Problem 77 - Chemistry: An Atoms-Focused Approach

Question

The freezing point of a solution prepared by dissolving 150 mg of caffeine in \(10.0 \mathrm{g}\) of camphor is lower by \(3.07^{\circ} \mathrm{C}\) than that of pure camphor \(\left(K_{f}=39.7^{\circ} \mathrm{C} / \mathrm{m}\right)\) What is the molar mass of caffeine? Elemental analysis of caffeine yields the following results: \(49.49 \%\) C, \(5.15 \% \mathrm{H}\) \(28.8796 \mathrm{N},\) and the remainder is \(\mathrm{O} .\) What is the molecular formula of caffeine?

Step-by-Step Solution

Answer

The step-by-step solution shows how to find the molar mass of caffeine using the freezing point depression data and determine the molecular formula of caffeine using the elemental analysis data. The calculated molecular formula of caffeine is C8H10N4O2.

Step 1: Convert mass of caffeine to grams

Convert the mass of caffeine from 150 mg to grams: \(150~\mathrm{mg} \cdot \dfrac{1~\mathrm{g}}{1000~\mathrm{mg}} = 0.150~\mathrm{g}\)

Step 2: Calculate the molality of the caffeine solution

Use the freezing point depression formula to calculate the molality of the solution, where the ΔTf is given as \(3.07^{\circ}\mathrm{C}\) and Kf is given as \(39.7^{\circ} \mathrm{C/m}\): Molality = \(\dfrac{\Delta T_f}{K_f} = \dfrac{3.07^{\circ}\mathrm{C}}{39.7^{\circ}\mbox{C/m}}\) = 0.0773 mol/kg

Step 3: Find the moles of caffeine

Use the formula molality = moles of solute/kg of solvent to find the moles of caffeine: Moles of caffeine = molality x mass of solvent (in kg) = 0.0773 mol/kg x 0.010 kg = \(7.73 \times 10^{-4}\) moles

Step 4: Find the molar mass of caffeine

Molar mass = mass of solute / moles of solute = 0.150 g / \(7.73 \times 10^{-4}\) moles = 194.1g/mol #Part 2: Finding the molecular formula of caffeine# Now we will use the elemental analysis data to determine the molecular formula of caffeine. The percentages for each element are provided: 49.49% C, 5.15% H, 28.8796% N, and the remainder is O.

Step 1: Convert percentages to grams

Assume that you have 100 g of caffeine, so the mass of each element will be the same as its percentage: Mass of C = 49.49 g Mass of H = 5.15 g Mass of N = 28.8796 g Mass of O = 100 - 49.49 - 5.15 - 28.8796 = 16.4804 g

Step 2: Convert grams to moles

To find the moles of each element, divide the mass of each element by its molar mass (C: 12.01 g/mol, H: 1.01 g/mol, N: 14.01 g/mol, O: 16.00 g/mol): Moles of C: \(\dfrac{49.49}{12.01}\) = 4.12 Moles of H: \(\dfrac{5.15}{1.01}\) = 5.10 Moles of N: \(\dfrac{28.8796}{14.01}\) = 2.06 Moles of O: \(\dfrac{16.4804}{16.00}\) = 1.03

Step 3: Find the empirical formula

Divide all the moles by the smallest value (1.03) to find the ratio of the elements: Ratio of C: \(\dfrac{4.12}{1.03}\) = 4 Ratio of H: \(\dfrac{5.10}{1.03}\) ≈ 5 Ratio of N: \(\dfrac{2.06}{1.03}\) = 2 Ratio of O: \(\dfrac{1.03}{1.03}\) = 1 Hence, the empirical formula is C4H5N2O.

Step 4: Calculate the empirical formula mass

Find the mass of the empirical formula using the molar mass of each element (C: 12.01 g/mol, H: 1.01 g/mol, N: 14.01 g/mol, O: 16.00 g/mol): Empirical formula mass = (4 x 12.01) + (5 x 1.01) + (2 x 14.01) + (1 x 16.00) = 97.06 g/mol

Step 5: Find the molecular formula

Divide the molar mass of caffeine (194.1 g/mol) by the empirical formula mass (97.06 g/mol) and multiply it by the empirical formula to find the molecular formula: Molecular formula = \(\dfrac{194.1}{97.06}\) x C4H5N2O ≈ 2 x C4H5N2O = C8H10N4O2 The molecular formula of caffeine is C8H10N4O2.

