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Determining an Empirical Formula by Combustion Analysis

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 1PE Chapter 3.15SE

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 1PE

Determining an Empirical Formula by Combustion Analysis Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of C02 and 0.306 g of H20. Determine the empirical formula of isopropyl alcohol. The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a 2.203-g sample of this compound produces 4.401 g CO2 and 1.802 g H2O. A separate experiment shows that it has a molar mass of 88.1 g/mol. Which of the following is the correct molecular formula for dioxane? (a) C2H4O (b) C4H4O2 (c) CH2 (d) C4H8O2.

Step-by-Step Solution:

Answer: Step 1 Combustion analysis can only determine the empirical formula of a compound; it cannot determine the molecular formula. However, other techniques can determine the molecular weight. Once we know this value, coupled with the empirical formulas, we can easily calculate what the molecular formula is. Empirical formula for a compound is the simplest integral ratio of all the number of atoms in a compound. For example the molecular formula of glucose is C6H12O6 but its empirical formula is CH2O . Thus, the empirical formula weight of this compound is the sum of atomic masses of one carbon atom, two hydrogen atoms and one oxygen atom.

Step 2 of 6

Chapter 3.15SE, Problem 1PE is Solved
Step 3 of 6

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Determining an Empirical Formula by Combustion Analysis