Determining the Empirical and Molecular Formulas of a Compound: A Step-by-Step Guide

In analytical chemistry, determining the empirical and molecular formulas of unknown compounds is a fundamental task. This guide will walk you through the process of solving a specific problem: analyzing a sample containing only CHO that produces certain amounts of CO2 and H2O upon combustion. This article is suitable for students, scientists, and anyone involved in analytical chemistry.

Understanding the Problem

We are given a compound containing carbon (C), hydrogen (H), and oxygen (O) that, when burned, produces carbon dioxide (CO2) and water (H2O). The goal is to determine the empirical and molecular formulas of this compound. To achieve this, we will follow a series of steps involving stoichiometric calculations and the application of the molar mass.

Strategy and Calculations

Step 1: Calculate the Mass of Carbon and Hydrogen

First, we will calculate the mass of carbon (C) and hydrogen (H) in the original sample from the given masses of CO2 and H2O.

Given mass of CO2 16.01 mg

Molar mass of CO2 44.01 g/mol

Moles of CO2 16.01 mg / 44.01 g/mol 0.3637 mmol

Moles of carbon (C) in CO2 0.3637 mmol

Mass of C 0.3637 mmol × 12.01 g/mol 4.37 mg

Given mass of H2O 4.37 mg

Molar mass of H2O 18.02 g/mol

Moles of H2O 4.37 mg / 18.02 g/mol 0.243 mmol

Moles of hydrogen (H) 0.243 mmol × 2 0.486 mmol

Mass of H 0.486 mmol × 1.01 g/mol 0.491 mg

Step 2: Calculate the Mass of Oxygen

Next, we will determine the mass of oxygen (O) in the original sample by subtracting the masses of carbon and hydrogen from the total mass of the sample.

Total mass of sample 10.68 mg

Mass of O 10.68 mg - 4.37 mg - 0.491 mg 5.819 mg

Step 3: Convert Mass to Moles for Each Element

Now, convert the mass of each element to moles:

Moles of C 4.37 mg / 12.01 g/mol 0.3637 mmol

Moles of H 0.491 mg / 1.01 g/mol 0.486 mmol

Moles of O 5.819 mg / 16.00 g/mol 0.3637 mmol

Step 4: Determine the Empirical Formula

To find the empirical formula, divide each mole value by the smallest mole value:

Dividing by 0.3637 mmol:

C: 0.3637 mmol / 0.3637 mmol 1

H: 0.486 mmol / 0.3637 mmol ≈ 1.33

O: 0.3637 mmol / 0.3637 mmol 1

Multiplying by 3 to get whole numbers:

C: 1 × 3 3

H: 1.33 × 3 ≈ 4

O: 1 × 3 3

Empirical formula: CH3O

Step 5: Determine the Molecular Formula

Calculate the empirical formula mass:

CH3O 12.01 3 × 1.01 16.00 88.03 g/mol

Given molar mass 176.1 g/mol

Multiplying factor 176.1 g/mol / 88.03 g/mol ≈ 2

Molecular formula: (CH3O)2 C2H6O2

Algebraic Method

For practice, an alternate approach using an algebraic method is also provided. Here, we assume the molecular formula to be CxHyOz and derive the values of x, y, and z based on the given data.

Using the given data, the derived values for x, y, and z are as follows:

x 6, y 8, z 6

Molecular formula: C6H8O6

Empirical formula: C3H4O3

Conclusion

In conclusion, the empirical formula of the compound is CH3O, and the molecular formula is C2H6O2. This method illustrates the rigorous calculations and logical steps required to determine the chemical composition of an unknown compound through sample combustion and stoichiometric analysis.