Problem 41 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic dibasic acid is used in volumetric analysis. The percentage of elements are \(\mathrm{C}=26.7 \%\), \(\mathrm{H}=2.2 \%\) and \(\mathrm{O}=71.1 \%\). Find the empirical formula of the acid. (a) \(\mathrm{CH}_{2} \mathrm{O}\) (b) \(\mathrm{CHO}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{4} \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formula is \(\text{CHO}_{2}\) (option b).

Step 1: Convert Percentage to Grams

Assume 100 grams of the compound for simplicity, so you have: - 26.7g of Carbon (\(\text{C}\))- 2.2g of Hydrogen (\(\text{H}\))- 71.1g of Oxygen (\(\text{O}\))

Step 2: Convert Grams to Moles

Use the atomic masses: \(\text{C} = 12 \text{ g/mol}\), \(\text{H} = 1 \text{ g/mol}\), \(\text{O} = 16 \text{ g/mol}\).Calculate moles:- Moles of \(\text{C} = \frac{26.7}{12} \approx 2.225\)- Moles of \(\text{H} = \frac{2.2}{1} = 2.2\)- Moles of \(\text{O} = \frac{71.1}{16} \approx 4.4437\)

Step 3: Determine the Mole Ratio

Divide each number of moles by the smallest number of moles (2.2 in this case):- C: \(\frac{2.225}{2.2} \approx 1.011\)- H: \(\frac{2.2}{2.2} = 1\)- O: \(\frac{4.4437}{2.2} \approx 2.019\)

Step 4: Formulate the Simplest Whole Number Ratio

The approximate mole ratio \(\text{C:H:O} = 1.011:1:2.019\) is close to the whole number ratio \(1:1:2\).Thus, the empirical formula is \(\text{CHO}_{2}\).

Step 5: Choose the Correct Answer

By comparing with the given options, the correct empirical formula of the acid is option (b) \(\text{CHO}_{2}\).

Key Concepts

Volumetric Analysis

Volumetric analysis is a technique in analytical chemistry that involves measuring the volume of a solution needed to react with a given amount of analyte. It's often used to determine the concentration or purity of substances in chemical reactions, especially in the field of organic chemistry.

When we are tasked with finding the empirical formula of an acid using elemental composition percentages, volumetric analysis can play a pivotal role. While standard volumetric analysis involves titrations, in this context, it helps us understand the proportions of each element present in the compound. This is achieved by converting percentages into moles, establishing a clear understanding of

• how the elements interact,
• how much of each element is required, and
• how they combine to form the empirical formula.

For this exercise, although straight volumetric titration isn't used, the concepts from volumetric analysis, such as transformation of weights into moles, are necessary steps in empirical formula determination.

Organic Chemistry

Organic chemistry focuses on the structure, properties, and reactions of organic compounds, which contain carbon. In the context of our exercise, this involves understanding the chemical behavior and structure of the organic dibasic acid, characterized by its empirical formula.

The empirical formula provides the simplest ratio of the elements within the compound, which is crucial for predicting reactivity and behavior in chemical reactions commonly explored in organic chemistry. Understanding this formula aids chemists in grasping:

• the fundamental architecture of the compound,
• how it might behave in chemical reactions, and
• potentially influencing the design of synthesis methods.

For example, knowing the empirical formula (\( \text{CHO}_2 \)) of an acid gives us clues into its possible stability and acidity. Organic chemistry endeavors often rely on such foundational data derived from elemental and empirical analysis to predict and direct chemical processes.

Mole Ratio

The mole ratio is a fundamental concept when it comes to chemical reactions and formula determination. It represents the ratio of moles of each element in a compound, reflecting the relative amount of each atomic species involved.

In our exercise, to determine the empirical formula, we calculated the moles of carbon, hydrogen, and oxygen based on their given percentages. After determining the moles of each element, the next step is to find the simplest whole number ratio of these moles.

• Dividing each element's mole value by the smallest number of moles ensures we obtain the simplest ratio possible.
• In the case presented, the smallest moles value was 2.2, allowing us to normalize each ratio, simplifying it to get approximately a 1:1:2 ratio.
• The ratio derived, despite being fractional initially, is rounded to the nearest whole number to provide clarity in the molecular structure of the compound.

Thus, the mole ratio helps in understanding how atoms assemble in a molecule, providing insight into the arrangement and stoichiometric composition.

Elemental Analysis

Elemental analysis is critical for determining the composition of a chemical sample. In the exercise context, it provides us with percentages of Carbon, Hydrogen, and Oxygen in the dibasic acid, allowing us to compute the empirical formula.

The process starts with converting these percents into mass, by assuming a sample size that simplifies calculations (usually 100 grams). Subsequently, it involves changing these masses into moles using atomic weights, giving insights into how much of each element integrates into the compound.

• Elemental analysis allows you to shift from qualitative assessment (types of elements present) to quantitative analysis (specific amounts of each element).
• This quantitative analysis is essential for accurately determining the compound's formula.
• The results, specifically the determined ratios, offer a window into the molecular framework that governs the compound's properties and behaviors.

Hence, elemental analysis serves as the backbone for formulating empirical formulas, ensuring that you have the vital data needed to understand any chemical compound deeply.