One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of \(1.50 \mathrm{~g}\) of fructose produced \(2.20 \mathrm{~g}\) of carbon dioxide and \(0.900 \mathrm{~g}\) of water. What is the empirical formula of fructose?

Stoichiometry is the section of chemistry that deals with the quantities of substances involved in chemical reactions. In the context of determining the empirical formula of a compound like fructose, stoichiometry involves using the mole concept to understand the ratios of atoms in the compound.

To find the empirical formula, we start by analyzing the combustion of fructose to produce carbon dioxide (CO2) and water (H2O). Since CO2 provides information about the carbon content, and H2O informs us about hydrogen, we use the mass of these products to determine the moles of carbon and hydrogen atoms in the original fructose sample.

Applying Stoichiometry in Empirical Formula Calculation

Using the mass of CO2 and water given from the combustion, and knowing their molar masses, we can calculate the number of moles of each, which in turn tells us the moles of carbon and hydrogen in the fructose. As fructose only contains carbon, hydrogen, and oxygen, we calculate the mass and thus moles of oxygen by the difference in mass from the original fructose sample. The empirical formula is found when these mole ratios are simplified to the smallest whole numbers, revealing the basic repeating unit of atoms in the molecule.

Combustion analysis is a laboratory method used to determine the elemental composition of a substance by burning it and analyzing the resulting products. In this case, fructose is burned to produce CO2 and H2O, where the mass of CO2 allows us to determine the amount of carbon, and the mass of H2O allows us to determine the amount of hydrogen in the sample.

Steps Involving Combustion Analysis

The steps to analyze the combustion of fructose begin with accurately measuring the mass of CO2 and H2O produced. It's essential to have precise measurements because the calculations for moles of carbon and hydrogen are directly dependent on these values. After finding the moles of carbon and hydrogen, the mass of oxygen is deduced from the initial mass of fructose, thus allowing for the complete characterization of the compound's composition.

The molar mass calculation is crucial in converting mass to moles, which is a pivotal step in determining the empirical formula. The molar mass, defined as the mass of one mole of a substance, is used as a conversion factor between grams and moles.

Importance of Molar Mass in Analysis

To calculate moles from masses of CO2 and H2O, we must use their molar masses. For CO2, the molar mass is the sum of the mass of one carbon atom and two oxygen atoms. Similarly, for H2O, we sum the mass of two hydrogen atoms with that of one oxygen atom. After calculating moles, we can determine the mass of these elements within the original sample. The final step in deducing the empirical formula involves dividing the determined moles of carbon, hydrogen, and oxygen by the smallest numbers of moles to achieve the simplest whole number ratio. Through the precise calculation of molar masses and conversions to moles, the empirical formula of fructose, CH2O, is revealed, reflecting its most basic atomic makeup.