DAT General Chemistry: Gas Density and Molar Mass of a Gas

DAT High-Yield Concept

Gas density is directly proportional to pressure and molar mass, but inversely proportional to absolute temperature: $d = \frac{P \mathcal{M}}{R T}$
Gas density is typically reported in $\pu{g L^{-1}}$ , unlike liquid or solid density which is expressed in $\pu{g mL^{-1}}$ .
Rearranging the density equation allows for the rapid determination of a gas's molar mass: $\mathcal{M} = \frac{d R T}{P}$
Always convert temperature to Kelvin ( $\pu{K}$ ) and ensure pressure and volume units match the ideal gas constant $R = \pu{0.08206 L atm mol^{-1} K^{-1}}$ .

Deriving Gas Density from the Ideal Gas Law

The ideal gas law relates pressure ( $P$ ), volume ( $V$ ), amount in moles ( $n$ ), and absolute temperature ( $T$ ) via $P V = n R T$ . Because moles can be expressed as mass divided by molar mass ( $n = \frac{m}{\mathcal{M}}$ ), we substitute this relationship into the ideal gas law:

P V = \left(\frac{m}{\mathcal{M}}\right) R T

Rearranging to isolate density ( $d = \frac{m}{V}$ ) yields the foundational gas density equation:

d = \frac{m}{V} = \frac{P \mathcal{M}}{R T}

Key Definition: Gas density (

d

) represents mass per unit volume for a gas, measured in

\pu{g L^{-1}}

. Because gas volume changes significantly with temperature and pressure, gas density must always be specified alongside environmental conditions.

Worked Example: Calculating Gas Density

Problem: Calculate the density of pure oxygen gas, $\ce{O2}$ , at $\pu{298 K}$ and $\pu{1.50 atm}$ .

Solution & Dimensional Analysis:

Identify molar mass: $\ce{O2}$ has a molar mass of $\mathcal{M} = \pu{32.00 g mol^{-1}}$ .
Apply the formula: $d = \frac{(\pu{1.50 atm})(\pu{32.00 g mol^{-1}})}{(\pu{0.08206 L atm mol^{-1} K^{-1}})(\pu{298 K})}$
Calculate: $d = \frac{48.00}{24.45388} = \pu{1.96 g L^{-1}}$ .

Common Exam Trap: Using atomic oxygen ( $\mathcal{M} = \pu{16.00 g mol^{-1}}$ ) instead of elemental diatomic oxygen ( $\ce{O2}$ , $\mathcal{M} = \pu{32.00 g mol^{-1}}$ ). Elemental gases such as $\ce{N2}$ , $\ce{O2}$ , and $\ce{F2}$ exist as homonuclear diatomics.

Determining Molar Mass from Gas Density

Measuring the density of an unknown gas provides a straightforward experimental route to determine its molar mass. Rearranging the gas density expression yields:

\mathcal{M} = \frac{d R T}{P}

This relationship is essential on the DAT for identifying unknown volatile liquids or elemental gases from vapor density measurements. (Note: The concept of molar volume at STP is covered in depth in its own dedicated lesson.)

Worked Example: Molar Mass Determination

Problem: An unknown gas displays a density of $\pu{2.10 g L^{-1}}$ at $\pu{300 K}$ and $\pu{0.950 atm}$ . What is the molar mass of the gas?

Solution:

Apply the rearranged equation: $\mathcal{M} = \frac{(\pu{2.10 g L^{-1}})(\pu{0.08206 L atm mol^{-1} K^{-1}})(\pu{300 K})}{\pu{0.950 atm}}$
Perform calculation: $\mathcal{M} = \frac{51.6978}{0.950} = \pu{54.4 g mol^{-1}}$ .

Common Exam Trap: Forgetting to convert temperature from Celsius to Kelvin ( $T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15$ ) or misplacing pressure in the denominator.

Mnemonic: Remember "Police Officers Matter = dirt"

(P \cdot \mathcal{M} = d \cdot R \cdot T)

. Solve for density (

d = \frac{P \mathcal{M}}{R T}

) or molar mass (

\mathcal{M} = \frac{d R T}{P}

) easily by reordering this single core phrase!

How Temperature and Pressure Change Gas Density

Gas density is highly sensitive to thermodynamic state variables:

Pressure Effect: Increasing pressure compresses the gas particles into a smaller volume, increasing mass per unit volume. Gas density is directly proportional to pressure ( $d \propto P$ ).
Temperature Effect: Increasing temperature increases kinetic energy, causing gas expansion. This larger volume reduces mass per unit volume. Gas density is inversely proportional to temperature ( $d \propto \frac{1}{T}$ ).
Molar Mass Effect: At identical temperature and pressure, heavier gases contain more mass per mole in the same volume. Comparing gases under identical conditions shows $\ce{CH4}$ < $\ce{C3H8}$ < $\ce{C4H10}$ in density.

Data about density of liquid and vapor phases of gases undes various temperatures for propane and n-butane. (Source: Zivenko Oleksii · CC BY-SA 4.0 · Wikimedia Commons)

Comparison of Variables Affecting Gas Density

Variable Change	Mathematical Relation	Effect on Gas Density ( $d$ )	Molecular Explanation
Increase Pressure ( $P \uparrow$ )	$d \propto P$	Increases	Gas molecules are compressed into a smaller volume.
Increase Temperature ( $T \uparrow$ )	$d \propto \frac{1}{T}$	Decreases	Thermal expansion increases container volume.
Higher Molar Mass ( $\mathcal{M} \uparrow$ )	$d \propto \mathcal{M}$	Increases	Equal moles occupy equal volumes, but heavier molecules add mass.

Critical Thinking Questions

Why does a hot air balloon rise in cool ambient air? Explain in terms of temperature, gas density, and buoyancy.
If container pressure is doubled while absolute temperature is halved, by what factor does the gas density change?
Two rigid containers of identical volume hold $\ce{He}$ and $\ce{CO2}$ at $\pu{25 degC}$ and $\pu{1.0 atm}$ . How do their densities compare, and why?

References

Sources returned by Google Search grounding during lesson verification.

Chapter 6.3: The Ideal Gas Law - Chemistry LibreTexts — chem.libretexts.org
Equation of State - NASA Glenn Research Center — grc.nasa.gov

Gas Density and Molar Mass of a Gas

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