Empirical Formula of an Organic Compound from Combustion Analysis

Understanding the empirical formula of an organic compound is crucial in determining its simplest whole number ratio of atoms. This article explains the process using a detailed example, demonstrating the step-by-step method to find the empirical formula from given combustion data. The example involves an organic compound that yields carbon dioxide (CO2) and water (H2O) upon combustion, with additional nitrogen present. This information can be vital for further analytical chemistry and material science studies.

Determining the Amounts of Each Element in the Compound

The first step is to determine the amounts of carbon (C), hydrogen (H), and nitrogen (N) in the sample based on the combustion products. This is approached by calculating the moles of each element present in the combustion products: carbon dioxide (CO2) and water (H2O).

Step 1: Determine Moles of Carbon and Hydrogen

First, convert the mass of CO2 to moles of carbon (C). The molar mass of CO2 is calculated as:

(44.01 text{ g/mol} 12.01 text{ g/mol C} 2 times 16.00 text{ g/mol O})

Then calculate the moles of CO2 produced:

( text{Moles of CO}_2 frac{1.1 text{ g}}{44.01 text{ g/mol}} approx 0.0250 text{ mol} )

Since each mole of CO2 contains 1 mole of carbon, the moles of carbon are:

( text{Moles of C} 0.0250 text{ mol} )

Next, convert the mass of H2O to moles of hydrogen (H). The molar mass of H2O is:

(18.02 text{ g/mol} 2 times 1.01 text{ g/mol H} 16.00 text{ g/mol O} )

Calculate the moles of H2O produced:

( text{Moles of H}_2text{O} frac{0.3 text{ g}}{18.02 text{ g/mol}} approx 0.0166 text{ mol} )

Since each mole of H2O contains 2 moles of hydrogen, the moles of hydrogen are:

( text{Moles of H} 2 times 0.0166 text{ mol} approx 0.0332 text{ mol} )

Step 2: Calculate the Mass of Carbon and Hydrogen

Calculate the mass of carbon and hydrogen from the moles determined previously:

( text{Mass of C} 0.0250 text{ mol} times 12.01 text{ g/mol} approx 0.300 text{ g} )

( text{Mass of H} 0.0332 text{ mol} times 1.01 text{ g/mol} approx 0.0335 text{ g} )

Step 3: Calculate the Mass of Nitrogen

The total mass of the compound is 0.45 g. To find the mass of nitrogen, subtract the combined mass of carbon and hydrogen from the total mass:

( text{Mass of N} 0.45 text{ g} - 0.300 text{ g} - 0.0335 text{ g} approx 0.1165 text{ g} )

The molar mass of nitrogen (N) is approximately 14.01 g/mol. Calculate the moles of nitrogen:

( text{Moles of N} frac{0.1165 text{ g}}{14.01 text{ g/mol}} approx 0.00831 text{ mol} )

Step 4: Determine the Mole Ratio for the Empirical Formula

Now, we have the moles of carbon (C), hydrogen (H), and nitrogen (N):

Moles of C: 0.0250 Moles of H: 0.0332 Moles of N: 0.00831

To find the simplest whole number ratio, divide each by the smallest number of moles (0.00831 mol):

C: (frac{0.0250}{0.00831} approx 3.01 approx 3) H: (frac{0.0332}{0.00831} approx 4.00 approx 4) N: (frac{0.00831}{0.00831} 1)

The mole ratio of C:H:N is 3:4:1, resulting in the empirical formula:

( text{C}_3text{H}_4text{N} )

Conclusion

The empirical formula of the compound is (text{C}_3text{H}_4text{N}). This process can be applied to various organic compounds where the amounts of carbon, hydrogen, and nitrogen can be derived from their combustion products. Understanding the empirical formula is essential for further chemical analysis and material characterization.