How to Find Moles of Gas Collected Over Water Using Vapour Pressure

You collect hydrogen gas over water and read the atmospheric pressure as \(101.3\ \mathrm{kPa}\). The gas volume is \(250.0\ \mathrm{mL}\), and the temperature is \(25.0^\circ\mathrm C\). Can you put \(101.3\ \mathrm{kPa}\) straight into the ideal gas equation to calculate the moles of hydrogen?

No. Your collection tube contains more than just hydrogen. Some water has evaporated into the tube, so the collected gas is actually a mixture of hydrogen and water vapour.

To find the moles of hydrogen, you must first subtract the water vapour pressure from the total pressure.

At \(25.0^\circ\mathrm C\), the supplied water vapour pressure is \(3.17\ \mathrm{kPa}\). Therefore:

\[ P_{\ce{H2}}=101.3-3.17 =98.13\ \mathrm{kPa} \]

You then use this corrected pressure in the ideal gas equation:

\[ \boxed{n=\frac{(P_{\mathrm{atm}}-P_{\ce{H2O}})V}{RT}} \]

This is the central method for finding moles of gas collected over water.

01Why you must subtract water vapour pressure

Imagine collecting hydrogen in an inverted measuring tube filled with water. The hydrogen enters through a delivery tube, pushes the water down, and occupies the space above it.

However, water molecules also escape from the liquid surface and enter the gas space. This process is called evaporation.

Eventually, the gas above the water contains both hydrogen molecules and water vapour molecules.

The important detail is that your pressure measurement includes contributions from both gases.

Dalton’s law of partial pressures tells us that the total pressure of a mixture of gases equals the sum of the individual partial pressures.

For hydrogen collected over water:

\[ P_{\mathrm{total}}=P_{\ce{H2}}+P_{\ce{H2O}} \]

Rearranging:

\[ \boxed{P_{\ce{H2}}=P_{\mathrm{total}}-P_{\ce{H2O}}} \]

Here:

• \(P_{\mathrm{total}}\) is the pressure of the entire gas mixture.
• \(P_{\ce{H2}}\) is the partial pressure of hydrogen.
• \(P_{\ce{H2O}}\) is the partial pressure of water vapour.

The same relationship applies if the collected gas is oxygen, nitrogen, carbon dioxide, or another gas.

What does partial pressure actually mean?

Think of the hydrogen and water vapour molecules as two groups sharing the same container.

Both groups move around independently and collide with the walls of the container. Each group contributes to the pressure measured against those walls.

The analogy is useful because it shows why the pressures add together. However, the particles do not need to form separate layers or occupy separate regions. Both gases are mixed throughout the available volume.

The pressure of the gas you want is therefore only part of the total measured pressure.

Why does atmospheric pressure equal total gas pressure?

In many HSC Chemistry questions, the water levels inside and outside the collection tube are equal.

When this happens, the pressure of the gas mixture inside the tube equals atmospheric pressure.

Therefore:

\[ P_{\mathrm{total}}=P_{\mathrm{atm}} \]

Substituting into Dalton’s law gives:

\[ \boxed{P_{\mathrm{gas}}=P_{\mathrm{atm}}-P_{\ce{H2O}}} \]

This is the equation you will use most often.

Important: You are subtracting the water vapour pressure from the total pressure, not from the gas volume.

02The method: finding moles of gas collected over water

Once you understand the pressure correction, the calculation is straightforward.

Use this sequence whenever a question asks you to calculate the moles of gas collected over water.

Step 1: Identify the temperature.

Find the temperature at which the gas was collected.

Convert it into kelvin:

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

Step 2: Find the water vapour pressure.

Use the supplied vapour pressure table or the value given in the question.

Make sure you use the value corresponding to the experimental temperature.

Step 3: Calculate the pressure of the dry gas.

If the water levels are equal:

\[ P_{\mathrm{gas}}=P_{\mathrm{atm}}-P_{\ce{H2O}} \]

If the question directly supplies the total pressure of the wet gas, use that instead of atmospheric pressure.

