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Manufacturers of headache remedies routinely claim that their own brands are more potent pain relievers than the competing brands. Their way of making the comparison is to compare the number of molecules in the standard dosage. Tylenol uses \(325 \mathrm{mg}\) of acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2}\right)\) as the standard dose, while Advil uses \(2.00 \times 10^{2} \mathrm{mg}\) of ibuprofen \(\left(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\right)\). Find the number of molecules of pain reliever in the standard doses of (a) Tylenol and (b) Advil.

Short Answer

Expert verified
Tylenol has \(1.29 \times 10^{21}\) molecules; Advil has \(5.84 \times 10^{20}\) molecules.

Step by step solution

01

Determine Molar Mass

First, calculate the molar mass of each compound. For acetaminophen (\( \mathrm{C}_8 \mathrm{H}_9 \mathrm{NO}_2 \)), the molar mass is calculated as follows:- Carbon (C): 8 atoms \( \times 12.01 \, \text{g/mol} = 96.08 \, \text{g/mol} \)- Hydrogen (H): 9 atoms \( \times 1.008 \, \text{g/mol} = 9.072 \, \text{g/mol} \)- Nitrogen (N): 1 atom \( \times 14.01 \, \text{g/mol} = 14.01 \, \text{g/mol} \)- Oxygen (O): 2 atoms \( \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)Adding these together, the molar mass of acetaminophen is \( 151.162 \, \text{g/mol} \). For ibuprofen (\( \mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_2 \)), the molar mass is:- Carbon (C): 13 atoms \( \times 12.01 \, \text{g/mol} = 156.13 \, \text{g/mol} \)- Hydrogen (H): 18 atoms \( \times 1.008 \, \text{g/mol} = 18.144 \, \text{g/mol} \)- Oxygen (O): 2 atoms \( \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)Adding these, the molar mass of ibuprofen is \( 206.274 \, \text{g/mol} \).
02

Convert Mass to Moles

Convert the mass of each medicament to moles using their molar masses. (a) For Tylenol:- Given mass = 325 mg = 0.325 g- Moles of acetaminophen = \( \frac{0.325 \, \text{g}}{151.162 \, \text{g/mol}} \approx 0.00215 \, \text{mol} \)(b) For Advil:- Given mass = 200 mg = 0.200 g- Moles of ibuprofen = \( \frac{0.200 \, \text{g}}{206.274 \, \text{g/mol}} \approx 0.000970 \, \text{mol} \)
03

Convert Moles to Molecules

Use Avogadro's number (\( 6.022 \times 10^{23} \) molecules/mol) to convert the moles of each compound to the number of molecules.(a) For Tylenol:- Number of molecules = \( 0.00215 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 1.29 \times 10^{21} \, \text{molecules} \)(b) For Advil:- Number of molecules = \( 0.000970 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 5.84 \times 10^{20} \, \text{molecules} \)

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Key Concepts

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

Molar Mass
Molar mass is a vital concept in chemistry that helps us understand the mass of a given compound's molecules. It is expressed in units of grams per mole (g/mol), and provides insight into the weight of one mole of substance, which is the fundamental quantity for chemical reactions and equations.

To determine the molar mass, we first sum up the atomic masses of all the atoms present in a molecule. For example, in acetaminophen, which has the molecular formula \( \text{C}_8 \text{H}_9 \text{NO}_2 \), we add up the masses of 8 carbon atoms, 9 hydrogen atoms, 1 nitrogen atom, and 2 oxygen atoms. Similarly, for ibuprofen, the masses of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms are summed. Each element has a specific atomic mass which can be found on the periodic table.

Knowing the molar mass enables the conversion of a substance's mass (such as from milligrams to grams) to moles. This step is essential for understanding the quantity of molecules or atoms involved in any given measurement — whether you're working in a laboratory or understanding the dosage in pharmaceuticals.
Avogadro's Number
Avogadro's number ( 6.022 × 10^{23} ) is a constant in chemistry crucial for converting between the amount of substance in moles and the number of particles — which can be atoms, molecules, or even smaller entities.

