Problem 73 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A compound having the empirical formula, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\), has a molecular weight of \(170 \pm 5\). The molecular formula of the compound is (a) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

The molecular formula of the compound is \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{2}\).

Step 1: Calculate Empirical Formula Mass

First, calculate the empirical formula mass of the compound \(\mathrm{C}_{3} \mathrm{H}_{4}\mathrm{O}\). Use the atomic masses for carbon (C, approximately 12 amu), hydrogen (H, approximately 1 amu), and oxygen (O, approximately 16 amu). The empirical formula mass is thus \(3 \times 12\) for carbon + \(4 \times 1\) for hydrogen + \(1 \times 16\) for oxygen.

Step 2: Determine the Molecular Mass

The problem provides the molecular mass of the compound as \(170 \pm 5\) amu. You should use this molecular mass to find out how many times the empirical formula mass fits into the molecular mass.

Step 3: Calculate the Ratio of Molecular Mass to Empirical Formula Mass

Divide the given molecular mass by the empirical formula mass to find the ratio, which will be the number of empirical formula units in the actual molecule.

Step 4: Determine the Molecular Formula

Multiply the subscripts in the empirical formula by the ratio found in step 3 to get the molecular formula of the compound.

Step 5: Compare with the Given Options

Check which of the given options is consistent with the molecular formula determined in step 4.

Key Concepts

Empirical Formula Mass

The empirical formula mass is an essential concept in chemistry, which serves as the foundation for understanding molecular compositions. It refers to the sum of the atomic masses of all the atoms in the empirical formula. An empirical formula represents the simplest whole-number ratio of elements in a compound, not the exact number of atoms like the molecular formula.

For instance, in the exercise, the empirical formula is given as \(\mathrm{C}_{3} \mathrm{H}_{4}\mathrm{O}\). To calculate the empirical formula mass, each element's atomic mass must be multiplied by its subscript in the formula: Carbon (C) has an atomic mass of roughly 12 amu, Hydrogen (H) 1 amu, and Oxygen (O) about 16 amu. Therefore, the empirical formula mass for this compound can be calculated as follows: \((3 \times 12) + (4 \times 1) + (16) = 56 \text{amu}\).

Molecular Weight Calculation

Molecular weight calculation is a key step in determining the molecular formula. The molecular weight, also known as molecular mass, is the weight of all atoms in a molecule and it is measured in atomic mass units (amu). It provides important information about the molecular size and how it influences reactions.

To calculate the molecular weight, the atomic weights of the constituent atoms are added together. In the given exercise, the molecular weight is provided as \((170 \pm 5)\) amu. With the empirical formula mass and the molecular weight at hand, the next step is to ascertain how many times the empirical unit fits into the molecular weight, which is fundamental to finding out the molecular formula of the compound.

Stoichiometry

Stoichiometry is the quantitative relationship between the amounts of reactants and products in a chemical reaction. It involves calculations concerning the masses of reactants and products, often relying on conversion factors derived from the balanced chemical equation and the principle of conservation of mass.

In our exercise, stoichiometry plays a crucial role in transitioning from the empirical formula to the molecular formula. By determining the ratio of the molecular mass (from the molecular weight calculation) to the empirical formula mass, one can understand how many empirical units make up a molecule. This ratio is fundamental to stoichiometry, as it allows for the precise scaling up from empirical to molecular formula, ensuring the conservation of mass and the correct representation of the substance involved.

Chemistry Competitive Exams

Students preparing for chemistry competitive exams need to be proficient in practical skills such as molecular formula determination. These exams test a student's understanding and application of chemical principles in problem-solving. Questions on empirical and molecular formulas are standard, as they test students' grasp of the fundamentals of chemical composition and stoichiometry.

In order to excel, students should practice a diverse range of problems, including those similar to our example. Understanding the process of calculating empirical formula mass, carrying out molecular weight calculation, and applying stoichiometry principles is critical performance in competitive exams. Moreover, quick and accurate application of these skills is often required under time constraints, which is why regular practice and a clear grasp of these concepts are indispensable.