Question

In an organic compound of molar mass \(108 \mathrm{~g} \mathrm{~mol}^{-1} \mathrm{C}\), \(\mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1: 3.5\) by weight. The molecular formula of the compound can be: (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{12} \mathrm{~N}_{2}\)

Short answer

(b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{N}_{2}\) is the molecular formula.

Step-by-step solution

Understanding the Ratio

The weight ratio of the elements in the compound is given as 9:1:3.5 for C:H:N. This means for every 9 parts of carbon, there is 1 part of hydrogen, and 3.5 parts of nitrogen by weight.

Determine the Molar Mass Proportion

We must find the amount of each element contributing to the molar mass of 108 g/mol. Let the weights of C, H, and N be represented as \(9x, x\), and \(3.5x\) respectively where \(x\) is a common factor. The equation becomes: \[ 9x + x + 3.5x = 108 \] This simplifies to: \[ 13.5x = 108 \]

Solve for x

To find \(x\), divide the total molar mass by the sum of the proportions: \[ x = \frac{108}{13.5} = 8 \] Thus, \(9x = 72\), \(x = 8\), and \(3.5x = 28\). So, the weights are 72 g of C, 8 g of H, and 28 g of N.

Calculate Moles of Each Element

Calculate the moles of each element using their atomic masses (C=12 g/mol, H=1 g/mol, N=14 g/mol):- Moles of C: \( \frac{72}{12} = 6 \)- Moles of H: \( \frac{8}{1} = 8 \)- Moles of N: \( \frac{28}{14} = 2 \)

Determine Molecular Formula

The ratio of moles for the elements gives us the subscripts for each atom in the molecular formula:- For C: 6 parts- For H: 8 parts- For N: 2 partsThus, the molecular formula is \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{N}_{2}\).

Key concepts

Molar Mass

Molar mass is a crucial concept in chemistry, particularly when dealing with molecular formula calculations. It represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To find the molar mass of an organic compound, we sum the atomic masses of all the elements present in the molecule. For example, carbon has an atomic mass of 12 g/mol, hydrogen 1 g/mol, and nitrogen 14 g/mol. The molar mass is essential for determining how much of each element contributes to the total weight of a molecule. In this exercise, the molar mass of the compound is given as 108 g/mol. This means that the sum of the masses of all the elements in the compound—carbon, hydrogen, and nitrogen—must equal 108 g/mol. Calculating this helps identify the correct molecular formula.

Organic Compounds

Organic compounds are characterized by the presence of carbon atoms bonded to hydrogen, nitrogen, oxygen, or other elements. They form the basis of all known life and are prevalent in many forms, from simple molecules to complex structures like proteins and DNA. Organic compounds are integral in a wide array of chemical processes. In this problem, we're dealing with an organic compound comprising carbon (C), hydrogen (H), and nitrogen (N). Understanding the composition of these compounds and how the elements interact in specific ratios is necessary to determine their molecular formulas accurately. This composition heavily influences properties such as their reactivity, polarity, and biological activity. By analyzing the given ratio of 9:1:3.5 for C:H:N, we can infer how these elements are combined in the compound, which, along with the molar mass, aids in deducing the molecular formula.

Elemental Ratio

Elemental ratio refers to the proportion of each element within a compound. It’s crucial for determining the empirical formula, which shows the simplest whole-number ratio of atoms in a compound. For this exercise, the ratio given is 9:1:3.5 by weight for carbon, hydrogen, and nitrogen respectively.To interpret these ratios, assume a base amount for easier calculation. Here, if we set a base of 9 parts for carbon, the corresponding amounts for hydrogen and nitrogen are 1 part and 3.5 parts by weight. This ratio allows us to assign a common factor ( \(x\)) for calculation to relate to the molar mass. Using the equation \(13.5x = 108\) g/mol, we solve for \(x\) to understand the contribution of each element to the compound's overall mass. Working through these ratios provides clarity on how the elements are distributed in the organic compound and helps to pinpoint the correct molecular formula.

Moles Calculation

Calculating moles is essential for converting a compound's mass to the number of particles or atoms it contains. The mole concept is a fundamental chemistry principle allowing chemists to count atoms by weighing them.In this exercise, we calculate the moles of each element by dividing the mass of each element by its respective atomic mass:

• Moles of Carbon (C): \(\frac{72}{12} = 6\)
• Moles of Hydrogen (H): \(\frac{8}{1} = 8\)
• Moles of Nitrogen (N): \(\frac{28}{14} = 2\)

This calculation provides the mole ratio, which directly translates to the subscripts in the molecular formula, yielding \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{N}_{2}\). Understanding how to accurately convert masses to moles through this process allows for the determination of the molecular structure of compounds—critical for various applications in chemistry and related fields.