Step 4: Convert the volume and pressure into compatible units.

For example, if you use pressure in kPa, convert the volume into litres and use the corresponding value of \(R\).

Step 5: Substitute into the ideal gas equation.

\[ PV=nRT \]

Rearrange to find the amount of gas:

\[ \boxed{n=\frac{PV}{RT}} \]

Remember that \(P\) must be the partial pressure of the gas you are calculating, not the total pressure of the wet mixture.

A useful pressure decision table

The last two situations are worth remembering. Subtracting water vapour pressure twice is just as incorrect as forgetting to subtract it.

03Temperature, pressure, and the gas constant \(R\)

Before attempting a calculation, check that your units match.

Temperature must be in kelvin

The ideal gas equation requires absolute temperature, measured in kelvin.

For example, if a gas is collected at \(25.0^\circ\mathrm C\):

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

Do not substitute \(25.0\) directly into the ideal gas equation.

Kelvin is used because gas behaviour depends on absolute temperature, rather than the arbitrary zero point of the Celsius scale.

If you need a refresher on how temperature, pressure, and volume relate, see How to Read Pressure-Volume and Temperature Gas Graphs.

Choose the correct value of \(R\)

The gas constant must match your pressure and volume units.

For most calculations in this article, we will use:

\[ R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}} \]

If the question supplies pressure in mmHg, you can either convert it into kPa or convert it into atm and use the corresponding value of \(R\).

The standard pressure conversion is:

\[ 760\ \mathrm{mmHg}=1\ \mathrm{atm} \]

\[ 1\ \mathrm{atm}=101.325\ \mathrm{kPa} \]

For more practice matching gas-law units, see Choosing R When Pressure Is in kPa and Volume Is in L.

Water vapour pressure depends on temperature

The water vapour pressure increases as temperature increases.

This means you cannot use the same water vapour pressure for every experiment.

The following are approximate illustrative values of saturated water vapour pressure:

These values are provided as examples of the type of data you may encounter. In an exam question, use the supplied table or specified value.

Why does this matter?

Suppose you collect gas at \(30^\circ\mathrm C\), but accidentally use the water vapour pressure for \(20^\circ\mathrm C\).

You would subtract too little pressure from the total. This would make the calculated pressure of your gas too high, causing you to overestimate the number of moles.

Always match the vapour pressure to the actual experimental temperature.

04What if the water levels are not equal?

So far, we have assumed that the water levels inside and outside the collection tube are equal.

This is an important experimental condition, not just a convenient drawing detail.

If the levels are different, the pressure of the collected gas is not necessarily equal to atmospheric pressure.

There are two possibilities:

• If the water level inside the collection tube is higher than the outside water level, the gas pressure inside the tube is lower than atmospheric pressure.
• If the water level inside the collection tube is lower than the outside water level, the gas pressure inside the tube is higher than atmospheric pressure.

The pressure difference comes from the weight of the water column.

The relationship is:

\[ \Delta P=\rho gh \]

where:

• \(\Delta P\) is the pressure difference in Pa.
• \(\rho\) is the density of water in \(\mathrm{kg\,m^{-3}}\).
• \(g\) is gravitational acceleration, approximately \(9.81\ \mathrm{m\,s^{-2}}\).
• \(h\) is the vertical difference between the water levels in metres.

For water, a height difference of \(10.0\ \mathrm{cm}\) produces approximately:

\[ \begin{aligned} \Delta P&=(1000)(9.81)(0.100)\\ &=981\ \mathrm{Pa}\\ &=0.981\ \mathrm{kPa} \end{aligned} \]

The sign of this correction depends on which water level is higher.

After correcting for the water-level difference, you must still subtract the water vapour pressure.

