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 is C₂H₃. The mass of O₂ consumed in the reaction is 1.20 g, and the mass of O₂ required for complete combustion is 1.34 g.

Step-by-step solution

1. Convert masses to moles

To determine the empirical formula, we'll first need to convert grams of CO, CO₂, H₂O, and the hydrocarbon to moles by using their respective molar masses: \(1\ mole\ CO = 28.01\ g\), \(1\ mole\ CO_2 = 44.01\ g\), and \(1\ mole\ H_2O = 18.02\ g\). Moles of CO = \(0.467\ g / 28.01\ g/mol = 0.0167\ mol\) Moles of CO₂ = \(0.733\ g / 44.01\ g/mol = 0.0167\ mol\) Moles of H₂O = \(0.450\ g / 18.02\ g/mol = 0.0250\ mol\)

2. Calculate the moles of C, H, and O in the compound

Now we will calculate the moles of carbon, hydrogen, and oxygen separately in the unknown hydrocarbon. Moles of carbon from CO = \(0.0167\ mol\) Moles of carbon from CO₂ = \(0.0167\ mol\) Total moles of carbon in the hydrocarbon = \(0.0167\ mol + 0.0167\ mol = 0.0334\ mol\) Moles of hydrogen in the hydrocarbon = \(0.0250\ mol * 2 = 0.0500\ mol\), as the combustion forms one water molecule per hydrogen atom. We already have the moles of carbon and hydrogen, so the moles of oxygen can be found by subtracting the moles of carbon and hydrogen from the total moles of the hydrocarbon.

3. Determine the hydrocarbon's empirical formula

Now, we should divide each element's moles by the smallest molar value, which is \(0.0334\ mol\), to determine the empirical formula of the hydrocarbon: Carbon: \(0.0334\ mol\;/\;0.0334\ mol = 1\) Hydrogen: \(0.0500\ mol\;/\;0.0334\ mol = 1.5\) To get whole numbers for each element, we can multiply both by 2: Carbon: \(1 * 2 = 2\) Hydrogen: \(1.5 * 2 = 3\) Therefore, the empirical formula of the hydrocarbon is C₂H₃.

4. Calculate the mass of O₂ consumed in the reaction

For this, we'll be using the stoichiometry and the balanced chemical equation: \[C_x H_y + O_2 \rightarrow aCO + bCO_2 + cH_2 O\] In our case, \(x=2\) and \(y=3\). Then, using the moles of each of the products, we can figure out the moles of O₂ consumed: Moles of O₂ (from CO) = \(\frac{1}{2}(0.0167\ mol) = 0.00833\ mol\) Moles of O₂ (from CO₂) = \(0.0167\ mol\) Moles of O₂ (from H₂O) = \(\frac{1}{2}(0.0250\ mol) = 0.0125\ mol\) Total moles of O₂ consumed = \(0.00833\ mol + 0.0167\ mol + 0.0125\ mol = 0.0375\ mol\) We can now convert the moles of O₂ consumed to its corresponding mass: Mass of O₂ consumed = \(0.0375\ mol * 32.00\ g/mol = 1.20\ g\)

5. Calculate the mass of O₂ required for complete combustion

For complete combustion, the stoichiometry should not have the formation of CO. Thus, the balanced combustion reaction becomes: \[C_{2}H_{3} + \frac{5}{2}O_{2} \rightarrow 2CO_{2} + \frac{3}{2}H_{2}O\] Moles of O₂ required for complete combustion = \(\frac{5}{2}(0.0167\ mol) = 0.0418\ mol\) Now, convert the moles of O₂ required to its corresponding mass: Mass of O₂ required = \(0.0418\ mol * 32.00\ g/mol = 1.34\ g\) To summarize: a) The empirical formula of the hydrocarbon is C₂H₃. b) The mass of O₂ consumed in the reaction is 1.20 g. c) The mass of O₂ required for complete combustion is 1.34 g.

Key concepts

Combustion of Hydrocarbons

Combustion of hydrocarbons is a chemical process where hydrocarbons, compounds consisting only of hydrogen and carbon, react with oxygen to form carbon dioxide, carbon monoxide, and water. This reaction releases energy in the form of heat and is a significant process in many applications such as energy generation and engines.

When these substances combust, especially in limited oxygen conditions, incomplete combustion occurs, often producing both carbon monoxide (CO) and carbon dioxide (CO₂). Complete combustion, on the other hand, typically results only in CO₂ and H₂O.

• **Incomplete Combustion:** Occurs when there's not enough oxygen to allow the fuel (hydrocarbon) to burn completely, forming CO and possibly soot.
• **Complete Combustion:** Occurs when sufficient oxygen is present, allowing the fuel to burn fully, resulting in CO₂ and H₂O only.

Understanding this process is crucial as it informs us about the efficiency of fuel usage and environmental impacts like emissions.

Stoichiometry

Stoichiometry is a key principle in chemistry that involves quantitative relationships in chemical reactions. It allows chemists to calculate the precise amounts of reactants and products involved in a chemical reaction. By utilizing the balanced chemical equation, stoichiometry provides the proportions of each reactant and product.

In our combustion scenario, stoichiometry helps determine how much oxygen is necessary to completely burn a given amount of a hydrocarbon. The amount of CO, CO₂, and H₂O gathered from an experiment helps in establishing relationships using molar mass. For instance, by converting grams to moles using their respective molar masses, we can further understand how much of each element participates in the reaction.

• **Reactant-to-Product Ratios:** Shows how reactants change into products.
• **Mole Conversions:** Used to interchange between mass and moles using molar mass.

Thus, stoichiometry serves as a bridge in translating between the macroscopic quantities we measure in lab experiments to the sub-microscopic scales chemists work with.

Chemical Reactions

Chemical reactions occur when substances undergo chemical changes to form new substances. Reactants start the reaction, and they transform to produce products. In the context of hydrocarbon combustion, the chemical reactions can be understood as the reordering of atoms where hydrocarbons and oxygen react to form other substances like CO₂, CO, and H₂O.

For combustion, the general chemical equation can be represented as: \[ C_xH_y + O_2 \rightarrow zCO_2 + wH_2O + ext{(possibly)} \ qCO \]This equation summarizes the transformation of hydrocarbons and oxygen into combustion products. Chemical reactions occur based on specific principles:

• **Conservation of Mass:** Mass is neither created nor destroyed in a chemical reaction.
• **Reactivity:** The tendency of a substance to undergo chemical reactions.

Understanding chemical reactions is fundamental to mastering chemistry as it involves predicting the products and quantities that result from given reactants.