Collecting a Gas over Water — Theory
In a pneumatic trough the gas you collect is a mixture of the desired (dry) gas and water vapor.
Dalton’s law says the total pressure equals the sum of partial pressures, so when the water levels inside
and outside the bottle are matched (same hydrostatic head) and the atmosphere is at barometric pressure \(P_\text{bar}\),
\[
P_\text{bar} \;=\; P_\text{gas} \;+\; P_{\ce{H2O}}
\qquad\Rightarrow\qquad
P_\text{gas} \;=\; P_\text{bar} - P_{\ce{H2O}} .
\]
With the ideal gas equation, moles of the dry gas are
\[
n \;=\; \frac{P_\text{gas} \, V}{R\,T}.
\]
If the water levels are not equal, a small hydrostatic correction is needed:
\(P_\text{bar} = P_\text{gas} + P_{\ce{H2O}} \pm \rho g\,\Delta h\) (sign depends on which level is higher).
This tool assumes levels are equal (most experiments adjust for this).
Procedure (what the calculator does)
- Convert units. Pressure to a consistent unit (internally Pa), \(T\to\mathrm{K}\), and \(V\) to a unit compatible with \(R\).
- Subtract water vapor. Compute the dry-gas pressure \(P_\text{gas}=P_\text{bar}-P_{\ce{H2O}}\).
- Use \(PV=nRT\). Find moles \(n=(P_\text{gas}V)/(RT)\).
- (Optional) Stoichiometry. If a reaction relates the collected gas to another species,
apply the factor \(\nu\) (mol target per mol gas): \(n_\text{target}=\nu n\), and \(m_\text{target}=n_\text{target}M\).
Water vapor pressure
\(P_{\ce{H2O}}\) depends strongly on temperature. You may enter it directly (from a table),
or estimate it with an Antoine fit. A few reference values:
| \(T\) (°C) | \(P_{\ce{H2O}}\) (kPa) | \(P_{\ce{H2O}}\) (mmHg) |
|---|
| 15 | 1.71 | 12.79 |
| 20 | 2.34 | 17.56 |
| 23 | 2.81 | 21.1 |
| 25 | 3.17 | 23.8 |
| 30 | 4.24 | 31.9 |
Worked example
Data: \(V=81.2\ \mathrm{mL}\) of \( \ce{O2} \) collected over water at \(23^{\circ}\mathrm{C}\),
\(P_\text{bar}=751\ \mathrm{mmHg}\), \(P_{\ce{H2O}}=21.1\ \mathrm{mmHg}\).
Find \(n(\ce{O2})\). Then, for the decomposition \(2\,\ce{Ag2O(s)}\to 4\,\ce{Ag(s)}+\ce{O2(g)}\),
find the mass of \(\ce{Ag2O}\) that decomposed.
\[
P_\text{gas} = 751 - 21.1 = 730\ \mathrm{mmHg}
\quad\Rightarrow\quad
P_\text{gas} = \frac{730}{760} = 0.961\ \mathrm{atm}
\]
\[
T = 23 + 273.15 = 296\ \mathrm{K},
\quad V = 81.2\ \mathrm{mL} = 0.0812\ \mathrm{L}
\]
\[
n = \frac{PV}{RT}
= \frac{(0.961\ \mathrm{atm})(0.0812\ \mathrm{L})}{(0.082057\ \mathrm{atm\,L\,mol^{-1}\,K^{-1}})(296\ \mathrm{K})}
\approx 3.21\times 10^{-3}\ \mathrm{mol}
\]
For the reaction, \( \nu = 2\ \mathrm{mol\ Ag_2O}/\mathrm{mol\ O_2} \):
\[
n_{\ce{Ag2O}} = \nu n = 2(3.21\times10^{-3}) = 6.42\times10^{-3}\ \mathrm{mol}
\]
With \(M(\ce{Ag2O})=231.7\ \mathrm{g\,mol^{-1}}\),
\[
m_{\ce{Ag2O}} = n_{\ce{Ag2O}} M \approx (6.42\times10^{-3})\times 231.7 \approx 1.49\ \mathrm{g}.
\]
Checks & common pitfalls
- Level the water. Ensure the water levels inside and outside the bottle are equal so \(P_\text{tot}=P_\text{bar}\).
Otherwise apply a hydrostatic correction.
- Use absolute temperature. Always convert °C to K before using \(PV=nRT\).
- Consistent \(R\). Match the gas constant to your pressure–volume units (e.g., \(0.082057\ \mathrm{atm\,L\,mol^{-1}\,K^{-1}}\)).
- Reasonableness. Increasing \(T\) or \(V\) at fixed \(n\) and \(P_\text{bar}\) changes \(n\) via \(PV=nRT\). If \(P_\text{gas}\le0\),
your inputs for \(P_\text{bar}\) and \(P_{\ce{H2O}}\) are inconsistent.
How the tool applies this
- Converts all inputs to consistent SI units and computes \(P_\text{gas}=P_\text{bar}-P_{\ce{H2O}}\).
- Uses \(n=(P_\text{gas}V)/(RT)\) to return moles of dry gas.
- Optionally multiplies by a user-supplied stoichiometric factor \(\nu\) and molar mass \(M\) to get a target amount and mass.
- Shows every conversion and arithmetic step in LaTeX, so you can follow or copy into your notes.
