Problem 90 - A Complete Resource Book in Chemistry for JEE Main

Question

On combustion, \(0.2 \mathrm{~g}\) of an organic compound containing carbon, hydrogen and oxygen yields \(0.147\) g carbon dioxide and \(0.12 \mathrm{~g}\) water. The percentage of carbon, hydrogen and oxygen respectively in the compound is (a) \(22.5,8.7\) and \(68.8\) (b) \(20,6.7\) and \(73.3\) (c) \(18.2,9.5\) and \(72.3\) (d) \(28.9,12.8\) and \(58.8\)

Step-by-Step Solution

Answer

The percentages of C, H, and O in the compound are 20%, 6.7%, and 73.3%, corresponding to option (b).

Step 1: Calculate the moles of carbon in carbon dioxide

To find the carbon content, we first determine the moles of carbon in the given carbon dioxide. Since the molar mass of CO2 is 44 g/mol and one mole of CO2 contains one mole of carbon, the moles of carbon are calculated as follows: \[\text{Moles of } CO_2 = \frac{0.147 \, \text{g}}{44 \, \text{g/mol}} = 0.003341 \, \text{mol} \]Thus, the moles of carbon in CO2 are also 0.003341 mol.

Step 2: Calculate the mass of carbon in carbon dioxide

The mass of carbon can be calculated using the moles of carbon from Step 1 and the atomic mass of carbon (12 g/mol):\[\text{Mass of C} = 0.003341 \, \text{mol} \times 12 \, \text{g/mol} = 0.0401 \, \text{g}\]

Step 3: Calculate the moles of hydrogen in water

Water has a molar mass of 18 g/mol, and each mole of water contains 2 moles of hydrogen. First, calculate the moles of water:\[\text{Moles of } H_2O = \frac{0.12 \, \text{g}}{18 \, \text{g/mol}} = 0.006667 \, \text{mol}\]The moles of hydrogen are:\[\text{Moles of H} = 2 \times 0.006667 \, \text{mol} = 0.013334 \, \text{mol}\]

Step 4: Calculate the mass of hydrogen in water

Using the moles of hydrogen calculated in Step 3 and the atomic mass of hydrogen (1 g/mol), we can find the mass:\[\text{Mass of H} = 0.013334 \, \text{mol} \times 1 \, \text{g/mol} = 0.013334 \, \text{g}\]

Step 5: Calculate the mass of oxygen in the compound

The total mass of the compound is 0.2 g. Subtract the mass of carbon and mass of hydrogen from this total to find the mass of oxygen:\[\text{Mass of O} = 0.2 \, \text{g} - 0.0401 \, \text{g} - 0.013334 \, \text{g} = 0.146566 \, \text{g}\]

Step 6: Calculate the percentage of each element

Convert the masses of carbon, hydrogen, and oxygen into percentages of the total mass:- Carbon percentage: \[ \frac{0.0401}{0.2} \times 100 = 20.05\% \]- Hydrogen percentage: \[ \frac{0.013334}{0.2} \times 100 = 6.667\% \]- Oxygen percentage: \[ \frac{0.146566}{0.2} \times 100 = 73.283\% \]

Step 7: Choose the closest match from the given options

From the calculated percentages: Carbon = 20.05%, Hydrogen = 6.667%, and Oxygen = 73.283%. These percentages closely match option (b) in the problem, which states 20%, 6.7%, and 73.3%.

Key Concepts

Empirical Formula Calculation

Understanding empirical formula calculation is crucial in chemistry to determine the simplest whole-number ratio of elements in a compound. It is derived from the proportion of elements, often found using data from combustion analysis. Here's why it's vital:

- **Simplifies chemical composition**: The empirical formula helps simplify the representation of a compound by showing the lowest ratio of its constituent atoms. This means, instead of stating exact numbers of atoms, we provide the simplest way in which they combine. - **Foundation for further calculations**: Once you have the empirical formula, you can use it to find molecular formulas if you know the molar mass of the compound. The molecular formula is a multiple of the empirical formula. - **Practical application**: In industrial applications, knowing the empirical formula is essential when working with unknown compounds, as it gives insight into potential reactions and properties. The process involves determining the masses or percentages of each element, converting them to moles, and then finding the simplest whole-number ratio. This establishes a framework for understanding the compound's basic structure.

Elemental Analysis

Elemental analysis is the process of determining the elemental composition of a substance. It plays a key role in organic chemistry to deduce the amounts of carbon, hydrogen, and oxygen in compounds through processes such as combustion analysis.

- **Combustion Analysis**: This is a common technique where the compound is burned, and products like carbon dioxide and water are measured. By knowing these amounts, students can deduce the quantities of carbon and hydrogen in the original compound. - **Importance for identification**: Conducting an elemental analysis is often the first step in identifying and characterizing unknown compounds. It helps chemists understand what a compound is composed of and how it might react. - **Calculative approach**: From a calculative perspective, this analysis converts measured masses (grams) to moles, which are then compared to ascertain the mass ratios of elements in the compound. Elemental analysis effectively ties together theory and practice, offering insight into the physical composition and chemical behavior of compounds.

Mole Calculations

A fundamental concept in chemistry, mole calculations, are built on the idea that the mole represents a quantity or 'count' of substance particles. This concept is pivotal because it allows chemists to relate the macro world of grams to the microscopic world of atoms and molecules.

- **Understanding moles**: A mole is defined as exactly 6.022 x 10²³ particles of a substance, which can be atoms, molecules, ions, or electrons. This is known as Avogadro's number.- **Conversion utility**: The mole allows for the conversion of mass (in grams) to moles, facilitating the calculation of atom or molecule counts. This is especially useful when determining the amounts involved in chemical reactions.- **Practical application**: In mole calculations, students often work with equations like \( ext{Moles} = \frac{ ext{mass in grams}}{ ext{molar mass in g/mol}} \). This equation helps determine how much of each element is present, providing a gateway to stoichiometry calculations.Mole calculations make it easier to balance chemical equations, understand reaction yields, and calculate concentrations, thus forming a core part of chemical education.