Question

In an organic compound of molar mass \(108 \mathrm{gm} \mathrm{mol}^{-1} \mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1:\) 3.5 by weight. 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) \(\mathrm{C}_{4} \mathrm{H}_{18} \mathrm{~N}_{3}\)

Short answer

The molecular formula of the compound is \textbf{(a) C6H8N2}.

Step-by-step solution

Determine the mass ratios of the elements

Given the compound consists of C, H, and N with mass ratios 9:1:3.5, calculate the mass of each element in a 100 g sample. For C: 9 g, for H: 1 g, for N: 3.5 g.

Calculate the moles of each element

Using the molar mass of each element, calculate the moles. Moles of C = 9 g / 12 g/mol, Moles of H = 1 g / 1 g/mol, Moles of N = 3.5 g / 14 g/mol.

Find the mole ratio

Divide the number of moles of each element by the smallest number obtained to get the simplest whole number ratio. Ratio of C:H:N is approximately 9:1:3.5 to 6:8:2.

Write the empirical formula

From the mole ratio, the empirical formula is CH(8/6)N(2/6). Simplify the subscripts to get the empirical formula C6H8N2.

Determine the molecular formula

Compare the empirical formula mass to the given molecular mass to determine the molecular formula. If the empirical formula mass matches the given molar mass, then the empirical formula is the molecular formula. The empirical formula mass of C6H8N2 is 6*12 + 8*1 + 2*14 = 108 g/mol, which matches the given molar mass.

Key concepts

Understanding Empirical and Molecular Formulas

Empirical formulas provide the simplest whole-number ratio of atoms in a compound, which is crucial for understanding the basic building blocks of a molecule. It does not necessarily represent the actual numbers of atoms in a molecule, rather it gives information about the relative proportions of different elements. For example, hydrogen peroxide's empirical formula is HO, showing that for every hydrogen atom, there is one oxygen atom. However, the molecular formula of hydrogen peroxide is H2O2, which represents the actual numbers of each type of atom in a molecule.

When solving problems related to empirical formulas, like the one provided, it's essential to convert the given mass percentages or ratios into moles since chemical reactions occur on a mole-to-mole basis. This is achieved by dividing the mass of each element by its respective atomic mass. Once the moles are calculated, identifying the smallest number of moles assists in creating a mole ratio which is then used to determine the empirical formula of the compound. It’s like finding the smallest common denominator in a fraction to simplify it.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, generally expressed in grams per mole (g/mol). It is a fundamental concept in chemistry that’s used to convert between mass and moles. As seen in our exercise, calculating the molar mass involves using the atomic masses of all atoms within a compound and summing them according to the number of each type of atom present.

For instance, the molar mass of carbon (C) is approximately 12 g/mol, hydrogen (H) is 1 g/mol, and nitrogen (N) is 14 g/mol. By using these values, along with your empirical formula, you can calculate the mass of the empirical formula unit. This value is then compared to the provided molar mass of the compound to determine if the empirical formula reflects the actual compound or if the molecular formula is a multiple of the empirical formula. In practice, this helps chemists to not only understand the composition of a compound but also to measure out how much of a chemical is needed for a reaction.

Stoichiometry and Its Role in Chemistry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's based on the conservation of mass and the concept of moles. The core idea is that molecules react in whole-number ratios, which are derived from the coefficients in a balanced chemical equation. These ratios allow chemists to predict the amounts of products that will be formed or the amounts of reactants needed for a reaction to proceed.

In our example, to deduce the correct molecular formula, stoichiometry principles were applied. After finding the mole ratios of the elements, these were used to write the empirical formula. As stoichiometry is essentially about numbers in balance, it's important to note that the molecular formula will always exist as a whole-number multiple of the empirical formula because compounds consist of discrete molecules, not fractions of them. The art and science of stoichiometry lie in balancing these ratios to reflect what occurs in the real world during chemical reactions.