Determining the Empirical and Molecular Formulas of a Carbon-Hydrogen-Oxygen Compound

When working with a compound composed of only carbon, hydrogen, and oxygen (C, H, and O), it is often necessary to determine its empirical and molecular formulas through combustion analysis. This process involves burning a known mass of the compound and measuring the mass of the resultant carbon dioxide (CO2) and water (H2O) produced. In this article, we detail the step-by-step method to find both the empirical and molecular formulas of such a compound, using the given example.

Understanding the Problem

We have a compound that undergoes complete combustion, producing CO2 and H2O. From the provided data, we know:

10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol.

The goal is to determine both the empirical and molecular formulas of the compound.

Strategy

The strategy involves several key steps:

Calculate the masses of carbon (C), hydrogen (H), and oxygen (O) in the compound using the combustion products. Convert these masses into moles for each element. Determine the empirical formula by finding the simplest whole-number ratio of moles for each element. Use the molar mass of the compound to find the molecular formula.

Calculations

Step 1: Calculate the Mass of Carbon (C) and Hydrogen (H)

First, determine the moles of CO2 produced:

[ text{Moles of CO}_2 frac{16.01 text{ mg}}{44.01 text{ g/mol}} 0.3637 text{ mmol} ]

The moles of carbon are the same as the moles of CO2, since:

[ text{Moles of C} 0.3637 text{ mmol} times 12.01 text{ g/mol} 4.37 text{ mg} ]

Next, determine the moles of H2O produced:

[ text{Moles of H}_2text{O} frac{4.37 text{ mg}}{18.02 text{ g/mol}} 0.243 text{ mmol} ]

The moles of hydrogen are twice the moles of H2O:

[ text{Moles of H} 0.243 text{ mmol} times 2 0.486 text{ mmol} times 1.01 text{ g/mol} 0.491 text{ mg} ]

Step 2: Calculate the Mass of Oxygen (O)

The mass of oxygen can be calculated by subtracting the masses of carbon and hydrogen from the total mass of the compound:

[ text{Mass of O} 10.68 text{ mg} - 4.37 text{ mg} - 0.491 text{ mg} 5.819 text{ mg} ]

Convert the mass of oxygen to moles:

[ text{Moles of O} frac{5.819 text{ mg}}{16.00 text{ g/mol}} 0.3637 text{ mmol} ]

Step 3: Determine the Empirical Formula

Divide each mole value by the smallest mole value (0.3637 mmol):

C: 0.3637 mmol / 0.3637 mmol 1 H: 0.486 mmol / 0.3637 mmol 1.33 (rounded to 1.33 for simplicity) 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

The empirical formula is CHO.

Step 4: Determine the Molecular Formula

Calculate the empirical formula mass:

[ 12.01 text{ g/mol} 1.01 text{ g/mol} 16.00 text{ g/mol} 39.02 text{ g/mol} ]

Divide the molar mass of the compound by the empirical formula mass:

[ frac{176.1 text{ g/mol}}{39.02 text{ g/mol}} 4.52 approx 4.5 ]

Multiplying the empirical formula by this factor:

[ text{CHO} times 4.52 text{C}_3text{H}_4text{O}_3 times 2 text{C}_6text{H}_8text{O}_6 ]

Therefore, the empirical formula is CHO and the molecular formula is C6H8O6.

Algebraic Method of Determining Molecular Formula

An algebraic approach can also be used to determine the molecular formula:

Let the molecular formula of the compound be CxHyOz. Using the balanced combustion equation:

CxHyOz 2left(frac{y}{2}-frac{z}{2}right)text{O}_2 rightarrow x text{CO}_2 frac{y}{2}text{H}_2text{O}

Calculate the moles of CO2 and H2O from their masses: 16.01 mg/44.01 g/mol 0.3637 mmol for CO2 4.37 mg/18.02 g/mol 0.243 mmol for H2O Mass of C, H, and O from burning 10.68 mg of the compound: 4.37 mg for C, 0.491 mg for H, 5.819 mg for O Calculate x and y by using the ratios from the combustion products: x 6, y 8 Use the molar mass equation to solve for z: 6 × 12.01 8 × 1.01 6 × 16.00 176.1 z 6 Therefore, the molecular formula is C6H8O6.

Conclusion

We have determined that the empirical formula and molecular formula of the compound are C3H4O3 and C6H8O6, respectively. Combustion analysis provides a powerful method to identify the composition of a compound from its combustion products.

Key Points

Combustion analysis is used to determine the empirical and molecular formulas of a compound. The empirical formula is the simplest whole-number ratio of the elements in the compound. The molecular formula is the empirical formula multiplied by a factor that gives the actual number of atoms of each element in the molecule. Molar masses are crucial for converting between mass and moles.