For example, if the inside water level is higher:

\[ P_{\mathrm{gas}} = P_{\mathrm{atm}}-\rho gh-P_{\ce{H2O}} \]

This correction is only needed when the question provides unequal water levels or a relevant pressure difference. Do not invent a correction when the question explicitly states that the levels are equal.

05Worked examples

Let’s apply the method to two different experimental situations.

06Ideal gas assumptions versus experimental corrections

There is an important distinction between correcting the pressure of a wet gas and accounting for imperfections in an experiment.

They are not the same thing.

For an ordinary HSC-style calculation, the ideal gas equation is generally applied under the assumption that the gas behaves ideally.

You also normally assume that:

• The gas has reached equilibrium with the water.
• The gas and water are at the same temperature.
• The supplied water vapour pressure represents the saturated vapour pressure.
• There is no significant gas leakage or dissolution unless the question states otherwise.

Subtracting water vapour pressure does not correct for non-ideal gas behaviour.

It simply removes the contribution of water vapour so that you can calculate the amount of the gas you actually want.

This distinction becomes particularly important when you use the calculated moles in a subsequent stoichiometry problem.

07Questions and solutions

The following questions move from a standard gas calculation towards more demanding problems involving experimental data, stoichiometry, and changing conditions.

Question 1

A student collects \(150.0\ \mathrm{mL}\) of argon gas over water at \(20.0^\circ\mathrm C\).

The atmospheric pressure is \(101.2\ \mathrm{kPa}\), and the water vapour pressure at this temperature is \(2.34\ \mathrm{kPa}\).

The water levels inside and outside the collection tube are equal.

Calculate the number of moles of argon collected.

Solution 1

The student collected \(6.08\times10^{-3}\ \mathrm{mol}\) of argon.

First, calculate the partial pressure of argon using Dalton’s law:

\[ \begin{aligned} P_{\ce{Ar}}&=P_{\mathrm{atm}}-P_{\ce{H2O}}\\ &=101.2-2.34\\ &=98.86\ \mathrm{kPa} \end{aligned} \]

Convert the volume into litres:

\[ V=0.1500\ \mathrm L \]

Convert the temperature into kelvin:

\[ T=20.0+273.15=293.15\ \mathrm K \]

Now substitute into the ideal gas equation:

\[ \begin{aligned} n&=\frac{PV}{RT}\\ &=\frac{(98.86)(0.1500)} {(8.314)(293.15)}\\ &=0.006084\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \boxed{n(\ce{Ar})=6.08\times10^{-3}\ \mathrm{mol}} \]

The important step was subtracting the water vapour pressure before using \(PV=nRT\).

Using the full atmospheric pressure would have overestimated the amount of argon.

Question 2

A student collects \(360.0\ \mathrm{mL}\) of a gas over water at \(27.5^\circ\mathrm C\).

The atmospheric pressure is \(748\ \mathrm{mmHg}\). The supplied water vapour pressure data are:

For this question, use linear interpolation to estimate the water vapour pressure at \(27.5^\circ\mathrm C\).

The water levels are equal.

Calculate the number of moles of gas collected, using \(R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\).

Solution 2

The collected sample contains approximately \(1.38\times10^{-2}\ \mathrm{mol}\) of gas.

There are three things to manage here: interpolation, pressure conversion, and temperature conversion.

First, estimate the water vapour pressure.

The temperature \(27.5^\circ\mathrm C\) is halfway between \(25.0^\circ\mathrm C\) and \(30.0^\circ\mathrm C\).

Using linear interpolation:

\[ \begin{aligned} P_{\ce{H2O}} &=\frac{3.17+4.24}{2}\\ &=3.705\ \mathrm{kPa} \end{aligned} \]

Next, convert the atmospheric pressure into kPa.

