Determining the Empirical Formula of a Compound from Combustion Analysis: A Step-by-Step Guide

Understanding how to determine the empirical formula of a compound from its combustion analysis is essential for any chemistry student. This process involves a series of calculations and conversions that help deduce the simplest whole-number ratio of the constituent elements. Let's walk through the steps using the provided example and some simplified calculations for clarity.

Introduction to Combustion Analysis

Combustion analysis is a method used to determine the empirical formula of an organic compound. By burning a sample of the compound in the presence of oxygen, we can measure the products: carbon dioxide (CO?), water (H?O), and, if applicable, other gases or residues. From these measurements, we can calculate the mass of each element in the original compound and subsequently determine its empirical formula.

Example Calculation

Let's consider a pure compound containing carbon (C), hydrogen (H), and chlorine (Cl) that was combusted in a carbon-hydrogen combustion analyzer. The combustion produced 0.400g of water (H?O) and 0.889g of carbon dioxide (CO?). Given the atomic weights of C (12 g/mol), H (1 g/mol), and Cl (35.5 g/mol), we can determine the empirical formula of the compound.

Step 1: Calculate the Moles of Each Element

First, we need to convert the masses of the products into moles. Let's assume we have 290g of CO?, 270g of H?O, and 66g of Cl? for simplicity.

CO?: Molar weight 44g/mol Moles of CO? 290g / 44g/mol 6.59 moles of C (since each mole of CO? contains one mole of C) H?O: Molar weight 18g/mol Moles of H?O 270g / 18g/mol 15 moles of H (since each mole of H?O contains two moles of H) Cl?: Given 66g of Cl?, the moles of Cl would be 66g / (2 * 35.5g/mol) 0.92 moles of Cl

Step 2: Determine the Simplest Whole-Number Ratio

Now, we need to find the simplest whole-number ratio of the moles of each element. Let's use the smallest number of moles (0.92 moles of Cl) as the reference:

C: 6.59 moles / 0.92 moles 7.18 (approximately 7 if rounded) H: 15 moles / 0.92 moles 16.30 (approximately 16 if rounded) C1: 0.92 moles / 0.92 moles 1

The simplest whole-number ratio of C:H:Cl is approximately 7:16:1.

Step 3: Verify with a Simplified Approach

Let's use a more simplified approach by considering 146.99g of the original compound:

C: 5.17g / 146.99g * 100 62.07 (percent by mass) H: 10.34g / 146.99g * 100 10.34 (percent by mass) O: 146.99g - 5.17g - 10.34g 131.48g, or 89.60 (percent by mass)

Using these mass percentages, calculate the moles of each element:

C: 5.17g / 12g/mol 0.43 moles H: 10.34g / 1g/mol 10.34 moles O: 131.48g / 16g/mol 8.22 moles

Divide each by the smallest number of moles (0.43) to find the simplest whole-number ratio:

C: 0.43 moles / 0.43 1 H: 10.34 moles / 0.43 24.04 (approximately 24 if rounded) O: 8.22 moles / 0.43 19.12 (approximately 19 if rounded)

The empirical formula is C?H??O??, which needs further refinement by considering the atomic weights of the known products.

Conclusion

The empirical formula of the compound can be derived by considering the simplest whole-number ratio of the elements found in the combustion products. This process involves detailed calculations and conversions, as demonstrated above. Understanding and practicing these steps are crucial for any student of chemistry, as they form the basis of many analytical techniques used in the field.

Additional Tips

Always double-check your calculations, especially with conversions and ratios. Use a periodic table to verify atomic and molar weights. Practice with different examples to build confidence and accuracy.

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