DAT General Chemistry: Gases Collected over Water

DAT High-Yield Concept

Gases generated in laboratory reactions are commonly collected by displacing water in an inverted container.
Gas collected over water is always a wet gas mixture containing both the target gas and evaporated water vapor (

A gas bubbled up into an inverted tube of water displaces the water but arrives carrying water vapor with it, so the tube holds a mixture rather than a dry sample. When the water level inside the tube is level with the water outside, the total pressure inside equals the atmospheric pressure, and by Dalton's law that total is the pressure of the collected gas plus the vapor pressure of water at that temperature. Subtracting the tabulated water vapor pressure gives the pressure of the gas itself, which is what belongs in the ideal gas law. The method only suits gases that are not appreciably soluble in water.

$\ce{P_{\text{total}} = P_{\text{dry gas}} + P_{\ce{H2O}}}$ ).
Water vapor pressure is determined only by temperature (it is independent of gas volume or total pressure).
Correction formula: $P_{\text{dry gas}} = P_{\text{total}} - P_{\ce{H2O}}$ .
Water levels are equalized inside and outside the gas-collecting tube to set the total inside pressure equal to the barometric pressure.

Gas Collection Mechanics

When a gas is generated by a chemical reaction in aqueous solution, it can be collected by drowning an inverted bottle or eudiometer filled with water into a trough of water. As the gas bubbles enter the tube, it displaces the water downward.

Because the gas bubbles travel through liquid water, water molecules continuously evaporate into the gas phase. The space above the water line fills with a gas mixture consisting of the target gas and water vapor. Under equilibrium conditions at a given temperature, the vapor pressure of water (

File:Water vapor pressure graph.jpg (Source: CC BY 2.5 · Wikimedia Commons)

) reaches a fixed value.

Dalton's Law Application:

The total pressure of the collected gas equals the sum of the partial pressure of the dry gas and the vapor pressure of water:

P_{\text{total}} = P_{\text{dry gas}} + P_{\ce{H2O}}

When the water levels inside and outside the collection tube are equalized, $P_{\text{total}} = P_{\text{atm}}$ .

Water-Vapor Pressure Correction Calculation

To perform stoichiometry or ideal gas law calculations ( $PV = nRT$ ), you must subtract the vapor pressure of water from the total atmospheric pressure. Using total pressure directly is a primary DAT trap that overestimates the moles of dry gas.

State	Pressure Composition	When Used in $PV = nRT$
Wet Gas	$P_{\text{total}} = P_{\text{gas}} + P_{\ce{H2O}}$	Never directly! Gives incorrect mole calculations.
Dry Gas	$P_{\text{dry gas}} = P_{\text{total}} - P_{\ce{H2O}}$	Always! Required to find true moles of produced gas.

DAT Mnemonic: "DRY gas = Deduct Required Y-vapor!"
Always subtract

P_{\ce{H2O}}

before plugging pressure into

PV = nRT

Worked Example

Problem: A sample of $\ce{O2(g)}$ is generated by the heating of potassium chlorate according to the balanced equation: $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$ The $\ce{O2(g)}$ is collected over water at $\pu{25.0 degC}$ under a total barometric pressure of $\pu{755.0 mmHg}$ . The volume of gas collected is $\pu{0.250 L}$ . The vapor pressure of water at $\pu{25.0 degC}$ is $\pu{23.8 mmHg}$ . Calculate the mass of dry $\ce{O2}$ gas produced.

Step 1: Calculate the partial pressure of dry $\ce{O2}$ .

P_{\ce{O2}} = P_{\text{total}} - P_{\ce{H2O}} = \pu{755.0 mmHg} - \pu{23.8 mmHg} = \pu{731.2 mmHg}

Step 2: Convert pressure to atmospheres and temperature to Kelvin.

P_{\ce{O2}} = \frac{\pu{731.2 mmHg}}{\pu{760.0 mmHg/atm}} = \pu{0.9621 atm}

T = \pu{25.0 degC} + 273.15 = \pu{298.15 K}

Step 3: Calculate moles of dry $\ce{O2}$ using the Ideal Gas Law.

n_{\ce{O2}} = \frac{P_{\ce{O2}} V}{R T} = \frac{(\pu{0.9621 atm})(\pu{0.250 L})}{(\pu{0.08206 L*atm/(mol*K)})(\pu{298.15 K})} = \pu{0.00983 mol}

Step 4: Calculate mass of $\ce{O2}$ (molar mass = $\pu{32.00 g/mol}$ ).

m_{\ce{O2}} = \pu{0.00983 mol} \times \pu{32.00 g/mol} = \pu{0.315 g}

DAT Exam Trap: Forgetting to subtract

P_{\ce{H2O}}

yields

P = \pu{755.0 mmHg}

(

\pu{0.9934 atm}

), which mistakenly calculates

\pu{0.325 g}

\ce{O2}

. Always check if the question specifies gas collected over water!

Critical Thinking Questions

If the temperature of the water bath increases during an experiment while total pressure remains constant, how does the partial pressure and mole fraction of the collected dry gas change?
Why must the water levels inside and outside the gas collection tube be made equal before reading the gas volume? What happens if the inner level is higher?
Why is gas collection over water unsuitable for gases like ammonia ( $\ce{NH3}$ ) or hydrogen chloride ( $\ce{HCl}$ )?

References

Sources returned by Google Search grounding during lesson verification.

patents.google.com — patents.google.com
9.12: Dalton's Law of Partial Pressures - Chemistry LibreTexts — chem.libretexts.org
Collection of Gases over Water - UCalgary Chemistry Textbook — chem.libretexts.org

Gases Collected over Water

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