Using \(760\ \mathrm{mmHg}=101.325\ \mathrm{kPa}\):

\[ \begin{aligned} P_{\mathrm{atm}} &=\frac{748}{760}\times101.325\\ &=99.73\ \mathrm{kPa} \end{aligned} \]

Now calculate the partial pressure of the collected gas:

\[ \begin{aligned} P_{\mathrm{gas}} &=99.73-3.705\\ &=96.02\ \mathrm{kPa} \end{aligned} \]

Convert the volume and temperature:

\[ V=0.3600\ \mathrm L \]

\[ T=27.5+273.15=300.65\ \mathrm K \]

Substitute into the ideal gas equation:

\[ \begin{aligned} n&=\frac{PV}{RT}\\ &=\frac{(96.02)(0.3600)} {(8.314)(300.65)}\\ &=0.01383\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \boxed{n=1.38\times10^{-2}\ \mathrm{mol}} \]

Notice that the water vapour pressure was interpolated before being subtracted.

Also notice that the atmospheric pressure was converted into kPa before subtracting the water vapour pressure. You cannot subtract pressures expressed in different units.

Question 3

A student collects \(300.0\ \mathrm{mL}\) of gas over water at \(20.0^\circ\mathrm C\).

The total pressure is \(101.3\ \mathrm{kPa}\), and the actual water vapour pressure is \(2.34\ \mathrm{kPa}\).

However, the student accidentally uses the water vapour pressure for \(30.0^\circ\mathrm C\), which is \(4.24\ \mathrm{kPa}\).

Calculate:

a) The correct number of moles of gas.

b) The number of moles calculated by the student.

c) The percentage error in the student’s result.

Explain why the error occurs.

Solution 3

The student underestimates the number of moles by approximately \(1.92\%\).

The mistake is using a water vapour pressure that is too high. This causes the student to subtract too much pressure from the total.

Part a: Calculate the correct number of moles.

The correct gas pressure is:

\[ \begin{aligned} P_{\mathrm{gas}}&=101.3-2.34\\ &=98.96\ \mathrm{kPa} \end{aligned} \]

The temperature is:

\[ T=293.15\ \mathrm K \]

The volume is:

\[ V=0.3000\ \mathrm L \]

Substitute:

\[ \begin{aligned} n_{\mathrm{correct}} &=\frac{(98.96)(0.3000)} {(8.314)(293.15)}\\ &=0.01218\ \mathrm{mol} \end{aligned} \]

Part b: Calculate the student’s result.

The student incorrectly uses \(4.24\ \mathrm{kPa}\) for water vapour pressure.

Their calculated gas pressure is:

\[ \begin{aligned} P_{\mathrm{wrong}}&=101.3-4.24\\ &=97.06\ \mathrm{kPa} \end{aligned} \]

Using this incorrect pressure:

\[ \begin{aligned} n_{\mathrm{wrong}} &=\frac{(97.06)(0.3000)} {(8.314)(293.15)}\\ &=0.01195\ \mathrm{mol} \end{aligned} \]

Part c: Calculate the percentage error.

Use the correct value as the reference:

\[ \%\text{ error} = \frac{|n_{\mathrm{wrong}}-n_{\mathrm{correct}}|} {n_{\mathrm{correct}}}\times100 \]

Substituting:

\[ \begin{aligned} \%\text{ error} &=\frac{|0.01195-0.01218|} {0.01218}\times100\\ &\approx1.92\% \end{aligned} \]

The student’s result is approximately \(1.92\%\) too low.

The physical reason is straightforward: at \(20.0^\circ\mathrm C\), the actual water vapour pressure is lower than at \(30.0^\circ\mathrm C\).

Using the higher-temperature value removes too much pressure from the gas mixture, so the calculated gas pressure and mole amount are both underestimated.

This is why matching the water vapour pressure to the experimental temperature is essential.

Question 4

A student reacts magnesium with excess hydrochloric acid to produce hydrogen gas.

The reaction is:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

The student uses \(0.120\ \mathrm g\) of magnesium and collects the hydrogen over water.

