Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A gaseous hydrocarbon gives upon combustion \(0.72 \mathrm{~g}\) of water and \(3.08 \mathrm{~g}\). of \(\mathrm{CO}_{2}\). The empirical formula of the hydrocarbon is : (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{7} \mathrm{H}_{8}\)

Step-by-Step Solution

Answer

The empirical formula of the hydrocarbon is \(\mathrm{C}_3\mathrm{H}_4\).

Step 1: Calculate moles of water produced

To find the moles of water, use the formula:\[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{\text{Mass of } \mathrm{H}_2\mathrm{O}}{\text{Molar mass of } \mathrm{H}_2\mathrm{O}} = \frac{0.72 \, \mathrm{g}}{18 \, \mathrm{g/mol}} = 0.04 \, \mathrm{mol} \]

Step 2: Determine moles of hydrogen in water

Each mole of water (\(\mathrm{H}_2\mathrm{O}\)) contributes 2 moles of hydrogen atoms. Thus, the moles of hydrogen are:\[ \text{Moles of } \mathrm{H} = 2 \times 0.04 \, \mathrm{mol} = 0.08 \, \mathrm{mol} \]

Step 3: Calculate moles of carbon dioxide produced

The moles of carbon dioxide are given by:\[ \text{Moles of } \mathrm{CO}_2 = \frac{\text{Mass of } \mathrm{CO}_2}{\text{Molar mass of } \mathrm{CO}_2} = \frac{3.08 \, \mathrm{g}}{44 \, \mathrm{g/mol}} = 0.07 \, \mathrm{mol} \]

Step 4: Determine moles of carbon in carbon dioxide

Each mole of \(\mathrm{CO}_2\) contains one mole of carbon. Therefore, the moles of carbon are:\[ \text{Moles of } \mathrm{C} = 0.07 \, \mathrm{mol} \]

Step 5: Obtain the empirical formula ratio

The ratio of moles of carbon to hydrogen is determined and simplified: \[ \text{Ratio (C:H)} = \frac{0.07}{0.07} : \frac{0.08}{0.07} = 1 : 1.14 \]This can be approximated to the simplest whole number ratio of 3:4 by multiplying both by 3.

Step 6: Determine empirical formula

Based on the simplified whole number ratio between carbon and hydrogen, the empirical formula is \(\mathrm{C}_3\mathrm{H}_4\).

Key Concepts

Gaseous Hydrocarbon Combustion

When a hydrocarbon combusts, it reacts with oxygen. This process results in the production of water (\(\mathrm{H}_2\mathrm{O}\)) and carbon dioxide (\(\mathrm{CO}_2\)). The complete combustion reaction for a hydrocarbon can be generalized as: \[ \text{Hydrocarbon} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]Gaseous hydrocarbons are common fuels, and understanding their combustion is crucial in chemistry. By measuring the amounts of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) produced, we can backtrack to determine specific details about the hydrocarbon, like its empirical formula.

Mole Calculations

Understanding and calculating moles is foundational in chemistry. A mole is a unit that measures the amount of a substance, and it's based on Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles/mol. This allows chemists to count atoms, ions, and molecules by weighing them. For example, to determine the moles of \(\mathrm{H}_2\mathrm{O}\) in our problem, we use its given mass (0.72 g) divided by its molar mass (18 g/mol): \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{0.72 \text{ g}}{18 \text{ g/mol}} = 0.04 \text{ mol} \]This calculation helps convey how many water molecules are involved in the reaction. Similarly, to find moles of \(\mathrm{CO}_2\), you divide its mass (3.08 g) by its molar mass (44 g/mol): \[ \text{Moles of } \mathrm{CO}_2 = \frac{3.08 \text{ g}}{44 \text{ g/mol}} = 0.07 \text{ mol} \]This allows for stoichiometric calculations, which we explore next.

Stoichiometry

Stoichiometry is the quantitative analysis of reactants and products in a chemical reaction. It uses the mole concept to calculate relationships and predict how much product forms from a given amount of reactants. In hydrocarbon combustion, stoichiometry helps us determine the amount of carbon and hydrogen in the original hydrocarbon.

• Each mole of \(\mathrm{H}_2\mathrm{O}\) contains 2 moles of hydrogen.
• Each mole of \(\mathrm{CO}_2\) contains 1 mole of carbon.

So, for our water calculation: \[ \text{Moles of } \mathrm{H} = 2 \times 0.04 \text{ mol } = 0.08 \text{ mol} \]And for carbon dioxide: \[ \text{Moles of } \mathrm{C} = 0.07 \text{ mol} \]This shows the mole relationship and is crucial for finding the empirical formula.

Empirical Formula Determination

The empirical formula of a compound gives the simplest whole number ratio of the elements present. To determine this, we use the mole calculations of carbon and hydrogen. In this exercise, we have:

• Moles of Carbon (C): 0.07 mol
• Moles of Hydrogen (H): 0.08 mol

The empirical formula is based on the ratio of these moles. The calculated ratio was initially 1:1.14, which approximates to a simpler whole number formula when multiplied by 3, yielding 3:4:\[ \text{Ratio (C:H)} = 3:4 \]Thus, the empirical formula becomes \(\mathrm{C}_3\mathrm{H}_4\). This process helps confirm the identity of the hydrocarbon, offering insights into its structure and properties.