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.

Step-by-step solution

Calculate the moles of carbon dioxide and water produced

Given that 2.20 g of carbon dioxide (CO₂) and 0.900 g of water (H₂O) are produced, we will first convert these masses into moles. Molar mass of CO₂ = 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol Molar mass of H₂O = 2 × 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol Moles of CO₂ = mass of CO₂ / molar mass of CO₂ = 2.20 g / 44.01 g/mol = 0.0500 mol Moles of H₂O = mass of H₂O / molar mass of H₂O = 0.900 g / 18.02 g/mol = 0.0500 mol

Determine the mass of carbon and hydrogen present in fructose

Knowing the moles of CO₂ and H₂O produced, we can calculate the mass of carbon (from CO₂) and hydrogen (from H₂O) in fructose. Mass of Carbon = moles of CO₂ × molar mass of C = 0.0500 mol × 12.01 g/mol = 0.600 g Mass of Hydrogen = moles of H₂O × (2 × molar mass of H) = 0.0500 mol × (2 × 1.01 g/mol) = 0.101 g (rounded)

Calculate the mass of oxygen present in fructose

The total mass of fructose is given as 1.50 g. We can determine the mass of oxygen in fructose by subtracting the masses of carbon and hydrogen from the total mass. Mass of Oxygen = 1.50 g - (0.600 g + 0.101 g) = 0.799 g (rounded)

Convert the masses of carbon, hydrogen, and oxygen into moles

Now, let's convert the masses of carbon, hydrogen, and oxygen to moles using their molar masses. Moles of Carbon = mass of Carbon / molar mass of C = 0.600 g / 12.01 g/mol = 0.0500 mol Moles of Hydrogen = mass of Hydrogen / molar mass of H = 0.101 g / 1.01 g/mol ≈ 0.100 mol Moles of Oxygen = mass of Oxygen / molar mass of O = 0.799 g / 16.00 g/mol ≈ 0.0500 mol

Find the ratio of the moles of each element and determine the empirical formula

To find the empirical formula, we will divide the moles of each element by the smallest number of moles, and then round each value to the nearest whole number. Ratio of C: 0.0500 mol / 0.0500 mol = 1 Ratio of H: 0.100 mol / 0.0500 mol = 2 Ratio of O: 0.0500 mol / 0.0500 mol = 1 So the empirical formula of fructose is CH₂O.

Key concepts

Combustion Analysis

Combustion analysis is an essential technique in chemistry used to determine the empirical formulas of organic compounds. It involves completely burning a sample and then analyzing the resultant products, typically carbon dioxide \((CO_2)\) and water \((H_2O)\). From these products, you can deduce the quantities of carbon, hydrogen, and sometimes, oxygen within the original compound. This method assumes that all the carbon in the compound converts to \(CO_2\) and all the hydrogen converts to \(H_2O\) during combustion. To utilize combustion analysis effectively, follow these steps:

• Measure the masses of \(CO_2\) and \(H_2O\) produced from the combustion.
• Use the molar masses of \(CO_2\) and \(H_2O\) to convert these masses into moles.
• Deducing the moles of carbon and hydrogen in the initial compound becomes straightforward since each mole of \(CO_2\) contains one mole of carbon atoms, and each mole of \(H_2O\) contains two moles of hydrogen atoms.
• By knowing the initial mass of the compound, you can estimate the mass of any oxygen present, completing the picture for compounds that contain only carbon, hydrogen, and oxygen.

Understanding combustion analysis aids in finding empirical formulas, which lays the groundwork for exploring a compound's molecular formula.

Stoichiometry

Stoichiometry is a foundational concept in chemistry, enabling you to calculate the amounts of substances involved in chemical reactions. It relates to the quantitative relationships found within a chemical equation. To apply stoichiometry in combustion analysis, and when determining empirical formulas, follow these simple steps:

• First, grasp the balanced chemical equation for the reaction.
• Identify the molar relationships between reactants and products within the equation. In a combustion reaction for instance, focus on how moles of the original compound relate to moles of \(CO_2\) and \(H_2O\) produced.
• Convert given masses to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
• Apply stoichiometric coefficients from the balanced equation to find the moles of unknown reactants or products.

The derived mole ratios provide a basis for understanding quantitative chemistry, essential for calculating empirical formulas. This approach is crucial for synthesizing new compounds and in analytical chemistry.

Molar Mass Calculation

Understanding and calculating molar mass is a significant step in chemistry when dealing with empirical formulas. The molar mass is the mass of one mole of a substance, expressed in grams per mole \((g/mol)\). It is calculated by summing the atomic masses of all the atoms in a molecule. This concept is used extensively in converting between the mass of a substance and the moles of that substance, as demonstrated in combustion analysis. Here's how to use molar masses effectively:

• Identify the molecular formula of the compound if available; if not, work with the empirical formula.
• Use a periodic table to find the atomic masses of the individual elements involved.
• Add the atomic masses according to the number of each type of atom to find the molar mass.
• To find moles from mass data, divide the mass of the substance by its molar mass.

Calculating molar mass allows you to connect the mass data collected during experiments to the number of molecules or formula units of a compound. This calculation is vital for accurate stoichiometric conversions and for determining empirical and molecular formulas in chemistry.