Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

\(500 \mathrm{~mL}\) of a hydrocarbon gas burnt in excess of oxygen, gave \(2500 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(3000 \mathrm{~mL}\) of water vapour (at constant \(\mathrm{T}\) and \(\mathrm{P}\) ). The formula of the hydrocarbon is (a) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{12}\) (d) \(\mathrm{CH}_{4}\)

Step-by-Step Solution

Answer

The formula of the hydrocarbon is \(\mathrm{C}_{5} \mathrm{H}_{12}\).

Step 1: Understanding the Reaction

When a hydrocarbon with formula \(C_xH_y\) combusts, it reacts with oxygen to produce carbon dioxide \((CO_2)\) and water \((H_2O)\). The balanced equation looks like this: \(C_xH_y + O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). We need to use the volumes given to find \(x\) and \(y\).

Step 2: Analyzing the Carbon Dioxide Production

According to the problem, \(2500\, mL\) of \(CO_2\) is produced. Since the stoichiometry of the reaction is 1:1 between moles of hydrocarbon carbon atoms and \(CO_2\) molecules, there are \(x = 2500/500 = 5\) carbon atoms in the formula (since the hydrocarbon burnt is originally \(500\, mL\)). Therefore, \(x = 5\).

Step 3: Analyzing the Water Production

The problem states that \(3000\, mL\) of \(H_2O\) vapor is produced. Because each molecule of \(H_2O\) contains two hydrogen atoms, and using the stoichiometry given, the ratio for water shows us that there are \(y/2 = 3000/500 = 6\). Solving \(y/2 = 6\) gives \(y = 12\). Thus, there are 12 hydrogen atoms in the hydrocarbon.

Step 4: Determining the Hydrocarbon Formula

Now that we have found \(x = 5\) and \(y = 12\), we conclude that the hydrocarbon formula is \(C_5H_{12}\).

Key Concepts

Chemical Stoichiometry

Chemical stoichiometry is the use of relationships in a balanced chemical reaction to calculate quantities of reactants and products. It involves the examination of the mole ratio between different compounds in a reaction. Let's take combustion as an example.

In a combustion reaction, a hydrocarbon reacts with oxygen to produce carbon dioxide and water. The balanced equation not only shows this transformation but offers the proportion of molecules involved.

When given a problem with specific volumes of gases produced, as in the exercise we examined, stoichiometry allows us to determine the molecular formula of the hydrocarbon gas.

Remember, stoichiometry relies on the conservation of mass. In chemical reactions, atoms are neither created nor destroyed. All atoms present in the reactants must be accounted for in the products. This principle is key to correctly solving stoichiometric problems.

Here is a handy reminder of stoichiometry's main points:

• Balance the chemical equation.
• Use mole ratios derived from the balanced equation.
• Calculate unknown quantities based on given information.

Combustion Reactions

Combustion reactions are characterized by a hydrocarbon reacting with oxygen to produce carbon dioxide and water. These reactions are exothermic, meaning they release heat.
In the context of hydrocarbon combustion, the general equation can be represented as:

• Hydrocarbons (fuel) + Oxygen → Carbon Dioxide + Water + Energy

Combustion reactions are predictable and easily balanced, which makes them a core topic in stoichiometry exercises.

The problem at hand involves burning a hydrocarbon gas. The entire process involves understanding that the carbon in carbon dioxide comes from the carbon atom within the hydrocarbon. Similarly, the hydrogen in water comes from the hydrogen atoms in the hydrocarbon.

Here's how to break down a combustion reaction:

• Identify how many carbon and hydrogen atoms are in the hydrocarbon.
• Use the moles of the products (CO₂ and H₂O) to back-calculate the composition of the reactant hydrocarbon.
• Ensure the entire equation is balanced according to the stoichiometric coefficients derived from the process.

Hydrocarbon Formulas

Hydrocarbon formulas represent the composition of hydrocarbons, which are organic compounds consisting entirely of hydrogen and carbon. The simplest type of hydrocarbons are alkanes, which follow the general formula \(C_nH_{2n+2}\).

When determining the formula of an unknown hydrocarbon, we look for clues in the reaction products. This step usually involves:

• Analyzing the number of carbon and hydrogen atoms transferred to the products, like CO₂ and H₂O.
• Using the reaction coefficients to ascertain the ratio of carbon to hydrogen in the hydrocarbon.

In the exercise, a hydrocarbon produced CO₂ and H₂O, leading us to calculate the number of carbon (\(x\)) and hydrogen (\(y\)) atoms. By comparing the volume of the products to the initial volume of the hydrocarbon, we derived the molecular formula of the compound.

Hydrocarbons are generally classified into categories based on the type of carbon bonding, such as:

• Alkanes (single bonds)
• Alkenes (at least one double bond)
• Alkynes (at least one triple bond)

The molecular formula determined also provides insight into the category of hydrocarbon based on the number of hydrogens relative to the carbons.