Molar Mass Determination — Ideal Gas Method
For a gas sample at pressure \(P\), volume \(V\), and absolute temperature \(T\), the ideal gas equation
\(PV=nRT\) connects macroscopic variables with the amount of substance \(n\). Because \(n=\dfrac{m}{M}\),
where \(m\) is the sample mass and \(M\) is the molar mass, substitution gives a direct formula for \(M\).
\[
PV = nRT,\quad n=\dfrac{m}{M}
\;\;\Longrightarrow\;\;
\boxed{\, M \;=\; \dfrac{m \cdot R \cdot T}{P \cdot V} \,}
\]
Dimensions & constants
With SI units \(P\,[\mathrm{Pa}],\,V\,[\mathrm{m^{3}}],\,T\,[\mathrm{K}],\,m\,[\mathrm{g}]\) and
\(R=8.314462618\ \mathrm{Pa\cdot m^{3}\cdot mol^{-1}\cdot K^{-1}}\), the result \(M\) comes out in \(\mathrm{g\cdot mol^{-1}}\).
Equivalent forms of \(R\) are useful when working directly in non-SI display units:
\[
R = 8.314462618\ \mathrm{Pa\cdot m^{3}\cdot mol^{-1}\cdot K^{-1}}
= 0.082057\ \mathrm{atm\cdot L\cdot mol^{-1}\cdot K^{-1}}
= 62.3637\ \mathrm{L\cdot Torr\cdot mol^{-1}\cdot K^{-1}}
\]
Unit toolkit (convert before calculating)
\[
1\ \mathrm{atm}=101325\ \mathrm{Pa},\quad
1\ \mathrm{bar}=10^{5}\ \mathrm{Pa},\quad
1\ \mathrm{mmHg}\approx 133.322\ \mathrm{Pa},\quad
1\ \mathrm{Torr}=133.322\ \mathrm{Pa},\quad
1\ \mathrm{psi}=6894.757\ \mathrm{Pa}
\]
\[
1\ \mathrm{L}=10^{-3}\ \mathrm{m^{3}},\qquad
1\ \mathrm{mL}=1\ \mathrm{cm^{3}}=10^{-6}\ \mathrm{m^{3}}
\]
\[
T\,(\mathrm{K}) \;=\; T\,(^\circ\mathrm{C}) + 273.15
\]
Two practical routes
1) Direct method (given \(P, V, T, m\))
- Convert \(P\to\mathrm{Pa}\), \(V\to\mathrm{m^{3}}\), \(T\to\mathrm{K}\), \(m\to\mathrm{g}\).
- Compute \(M=\dfrac{m \cdot R \cdot T}{P \cdot V}\).
- Report \(M\) in \(\mathrm{g\cdot mol^{-1}}\) with appropriate significant figures.
2) Weighing method (find \(V\) and \(m\) from masses)
When a calibrated gas volume is not given, determine the vessel volume from water
and the gas mass by difference, then use the same formula for \(M\).
\[
V_{\text{vessel}} \;=\; \dfrac{m_{\text{water}}}{\rho_{\text{water}}}
\quad\text{with}\quad
m_{\text{water}} = m_{\text{(vessel+water)}} - m_{\text{empty}}
\]
\[
m_{\text{gas}} \;=\; m_{\text{(vessel+gas)}} - m_{\text{empty}}
\]
\[
M \;=\; \dfrac{m_{\text{gas}} \cdot R \cdot T}{P \cdot V_{\text{vessel}}}
\]
Worked example (numbers as in the standard propylene case)
Data: \(m=0.1654\ \mathrm{g}\), \(T=24.0^\circ\mathrm{C}\), \(P=740.3\ \mathrm{mmHg}\), \(V=98.41\ \mathrm{mL}\).
\[
\begin{aligned}
T &= 24.0 + 273.15 = 297.2\ \mathrm{K} \\
P &= 740.3\ \mathrm{mmHg}\times \dfrac{1\ \mathrm{atm}}{760\ \mathrm{mmHg}}
= 0.9741\ \mathrm{atm} \\
V &= 98.41\ \mathrm{mL}\times 10^{-3} = 0.09841\ \mathrm{L}
\end{aligned}
\]
\[
M \;=\;
\dfrac{0.1654\ \mathrm{g}\;\cdot\;0.082057\ \mathrm{atm\cdot L\cdot mol^{-1}\cdot K^{-1}}\;\cdot\;297.2\ \mathrm{K}}
{0.9741\ \mathrm{atm}\;\cdot\;0.09841\ \mathrm{L}}
\;\approx\; 42.08\ \mathrm{g\cdot mol^{-1}}
\]
Assumptions & limitations
- Ideal behavior: deviations increase at high \(P\) or very low \(T\). For improved accuracy, a real-gas equation (e.g., van der Waals) or a compressibility factor \(Z\) can be used:
\[
PV = n Z R T \;\Rightarrow\; M = \dfrac{m \cdot Z \cdot R \cdot T}{P \cdot V}.
\]
- Dry gas: if water vapor is present, use the dry-gas pressure \(P_{\text{dry}}=P_{\text{total}}-P_{\mathrm{H_2O}}\).
- Absolute pressure: ensure pressure is not gauge-only; convert to absolute before use.
- Temperature uniformity: \(T\) must be the gas temperature in the vessel.
Quality checks
- Units cancel properly: verify that \(M\) ends in \(\mathrm{g\cdot mol^{-1}}\).
- Order of magnitude: most small organic gases fall in \(30\)–\(80\ \mathrm{g\cdot mol^{-1}}\); large discrepancies suggest a unit error.
- Significant figures: match the least precise input (the calculator allows 2–8 s.f.).
Common pitfalls
- Forgetting to convert \(^\circ\mathrm{C}\) to \(\mathrm{K}\).
- Mixing \(R\) with incompatible units (e.g., using \(R=0.082057\) with \(P\) in \(\mathrm{Pa}\)).
- Using gauge pressure or including water vapor pressure inadvertently.
- Rounding too early; keep full precision until the final step.