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

(a) The empirical formula of the hydrocarbon is C2H3. (b) The grams of O2 used in the reaction: 1.069 g. (c) The grams of O2 required for complete combustion: 1.334 g.

Step-by-step solution

Find the moles of CO, CO2, and H2O produced

To find the moles of CO, CO2, and H2O produced, we will divide the given masses by their respective molar masses: * Moles of CO: \(\frac{0.467}{28.01}\) = 0.0167 mol * Moles of CO2: \(\frac{0.733}{44.01}\) = 0.0167 mol * Moles of H2O: \(\frac{0.450}{18.02}\) = 0.0250 mol

Determine the moles of C and H in the hydrocarbon

Now, we will determine the moles of C and H in the hydrocarbon: * Moles of C: \(0.0167\,\text{mol (from CO)} + 0.0167\,\text{mol (from CO2)}\) = 0.0334 mol * Moles of H: \(2\times0.0250\,\text{mol (from H2O)}\) = 0.0500 mol

Calculate the empirical formula of the hydrocarbon

To find the empirical formula of the hydrocarbon, we will find the smallest whole number ratio between the moles of C and H: * Moles of C : Moles of H = \(\frac{0.0334}{0.0334}\) : \(\frac{0.0500}{0.0334}\) = 1 : 1.5 Since we need whole numbers, we can multiply both sides by 2: * Moles of C : Moles of H = 2 : 3 So, the empirical formula of the hydrocarbon is C2H3.

Calculate the grams of O2 used in the reaction

We will now find the grams of O2 used in the reaction. First, let's write down the balanced chemical equation for the incomplete combustion of C2H3. C2H3 + O2 -> CO + CO2 + H2O Next, we will determine the moles of O2 used in the reaction considering the moles of CO and CO2: * Moles of O2: \(0.0167\,\text{mol (for CO)} + 0.0167\,\text{mol (for CO2)}\) = 0.0334 mol Finally, we will convert the moles of O2 to grams: * Grams of O2: \(0.0334\,\text{mol}\times 32.00\,\frac{g}{\text{mol}}\) = 1.069 g Therefore, 1.069 g of O2 were used in the reaction.

Calculate the grams of O2 required for complete combustion

To calculate the grams of O2 required for complete combustion of C2H3, let's write down the balanced chemical equation for the complete combustion: C2H3 + \(\frac{5}{2}\)O2 -> 2CO2 + \(\frac{3}{2}\)H2O Now, we can calculate the moles of O2 required for complete combustion using the stoichiometry of the balanced equation: * Moles of O2 required: \(\frac{5}{2}\times 0.0167 \,\text{mol (from C2H3)}\) = 0.0417 mol Finally, we will convert the moles of O2 required to grams: * Grams of O2 required: \(0.0417\,\text{mol}\times 32.00\,\frac{g}{\text{mol}}\) = 1.334 g Therefore, 1.334 g of O2 would have been required for complete combustion of C2H3. To summarize, (a) The empirical formula of the hydrocarbon is C2H3. (b) The grams of O2 used in the reaction: 1.069 g. (c) The grams of O2 required for complete combustion: 1.334 g.