Problem 94 - Introductory Chemistry Atoms First

Question

A compound used as an insecticide that contains only \(C, H\), and \(C l\) is subjected to combustion analysis, yielding \(24.78 \% \mathrm{C}\) and \(2.08 \% \mathrm{H}\). (a) What is the empirical formula for this compound? (b) What is the molecular formula if its actual formula is four times the mass of its empirical formula?

Step-by-Step Solution

Answer

The empirical formula for the compound is CHCl, and the molecular formula is C₄H₄Cl₄.

Step 1: Calculate the moles of each element

First, we need to assume a sample mass of the compound, say 100 grams. This way, the given percentages can convert to the mass of each element in grams. Then, we will convert the mass to moles using the atomic masses of carbon, hydrogen, and chlorine. - Mass of Carbon (C) = 24.78 g - Mass of Hydrogen (H) = 2.08 g - Mass of Chlorine (Cl) = 100 g - (24.78 g + 2.08 g) = 73.14 g Now, we will convert the masses to moles by dividing the mass by the respective atomic masses of the elements. - Moles of C = 24.78 g / 12.01 g/mol ≈ 2.065 moles - Moles of H = 2.08 g / 1.008 g/mol ≈ 2.063 moles - Moles of Cl = 73.14 g / 35.45 g/mol ≈ 2.063 moles

Step 2: Determine the empirical formula

To find the empirical formula, we need to express the ratio of the moles of the elements in the compound as the smallest possible whole numbers. Divide the moles of each element by the smallest number of moles, then round it to the nearest whole number. - Ratio of C atoms: 2.065 moles / 2.063 ≈ 1 - Ratio of H atoms: 2.063 moles / 2.063 ≈ 1 - Ratio of Cl atoms: 2.063 moles / 2.063 ≈ 1 Thus, the empirical formula of the compound is CHCl.

Step 3: Calculate the molecular formula

The problem states that the molecular formula has a mass four times larger than the empirical formula. To find the molecular formula, first, calculate the molar mass of the empirical formula. - Molar mass of CHCl = 12.01 g/mol (C) + 1.008 g/mol (H) + 35.45 g/mol (Cl) ≈ 48.47 g/mol Now, determine the molar mass of the molecular formula, which is four times the mass of the empirical formula. - Molar mass of molecular formula = 4 × 48.47 g/mol ≈ 193.88 g/mol Since the molecular formula has a mass four times larger than the empirical formula, the number of atoms of each element in the molecular formula will be four times that in the empirical formula. - Molecular formula = C₄H₄Cl₄ So, the molecular formula of the compound is C₄H₄Cl₄.

Key Concepts

Combustion Analysis

Combustion analysis is a laboratory technique used to determine the empirical formula of a compound, especially those containing carbon and hydrogen. During combustion, the compound is burned in an excess of oxygen. Rather than determining the direct amounts of each element, you measure the carbon dioxide and water produced, as these can be connected back to the quantities of carbon and hydrogen in the original sample.
In this specific exercise, combustion analysis is used to determine the percentage composition of carbon, hydrogen, and indirectly chlorine, within the insecticide compound. The percentages are crucial for calculating the empirical formula, as they represent the relative masses of each element present. By knowing the percentages of each element, one can infer the simplest whole number molar ratio in which these atoms combine in the compound.

Molecular Formula

The molecular formula of a compound indicates the actual number of each type of atom in a molecule, highlighting the total elements and their proportions. In many cases, it is a multiple of the empirical formula, which represents the smallest whole number ratio.

• Empirical Formula - Represents the simplest ratio, used as a baseline to calculate the molecular formula.
• Molar Mass - The molecular formula's relation with molar mass provides a way to ascertain how many times the empirical formula should be multiplied.

In the exercise, the provided information about the compound's empirical formula (CHCl) and its relation to the molecular formula, which is four times the mass of the empirical formula, guides solving its true molecular makeup. Calculating the mass of the empirical unit and then multiplying by four leads to the conclusion that the molecular formula is C₄H₄Cl₄.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in chemical reactions. It is the tool used to calculate amounts, often moles or masses, based on balanced chemical equations.
To solve problems involving stoichiometry, one often needs:

• The balanced chemical equation
• Molar ratios from the equation
• Molar masses of reactants and products

In the context of the exercise, understanding stoichiometry is essential for converting percentage mass into moles (by using atomic masses). Each calculation step is based on stoichiometry principles - such as converting mass to moles and then using mole ratios to determine empirical formulas.

Chemical Formulas

Chemical formulas represent the composition of substances using symbols for the elements and subscript numbers to denote the number of atoms of each element. Understanding the distinction between empirical and molecular formulas is vital:

• Empirical Formula - The simplest integer ratio of the elements in a compound. It's used as a stepping stone to determine the more detailed molecular formula.
• Molecular Formula - Exact count of atoms in a molecule, which might require additional information like molar mass to determine from the empirical formula.

In this exercise, mastering the interpretation and conversion between these chemical formulas enables determination of the compound's true nature. Ultimately, this culminates in understanding the entire molecular structure and composition expressed in the correct molecular formula.