Problem 68 - A Complete Resource Book in Chemistry for JEE Main

Question

If a compound on analysis was found to contain \(\mathrm{C}=\) \(18.5 \%, \mathrm{H}=1.55 \%, \mathrm{Cl}=55.04 \%\) and \(\mathrm{O}=24.81 \%\) then its empirical formula is (a) \(\mathrm{CHClO}\) (b) \(\mathrm{CH}_{2} \mathrm{ClO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OCl}\) (d) \(\mathrm{ClC}_{2} \mathrm{H}_{5} \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formula is \(\text{CHClO}\), so option (a) is correct.

Step 1: Calculate moles of each element

To find the empirical formula, begin by assuming you have 100 g of the compound, so each percentage translates directly to grams:- C: 18.5 g- H: 1.55 g- Cl: 55.04 g- O: 24.81 gCalculate the moles of each element using their atomic masses:For Carbon:\[\text{Moles of C} = \frac{18.5}{12.01} \approx 1.54 \] For Hydrogen:\[\text{Moles of H} = \frac{1.55}{1.01} \approx 1.53 \] For Chlorine:\[\text{Moles of Cl} = \frac{55.04}{35.45} \approx 1.55 \] For Oxygen:\[\text{Moles of O} = \frac{24.81}{16.00} \approx 1.55 \]

Step 2: Determine simplest mole ratio

To find the simplest ratio, divide each element's mole number by the smallest number of moles computed:\[\text{Smallest number of moles} = 1.53 \] Carbon:\[\frac{1.54}{1.53} \approx 1 \] Hydrogen:\[\frac{1.53}{1.53} = 1 \] Chlorine:\[\frac{1.55}{1.53} \approx 1 \] Oxygen:\[\frac{1.55}{1.53} \approx 1 \]

Step 3: Write and verify the empirical formula

Since the moles of each element rounded to whole numbers remain 1, the empirical formula is determined based on these whole number ratios: \[\text{Empirical Formula} = \text{CHClO} \]The calculation shows that simply rounding each element's ratio gives us one atom of each, confirming the empirical formula is indeed \(\text{CHClO}\).

Key Concepts

Elemental Analysis

Elemental analysis is the process of determining the percentage composition of elements within a compound. In our example compound, the analysis revealed the following composition: 18.5% for Carbon (C), 1.55% for Hydrogen (H), 55.04% for Chlorine (Cl), and 24.81% for Oxygen (O). By assuming a total sample mass of 100 grams, we can directly interpret these percentages as grams of each element. This conversion simplifies our calculations as it allows us to focus directly on comparing the relative amounts of each element based on these fixed masses.

• This analytic approach is especially useful in determining the empirical formula, which is a critical step in understanding a compound's basic chemical makeup.
• The percentage composition provides insight into the compound's structural characteristics without needing to identify its molecular formula.

Mole Ratio

The mole ratio is a fundamental concept in chemistry that reflects the relative number of moles of each element present in a compound. After converting the mass of each element into moles using their respective atomic masses, we calculated the following: - Carbon: approximately 1.54 moles - Hydrogen: approximately 1.53 moles - Chlorine: approximately 1.55 moles - Oxygen: approximately 1.55 moles Understanding the mole ratio involves simplifying these mole values to the smallest possible whole numbers to express how each element relates to the others. By dividing each mole quantity by the smallest number of moles (in this case, 1.53), we arrive at a simple 1:1:1:1 ratio for each element, leading us to construct the empirical formula. These ratios are crucial because they provide the foundational information to express the simplest representation of a compound's constituent elements.

Chemical Composition

Chemical composition refers to the types and amounts of elements present in a compound. It's all about understanding what makes up the compound at the most basic level. In the exercise, we started with given percentages for C, H, Cl, and O, which led us to weight these against their molar masses.

• Understanding chemical composition through empirical formulas provides insights into the molecular structure and reactivity patterns of compounds.
• This basic understanding is crucial for any further qualitative or quantitative analysis, influencing everything from predicting reactions to understanding material properties.

By using the concept of chemical composition, we can piece together what elements are in the compound, how much of each there is, and in doing so, gain insights into potential chemical properties and behaviors.

Stoichiometry

Stoichiometry is the branch of chemistry that focuses on the quantitative aspects of chemical formulas and reactions. It provides the arithmetic framework for deducing relationships between reactants and products in a chemical reaction. While working through our exercise, stoichiometry allows us to systematically use the atomic masses of elements at hand and their ratios to determine the form and amount of compounds produced or needed in reactions. In terms of our empirical formula calculation, stoichiometry served as the method by which the proportions translated into a useful molecular framework with a 1:1:1:1 elemental ratio, leading to the empirical formula CHClO.

• Through stoichiometry, chemists can determine the amounts of substances required or produced, making it essential for laboratory settings or industrial processes.
• It's vital in extrapolating beyond mere ratios to making predictions about chemical products and reactions based on the empirical formula.

Ultimately, it is stoichiometry that unites elemental analysis into a coherent, predictive model for chemical interactions and transformations.