Question

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Short answer

The empirical formula of the hydrocarbon compound is \(\mathrm{C}_{2}\mathrm{H}_{3}\). In the incomplete combustion reaction, 1.336 g of \(\mathrm{O}_{2}\) were used. For complete combustion, 3.741 g of \(\mathrm{O}_{2}\) would have been required.

Step-by-step solution

Finding moles of CO, CO2 and H2O

Use the molar mass of \(\mathrm{CO}\), \(\mathrm{CO}_{2}\), and \(\mathrm{H}_{2}\mathrm{O}\) to find moles: Moles of \(\mathrm{CO} = \frac{0.467 g}{28.01 g/mol} = 0.0167 \, mol\) Moles of \(\mathrm{CO}_{2} = \frac{0.733 g}{44.01 g/mol} = 0.0167 \, mol\) Moles of \(\mathrm{H}_{2}\mathrm{O} = \frac{0.450 g}{18.02 g/mol} = 0.0250 \, mol\)

Writing down the balanced equation for the combustion of the hydrocarbon

Let's assume the hydrocarbon to be \(\mathrm{C}_{x}\mathrm{H}_{y}\): I. Incomplete combustion: \(\mathrm{C}_{x}\mathrm{H}_{y} + \alpha \mathrm{O}_{2} \rightarrow 0.0167 \, mol\,\mathrm{CO} + 0.0167 \, mol\,\mathrm{CO}_{2} + 0.0250 \, mol\,\mathrm{H}_{2}\mathrm{O}\)

Converting moles into atoms of Carbon and Hydrogen

From the moles of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\): Carbon atoms = \(0.0167 \, mol\,\mathrm{CO} + 0.0167 \, mol\,\mathrm{CO}_{2} = 0.0334 \, mol\,\mathrm{C}\) From the moles of \(\mathrm{H}_{2}\mathrm{O}\): Hydrogen atoms = \(2 \times 0.0250 \, mol\,\mathrm{H}_{2}\mathrm{O} = 0.0500 \, mol\,\mathrm{H}\)

Determine the empirical formula

We have: 0.0334 mol of Carbon 0.0500 mol of Hydrogen Divide by the lowest mole quantity: \(\frac{0.0334}{0.0334} = 1 \; (Carbon)\) \(\frac{0.0500}{0.0334} = 1.5 \; (Hydrogen)\) Since we need integer values for the empirical formula, let's multiply by 2: \(1 \times 2 = 2 \; (Carbon)\) \(1.5 \times 2 = 3 \; (Hydrogen)\) So, the empirical formula is \(\mathrm{C}_{2}\mathrm{H}_{3}\). (b) Calculate the grams of \(\mathrm{O}_{2}\) used in the reaction:

Finding moles of O2 used

From the balanced incomplete combustion equation: \(\mathrm{C}_{2}\mathrm{H}_{3} + \alpha \mathrm{O}_{2} \rightarrow 0.0167 \, mol\,\mathrm{CO} + 0.0167 \, mol\,\mathrm{CO}_{2} + 0.0250 \, mol\,\mathrm{H}_{2}\mathrm{O}\) For every mole of \(\mathrm{CO}_{2}\) formed, 1 mole of \(\mathrm{CO}\) and 0.5 mole of \(\mathrm{H}_{2}\mathrm{O}\) are produced. This means that 1 mole of hydrocarbon (\(\mathrm{C}_{2}\mathrm{H}_{3}\)) reacts with 2.5 moles of \(\mathrm{O}_{2}\) to produce 0.0334 moles of Carbon and 0.0250 moles of Hydrogen. Therefore, \(\alpha = 2.5\)

Calculate the grams of O2

Moles of \(\mathrm{O}_{2}\) used: \(2.5 \times 0.0167 = 0.04175 \, mol\) Grams of \(\mathrm{O}_{2}\) used: \(0.04175 \, mol \times 32.00 g/mol = 1.336 g\) (c) Calculate the grams of \(\mathrm{O}_{2}\) required for complete combustion:

Write the balanced equation for complete combustion

The complete combustion equation: \(\mathrm{C}_{2}\mathrm{H}_{3} + 3.5 \mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + \dfrac32 \mathrm{H}_{2}\mathrm{O}\)

Find moles of O2 needed for complete combustion

For complete combustion, 0.0334 moles of hydrocarbon require 3.5 moles of \(\mathrm{O}_{2}\).

Calculate the grams of O2 required

Moles of \(\mathrm{O}_{2}\) required: \(0.0334 \, mol \times 3.5 = 0.1169 \, mol\) Grams of \(\mathrm{O}_{2}\) required: \(0.1169 \, mol \times 32.00 g/mol = 3.741 g\)