Problem 34 - Chemistry and Chemical Reactivity

Question

Azulene is a beautiful blue hydrocarbon. If \(0.106 \mathrm{g}\) of the compound is burned in oxygen, \(0.364 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0596 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are isolated. (a) What is the empirical formula of azulene? (b) If a separate experiment gave \(128.2 \mathrm{g} / \mathrm{mol}\) as the molar mass of the compound, what is its molecular formula?

Step-by-Step Solution

Answer

The empirical formula could not be conclusively derived; molecular interpretations suggest reassess based traditional approximations primary tasks.

Step 1: Determine Moles of Carbon and Hydrogen

First, find the moles of carbon in \(0.364\, \text{g} \) of \(\text{CO}_2\). The molar mass of \(\text{CO}_2\) is \(44.01 \text{g/mol}\), and there is one carbon atom per molecule. Thus:\[\text{moles of C} = \frac{0.364}{44.01} \approx 0.00827 \text{ mol}\] Do the same for hydrogen in \(0.0596\, \text{g} \) of \(\text{H}_2\text{O}\). The molar mass of \(\text{H}_2\text{O}\) is \(18.02 \text{g/mol}\), and there are two hydrogen atoms per molecule. Thus:\[\text{moles of H} = 2 \times \frac{0.0596}{18.02} \approx 0.00662 \text{ mol}\]

Step 2: Mass of Carbon and Hydrogen

Calculate the mass of carbon and hydrogen using their moles and atomic masses. - For Carbon: \(0.00827 \text{ mol} \times 12.01 \text{ g/mol} = 0.0993 \text{ g}\)- For Hydrogen: \(0.00662 \text{ mol} \times 1.008 \text{ g/mol} = 0.00667 \text{ g}\)

Step 3: Determine Mass of Oxygen in Azulene

Subtract the mass of carbon and hydrogen from the original mass of azulene (0.106 g):\[0.106 - (0.0993 + 0.00667) = 0.00003 \text{ g of Oxygen}\]

Step 4: Determine Moles of Oxygen

Calculate the moles of oxygen using the mass found:\[moles\ of\ O = \frac{0.00003}{16.00}\approx 0.00000188 \text{ mol}\]

Step 5: Calculate Empirical Formula Ratio

Find the simplest ratio of moles of C, H, and O:- C: \(0.00827 / 0.00000188 \approx 4400\)- H: \(0.00662 / 0.00000188 \approx 3500\)- O: \(0.00000188 / 0.00000188 = 1\) Since the calculated ratio doesn't provide meaningful small whole numbers, recheck calculations, usually empirical is found through adjusting known traditional process with correct masses.

Step 6: Revised Empirical Formula Calculation

After verifying steps, and assuming simplest initial result or different error assumptions: Assume error only on adjusted parts: different ratios c and h repeat empirical such as initial approaches different counting, etc.

Step 6: Determine Molecular Formula

Use the molar mass to find the molecular formula. Assume correct empirical findings, next can use initial guidance empirically:E.g. Assume small empirical part, repeat base smallest, then drive: \(128.2 \text{ g/mol}/ 13\text{unit base molarity}\), find general small multiples to fill.

Key Concepts

Combustion Analysis

Combustion analysis is a fundamental technique in chemistry used to determine the elemental composition of a compound. When a substance, such as a hydrocarbon, is burned in oxygen, it typically produces carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)). By measuring the masses of these products, one can deduce the amounts of carbon and hydrogen in the original compound.

Here’s how it works:

• Measure Masses: In the given problem, the masses of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) were recorded after burning azulene.
• Calculate Moles: Using the known molar masses of \( \text{CO}_2 \) (44.01 g/mol) and \( \text{H}_2\text{O} \) (18.02 g/mol), determine the moles of carbon and hydrogen in these products.
• Translate to Elements: Recognize that each molecule of \( \text{CO}_2 \) contains one carbon atom, while each molecule of \( \text{H}_2\text{O} \) contains two hydrogen atoms.

Through these calculations, the elemental composition of the original hydrocarbon can be pieced together, forming the basis for determining its empirical formula.

Molar Mass Determination

The determination of the molar mass of a compound is crucial for finding its molecular formula. Knowing just the empirical formula gives the simplest ratio of atoms present in a compound, but additional data, like molar mass, helps fully define the compound.

So how is molar mass determined? Here are steps to make it clear:

• Experimental Methods: In some experiments, like the one with azulene, separate tests give a measurement of the molar mass using methods such as mass spectrometry or cryoscopy.
• From Empirical to Molecular: Once the molar mass is known, it is compared with the empirical formula mass. If the molar mass equals the empirical formula mass, the empirical and molecular formulas are the same. If not, multiply the empirical formula by an integer to match the molar mass.

For blue azulene in the problem, after determining the empirical formula, using the provided molar mass of 128.2 g/mol helps ascertain the complete molecular formula.

Hydrocarbon Compounds

Hydrocarbons are compounds solely composed of hydrogen and carbon atoms and are divided into several categories, such as alkanes, alkenes, and aromatic hydrocarbons. Each type has unique structural characteristics, influencing its chemical properties.

Why hydrocarbons are distinctive:

• Structure: Alkanes are saturated with single bonds, while alkenes contain at least one double bond. Aromatic hydrocarbons, like azulene, have distinct ring structures with electrons shared differently.
• Properties: The saturation level and structure affect stability, reactivity, and color. Azulene is noted particularly for its vibrant blue color, unusual among hydrocarbons.
• Combustion: Hydrocarbons are primarily burned for energy, producing \( \text{CO}_2 \) and \( \text{H}_2\text{O} \), fundamental for analyses.

Understanding hydrocarbons' diverse structures and behaviors is vital for patterns seen in combustion analysis and subsequent analyses.

Chemical Stoichiometry

Chemical stoichiometry is a key mathematical aspect of chemistry that deals with the relationships between reactants and products in chemical reactions. It's like a recipe, ensuring the reactants are balanced to predict the amounts of products formed or reactants consumed.

In stoichiometry:

• Balanced Equations: Chemical equations must be balanced with equal numbers of each element on both sides. For combustion, it typically involves balancing hydrocarbons with oxygen to produce \( \text{CO}_2 \) and \( \text{H}_2\text{O} \).
• Mole Ratios: Stoichiometry uses mole ratios from balanced equations to convert between masses, moles, and individual molecules.
• Yield and Reactions: By using stoichiometry principles, one can calculate theoretical yields, compare them to actual yields, and determine limiting reactants.

In the context of azulene, chemical stoichiometry helps determine its empirical and molecular formula by interpreting the outcome of its combustion reaction.