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: When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.460 g of a particular hydrocarbon was burned in air, 0.477 g of CO, 0.749 g of CO2, and 0.460 g of H2O were formed.What is the empirical formula of the compound?

Solution: When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.460 g of a particular hydrocarbon was burned in air, 0.477 g of CO, 0.749 g of CO2, and 0.460 g of H2O were formed

Problem

When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.460 g of a particular hydrocarbon was burned in air, 0.477 g of CO, 0.749 g of CO2, and 0.460 g of H2O were formed.

What is the empirical formula of the compound?

Solution

We have to determine the molecular formula of a hydrocarbon, 0.460 g of which 0.477 g CO, 0.749 g CO2 and 0.460 g H2O upon combustion.


Empirical formula is a chemical formula that gives the relative proportions of elements in a compound but not the actual number of atoms of elements in a compound.


We will calculate the empirical formula through these steps:

Step 1: Find the mass of C and H using the given masses of CO, CO2 and H2O.

Step 2: Find the percentage of C and H in the compound.

Step 3: Dividing the percentages of C and H by their average atomic masses.

Step 4: Dividing the obtained numbers by the smallest number among them and rounding off to the nearest whole numbers to get the empirical formula.


Solution BlurView Complete Written Solution