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?

Using the molar masses of CO (\(28.01\mathrm{~g/mol}\)), CO2 (\(44.01\mathrm{~g/mol}\)), and H2O (\(18.02\mathrm{~g/mol}\)), let's calculate the amount of moles of each compound formed. moles of CO: \(\dfrac{0.467\mathrm{~g}}{28.01\mathrm{~g/mol}} = 0.0167\mathrm{~mol}\) moles of CO2: \(\dfrac{0.733\mathrm{~g}}{44.01\mathrm{~g/mol}} = 0.0167\mathrm{~mol}\) moles of H2O: \(\dfrac{0.450\mathrm{~g}}{18.02\mathrm{~g/mol}} = 0.025\mathrm{~mol}\)

There are two sources of carbon in the products: CO and CO2. Both of the carbon moles in CO and CO2 come from the hydrocarbon. Carbon: \(0.0167+0.0167=0.0334\mathrm{~mol}\) The moles of hydrogen in the hydrocarbon come from the water produced: Hydrogen: \(2\times0.025=0.050\mathrm{~mol}\) Oxygen: Total oxygen used in the reaction - Oxygen used to form CO and CO2 Total oxygen used in the reaction comes from the moles of CO and CO2 in the products and their oxygen content: \(0.0167+2\times0.0167=0.05\mathrm{~mol}\) So the empirical formula is C\(_{0.0334}\)H\(_{0.050}\)O\(_{0.0167}\). To obtain the simplest whole-number ratio of elements in the compound, we will divide each subscript by the smallest subscript: C: \(\dfrac{0.0334}{0.0167} = 2\) H: \(\dfrac{0.050}{0.0167} = 3\) O: \(\dfrac{0.0167}{0.0167} = 1\) The empirical formula of the hydrocarbon is C2H3O. (a) Empirical formula: C2H3O

To find the grams of O2 used, lets first find the moles of O2: moles of O2: \(0.05\mathrm{~mol}\) (from the previous calculations) Now, using the molar mass of O2 (\(32.00\mathrm{~g/mol}\)), we can find the grams of O2 used: grams of O2: \(0.05\mathrm{~mol} \times 32.00\mathrm{~g/mol} = 1.6\mathrm{~g}\) (b) Grams of O2 used in the reaction: 1.6 g

For complete combustion, the hydrocarbon would react with oxygen to produce only CO2 and H2O. The equation for the complete combustion would be: C2H3O + \(x\)O2 \(\rightarrow\) 2CO2 + \(1.5\)H2O To find the value of x, we'll balance the oxygen atoms: 2x = 4 + 3 x = 3.5 The balanced equation for complete combustion would be: C2H3O + 3.5O2 \(\rightarrow\) 2CO2 + \(1.5\)H2O Now, we can calculate the grams of O2 required for complete combustion: moles of O2: \(3.5 \times 0.0167\mathrm{~mol} = 0.05845\mathrm{~mol}\) grams of O2: \(0.05845\mathrm{~mol} \times 32.00\mathrm{~g/mol} = 1.87\mathrm{~g}\) (c) Grams of O2 required for complete combustion: 1.87 g