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) \(\mathrm{CH}_{4}\)

Short answer

The formula of the hydrocarbon is \( C_5H_{12} \).

Step-by-step solution

Write the General Combustion Reaction

The general formula for a hydrocarbon is \( C_xH_y \). Upon complete combustion in the presence of oxygen, the hydrocarbon will produce carbon dioxide \( CO_2 \) and water vapor \( H_2O \). The balanced chemical equation is: \[ C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \]

Analyze the Volume Ratios

According to Avogadro's Law, the volumes of gases involved in reactions are proportional to the number of molecules (at constant temperature and pressure). Here, \(500\ \mathrm{mL}\) of the hydrocarbon produces \(2500\ \mathrm{mL}\) of \( CO_2 \) and \(3000\ \mathrm{mL}\) of \( H_2O \). This gives us the molar ratio: \[ 500\ \mathrm{mL}\ C_xH_y \rightarrow 2500\ \mathrm{mL}\ CO_2 + 3000\ \mathrm{mL}\ H_2O \] Thus, \(1\ \text{mole of}\ C_xH_y\) produces \(5\ \text{moles of}\ CO_2\) and \(6\ \text{moles of}\ H_2O\).

Relate Volumes to Chemical Formula

From the balanced equation, \(x = 5\), which means there are 5 carbon atoms in the hydrocarbon's formula, given there are 5 moles of \( CO_2 \) produced per mole of hydrocarbon. Similarly, solving for \(y\) considering there are \(6\) hydrogen molecules per hydrocarbon, gives \(y = 12\) since 1 mole of \( H_2O \) contains \(2\) hydrogen atoms. Therefore, the hydrocarbon formula is \( C_5H_{12} \).

Choose the Correct Option

The hydrocarbon formula calculated is \( C_5H_{12} \), which is option (c). Therefore, the correct answer for the hydrocarbon that results in these combustion volumes is \( C_5H_{12} \).

Key concepts

Avogadro's Law

Avogadro's Law is a fundamental principle in chemistry stating that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. This concept is vital when working with gases in chemical reactions.
In the given problem, the hydrocarbon is burned in the presence of oxygen. The reaction yields carbon dioxide and water vapor. Since the problem states that conditions of temperature and pressure are constant, we can use Avogadro's Law. This allows us to equate the volumes directly to the number of moles.
For instance, when we see that 500 mL of hydrocarbon yields 2500 mL of CO₂ and 3000 mL of H₂O, we can directly interpret the volumes as a molar ratio. This is why the moles of hydrocarbon relate to the moles of products formed directly, aiding in determining the capacity of each component in the reaction.

Balanced Chemical Equation

A balanced chemical equation is essential in understanding chemical reactions because it shows the exact proportions of reactants and products. This involves ensuring the same number of each type of atom appears on both sides of the equation.
When we consider the combustion of hydrocarbons, a general equation is given by: \[ C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \]This equation needs to be balanced, meaning both the reactant and product sides have the same number of carbon, hydrogen, and oxygen atoms.

• 'C_xH_y' represents hydrocarbon molecules.
• The coefficient in front of \(O_2\) shows the necessary oxygen moles for combustion per mole of hydrocarbon.
• The products \(xCO_2\) and \(\frac{y}{2}H_2O\) indicate the resulting molecules from the reaction.

Balancing such an equation ensures that all atoms are conserved during the reaction, following the Law of Conservation of Mass.

Volume Ratios

In chemical reactions involving gases, particularly those examined using Avogadro's Law, understanding volume ratios is crucial. Volume ratios are the coefficients in a balanced chemical equation interpreted as the ratio of volumes for gaseous reactants and products involved in the reaction.
In the scenario provided, the ratios were described using volumes:

• 500 mL of hydrocarbon (\(C_xH_y\))
• 2500 mL of carbon dioxide (\(CO_2\))
• 3000 mL of water vapor (\(H_2O\))

These volumes directly translate to a molar ratio, as per Avogadro's principle. From this, it follows that 1 mole of the hydrocarbon results in 5 moles of carbon dioxide and 6 moles of water.
This technique allows scientists to determine the stoichiometry of a reaction, helping to derive the formula of unknown hydrocarbons when given volumes of gases produced or consumed. In this problem, volume ratios were essential in deriving \(C_5H_{12}\) as the formula of the hydrocarbon.