Question

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?

Short answer

The empirical formula of fructose is CH₂O, derived from the ratio of moles of Carbon, Hydrogen, and Oxygen atoms (1:2:1) in the simplest whole number form.

Step-by-step solution

Find moles of carbon (C)

The complete combustion of fructose produces carbon dioxide (CO₂), which contains one carbon atom. To find the moles of carbon, we need to know the mass and molar mass of carbon dioxide produced. The mass of carbon dioxide produced is 2.20 g, and the molar mass of carbon dioxide is 44.01 g/mol (12.01 g/mol for carbon and 16.00 g/mol each for two oxygen atoms). Now, we can calculate the moles of Carbon: moles of CO₂ = \( \frac{2.20 g}{44.01 g/mol} = 0.0500 \ mol \) Since there is only one carbon atom per molecule of CO₂, the number of moles of carbon is 0.0500 mol.

Find moles of hydrogen (H)

The complete combustion of fructose produces water (H₂O), which contains two hydrogen atoms. To find the moles of hydrogen, we need to know the mass and molar mass of water produced. The mass of water produced is 0.900 g, and the molar mass of water is 18.02 g/mol (1.01 g/mol each for two hydrogen atoms and 16.00 g/mol for oxygen atoms). Now, we can calculate the moles of Hydrogen: moles of H₂O = \( \frac{0.900 g}{18.02 g/mol} = 0.050 \ mol \) Since there are two hydrogen atoms per molecule of H₂O, the number of moles of hydrogen is moles of H = 0.0500 mol × 2 = 0.100 mol

Calculate mass of oxygen (O)

The mass of fructose is given as 1.50 g. We've calculated the mass of carbon and hydrogen, so now we can find the mass of oxygen: mass of Oxygen = mass of Fructose - mass of Carbon - mass of Hydrogen mass of Oxygen = 1.50 g - (0.0500 mol × 12.01 g/mol) - (0.100 mol × 1.01 g/mol) mass of Oxygen = 1.50 g - 0.600 g - 0.101 g = 0.799 g

Find moles of oxygen (O)

Now, we can calculate the moles of oxygen by dividing the mass of oxygen by its molar mass (16.00 g/mol): moles of O = \( \frac{0.799 g}{16.00 g/mol} = 0.050 \ mol \)

Find the empirical formula

To find the empirical formula, we need to determine the simplest whole number ratio of moles of Carbon, Hydrogen, and Oxygen atoms. moles of C : moles of H : moles of O = 0.0500 : 0.100 : 0.0500 Divide all the mole ratios by the smallest mole number (0.0500 mol) to determine the simplest whole number ratio. C : H : O = 1 : 2 : 1 The empirical formula of fructose is CH₂O.

Key concepts

Combustion Reaction

A combustion reaction is a chemical process where a substance combines with oxygen to release energy in the form of heat and light. In the case of fructose, a combustion reaction involves burning fructose in the presence of oxygen to produce carbon dioxide

• (CO₂)
• water (H₂O)

This type of reaction is crucial in determining the empirical formula of substances like fructose by analyzing the products of the combustion.
By examining the amount and types of substances produced from a complete combustion reaction, we can gain insights into the composition of the original compound. Such reactions are frequently used in the laboratory to deduce the elemental makeup of organic compounds.

Molar Mass

Molar mass is defined as the mass of one mole of a substance, often expressed in grams per mole (g/mol). It provides a bridge between the mass of a substance and the number of moles, which is essential when converting between mass and moles in calculations. To determine the molar mass:

• Identify each element in the compound.
• Multiply the atomic mass of each element (found on the periodic table) by the number of times the element appears in the compound.
• Sum these values to find the total molar mass of the compound.

For example, in carbon dioxide (CO₂), the molar mass is calculated by adding the molar mass of one carbon atom (12.01 g/mol) with the molar masses of two oxygen atoms (16.00 g/mol each), resulting in 44.01 g/mol for CO₂.
Understanding molar masses is vital for chemists when calculating the quantities of reactants and products involved in a chemical reaction.

Chemical Compounds

Chemical compounds are substances formed from two or more elements that are chemically bonded together in fixed proportions. These compounds can be identified by their chemical formulas, which denote the types and numbers of atoms present.
In organic chemistry, common compounds include hydrocarbons and carbohydrates such as fructose. Chemical compounds such as fructose can be analyzed to determine their empirical formulas using combustion data.

• The empirical formula represents the simplest whole-number ratio of the elements in a compound.
• Understanding the composition of chemical compounds is essential for applications ranging from material synthesis to pharmaceutical development.

By knowing the chemical composition, scientists can predict the properties and potential reactions of these substances.

Mole Calculation

Mole calculations allow chemists to relate masses of substances with the number of particles (atoms or molecules) involved in a chemical reaction. The concept of the mole serves as a handy conversion factor, equal to Avogadro's number (

• \(6.022 imes 10^{23}\).

It is used in translating between the mass of a substance and the number of moles.
These calculations involve:

• Determining the number of moles from mass using the formula:
• \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)

• Applying stoichiometric factors to relate the moles of one substance with another.

Mole calculations are foundational in laboratory practices for quantifying reactants and products, ensuring reactions proceed with the correct proportions, and analyzing experimental data.

Fructose

Fructose is a simple sugar that is central to many biological processes. As a carbohydrate, it serves as an energy source and is primarily found in fruits. Its chemical formula is often written as C₆H₁₂O₆, indicating its composition of carbon, hydrogen, and oxygen.
To determine its empirical formula—which gives the simplest whole-number ratio of its elements—scientists perform combustion reactions and analyze the resulting products.

• The complete combustion of fructose produces carbon dioxide and water, which are used to calculate the amount of carbon, hydrogen, and oxygen originally present in fructose.
• The empirical formula derived aligns with its general carbohydrate structure—providing insights into its base composition.

This analysis helps underline the fundamental building blocks of more complex carbohydrates and illuminates fructose's role in both chemistry and biology.