The experimental data are:

The water levels are equal.

a) Calculate the actual number of moles of hydrogen collected.

b) Calculate the theoretical number of moles of hydrogen that could be produced from the magnesium.

c) Calculate the percentage of the theoretical amount that was collected.

d) Explain why the result does not, by itself, prove that the chemical reaction had a percentage yield equal to your calculated collection percentage.

Solution 4

The experiment collected approximately \(0.00463\ \mathrm{mol}\) of hydrogen, representing an apparent collection recovery of \(93.7\%\) of the theoretical amount.

Part a: Calculate the actual moles of hydrogen collected.

First, correct the pressure:

\[ \begin{aligned} P_{\ce{H2}}&=99.4-2.81\\ &=96.59\ \mathrm{kPa} \end{aligned} \]

Convert the volume:

\[ V=0.118\ \mathrm L \]

Convert the temperature:

\[ T=23.0+273.15=296.15\ \mathrm K \]

Substitute:

\[ \begin{aligned} n(\ce{H2}) &=\frac{PV}{RT}\\ &=\frac{(96.59)(0.118)} {(8.314)(296.15)}\\ &=0.004629\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \boxed{n(\ce{H2})=0.00463\ \mathrm{mol}} \]

Part b: Calculate the theoretical amount of hydrogen.

The balanced equation is:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

The mole ratio between magnesium and hydrogen is \(1:1\).

Therefore, one mole of magnesium produces one mole of hydrogen, assuming the reaction goes to completion.

Calculate the initial moles of magnesium:

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

\[ \begin{aligned} n(\ce{Mg}) &=\frac{0.120}{24.3}\\ &=0.004938\ \mathrm{mol} \end{aligned} \]

Since the mole ratio is \(1:1\):

\[ \boxed{n_{\mathrm{theoretical}}(\ce{H2})=0.00494\ \mathrm{mol}} \]

Part c: Calculate the percentage collected.

Use:

\[ \%\text{ recovery} = \frac{n_{\mathrm{actual}}} {n_{\mathrm{theoretical}}}\times100 \]

Substitute:

\[ \begin{aligned} \%\text{ recovery} &=\frac{0.004629}{0.004938}\times100\\ &=93.7\% \end{aligned} \]

Therefore:

\[ \boxed{\%\text{ recovery}=93.7\%} \]

Part d: Explain the experimental limitation.

A collection recovery of \(93.7\%\) does not necessarily mean that the chemical reaction itself had a \(93.7\%\) yield.

Some hydrogen may have escaped through leaks, dissolved in the water, or remained in the apparatus rather than entering the measuring tube.

The magnesium may also have reacted incompletely, or the experimental measurements may contain errors.

These possibilities produce a similar observation: less hydrogen is measured than the theoretical amount.

The pressure correction removes the contribution of water vapour, but it cannot recover hydrogen that was never measured.

The correct conclusion is that the experiment collected \(93.7\%\) of the theoretical amount of hydrogen. Further evidence would be needed to identify why the measured amount was lower.

Question 5

A student traps a fixed amount of dry nitrogen gas in an inverted graduated tube over water.

The student measures the gas under two different sets of conditions. At both measurements, the water levels inside and outside the tube are equal.

The following data are recorded:

Assume:

• The amount of dry nitrogen remains constant.
• The nitrogen behaves ideally.
• The gas is saturated with water vapour at both temperatures.
• No nitrogen escapes or dissolves significantly.
• Each volume measurement has an uncertainty of approximately \(1\ \mathrm{mL}\).

A student argues that the gas volume should increase only according to Charles’s law. They calculate a final volume of approximately \(206.8\ \mathrm{mL}\) and conclude that the measured \(215.0\ \mathrm{mL}\) must mean additional nitrogen entered the tube.

Is the student’s conclusion correct?

Calculate the expected volume in State B and explain why the simple Charles’s law prediction is insufficient.

Solution 5

The student’s conclusion is incorrect. The measured volume of \(215.0\ \mathrm{mL}\) is consistent with the same amount of nitrogen remaining in the tube.

