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Chapter 3: Problem 102

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

Expert verified
(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

01

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
02

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
03

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.
04

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.
05

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.

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Most popular questions from this chapter

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and \(1.00\) \(\mathrm{mol} \mathrm{} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2}\), molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), molar mass \(=88 \mathrm{~g} / \mathrm{mol}\)

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), a component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.50 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(15.0 \mathrm{gal}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) (the capacity of an average fuel tank)? (d) How many grams of \(\mathrm{CO}_{2}\) are produced when \(15.0 \mathrm{gal}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) are combusted?

(a) What is the mass, in grams, of \(1.223\) mol of iron (III) sulfate? (b) How many moles of ammonium ions are in \(6.955 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(1.50 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) ? (d) What is the molar mass of diazepam (Valium \()\) if \(0.05570 \mathrm{~mol}\) has a mass of \(15.86 \mathrm{~g}\) ?

Calculate the percentage by mass of oxygen in the following compounds: (a) morphine, \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\); (b) \(\mathrm{co}\) deine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\) (c) cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\); (d) tetracycline, \(\mathrm{C}_{22} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{8} ;\) (e) digitoxin, \(\mathrm{C}_{41} \mathrm{H}_{64} \mathrm{O}_{13}\); (f) vancomycin, \(\mathrm{C}_{66} \mathrm{H}_{75} \mathrm{Cl}_{2} \mathrm{~N}_{9} \mathrm{O}_{24}\)
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