Problem 142 - A Complete Resource Book in Chemistry for JEE Main

Question

In an organic compound of molar mass is \(108 \mathrm{~g} \mathrm{~mol}^{-1}\). \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 35\) by weight in it. Molecular formula can be (a) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{~N}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{18} \mathrm{~N}_{3}\)

Step-by-Step Solution

Answer

The molecular formula is (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{N}_{3}\).

Step 1: Determine the Mass Contribution of Each Element

Given the compound's molar mass is 108 g/mol and the elements maintain a weight ratio of 9:1:35 for C:H:N, we find the actual mass of each element in 108 g:- Carbon (C): \(\frac{9}{9+1+35} \times 108 = 24 \text{ g} \)- Hydrogen (H): \(\frac{1}{9+1+35} \times 108 = 2.7 \text{ g} \)- Nitrogen (N): \(\frac{35}{9+1+35} \times 108 = 81 \text{ g} \)

Step 2: Calculate the Moles of Each Element

Next, convert the masses to moles using their atomic masses: C = 12 g/mol, H = 1 g/mol, N = 14 g/mol.- Moles of Carbon (C): \(\frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles} \)- Moles of Hydrogen (H): \(\frac{2.7 \text{ g}}{1 \text{ g/mol}} = 2.7 \text{ moles} \)- Moles of Nitrogen (N): \(\frac{81 \text{ g}}{14 \text{ g/mol}} = 5.79 \text{ moles} \)

Step 3: Determine Simplest Whole Number Ratio of Moles

Find the simplest whole number ratio of moles for C, H, and N by dividing by the smallest number of moles.- C: \(\frac{2}{2} = 1 \)- H: \(\frac{2.7}{2} = 1.35 \)- N: \(\frac{5.79}{2} = 2.895 \)Rounding to nearest whole number gives C:H:N = 1:1:3.

Step 4: Compare with Possible Molecular Formulas

Using the molecular formula from the ratio C:H:N = 1:1:3, compare it to possible molecules:- (a) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) doesn't fit because of incorrect H and N proportions.- (b) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{~N}\) doesn't match with N proportion.- (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) fits the C:H:N = 1:1:3 ratio.- (d) \(\mathrm{C}_{4} \mathrm{H}_{18} \mathrm{~N}_{3}\) doesn't fit the C ratio.

Key Concepts

Molar Mass

Understanding molar mass is a crucial part of chemistry, especially when you're dealing with molecular formulas. Molar mass is the weight of one mole (6.022 \(\times\) 10\(^23\) particles) of any substance. For example, the molar mass of carbon (C) is 12 g/mol. Knowing the molar mass allows chemists to convert between mass and moles — an essential aspect of many chemical calculations. This makes it easier to figure out how much of each element is present in a sample.

When dealing with compounds, you need to sum up the atomic masses of the component elements. For example, in a compound like water \( \text{(H}_2\text{O)} \), you would add the masses of two hydrogen atoms and one oxygen atom to get the molar mass of water. By understanding and using molar mass, you can start to piece together how much of a substance is present in any given sample, which is especially useful when working with reactions.

Elemental Composition

Elemental composition is about figuring out what elements are in a compound and in what ratio. In this exercise, the weight ratio of carbon, hydrogen, and nitrogen is crucial. An elemental composition expressed in percentages or ratios gives insight into how much of each element makes up the compound.

• Weight Ratio: Here, carbon, hydrogen, and nitrogen are in a 9:1:35 ratio. This tells us their relative masses in the compound.
• Molar Ratio: Once you have the weight of each element, use the atomic masses (like C = 12 g/mol) to find how many moles of each are present.

Breaking down into moles is important because it provides the number of atoms, offering a clearer picture of the compound's true composition. The goal is to find the simplest whole number ratio, reflecting the compound's actual structure.

Stoichiometry

Stoichiometry is the heart of chemistry, bridging the gap between reactants and products. It's the math behind chemical reactions, ensuring that atoms are conserved. In terms of this exercise, stoichiometry helps identify the molecular formula once you have the elements and their molar ratios.

Based on the calculated moles: carbon (C) is 2 moles, hydrogen (H) is 2.7 moles, and nitrogen (N) is 5.79 moles. When simplified to their simplest whole-number ratio, the stoichiometry here points to a molecular formula that aligns with specific multiple values, like a 1:1:3 ratio. This work allows proper matching with possible chemical formulas, ensuring that the chemical composition is accurate and authentically represents the compound's nature.

Organic Chemistry

Organic chemistry focuses on carbon-containing compounds, essential to life. In this context, organic chemistry involves finding the practical applications of molecular formulas.

• Molecular Formulas: From the given exercise, the goal is to work out the correct formula using understood ratios and molar masses.
• Compounds: Organic compounds like those discussed are critical in many fields, from pharmaceuticals to materials science, making this topic integral to various scientific pursuits.

While many compounds can fit initial criteria, only a correctly matched formula reflects actual molecular structure. Understanding the fundamentals of organic chemistry aids in building a more comprehensive picture of these complex molecules.