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Chapter 12: Problem 56

Caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\), is a stimulant found in tea and coffee. A sample of the substance was dissolved in \(45.0 \mathrm{~g}\) of chloroform, \(\mathrm{CHCl}_{3},\) to give a \(0.0946 \mathrm{~m}\) solution. How many grams of caffeine were in the sample?

Short Answer

Expert verified
The sample contained approximately 0.827 grams of caffeine.

Step by step solution

01

Understand the Meaning of Molality

Molality (m) is defined as the number of moles of solute per kilogram of solvent. For our problem, the solution has a molality of 0.0946 m and chloroform is the solvent.
02

Calculate the Moles of Caffeine

Since we know the molality (0.0946 mol/kg) and the mass of chloroform (45.0 g = 0.045 kg), we can calculate the moles of caffeine using the formula: \[ ext{moles of caffeine} = ext{molality} \times ext{mass of solvent (in kg)} = 0.0946 \times 0.045 \approx 0.004257 ext{ moles} \]
03

Determine the Molar Mass of Caffeine

The molar mass of caffeine, \( \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} \), can be calculated as follows: - Carbon (C): 12.01 g/mol \( \times 8 = 96.08 \text{ g/mol} \)- Hydrogen (H): 1.01 g/mol \( \times 10 = 10.1 \text{ g/mol} \)- Nitrogen (N): 14.01 g/mol \( \times 4 = 56.04 \text{ g/mol} \)- Oxygen (O): 16.00 g/mol \( \times 2 = 32.00 \text{ g/mol} \)Adding these gives: \\[ ext{Molar mass of caffeine} = 96.08 + 10.1 + 56.04 + 32.00 = 194.22 \text{ g/mol} \]
04

Calculate the Mass of Caffeine

Now that we know the moles of caffeine and its molar mass, we can calculate the mass of caffeine in the sample using the formula: \\[ \text{mass of caffeine} = \text{moles of caffeine} \times \text{molar mass of caffeine} = 0.004257 \times 194.22 \approx 0.827 \text{ g} \]

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Key Concepts

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

Moles of Solute
Understanding the concept of moles is fundamental to grasping how solutions are measured and created in chemistry. In the context of the given problem, the solute is caffeine. Moles of solute represents the amount of caffeine that is present in the solution. Here, we use the formula related to molality, which is defined as the number of moles of solute per kilogram of solvent. The relationship can be expressed as:\[ \text{moles of solute} = \text{molality} \times \text{mass of solvent (in kg)} \]For instance, given a molality of 0.0946 mol/kg and the mass of chloroform as 45.0 grams (which converts to 0.045 kg when divided by 1000), the moles of caffeine can be calculated as:\[ \text{moles of caffeine} = 0.0946 \times 0.045 \approx 0.004257 \text{ moles} \]This step involves multiplying the known molality by the kilogram weight of the solvent to estimate the moles of caffeine dissolved in the chloroform.
Molar Mass Calculation
Molar mass is a crucial concept in chemistry, especially when converting between mass and moles. It is defined as the mass of one mole of a given substance and is typically expressed in grams per mole (g/mol). To calculate the molar mass of caffeine, we look at the chemical formula \( \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{O}_{2} \), which indicates the elements and their counts within a molecule:
  • Carbon \((C)\) appears 8 times, with each atom having an atomic mass of 12.01 g/mol. Thus, contributing \(8 \times 12.01 = 96.08 \text{ g/mol}\).
  • Hydrogen \((H)\) appears 10 times, contributing \(10 \times 1.01 = 10.1 \text{ g/mol}\).
  • Nitrogen \((N)\) appears 4 times, contributing \(4 \times 14.01 = 56.04 \text{ g/mol}\).
  • Oxygen \((O)\) appears 2 times, contributing \(2 \times 16.00 = 32.00 \text{ g/mol}\).
By summing these contributions, we find the molar mass of caffeine:\[ 96.08 + 10.1 + 56.04 + 32.00 = 194.22 \text{ g/mol} \]
Mass of Solute Calculation
Once we have determined the moles of solute (caffeine) and its molar mass, calculating the mass of the solute in the solution becomes straightforward. The formula to determine the mass of the solute from its moles and molar mass is:\[ \text{mass of solute} = \text{moles of solute} \times \text{molar mass of solute} \]In the given exercise, the moles of caffeine have already been calculated as approximately 0.004257 moles. With the molar mass of caffeine known to be 194.22 g/mol, the mass of caffeine is:\[ \text{mass of caffeine} = 0.004257 \times 194.22 \approx 0.827 \text{ g} \]Thus, about 0.827 grams of caffeine is present in the sample dissolved in the 45 grams of chloroform, demonstrating a clear pathway from molality through moles to mass of the solute.

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Most popular questions from this chapter

The osmotic pressure of blood at \(37^{\circ} \mathrm{C}\) is 7.7 atm. A solution that is given intravenously must have the same osmotic pressure as the blood. What should be the molarity of a glucose solution to give an osmotic pressure of \(7.7 \mathrm{~atm}\) at \(37^{\circ} \mathrm{C} ?\)

Which of the following ions would be expected to have the greater energy of hydration, \(\mathrm{Mg}^{2+}\) or \(\mathrm{Al}^{3+}\) ?

A sample of an ionic solid is dissolved in \(1.00 \mathrm{~kg}\) of water. The freezing point of the water is \(-0.01^{\circ} \mathrm{C}\). If three times the mass of ionic solid is dissolved in \(1.00 \mathrm{~kg}\) of water and the resulting freezing point of the solution is \(-0.09^{\circ} \mathrm{C},\) which of the following would be possible formulas for the solid: \(\mathrm{MX}, \mathrm{MX}_{2}\), or \(\mathrm{MX}_{3}\), where \(\mathrm{M}\) represents a metal cation and \(\mathrm{X}\) an anion with a l-charge? Explain your reasoning.

Equal numbers of moles of two soluble, substances, substance A and substance \(\mathrm{B}\), are placed into separate 1.0 - \(\mathrm{L}\) samples of water. a. The water samples are cooled. Sample A freezes at \(-0.50^{\circ} \mathrm{C},\) and Sample \(\mathrm{B}\) freezes at \(-1.00^{\circ} \mathrm{C}\). Explain how the solutions can have different freezing points. b. You pour \(500 \mathrm{~mL}\) of the solution containing substance B into a different beaker. How would the freezing point of this 500 -mL portion of solution \(\mathrm{B}\) compare to the freezing point of the 1.0 - \(\mathrm{L}\) sample of solution \(\mathrm{A}\) ? c. Calculate the molality of the solutions of \(\mathrm{A}\) and \(\mathrm{B}\). Assume that \(i=1\) for substance \(\mathrm{A}\). d. If you were to add an additional \(1.0 \mathrm{~kg}\) of water to solution \(\mathrm{B}\), what would be the new freezing point of the solution? Try to write an answer to this question without using a mathematical formula.

The solubility of carbon dioxide in water is \(0.161 \mathrm{~g}\) \(\mathrm{CO}_{2}\) in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) A soft drink is carbonated with carbon dioxide gas at 5.50 atm pressure. What is the solubility of carbon dioxide in water at this pressure?
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