The concept is simple: one mole of any substance contains exactly 6.022 × 10^{23} of its particles. Named after the Italian scientist Amedeo Avogadro, this number helps bridge the microscopic world of atoms and molecules with the macroscopic world we experience.

When you have calculated the number of moles from the mass using the molar mass, Avogadro's number allows you to then find out how many molecules you have. For example, in the context of pharmaceutical chemistry, by knowing the mass of acetaminophen in Tylenol and its molar mass, we can determine the moles present, and subsequently, using Avogadro's number, the total number of acetaminophen molecules in a given dose.

This conversion is highly beneficial in crafting specific chemical formulations and understanding how various drugs compare on a molecular level.
Pharmaceutical Chemistry
Pharmaceutical chemistry involves the application of chemical techniques to the development and manufacturing of pharmaceutical substances. This field combines principles from molecular calculations, including molar mass and Avogadro's number, to ensure the efficacy and safety of medicines.

In the context of a simple exercise, such as comparing the number of molecules in different doses of pain relievers, pharmaceutical chemistry helps in identifying which compound, and therefore which drug, might be more effective based on its molecular content.
  • Understanding molar mass ensures the correct dosage of chemicals is administered in medicine.
  • Avogadro's number aids in calculating the precise number of molecules present, influencing potency comparisons.
  • These calculations help ascertain the pharmacodynamics of drugs, or how a drug acts on the body.
With this knowledge, chemists can fine-tune drug formulations precisely, leading to safer and more effective medicines. Understanding these principles is key to developing therapies that can treat a wide array of health issues effectively.

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Most popular questions from this chapter

A gas fills the right portion of a horizontal cylinder whose radius is \(5.00 \mathrm{~cm} .\) The initial pressure of the gas is \(1.01 \times 10^{5} \mathrm{~Pa}\). A frictionless movable piston separates the gas from the left portion of the cylinder, which is evacuated and contains an ideal spring, as the drawing shows. The piston is initially held in place by a pin. The spring is initially unstrained, and the length of the gas-filled portion is \(20.0 \mathrm{~cm}\). When the pin is removed and the gas is allowed to expand, the length of the gas-filled chamber doubles. The initial and final temperatures are equal. Determine the spring constant of the spring.

Multiple-Concept Example 4 deals with the concepts used in this problem. Conceptual Example 3 is also pertinent. A bubble, located \(0.200 \mathrm{~m}\) beneath the surface in a glass of beer, rises to the top. The air pressure at the top is \(1.01 \times 10^{5} \mathrm{~Pa}\). Assume that the density of beer is the same as that of fresh water. If the temperature and number of moles of \(\mathrm{CO} 2\) remain constant as the bubble rises, find the ratio of its volume at the top to that at the bottom.

Suppose that a tank contains \(680 \mathrm{~m}^{3}\) of neon at an absolute pressure of \(1.01 \times 10^{5} \mathrm{~Pa}\). The temperature is changed from 293.2 to \(294.3 \mathrm{~K}\). What is the increase in the internal energy of the neon?

The temperature of an ideal gas is doubled while the volume is kept constant. Does the absolute pressure of the gas double when the temperature that doubles is (a) the Kelvin temperature and (b) the Celsius temperature? Explain. Problem Determine the ratio \(P_{2} / P_{1}\) of the final pressure \(P_{2}\) to the initial pressure \(P_{1}\) when the temperature rises (a) from 35.0 to \(70.0 \mathrm{~K}\) and \((\mathrm{b})\) from 35.0 to \(70.0{ }^{\circ} \mathrm{C}\). Check to see that your answers are consistent with your answers to the Concept Questions.

The diffusion constant for the amino acid glycine in water is \(1.06 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). In a 2.0 \(\mathrm{cm}\) -long tube with a cross-sectional area of \(1.5 \times 10^{-4} \mathrm{~m}^{2}\), the mass rate of diffusion is \(m / t=4.2 \times 10^{-14} \mathrm{~kg} / \mathrm{s}\), because the glycine concentration is maintained at a value of \(8.3 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{3}\) at one end of the tube and at a lower value at the other end. What is the lower concentration?

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