Textbook Question
Combustion analysis of naphthalene, a hydrocarbon used in mothballs, produces 8.80 g CO2 and 1.44 g H2O. Calculate the empirical formula of naphthalene.
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Ch.3 - Molecules and Compounds
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217
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Table of contents
- Ch.1 - Matter, Measurement & Problem Solving90
- Ch.2 - Atoms & Elements94
- Ch.3 - Molecules and Compounds103
- Ch.4 - Chemical Reactions and Chemical Quantities46
- Ch.5 - Introduction to Solutions and Aqueous Solutions65
- Ch.6 - Gases100
- Ch.7 - Thermochemistry85
- Ch.8 - The Quantum-Mechanical Model of the Atom51
- Ch.9 - Periodic Properties of the Elements79
- Ch.10 - Chemical Bonding I: The Lewis Model70
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory68
- Ch.12 - Liquids, Solids & Intermolecular Forces44
- Ch.13 - Solids & Modern Materials42
- Ch.14 - Solutions77
- Ch.15 - Chemical Kinetics88
- Ch.16 - Chemical Equilibrium57
- Ch.17 - Acids and Bases121
- Ch.18 - Aqueous Ionic Equilibrium126
- Ch.19 - Free Energy & Thermodynamics80
- Ch.20 - Electrochemistry87
- Ch.21 - Radioactivity & Nuclear Chemistry61
- Ch.22 - Organic Chemistry114
Chapter 3, Problem 105
Combustion analysis of a hydrocarbon produces 33.01 g CO2 and 13.51 g H2O. Calculate the empirical formula of the hydrocarbon.
Verified step by step guidance1
Determine the moles of carbon in CO2 by using the molar mass of CO2 (44.01 g/mol) and the fact that each mole of CO2 contains one mole of carbon.
Calculate the moles of hydrogen in H2O by using the molar mass of H2O (18.02 g/mol) and the fact that each mole of H2O contains two moles of hydrogen.
Convert the moles of carbon and hydrogen to a mole ratio by dividing each by the smallest number of moles calculated in the previous steps.
If necessary, multiply the mole ratio by a whole number to obtain the smallest whole number ratio of carbon to hydrogen.
Write the empirical formula using the whole number ratio of carbon to hydrogen.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Combustion Analysis
Combustion analysis is a technique used to determine the composition of a hydrocarbon by burning it in excess oxygen. The products of this reaction are typically carbon dioxide (CO2) and water (H2O). By measuring the masses of these products, one can infer the amounts of carbon and hydrogen in the original hydrocarbon, which is essential for calculating its empirical formula.
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Combustion of Non-Hydrocarbons
Empirical Formula
The empirical formula of a compound represents the simplest whole-number ratio of the elements present in that compound. It is derived from the moles of each element calculated from the mass of the products obtained in combustion analysis. This formula is crucial for understanding the basic composition of the hydrocarbon before determining its molecular structure.
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02:26Empirical vs Molecular Formula
Mole Concept
The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of particles it contains. One mole of any substance contains Avogadro's number (approximately 6.022 x 10^23) of entities, whether atoms, molecules, or ions. In combustion analysis, converting the mass of CO2 and H2O to moles allows for the determination of the number of carbon and hydrogen atoms in the hydrocarbon, which is necessary for calculating the empirical formula.
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Related Practice
Textbook Question
From the given molar mass and empirical formula of several compounds, find the molecular formula of each compound. a. C4H9, 114.22 g/mol
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Textbook Question
Tartaric acid is the white, powdery substance that coats tart candies such as Sour Patch Kids. Combustion analysis of a 12.01-g sample of tartaric acid—which contains only carbon, hydrogen, and oxygen—produces 14.08 g CO2 and 4.32 g H2O. Determine the empirical formula of tartaric acid.
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Textbook Question
The foul odor of rancid butter is due largely to butyric acid, a compound containing carbon, hydrogen, and oxygen. Combustion analysis of a 4.30-g sample of butyric acid produces 8.59 g CO2 and 3.52 g H2O. Determine the empirical formula of butyric acid.
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Textbook Question
A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the empirical formula of phosphorus selenide.
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Textbook Question
From the given molar mass and empirical formula of several compounds, find the molecular formula of each compound. b. CCl, 284.77 g/mol
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