Key Concepts

Molar Mass Calculation

Calculating the molar mass is an essential skill in chemistry. It helps us determine the mass of a given chemical element or compound per mole of particles, usually in grams per mole (g/mol). In our exercise, the molar mass of caffeine is calculated using the freezing point depression method. This approach involves lowering the freezing point of a solvent by adding a solute, which in this case is caffeine.

Using the equation for freezing point depression: \[\Delta T_f = i \cdot K_f \cdot m\]where:

• \(\Delta T_f\) is the decrease in freezing point
• \(i\) is the van't Hoff factor (equal to 1 for non-electrolytes)
• \(K_f\) is the freezing point depression constant of the solvent
• \(m\) is the molality of the solution

Here, we calculated the molality as 0.0773 mol/kg using the given data (\(\Delta T_f = 3.07^{\circ}\mathrm{C}\) and \(K_f = 39.7^{\circ} \mathrm{C/m}\)). From the molality, the moles of caffeine were found, and subsequently, its molar mass was determined to be 194.1 g/mol. This is a vital concept because it allows us to understand how many grams one mole of the substance weighs.

Empirical Formula

The empirical formula represents the simplest, whole-number ratio of elements in a compound. It gives us a basic formula that broadly outlines the composition without revealing the exact number of atoms. To determine the empirical formula, we make use of elemental analysis data that tells us the composition by percentage of each element in a compound.

For caffeine, the percentages given were: 49.49% Carbon, 5.15% Hydrogen, 28.8796% Nitrogen, and the remainder in Oxygen. Assuming a 100 g sample, we directly convert these percentages to grams. Using the atomic masses of each element:

• Carbon: 12.01 g/mol
• Hydrogen: 1.01 g/mol
• Nitrogen: 14.01 g/mol
• Oxygen: 16.00 g/mol

We convert each element's mass to moles and then find the simplest ratio by dividing by the smallest number of moles. This gives us an empirical formula for caffeine as \( \text{C}_4\text{H}_5\text{N}_2\text{O} \). This step is crucial because it lays down the foundation for finding the molecular formula.

Molecular Formula

The molecular formula reveals the actual number of atoms in each element present in a compound, not just the ratio. Unlike the empirical formula, it reflects the true composition of the molecule.

In our example, the molecular formula of caffeine was determined after calculating the empirical formula. The next step was to find the empirical formula mass, which turned out to be 97.06 g/mol. Knowing the molar mass of caffeine from the molar mass calculation (194.1 g/mol), we used the following relation:\[\text{Molecular formula} = \left(\frac{\text{Molar mass}}{\text{Empirical formula mass}}\right) \times \text{Empirical formula}\]This calculation gave us a factor of approximately 2, indicating that the molecular formula is twice that of the empirical formula. Therefore, the molecular formula of caffeine is \( \text{C}_8\text{H}_{10}\text{N}_4\text{O}_2\). The molecular formula is important as it helps in determining structural formulas that will tell us more detailed information on molecular shape and function.

Elemental Analysis

Elemental analysis is a process that provides us with the elemental composition of a compound. It's an essential technique used heavily in chemistry to determine what elements are present in a compound and in what percentages.

This analysis involves a few key steps:

• Measuring the proportion of each element by mass within a sample.
• Typically conducted through combustion analysis or other laboratory methods to gather data.
• The results are often given in percentages, as seen in our caffeine example: Carbon (49.49%), Hydrogen (5.15%), Nitrogen (28.8796%), and the rest Oxygen.

Using this data, we can progress towards determining the empirical formula. Essentially, elemental analysis forms the groundwork for understanding the molecular framework of the substance in question and is a fundamental tool for chemists in identifying unknown compounds or verifying a substance's composition.