All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: In a given diffusion apparatus, 15.0 mL of HBr gas diffuses in 1.0 min. In the same apparatus and under the same conditions, 20.3 mL of an unknown gas diffuses in 1.0 min. The unknown gas is a hydroca

Solution: In a given diffusion apparatus, 15.0 mL of HBr gas diffuses in 1.0 min. In the same apparatus and under the same conditions, 20.3 mL of an unknown gas diffuses in 1.0 min. The unknown gas is a hydroca

Problem

In a given diffusion apparatus, 15.0 mL of HBr gas diffuses in 1.0 min. In the same apparatus and under the same conditions, 20.3 mL of an unknown gas diffuses in 1.0 min. The unknown gas is a hydrocarbon.

Find its molecular formula.

Solution

Recall: Graham's Law of Effusion allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.

This means that when comparing two gases:

Let's designate HBr as gas 1 and the unknown as gas 2. We're given:
rateHBr = 15.0 mL/min                  MHBr = 1.01 g/mol H + 79.90 g/mol Br = 80.91 g/mol
rateunk = 20.3 mL/min                  Munk = ?

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