The predicted volume is approximately \(215.4\ \mathrm{mL}\).

The important complication is that Charles’s law assumes constant pressure. Here, the partial pressure of nitrogen changes between the two measurements.

Step 1: Calculate the nitrogen partial pressure in each state.

In State A:

\[ \begin{aligned} P_{\ce{N2},1} &=100.0-2.34\\ &=97.66\ \mathrm{kPa} \end{aligned} \]

In State B:

\[ \begin{aligned} P_{\ce{N2},2} &=98.0-4.24\\ &=93.76\ \mathrm{kPa} \end{aligned} \]

Notice that the nitrogen pressure decreases from \(97.66\) to \(93.76\ \mathrm{kPa}\).

This happens because atmospheric pressure decreases while water vapour pressure increases.

Step 2: Convert the temperatures into kelvin.

For State A:

\[ T_1=20.0+273.15=293.15\ \mathrm K \]

For State B:

\[ T_2=30.0+273.15=303.15\ \mathrm K \]

Step 3: Choose the correct gas relationship.

The amount of dry nitrogen is constant, and we are comparing two different temperatures, pressures, and volumes.

Therefore, we use the combined gas relationship:

\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]

Crucially, both pressures in this equation must be the partial pressures of nitrogen, not the total pressures of the wet gas mixtures.

Rearrange to find the final volume:

\[ V_2=\frac{P_1V_1T_2}{P_2T_1} \]

Substitute:

\[ \begin{aligned} V_2&= \frac{(97.66)(200.0)(303.15)} {(93.76)(293.15)}\\ &\approx215.4\ \mathrm{mL} \end{aligned} \]

Therefore:

\[ \boxed{V_2\approx215.4\ \mathrm{mL}} \]

Step 4: Compare the predicted and measured volumes.

The predicted volume is \(215.4\ \mathrm{mL}\), while the measured volume is \(215.0\ \mathrm{mL}\).

The difference is:

\[ 215.4-215.0=0.4\ \mathrm{mL} \]

This is smaller than the stated measurement uncertainty of approximately \(1\ \mathrm{mL}\).

The two results are therefore consistent with the same amount of nitrogen remaining in the collection tube.

Step 5: Explain why Charles’s law gives the wrong prediction.

The student’s calculation assumes that the nitrogen partial pressure remains constant.

Charles’s law states that volume is directly proportional to absolute temperature when pressure and amount of gas are constant.

Using that assumption:

\[ \begin{aligned} V_2&=V_1\frac{T_2}{T_1}\\ &=200.0\times\frac{303.15}{293.15}\\ &\approx206.8\ \mathrm{mL} \end{aligned} \]

However, the nitrogen partial pressure does not remain constant.

It decreases from \(97.66\ \mathrm{kPa}\) to \(93.76\ \mathrm{kPa}\). A lower partial pressure allows the same amount of nitrogen to occupy a larger volume at the new temperature.

There is another subtle detail.

Although the amount of nitrogen is fixed, the amount of water vapour in the tube is not fixed. As the temperature increases, more water evaporates into the gas mixture.

The total amount of gas in the tube therefore changes, even though the amount of nitrogen does not.

This is why applying Charles’s law directly to the measured wet-gas volume gives the wrong prediction.

The correct comparison must track the nitrogen partial pressure at each temperature.

08Where this method leads next

Once you can calculate moles of gas collected over water, you can use that amount in a balanced chemical equation to determine the quantities of reactants consumed or products formed.

For example, you could collect hydrogen from magnesium reacting with hydrochloric acid, correct its pressure for water vapour, calculate its moles, and then use the mole ratio to determine the amount of magnesium that reacted.

The important habit is to separate the two jobs:

1. Correct the experimental gas pressure to find the partial pressure of the gas you want.
2. Use the ideal gas equation to calculate its amount in moles.

Get those two steps right, and gas collection calculations become much more manageable.