Problem 91 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

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

Step 1: Determine Moles of Carbon

First, we calculate the moles of carbon in the carbon dioxide (CO_2). The molecular weight of CO_2 is 44.01 \mathrm{g/mol\}:\[ \text{moles of C} = \frac{0.257 \mathrm{g \ CO}_2}{44.01 \mathrm{g/mol}} \approx 0.00584 \mathrm{mol} \]Since each \(\mathrm{CO}_2\) molecule contains one atom of C, this is the number of moles of C.

Step 2: Determine Moles of Hydrogen

Next, calculate the moles of hydrogen from the water (H_2O). The molecular weight of H_2O is 18.02 \mathrm{g/mol\}:\[ \text{moles of H} = \frac{0.0350 \mathrm{g \ H}_2\mathrm{O}}{18.02 \mathrm{g/mol}} \approx 0.00194 \mathrm{mol} \]Since each \(\mathrm{H}_2O\) molecule contains two hydrogen atoms, the moles of H are twice this number:\[ \text{moles of H} = 2 \times 0.00194 \approx 0.00388\ \mathrm{mol}\].

Step 3: Determine Mass of Oxygen in Compound

The total initial mass of the compound is \(0.105 \mathrm{g}\). First calculate the mass of carbon and hydrogen:\[ \text{mass of C} = 0.00584 \ \mathrm{mol} \times 12.01\ \mathrm{g/mol} \approx 0.0701 \mathrm{g}\]\[ \text{mass of H} = 0.00388 \ \mathrm{mol} \times 1.008\ \mathrm{g/mol} \approx 0.00391 \mathrm{g}\]The mass of oxygen is obtained by subtracting the masses of carbon and hydrogen from the total mass:\[ \text{mass of O} = 0.105 \mathrm{g} - (0.0701 \mathrm{g} + 0.00391 \mathrm{g}) \approx 0.031 \mathrm{g}\]

Step 4: Determine Moles of Oxygen

Calculate the moles of oxygen using its molar mass, \(16.00 \mathrm{g/mol}\):\[ \text{moles of O} = \frac{0.031 \mathrm{g}}{16.00 \mathrm{g/mol}} \approx 0.00194 \mathrm{mol}\]

Step 5: Determine Empirical Formula

Find the mole ratio by dividing each by the smallest number of moles:\[ \text{C ratio} = \frac{0.00584}{0.00194} \approx 3\]\[ \text{H ratio} = \frac{0.00388}{0.00194} \approx 2\]\[ \text{O ratio} = \frac{0.00194}{0.00194} \approx 1\]So, the empirical formula is \(\mathrm{C}_3\mathrm{H}_2\mathrm{O}\).

Key Concepts

Combustion Analysis

Combustion analysis is a crucial method employed in chemistry to determine the composition of an unknown organic compound. This analysis involves burning the compound in the presence of oxygen and analyzing the resulting combustion products, typically carbon dioxide ( CO_2 ) and water ( H_2O ). By measuring the masses of CO_2 and H_2O produced, you can infer the initial quantities of carbon and hydrogen in the compound.
Although oxygen is often present in the original compound, its amount is usually determined indirectly by deducting the masses of carbon and hydrogen from the total mass. To summarize, key steps in combustion analysis include:

• Measuring the amount of carbon in carbon dioxide.
• Measuring the amount of hydrogen in water.
• Calculating the mass of oxygen through deduction.

Understanding these principles is essential to grasping the full picture of how combustion analysis works in establishing the empirical formula of organic compounds.

Moles Calculation

The concept of moles is fundamental to chemistry, often serving to bridge the microscopic world of atoms and molecules with the macroscopic world we observe. One mole equals Avogadro's number of particles, approximately 6.022 imes 10^{23} of them.
In the context of combustion analysis, calculating moles allows us to translate mass measurements (in grams) into actual particle counts for comparison to deduce the sample's composition.Here's a concise guide to moles calculations in chemical analysis:

• Determine the molecular weight of the substance you are interested in (like CO_2 or H_2O).
• Calculate the moles by using the formula:
  \[\text{moles} = \frac{\text{mass in grams}}{\text{molar mass}}.\]
• For elements that occur in a compound, adjust for stoichiometry. For example, in H_2O, one molecule contains two atoms of hydrogen.

These mole calculations are essential in determining the ratios necessary for deriving the empirical formula—representing the simplest whole number ratio of elements in a compound.

Organic Compounds

Organic compounds are primarily composed of carbon atoms, typically bonded with hydrogen, oxygen, and sometimes other elements like nitrogen or sulfur.
Carbon's ability to form stable bonds with many elements, including itself, allows for a vast diversity of organic molecules from simple to highly complex structures. Understanding organic compounds involves knowing the following:

• Organic compounds always contain carbon-hydrogen ( C-H ) bonds.
• The variety of possible combinations means that each compound can have multiple forms (isomers) sharing the same molecular formula but differing structures.
• In combustion analysis, identifying the precise count of carbon, hydrogen, and oxygen atoms is critical for defining the empirical formula.

In essence, when working with organic compounds, you're often tasked with understanding both their molecular structure and their empirical representation. This understanding is integral when conducting combustion analysis as it informs both the qualitative and quantitative composition of the compound in question.