Ch.3 - Chemical ReactionsWorksheetSee all chapters
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 contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?

Solution: A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?

Problem

A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?

Solution
  • Determine the empirical formula by calculating the moles of C, H and N present in the compound. This can be calculated using the provided mass of CO2 and H2O
  • We can directly assume that all the C in the compound are converted to CO2 while the Hs go to H2O
  • Establish a relationship using stoichiometry to get the moles of C and O. 
  • Mass of N (eventually moles) can be calculated once we already have the mass of C and H
  • 1 mole CO2 contains 1 mole C while H2O contains 2 moles of H. Calculation of mass and mole of C will appear as:

molar mass of CO2

C - 12.01 (1) = 12.01

O - 16 (2) = 32

12.01 + 32 = 44.01 g/mol

molar mass of H2O:

H - 1.01 (2) = 2.02

O - 16 (1) = 16

16 + 2.02 = 18.02 g/mol

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