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A compound that contains only \(\mathrm{C}\) and \(\mathrm{H}\) was burned in excess \(\mathrm{O}_{2}\) to give \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). When \(0.270 \mathrm{~g}\) of the compound was burned, the amount of \(\mathrm{CO}_{2}\) formed reacted completely with \(20.0\) \(\mathrm{mL}\) of \(2.00 \mathrm{M} \mathrm{NaOH}\) solution according to the equation $$ 2 \mathrm{OH}^{-}(a q)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ When \(0.270 \mathrm{~g}\) of the compound was dissolved in \(50.0 \mathrm{~g}\) of camphor, the resulting solution had a freezing point of \(177.9^{\circ} \mathrm{C}\). [Pure camphor freezes at \(179.8^{\circ} \mathrm{C}\) and has \(\left.K_{\mathrm{f}}=37.7\left({ }^{\circ} \mathrm{C} \cdot \mathrm{kg}\right) / \mathrm{mol} .\right]\) (a) What is the empirical formula of the compound? (b) What is the molecular weight of the compound? (c) What is the molecular formula of the compound?

Step by step solution

01

Determine the moles of CO2 formed

Given that 20.0 mL of 2.00 M NaOH is used, calculate the moles of OH鈦� available: \[ \text{moles of OH}^- = 0.0200 \text{ L} \times 2.00 \frac{\text{mol}}{\text{L}} = 0.0400 \text{ mol} \] From the balanced equation, 2 moles of OH鈦� react with 1 mole of CO2, so the moles of CO2 can be determined as follows: \[ \text{moles of CO}_2 = \frac{0.0400 \text{ mol OH}^-}{2} = 0.0200 \text{ mol CO}_2 \]

02

Calculate the grams of carbon in the compound

Since 1 mole of CO2 contains 1 mole of carbon, the moles of carbon are equal to the moles of CO2. The mass of carbon is given by: \[ \text{mass of } C = 0.0200 \text{ mol} \times 12.01 \frac{\text{g}}{\text{mol}} = 0.240 \text{ g} \]

03

Calculate the grams of hydrogen in the compound

The original mass of the compound is 0.270 g, and the mass of carbon is 0.240 g. The mass of hydrogen is given by: \[ \text{mass of } H = 0.270 \text{ g} - 0.240 \text{ g} = 0.030 \text{ g} \] Thus, moles of hydrogen are: \[ \text{moles of } H = \frac{0.030 \text{ g}}{1.008 \text{ g/mol}} = 0.02976 \text{ mol} \]

04

Determine the empirical formula

Calculate the mole ratio of hydrogen to carbon:\[ \text{Mole ratio of } C = \frac{0.0200}{0.0200} = 1 \]\[ \text{Mole ratio of } H = \frac{0.02976}{0.0200} \approx 1.49 \]Rounding to the nearest whole number:For hydrogen, round 1.49 to 1.5 and multiply both the values by 2 to get whole numbers:\[ \text{Empirical formula} = \text{C}_2\text{H}_3 \]

05

Find the molar mass by freezing point depression

Freezing point change \(\Delta T = 179.8^{\circ} \text{C} - 177.9^{\circ} \text{C} = 1.9^{\circ} \text{C}\).Using the freezing point depression formula:\[ \Delta T = i \cdot K_f \cdot m \Rightarrow m = \frac{\Delta T}{K_f} = \frac{1.9^{\circ} \text{C}}{37.7^{\circ} \text{C} \cdot \text{mol}^{-1}} = 0.0504 \text{ mol/kg} \]Since 0.270 g (0.000270 kg) of the compound is dissolved, the moles of solute are:\[ 0.0504 \times 0.0500 \text{ kg} = 0.00252 \text{ mol} \]Molar mass \(M\) is:\[ M = \frac{0.270 \text{ g}}{0.00252 \text{ mol}} = 107.14 \text{ g/mol} \]

06

Determine the molecular formula

The empirical formula mass is the molecular weight of C鈧侶鈧�:\[ \text{empirical mass} = 2(12.01) + 3(1.008) = 27.034 \text{ g/mol} \]Find the ratio of molecular weight to empirical formula weight:\[ \text{molar mass ratio} = \frac{107.14 \text{ g/mol}}{27.034 \text{ g/mol}} \approx 4 \]Therefore, the molecular formula is 4 times the empirical formula:\[ \text{Molecular formula} = \text{C}_{8}\text{H}_{12} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Analysis

Combustion analysis is a technique used to determine the empirical formula of a compound containing carbon and hydrogen. When such a compound is burned in the presence of excess oxygen, it produces carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). The amount of \(\text{CO}_2\) produced is directly related to the amount of carbon in the compound. Similarly, the amount of \(\text{H}_2\text{O}\) gives information about the hydrogen content.

• In the given example, the compound was burned to yield \(0.0200\) mol of \(\text{CO}_2\).
• This allows us to calculate the mass of carbon, knowing that each mole of \(\text{CO}_2\) contains one mole of carbon atoms.

By analyzing both the mass of carbon and the mass of the compound, chemists can deduce the amount of hydrogen present. This balance helps in finding the empirical formula.

Freezing Point Depression Method

Freezing point depression is a colligative property that allows for the determination of molecular weight or molarity of a solute in a solvent. When a solute dissolves in a solvent, it lowers the freezing point of the solution compared to the pure solvent.

• The change in freezing temperature (\(\Delta T\)) is used in calculations, where \(\Delta T = 179.8^{\circ} C - 177.9^{\circ} C = 1.9^{\circ} C\).
• Using the formula \(\Delta T = i \cdot K_f \cdot m\), the molality (\(m\)) can be calculated.
• The constant \(K_f\) is specific to the solvent, which in this case is camphor, having the value \(37.7^{\circ} C \cdot kg/mol\).

This method allows us to find the molar mass of a solute, enhancing our understanding of its molecular structure, as seen with the compound's molar mass calculation.

Chemical Formula Calculation

Determining the chemical formula involves using both empirical data and molecular weight to find the compound's actual structure. This is accomplished by establishing the empirical formula first, then adjusting it to reflect the molecular weight.

• First, calculate the empirical formula, which in this example is \(\text{C}_2\text{H}_3\).
• Then, use the calculated empirical mass and compare it against the experimentally determined molecular mass (e.g., \(107.14\) g/mol).
• The ratio of molecular weight to empirical weight tells how many empirical units exist in the molecule.

Understanding and calculating the chemical formula allows scientists to comprehend the precise number of each atom in a particle and predict its chemical behavior.

Stoichiometry in Chemistry

Stoichiometry is the study of quantitative relationships in chemical reactions. It plays a vital role in calculating the balanced and exact proportions of reactants and products in a reaction.

• In a combustion reaction, stoichiometry allows us to understand the conversion of the original compound into \(\text{CO}_2\) and \(\text{H}_2\text{O}\).
• The balanced equation helps in relating the moles of reactants and products, as seen in the equation used in \(\text{NaOH}\) reaction to find \(\text{CO}_2\).
• Stoichiometry not only helps in analyzing reactions but also aids in the synthesis and decomposition processes, serving as an essential tool for chemists.

With stoichiometry, you can accurately predict the amount needed for any reaction, ensuring efficiency and precision in chemical experimentation.