Question

An organic compound was found to contain only $\mathrm{C},\mathrm{H},$ and Cl. When a $1.50-\mathrm{g}$ sample of the compound was completely combusted in air, 3.52 $\mathrm{g}$ of ${\mathrm{CO}}_2$ was formed. In a separate experiment, the chlorine in a $1.00-\mathrm{g}$ sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.

Short answer

The empirical formula of the organic compound containing only carbon, hydrogen, and chlorine is C8H50Cl.

Step-by-step solution

Calculate the moles of carbon in CO2

Given that 3.52 g of CO2 is produced from combustion of the 1.50 g sample, we can calculate the moles of carbon in CO2 using the molar mass of carbon (12.01 g/mol) and CO2 (44.01 g/mol): Moles of carbon = (3.52 g CO2) x (1 mol CO2 / 44.01 g CO2) x (1 mol C / 1 mol CO2) = 0.0800 mol C

Calculate the moles of chlorine in AgCl

We are given that 1.27 g of AgCl is produced from the 1.00 g sample of the compound. We can calculate the moles of chlorine in AgCl using the molar mass of chlorine (35.45 g/mol) and AgCl (143.32 g/mol): Moles of chlorine = (1.27 g AgCl) x (1 mol AgCl / 143.32 g AgCl) x (1 mol Cl / 1 mol AgCl) = 0.00999 mol Cl

Calculate the moles of hydrogen

We have the mass of the original compound (1.50 g) and the moles of carbon and chlorine calculated in Steps 1 and 2. Now, we can find the mass of hydrogen and calculate the moles of hydrogen using the molar mass of hydrogen (1.01 g/mol): Mass of hydrogen = 1.50 g (total mass) - (0.0800 mol C x 12.01 g/mol C) - (0.00999 mol Cl x 35.45 g/mol Cl) = 0.506 g H Moles of hydrogen = (0.506 g H) / (1.01 g/mol H) = 0.501 mol H

Find the empirical formula

We have the moles of every element: carbon (0.0800 mol), hydrogen (0.501 mol), and chlorine (0.00999 mol). To find the empirical formula, we'll divide all the calculated moles by their lowest common multiple: C: (0.0800 mol) / (0.00999 mol) ≈ 8 H: (0.501 mol) / (0.00999 mol) ≈ 50 Cl: (0.00999 mol) / (0.00999 mol) = 1 Therefore, the empirical formula of the compound is C8H50Cl.

Key concepts

Organic Compounds: The Building Blocks of Life

Organic compounds are a fascinating category of chemical substances that include both natural and synthetic creations. They are mainly composed of carbon atoms, often paired with hydrogen, oxygen, nitrogen, and other elements. These compounds are the foundation for life as we know it, playing crucial roles in the structure and function of living organisms. When we discuss organic compounds, some common examples include proteins, carbohydrates, and fats. These substances are vital to numerous biological processes and are studied extensively in chemistry and biology. A unique feature of organic compounds is the ability of carbon atoms to form stable covalent bonds with each other, creating diverse and complex molecular structures. This versatility allows for an incredible variety of compounds with different properties and functions. In the exercise you are tackling, you need to determine the empirical formula of an organic compound, demonstrating its composition in terms of the simplest whole number ratio of elements present. This is where techniques like combustion analysis come into play.

Combustion Analysis: Unlocking the Composition of Compounds

Combustion analysis is a clever technique chemists use to ascertain the elemental makeup of organic compounds. It involves burning the compound and analyzing the resulting combustion products to deduce the compound's original composition. Here's how it typically works:

• The organic compound is combusted in the presence of oxygen, where the carbon is converted to carbon dioxide ( CO_2 ).
• Hydrogen present in the compound is converted to water ( H_2O ), and if there's nitrogen, it might form nitrogen gas ( N_2 ) or other nitrogen oxides.
• By measuring the mass of these combustion products, the quantities of carbon, hydrogen, and other elements can be calculated.

In your exercise, combustion analysis was used to measure the amount of carbon in the compound. This data, combined with information from another analysis (conversion of chlorine to silver chloride), helps chemists piece together the empirical formula, offering a window into the compound's foundational building blocks.

Mole Calculations: A Key Tool in Chemistry

Understanding the mole concept is central in chemistry, especially when you aim to understand the quantitative aspects of chemical reactions and compounds. A 'mole' is a unit that measures the amount of substance, often used to simplify large numbers of molecules or atoms by providing a convenient counting method. Specifically, one mole is equivalent to approximately 6.022 x 10^23 entities (Avogadro's number). Mole calculations allow chemists to relate microscopic atomic masses to macroscopic amounts we can measure and observe. In the context of your exercise, you calculated moles by:

• Using the mass of the carbon dioxide produced to find the moles of carbon in the original compound.
• Using the mass of silver chloride to determine the moles of chlorine.
• Subtracting the mass contributions of carbon and chlorine from the total, allowing the calculation of moles of hydrogen.

These calculations formed the basis for determining the empirical formula, by normalizing the mole ratios to find the simplest integer ratio of the elements present in the compound. Practicing mole calculations helps you become fluent in translating raw data into a tangible understanding of chemical compositions and reactions.