathrm A}=31.3\ \mathrm{g\,mol^{-1}} \]

The result is too large because the calculation combines the mass of the entire mixture with the amount of dry gas only.

Why Student B is incorrect

Student B uses both the total pressure and total mass.

This is internally consistent if the goal is to calculate the average molar mass of the entire gas mixture.

However, it does not give the molar mass of the unknown dry gas.

The calculation gives approximately:

\[ M_{\mathrm B}=30.3\ \mathrm{g\,mol^{-1}} \]

This is the average molar mass of the unknown gas and water vapour together.

The distinction is subtle but important: the equation does not automatically distinguish the gases in a mixture. You must first decide which gas or mixture you are calculating the molar mass for, then make sure both the mass and pressure correspond to it.

08Where to go next

The central method is straightforward once your measurements are consistent:

\[ \boxed{M=\frac{mRT}{PV}} \]

The harder part is deciding what the measurements actually represent.

Before substituting, check that:

• The mass belongs to the gas you are investigating.
• The pressure is absolute and, if necessary, corrected for water vapour.
• The volume includes all the gas being measured.
• The temperature is in kelvin.
• The gas constant matches your units.
• All measurements describe the same state of the sample.

Once you can calculate the amount of gas reliably, you can use that information in chemical reaction calculations. The next useful connection is Gas Volumes in Chemical Reactions for HSC Chemistry, where pressure, volume, and temperature measurements help determine the quantities of substances involved in a reaction.