Calculate The Empirical Formula For Naphthalene

Calculating the Empirical Formula for Naphthalene: A Step-by-Step Guide

Naphthalene, a common aromatic hydrocarbon found in mothballs, provides an excellent example for understanding empirical formula calculations. The empirical formula represents the simplest whole-number ratio of atoms in a compound. While the molecular formula shows the actual number of atoms, the empirical formula provides a fundamental understanding of the constituent elements' proportions. This article will guide you through a detailed calculation of naphthalene's empirical formula, explaining each step and incorporating relevant organic chemistry concepts.

Understanding Empirical and Molecular Formulas

Before diving into the calculation, it's crucial to grasp the difference between empirical and molecular formulas.

• Empirical Formula: The simplest whole-number ratio of atoms in a compound. For example, the empirical formula for glucose (C₆H₁₂O₆) is CH₂O.

• Molecular Formula: Shows the actual number of atoms of each element present in a molecule. Glucose's molecular formula is C₆H₁₂O₆.

Determining the Empirical Formula of Naphthalene: The Combustion Analysis Method

The most common method for determining the empirical formula of organic compounds like naphthalene is combustion analysis. This technique involves burning a known mass of the compound in excess oxygen. The products of the combustion reaction – carbon dioxide (CO₂) and water (H₂O) – are then carefully collected and weighed. From the masses of CO₂ and H₂O, we can determine the mass of carbon and hydrogen in the original sample of naphthalene.

Let's assume, for the purpose of this example, that the combustion analysis of a 1.000 g sample of naphthalene yielded 3.26 g of CO₂ and 0.68 g of H₂O.

Step-by-Step Calculation:

Step 1: Calculate the Moles of Carbon Dioxide (CO₂) and Water (H₂O)

First, we need to convert the masses of CO₂ and H₂O to moles using their respective molar masses.

• 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: (3.26 g CO₂) / (44.01 g/mol) = 0.0741 mol CO₂

• Moles of H₂O: (0.68 g H₂O) / (18.02 g/mol) = 0.0377 mol H₂O

Step 2: Calculate the Moles of Carbon (C) and Hydrogen (H)

Since each mole of CO₂ contains one mole of carbon, the number of moles of carbon in the naphthalene sample is equal to the number of moles of CO₂.

• Moles of C: 0.0741 mol CO₂ = 0.0741 mol C

Similarly, each mole of H₂O contains two moles of hydrogen.

• Moles of H: 0.0377 mol H₂O * 2 mol H/mol H₂O = 0.0754 mol H

Step 3: Determine the Mole Ratio of Carbon (C) to Hydrogen (H)

To find the simplest whole-number ratio, divide the number of moles of each element by the smallest number of moles.

• Mole ratio of C: 0.0741 mol C / 0.0741 mol = 1.00
• Mole ratio of H: 0.0754 mol H / 0.0741 mol = 1.02

The ratio is approximately 1:1. However, we need whole numbers for the empirical formula. The slight discrepancy (1.02) is likely due to experimental error. We can round this to the nearest whole number, giving a ratio of 1:1.

Step 4: Write the Empirical Formula

Based on the mole ratio, the empirical formula for naphthalene is CH.

Refining the Empirical Formula: Considering the Molecular Weight

The empirical formula (CH) provides the simplest whole-number ratio. However, this isn't the true molecular formula of naphthalene. To find the molecular formula, we need additional information, specifically the molecular weight (molar mass) of naphthalene.

Let's assume the molar mass of naphthalene is experimentally determined to be approximately 128 g/mol.

Step 5: Determine the Molecular Formula

1. Calculate the empirical formula weight: The empirical formula weight of CH is 12.01 g/mol (C) + 1.01 g/mol (H) = 13.02 g/mol.

2. Determine the ratio between the molecular weight and the empirical formula weight: 128 g/mol (molecular weight) / 13.02 g/mol (empirical formula weight) ≈ 9.83

3. Round the ratio to the nearest whole number: This ratio is approximately 10.

4. Multiply the subscripts in the empirical formula by the whole-number ratio: The empirical formula CH multiplied by 10 gives us C₁₀H₁₀. This is the molecular formula of naphthalene.

Understanding the Discrepancy and Error Analysis

In real-world combustion analysis, slight discrepancies in the results are expected due to experimental errors. These errors can arise from:

• Incomplete combustion: Some of the sample might not completely burn, leading to lower than expected CO₂ and H₂O yields.
• Leakage in the apparatus: Loss of CO₂ or H₂O during the collection process can also skew the results.
• Measurement errors: Inaccuracies in weighing the sample and products can introduce errors.

Therefore, slight adjustments and rounding are often necessary when determining the empirical formula from experimental data.

Applications of Empirical Formula Calculations

Understanding how to calculate empirical formulas is crucial in many areas of chemistry, including:

• Qualitative and quantitative analysis: Determining the composition of unknown compounds.
• Synthesis of new compounds: Predicting the stoichiometry of chemical reactions.
• Material science: Characterizing the composition of materials.

Conclusion

Calculating the empirical formula of naphthalene, or any organic compound, using combustion analysis requires careful attention to detail and understanding of stoichiometry. The process involves several steps, from converting masses to moles, determining mole ratios, and ultimately obtaining the empirical formula. While the initial calculation may yield a slightly imperfect ratio, understanding experimental error and making reasonable adjustments are crucial for deriving a meaningful and accurate result. This process, combined with the knowledge of the molecular weight, allows the determination of the correct molecular formula. This detailed explanation aims to equip you with the necessary knowledge and confidence to perform similar calculations for various compounds. Remember, practice is key to mastering this essential skill in chemistry.