Problem 79 - Chemistry The Molecular Science

Question

Quinone, which is used in the dye industry and in chemical photography, is an organic compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} .\) Determine the empirical formula if \(0.105 \mathrm{~g}\) of the compound gives \(0.257 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0350 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) when burned completely.

Step-by-Step Solution

Answer

The empirical formula of the compound is \(\mathrm{C}_3\mathrm{H}_2\mathrm{O}\).

Step 1: Calculate Moles of Carbon in CO2

First, determine the number of moles of carbon in the CO2 produced. Given that the molar mass of CO2 is approximately 44.01 g/mol, the moles of carbon can be obtained from the mass of CO2 produced:\[m_{CO_2} = 0.257 \, ext{g}, \quad MM_{CO_2} = 44.01 \, \text{g/mol}\]Given the formula for CO2, each mole of CO2 contains one mole of carbon:\[moles\,of\,C = \frac{0.257}{44.01} = 0.00584 \, \text{mol}\]

Step 2: Calculate Moles of Hydrogen in H2O

Now, find the moles of hydrogen from the mass of water. Each mole of water (H2O) has two moles of hydrogen atoms:\[m_{H_2O} = 0.0350 \, \text{g}, \quad MM_{H_2O} = 18.02 \, \text{g/mol}\]Calculate the moles of water, then multiply by 2 for hydrogen:\[moles\,of\,H = 2 \times \left(\frac{0.0350}{18.02}\right) = 0.00388 \, \text{mol}\]

Step 3: Determine Mass of Oxygen in Quinone

To find the mass of oxygen in the quinone, note the total initial mass of the compound and the masses of carbon and hydrogen:- Total mass of quinone = 0.105 g- Mass of carbon = moles of carbon × molar mass C = 0.00584 × 12.01 g/mol- Mass of hydrogen = moles of hydrogen × molar mass H = 0.00388 × 1.01 g/molCalculate these:\[\text{Mass of C} = 0.00584 \times 12.01 = 0.0701 \, \text{g}, \quad \text{Mass of H} = 0.00388 \times 2.02 = 0.00392 \, \text{g}\]Therefore, the mass of oxygen in quinone:\[\text{Mass of O} = 0.105 - (0.0701 + 0.00392) = 0.0310 \, \text{g}\]

Step 4: Calculate Moles of Oxygen

Finally, compute the moles of oxygen found in quinone:Given the molar mass of oxygen is 16.00 g/mol:\[moles\,of\,O = \frac{0.0310}{16.00} = 0.00194 \, \text{mol}\]

Step 5: Calculate Empirical Formula

The empirical formula is derived from the ratio of moles of C, H, and O. Calculate the simplest whole-number ratio:- C: 0.00584 mol- H: 0.00388 mol- O: 0.00194 molDividing each by the smallest value (0.00194 mol):\[\text{Ratio of C} = \frac{0.00584}{0.00194} \approx 3 ewline\text{Ratio of H} = \frac{0.00388}{0.00194} \approx 2 ewline\text{Ratio of O} = \frac{0.00194}{0.00194} = 1\]Thus, the empirical formula is \(\mathrm{C}_3\mathrm{H}_2\mathrm{O}\).

Key Concepts

Combustion Analysis

In chemistry, combustion analysis is a method used to determine the elemental composition of an organic compound. This technique is particularly useful when dealing with compounds that contain carbon, hydrogen, and sometimes other elements such as oxygen. When an organic compound undergoes combustion, it reacts with oxygen to produce carbon dioxide (CO\(_2\)) and water (H\(_2\)O). By measuring the amounts of these combustion products, one can deduce the proportions of carbon and hydrogen in the original compound. In our given problem, quinone, an organic compound, undergoes combustion to yield specific amounts of CO\(_2\) and H\(_2\)O. By converting the masses of these products to moles using their respective molar masses, we can determine the moles of carbon and hydrogen that were initially present in the compound. This detailed process forms the basis of combustion analysis, which is a key technique in determining empirical formulas of organic substances.

Organic Compounds

Organic compounds are chemical compounds that primarily consist of carbon and hydrogen, often containing additional elements such as oxygen, nitrogen, and sulfur. Quinone, in this specific example, is an organic compound composed solely of carbon (C), hydrogen (H), and oxygen (O). These compounds form a vast array of structures and are foundational to chemical industries, biological systems, and more. The carbon atoms in organic compounds can create extensive networks due to their ability to form up to four covalent bonds, allowing for various configurations like chains and rings. When such compounds undergo a reaction, like combustion, analyzing the output allows for the determination of the substance's empirical formula, offering insights into its simplest structural model. Understanding the nature of organic compounds is essential in various fields, including chemistry, pharmacology, and materials science.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. In the context of this exercise, it's used to convert the mass of the combustion products (CO\(_2\) and H\(_2\)O) back to the amounts of carbon and hydrogen originally present in the quinone. The steps involve using the molar masses of these compounds to find the moles of carbon and hydrogen, as we need exact ratios to determine the compound's empirical formula. First, we convert the mass of CO\(_2\) to moles to find the moles of carbon since each molecule of CO\(_2\) contains one atom of carbon. Next, we perform a similar conversion for H\(_2\)O to find moles of hydrogen, remembering each molecule of H\(_2\)O contains two hydrogen atoms. Finally, stoichiometry helps in determining the number of moles of oxygen directly present by subtracting the computed masses of carbon and hydrogen from the total mass of the compound. Simplifying the ratios of these moles gives us the empirical formula C\(_3\)H\(_2\)O, representing the simplest integral ratio of elements in quinone.