Problem 88 - Chemistry: The Central Science

Question

Vanillin, the dominant flavoring in vanilla, contains \(\mathrm{C}\), \(\mathrm{H}\), and \(\mathrm{O} .\) When \(1.05 \mathrm{~g}\) of this substance is completely combusted, \(2.43 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.50 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

Step-by-Step Solution

Answer

The empirical formula of vanillin is C₃H₃O based on the mole ratio 3:3:1 determined from the combustion data.

Step 1: Calculate the moles of CO₂ and H₂O formed after combustion

To find the moles, we will use the formula: moles = mass / molar mass Molar mass of CO₂ = 12.01 (C) + 2 × 16.00 (O) = 44.01 g/mol Molar mass of H₂O = 2 × 1.01 (H) + 16.00 (O) = 18.02 g/mol Moles of CO₂ = (2.43 g) / (44.01 g/mol) = 0.0552 mol Moles of H₂O = (0.50 g) / (18.02 g/mol) = 0.0277 mol

Step 2: Relate the moles of CO₂ and H₂O to moles of C and H, respectively, in vanillin

One mole of CO₂ contains one mole of C, and one mole of H₂O contains two moles of H. Moles of C = 0.0552 mol (from CO₂) Moles of H = 2 × 0.0277 mol (from H₂O) = 0.0554 mol

Step 3: Calculate the mass of O in vanillin, and then calculate its moles

First, find the mass of C and H in vanillin. Mass of C = (0.0552 mol) × (12.01 g/mol) = 0.663 g Mass of H = (0.0554 mol) × (1.01 g/mol) = 0.056 g Mass of O in vanillin = Initial mass of vanillin - Mass of C - Mass of H Mass of O = 1.05 g - 0.663 g - 0.056 g = 0.331 g Moles of O = Mass of O / Molar mass of O Moles of O = 0.331 g / 16.00 g/mol = 0.0207 mol

Step 4: Calculate the mole ratio of C, H, and O in the compound

Divide the moles of each element by the smallest number of moles to get the mole ratio. C:H:O = 0.0552 : 0.0554 : 0.0207 Find the smallest number of moles Min. moles = 0.0207 Mole ratio = (0.0552/0.0207) : (0.0554/0.0207) : (0.0207/0.0207) = 2.67 : 2.68 : 1 Due to experimental errors, we consider all of these whole numbers. So, we have: Mole ratio = 3 : 3 : 1

Step 5: Determine the empirical formula of vanillin based on the mole ratio obtained

With the mole ratio obtained, we can write the empirical formula for vanillin as C₃H₃O.

Key Concepts

Combustion Analysis

Combustion analysis is an important method used to determine the empirical formula of a compound. This technique involves burning a sample completely in the presence of oxygen, which results in the formation of carbon dioxide (CO₂) and water (H₂O). By measuring the masses of these products, the amounts of carbon and hydrogen in the original compound can be determined. This process is particularly useful for compounds containing carbon, hydrogen, and oxygen.

In the case of vanillin, burning 1.05 g results in 2.43 g of CO₂ and 0.50 g of H₂O, allowing us to ascertain the amount of carbon and hydrogen present. Through calculations, we find the moles of CO₂ and H₂O, which correspond to the moles of carbon and hydrogen originally present in the vanillin. This initial step is crucial for determining the elemental composition of the compound from which the empirical formula is derived.

Molar Mass Calculation

Calculating molar mass is essential when working on combustion analysis and determining empirical formulas. The molar mass is the mass of one mole of a substance, and it serves as a bridge between the mass of a substance and the number of moles.

For carbon dioxide (CO₂), the molar mass is calculated by adding the atomic mass of carbon (12.01 g/mol) and twice the atomic mass of oxygen (16.00 g/mol), resulting in a molar mass of 44.01 g/mol. The molar mass of water (H₂O) is calculated similarly; it involves adding twice the atomic mass of hydrogen (1.01 g/mol) and the atomic mass of oxygen (16.00 g/mol), resulting in a molar mass of 18.02 g/mol.

By using these molar masses, we can determine the number of moles of CO₂ and H₂O produced in the combustion of vanillin, crucial for further calculations in finding the empirical formula.

Elemental Composition

Understanding elemental composition is key to finding the empirical formula of a compound. The elemental composition refers to the amount (by mass or moles) of each element present in a compound. For vanillin, this involves determining how much carbon, hydrogen, and oxygen it contains.

Once we have the moles of carbon and hydrogen, we also need the moles of oxygen. The oxygen is not directly measured from combustion products like carbon and hydrogen. Instead, it's determined by subtracting the mass of carbon and hydrogen from the initial mass of vanillin, giving us the mass of oxygen. We then convert this mass of oxygen to moles by dividing by the atomic mass of oxygen.

This collected data provides us all the necessary values to calculate the elemental composition, which ultimately helps us in deriving the empirical formula of vanillin.

Chemical Stoichiometry

Chemical stoichiometry involves understanding and using relationships between amounts of reactants and products in chemical reactions. This concept is fundamental when transitioning from experimental data to empirical formulas.

In our vanillin example, stoichiometry allows us to use the calculated moles of carbon, hydrogen, and oxygen to form mole ratios. These mole ratios highlight the simplest whole number relationships between the atoms in a compound.

After determining these relationships from combustion data, it's necessary to find the lowest number of moles and divide the moles of each element by this minimum number. This calculation results in the empirical formula, the simplest representation of the compound, depicted as C₃H₃O for vanillin in our case.

Understanding chemical stoichiometry thus not only simplifies complex chemical information but also helps us derive critical information about the composition of compounds.