Ideal Gas Applications – Molar Mass & Density

Molar Mass & Density Problems (Chapter 19)

Fundamental Relations & Mind-Set

• Core gas law used in every example: $PV = nRT$
• Derived expressions
  • Moles: $n = \frac{PV}{RT}$
  • Molar mass: $M = \frac{m}{n}$ where (m) = mass in grams
  • Density: $\rho = \frac{m}{V}$
• Instructor’s philosophy
  • Minimize memorization; derive from $PV = nRT$ instead of memorizing separate “molar–mass” or “density” versions.
  • Begin every problem by explicitly stating the quantity sought (molar mass, density, etc.) and its defining equation.

Multiple-Choice Pitfalls ("Distractor" Strategy)

• Test writers often place an intermediate result as option A.
• Rule of thumb:
  • If the first answer looks too easy, re-trace your steps.
  • Make sure you finish all algebra before selecting.

Example 1 – Molar Mass of an Unknown Gas

• Given
  • Mass: 52.6 g
  • Pressure: 0.984 atm
  • Volume: 75 L
  • Temperature: 294 K (already in Kelvin ⇒ no conversion)
• Objective: Find $M = \frac{m}{n}$
• Steps
  1. Compute moles from $n = \frac{PV}{RT}$
  • $n = \frac{0.984\,\text{atm} \times 75\,\text{L}}{0.0821\,\frac{\text{L·atm}}{\text{mol·K}} \times 294\,\text{K}} = 3.06\,\text{mol}$
  1. Plug into molar-mass formula
  • $M = \frac{52.6\,\text{g}}{3.06\,\text{mol}} = 17.2\,\text{g·mol}^{-1}$
• Common error: Selecting 3.06 g·mol⁻¹ (the moles) as final answer.

Intrinsic vs Extrinsic Properties

• Density is intrinsic (independent of sample size).
• Choose any convenient sample; instructor defaults to 1 mol for gases.
• Extrinsic quantities (mass, volume, moles) do scale with sample size.

Example 2 – Density of O₂ at STP

• Given: $T = 273\,\text{K},\; P = 1\,\text{atm}$ (Standard Temperature & Pressure)
• Shortcut at STP
  • 1 mol of any ideal gas occupies $22.4\,\text{L}$
• Procedure
  1. Assume 1 mol; obtain mass from periodic table:
  • $m = 31.998\,\text{g}$
  1. Use shortcut volume:
  • $V = 22.4\,\text{L}$
  1. Density:
  • $\rho = \frac{31.998\,\text{g}}{22.4\,\text{L}} = 1.43\,\text{g·L}^{-1}$
• Verification with PV = nRT (no shortcut)
  • $V = \frac{nRT}{P} = \frac{1\,\text{mol} \times 0.0821 \times 273}{1} = 22.413\,\text{L}$
  • Confirms shortcut value.

Example 3 – Density of SO₃ at 294 K & 1.14 atm (Non-STP)

• Given
  • $T = 294\,\text{K},\; P = 1.14\,\text{atm}$
• Assume 1 mol sample (because density is intrinsic)
  • Molar mass of SO₃:
  • $M = 32.06 + 3\times15.998 = 80.05\,\text{g·mol}^{-1}$
• Find volume with PV = nRT
  • $V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 294}{1.14} = 21.2\,\text{L}$
• Density
  • $\rho = \frac{80.05\,\text{g}}{21.2\,\text{L}} = 3.78\,\text{g·L}^{-1}$
• Trap: 21.2 L often appears as a distractor.

Gas Constant & Units Reminder

• Use $R = 0.0821\,\frac{\text{L·atm}}{\text{mol·K}}$ only when pressure is in atmospheres and volume in liters.
• Convert units first if data are given in mm Hg, kPa, or °C.

Checks for STP vs Non-STP

• STP definition in this course:
  • $T = 273\,\text{K}$
  • $P = 1\,\text{atm}$
• At STP: directly use $22.4\,\text{L·mol}^{-1}$ to bypass PV = nRT.
• At non-STP: must apply the full ideal-gas equation.

Key Takeaways for Exams

• Always write down what you need ((M) or (\rho)) before computing.
• Finish every algebraic step before looking at the answer set.
• Remember intrinsic nature of density ⇒ 1 mol assumption is legal and simplifies work.
• Verify STP conditions; use the 22.4 L shortcut only when applicable.
• Watch for unit consistency and clever distractors (especially option A).

Looking Ahead

• Next lecture covers gas-law stoichiometry.
• A review video will follow to consolidate Chapter 19.