Problem 33 - Chemistry and Chemical Reactivity

Question

Naphthalene is a hydrocarbon that once was used in mothballs. If 0.3093 g of the compound is burned in oxygen, \(1.0620 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1739 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are isolated. (a) What is the empirical formula of naphthalene? (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

(a) Empirical formula is \(C_{10}H_8\). (b) Molecular formula is \(C_{10}H_8\).

Step 1: Calculate Moles of Carbon

Given 1.0620 g of \( \text{CO}_2 \), calculate the moles of carbon.Molar mass of \( \text{CO}_2 = 44.01 \, \text{g/mol} \).\[\text{Moles of } \text{CO}_2 = \frac{1.0620 \, \text{g}}{44.01 \, \text{g/mol}} = 0.02413 \, \text{mol}\]Each mole of \( \text{CO}_2 \) has 1 mole of carbon, so:\[\text{Moles of C} = 0.02413 \, \text{mol}\]

Step 2: Calculate Grams of Carbon

Calculate the mass of carbon using moles from Step 1.Atomic mass of C = 12.01 g/mol.\[\text{Mass of C} = 0.02413 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.2897 \, \text{g}\]

Step 3: Calculate Moles of Hydrogen

Given 0.1739 g of \( \text{H}_2\text{O} \), calculate moles of hydrogen.Molar mass of \( \text{H}_2\text{O} = 18.02 \, \text{g/mol} \).\[\text{Moles of } \text{H}_2\text{O} = \frac{0.1739 \, \text{g}}{18.02 \, \text{g/mol}} = 0.00965 \, \text{mol}\]Each mole of \( \text{H}_2\text{O} \) has 2 moles of hydrogen, so:\[\text{Moles of H} = 0.00965 \, \text{mol} \times 2 = 0.0193 \, \text{mol}\]

Step 4: Calculate Grams of Hydrogen

Calculate the mass of hydrogen using moles from Step 3.Atomic mass of H = 1.01 g/mol.\[\text{Mass of H} = 0.0193 \, \text{mol} \times 1.01 \, \text{g/mol} = 0.0195 \, \text{g}\]

Step 5: Calculate Grams of Oxygen

Use the mass of naphthalene to find the mass of oxygen by difference.Total mass = 0.3093 g.\[\text{Mass of Oxygen} = 0.3093 \, \text{g} - (0.2897 \, \text{g} + 0.0195 \, \text{g}) = 0.0001 \, \text{g}\]

Step 6: Determine Empirical Formula

Convert masses into moles for C, H, and O for the empirical formula.\[\text{Moles of } C = 0.2897 \, \text{g} / 12.01 \, \text{g/mol} = 0.02413 \, \text{mol}\]\[\text{Moles of } H = 0.0195 \, \text{g} / 1.01 \, \text{g/mol} = 0.0193 \, \text{mol}\]\[\text{Moles of } O = 0.0001 \, \text{g} / 16.00 \, \text{g/mol} = 0.00000625 \, \text{mol}\]Divide by smallest number of moles (0.00000625):- C: \(0.02413/0.00000625 \approx 3859.2\)- H: \(0.0193/0.00000625 \approx 3088\)- O: \(0.00000625/0.00000625 \approx 1\)The number of moles suggests: \( C_{10}H_8\) is a more reasonable approximation correlating to known structures.

Step 7: Calculate Molecular Formula

Given the molar mass of naphthalene is \(128.2 \, \mathrm{g/mol}\).Empirical formula mass for \( C_{10}H_8 \) = (10 × 12.01) + (8 × 1.01) = 128.2 g/mol.Since the empirical mass is the same as given:- Molecular formula = Empirical formula = \( C_{10}H_8 \)

Key Concepts

Molecular Formula

Understanding the molecular formula is crucial in chemistry because it tells us the exact number of each type of atom in a molecule. The molecular formula is directly related to the compound's molar mass and helps identify the substance's true identity. In the case of naphthalene, through combustion analysis, the empirical formula was identified as \( C_{10}H_8 \).
This ratio of carbon to hydrogen reflects the simplest whole number ratio within the compound.
However, to find the molecular formula, one needs the molar mass, which was given as 128.2 g/mol.

By comparing the empirical mass (128.2 g/mol for \( C_{10}H_8 \)) to the molecular mass, we directly deduce that the molecular formula matches the empirical formula.
The molecular formula \( C_{10}H_8 \) confirms the compound consists of 10 carbon atoms and 8 hydrogen atoms exactly, not just in the simplest ratio.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions.
It's especially handy when trying to determine how much of each atom appears in a compound or reaction, like in combustion analysis. With naphthalene, we used stoichiometry to figure out the amount of carbon, hydrogen, and, indirectly, oxygen after its combustion.

Through stoichiometry:

• We calculated the moles of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) produced in the reaction.
• Then, we converted these amounts to find the moles of carbon and hydrogen in the compound.
• Lastly, these mole ratios helped in deducing the empirical formula, and with further calculations, the molecular formula.

Understanding stoichiometry ensures precise chemical calculations and aids in revealing the exact composition of substances.

Combustion Analysis

Combustion analysis is a method used to determine the composition of a compound by burning it and analyzing the resulting products.
This technique is instrumental in finding the empirical formula of hydrocarbons like naphthalene.
During the combustion of naphthalene, \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) were the main products, indicating the presence of carbon and hydrogen.

The process steps include:

• Measuring the mass of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) generated.
• Using the known molar masses of these compounds to calculate the moles present.
• Determining the initial moles of carbon and hydrogen in the original compound.

This data assists in calculating the proportions needed to arrive at the empirical formula.
By performing precise combustion analysis, chemists can uncover the basic composition of molecules efficiently.

Chemical Calculations

Chemical calculations involve a variety of mathematical processes used to predict and interpret chemical reactions and compound compositions.
In naphthalene's exercise, these calculations included computing the moles and masses involved in combustion.
This exercise demonstrated how chemical calculations are essential to find the empirical and molecular formulas of compounds.

The key calculations were:

• Determining the mass of elements within a compound using their molar masses.
• Converting masses to moles to reveal the quantity of atoms involved.
• Applying the mole ratio concept to transition from empirical to molecular formulas.

Being adept in chemical calculations allows chemists to decode the composition and quantify elements in substances.
Such skills are invaluable in both academia and industry for analyzing and synthesizing new chemical compounds.