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Ch.3 - Mass Relationships in Chemical Reactions
Chapter 3, Problem 14

Combustion analysis is performed on 0.50 g of a hydrocar-bon and 1.55 g of CO2, and 0.697 g of H2O are produced. The mass spectrum for the hydrocarbon is provided below. What is the molecular formula? (LO 3.12 and 3.13)
(a) C5H11 (b) C8H18 (c) C11H10 (d) C10H22

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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 and measuring the amounts of carbon dioxide (CO2) and water (H2O) produced. The mass of CO2 produced indicates the amount of carbon in the hydrocarbon, while the mass of H2O indicates the amount of hydrogen. This data is crucial for calculating the empirical formula of the compound.
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Empirical vs. Molecular Formula

The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula indicates the actual number of atoms of each element in a molecule. To find the molecular formula from the empirical formula, one must know the molar mass of the compound, which can be determined through mass spectrometry or other methods. Understanding the distinction between these formulas is essential for accurately identifying the compound.
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Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the conservation of mass. In combustion analysis, stoichiometric relationships allow for the conversion of the masses of CO2 and H2O produced into moles of carbon and hydrogen, respectively. This quantitative approach is fundamental for determining the empirical formula and ultimately the molecular formula of the hydrocarbon.
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