Question

An organic compound was found to contain only $\mathrm{C},\mathrm{H},$ and $\mathrm{Cl}$. 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 $1.00-g$ sample of the compound was converted to $1.27\mathrm{g}$ of AgCl. Determine the empirical formula of the compound.

Short answer

Based on the given data, the empirical formula of the organic compound containing C, H, and Cl is $C_2{H}_{3}Cl$.

Step-by-step solution

Calculate moles of Carbon

First, convert the mass of CO2 produced to moles of Carbon using the molecular weight of CO2. Since the molecular weight of CO2 is 44.01 g/mol (12.01 g/mol for C and 32.00 g/mol for O2), and each CO2 molecule contains one atom of carbon, we can find the moles of carbon as follows: $\text{moles of Carbon}=\frac{3.52\text{g CO}2}{44.01\text{g/mol}}$

Calculate moles of Chlorine

Next, convert the mass of AgCl produced to moles of Chlorine using the molecular weight of AgCl. Since the molecular weight of AgCl is 143.32 g/mol (107.87 g/mol for Ag and 35.45 g/mol for Cl), and each AgCl molecule contains one atom of Chlorine, we can find the moles of Chlorine as follows: $\text{moles of Chlorine}=\frac{1.27\text{g AgCl}}{143.32\text{g/mol}}$

Calculate the mass of Hydrogen

Now, we will calculate the mass of Hydrogen present in the 1.50-g sample by subtracting the masses of Carbon and Chlorine from the sample mass (1.50 g). $\text{mass of Hydrogen}=\text{mass of sample}-\text{mass of Carbon}-\text{mass of Chlorine}$

Calculate moles of Hydrogen

Next, convert the mass of Hydrogen calculated in Step 3 to moles of Hydrogen using the molecular weight of H (1.01 g/mol). $\text{moles of Hydrogen}=\frac{\text{mass of Hydrogen}}{1.01\text{g/mol}}$

Determine Mole Ratios of C, H, and Cl

Now that we have the moles of each element, we can find the simplest whole number ratio (mole ratio) of the elements by dividing each value by the smallest number of moles among C, H, and Cl.

Determine the Empirical Formula

Using the mole ratios obtained in Step 5, we can determine the empirical formula of the organic compound. The compound consists of C, H, and Cl in the simplest whole number ratios found in Step 5. This will give us the empirical formula of the compound.

Key concepts

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of an organic compound. This method is particularly useful for substances that contain carbon, hydrogen, and sometimes other elements such as oxygen or chlorine. The process involves burning the organic compound in the presence of oxygen, which results in the formation of carbon dioxide and water, among other products, depending on the components present. To perform a combustion analysis:

• Burn a known mass of the organic compound completely in air or pure oxygen.
• Measure the mass of carbon dioxide and water produced from the combustion.

From these measurements, you can determine the amounts of carbon and hydrogen in the original compound. In our specific exercise, the identified product was carbon dioxide, which indicates the presence of carbon. The number of moles of carbon can be found from carbon dioxide using its molar mass. This allows us to track back how much carbon was present in the initial sample of the compound.

Organic Chemistry

Organic chemistry is the study of the structure, properties, composition, reactions, and synthesis of carbon-containing compounds. This field covers a vast number of compounds and is essential due to the carbon atom's ability to form various kinds of structures, including chains, rings, and intricate framework structures. In our combustion analysis problem, we're dealing with an organic compound composed of carbon, hydrogen, and chlorine.

• Carbon forms the backbone of organic compounds.
• Hydrogen, the simplest element, often bonds with carbon to form hydrocarbons.
• Elements like chlorine can substitute hydrogen to form various functional groups, altering the compound's properties.

The study of such compounds allows chemists to determine their molecular structures and predict their behavior and reactivity, which is crucial for designing new materials, drugs, and polymers.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It is based on the conservation of mass and the principle that matter cannot be created or destroyed. In the problem, stoichiometry is used to determine the empirical formula of the organic compound by analyzing the amounts of carbon, hydrogen, and chlorine. Here's how stoichiometry was applied:

• Calculate the moles of each element (carbon, chlorine, and hydrogen) using their respective outputs (CO2 and AgCl from the problem statement).
• Find the mole ratios by dividing the number of moles of each element by the smallest number of moles calculated.
• Use these simplest whole number ratios to derive the empirical formula of the compound.

Stoichiometry thus serves as an essential tool for converting mass-to-moles and comparing these values to find the mole ratios necessary to define the compound's simplest formula. This understanding allows chemists to predict the outcomes of reactions and design chemical syntheses effectively.