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: A compound has the empirical formula CHCl. A 256-mL flask, at 373 K and 750. torr, contains 0.800 g of the gaseouscompound. Give the molecular formula.

Problem

A compound has the empirical formula CHCl. A 256-mL flask, at 373 K and 750. torr, contains 0.800 g of the gaseous
compound. Give the molecular formula.

Solution
  • We need to find the molar mass of the empirical formula and the actual gas since this will tell us how many atoms of C, H and Cl are present in the gas and the actual molecular formula of the gas
  • Molar mass of CHCl will appear as:

C - 12.01 (1) = 12.01

H - 1.01 (1) = 1.01

Cl - 35.45 (1) = 35.45

12.01 + 1.01 + 35.45 = 48.47 g/mol

  • For the actual molecular formula, we will be using the ideal gas equation and use the conditions provided to find the moles
  • Calculating for moles:
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